Toniolo and Linder, Equation (10+)

Percentage Accurate: 53.7% → 89.2%

Time: 18.3s

Alternatives: 22

Speedup: 24.8×

• 26.8% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

\[\begin{array}{l} \\ \frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \end{array} \]

FPCore

(FPCore (t l k)
 :precision binary64
 (/
  2.0
  (*
   (* (* (/ (pow t 3.0) (* l l)) (sin k)) (tan k))
   (+ (+ 1.0 (pow (/ k t) 2.0)) 1.0))))

C

double code(double t, double l, double k) {
	return 2.0 / ((((pow(t, 3.0) / (l * l)) * sin(k)) * tan(k)) * ((1.0 + pow((k / t), 2.0)) + 1.0));
}

Fortran

real(8) function code(t, l, k)
    real(8), intent (in) :: t
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    code = 2.0d0 / (((((t ** 3.0d0) / (l * l)) * sin(k)) * tan(k)) * ((1.0d0 + ((k / t) ** 2.0d0)) + 1.0d0))
end function

Java

public static double code(double t, double l, double k) {
	return 2.0 / ((((Math.pow(t, 3.0) / (l * l)) * Math.sin(k)) * Math.tan(k)) * ((1.0 + Math.pow((k / t), 2.0)) + 1.0));
}

Python

def code(t, l, k):
	return 2.0 / ((((math.pow(t, 3.0) / (l * l)) * math.sin(k)) * math.tan(k)) * ((1.0 + math.pow((k / t), 2.0)) + 1.0))

Julia

function code(t, l, k)
	return Float64(2.0 / Float64(Float64(Float64(Float64((t ^ 3.0) / Float64(l * l)) * sin(k)) * tan(k)) * Float64(Float64(1.0 + (Float64(k / t) ^ 2.0)) + 1.0)))
end

MATLAB

function tmp = code(t, l, k)
	tmp = 2.0 / (((((t ^ 3.0) / (l * l)) * sin(k)) * tan(k)) * ((1.0 + ((k / t) ^ 2.0)) + 1.0));
end

Wolfram

code[t_, l_, k_] := N[(2.0 / N[(N[(N[(N[(N[Power[t, 3.0], $MachinePrecision] / N[(l * l), $MachinePrecision]), $MachinePrecision] * N[Sin[k], $MachinePrecision]), $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision] * N[(N[(1.0 + N[Power[N[(k / t), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Initial Program: 53.7% accurate, 1.0× speedup

Math

\[\begin{array}{l} \\ \frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \end{array} \]

FPCore

(FPCore (t l k)
 :precision binary64
 (/
  2.0
  (*
   (* (* (/ (pow t 3.0) (* l l)) (sin k)) (tan k))
   (+ (+ 1.0 (pow (/ k t) 2.0)) 1.0))))

C

double code(double t, double l, double k) {
	return 2.0 / ((((pow(t, 3.0) / (l * l)) * sin(k)) * tan(k)) * ((1.0 + pow((k / t), 2.0)) + 1.0));
}

Fortran

real(8) function code(t, l, k)
    real(8), intent (in) :: t
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    code = 2.0d0 / (((((t ** 3.0d0) / (l * l)) * sin(k)) * tan(k)) * ((1.0d0 + ((k / t) ** 2.0d0)) + 1.0d0))
end function

Java

public static double code(double t, double l, double k) {
	return 2.0 / ((((Math.pow(t, 3.0) / (l * l)) * Math.sin(k)) * Math.tan(k)) * ((1.0 + Math.pow((k / t), 2.0)) + 1.0));
}

Python

def code(t, l, k):
	return 2.0 / ((((math.pow(t, 3.0) / (l * l)) * math.sin(k)) * math.tan(k)) * ((1.0 + math.pow((k / t), 2.0)) + 1.0))

Julia

function code(t, l, k)
	return Float64(2.0 / Float64(Float64(Float64(Float64((t ^ 3.0) / Float64(l * l)) * sin(k)) * tan(k)) * Float64(Float64(1.0 + (Float64(k / t) ^ 2.0)) + 1.0)))
end

MATLAB

function tmp = code(t, l, k)
	tmp = 2.0 / (((((t ^ 3.0) / (l * l)) * sin(k)) * tan(k)) * ((1.0 + ((k / t) ^ 2.0)) + 1.0));
end

Wolfram

code[t_, l_, k_] := N[(2.0 / N[(N[(N[(N[(N[Power[t, 3.0], $MachinePrecision] / N[(l * l), $MachinePrecision]), $MachinePrecision] * N[Sin[k], $MachinePrecision]), $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision] * N[(N[(1.0 + N[Power[N[(k / t), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] + 1.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Alternative 1: 89.2% accurate, 0.5× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 1.6 \cdot 10^{-213}:\\ \;\;\;\;\frac{2}{{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t\_m}{\cos k}}\right)}^{2}}\\ \mathbf{elif}\;t\_m \leq 6.9 \cdot 10^{-65}:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t\_m}\right)\right) \cdot \frac{{t\_m}^{1.5}}{\ell}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\left(\frac{t\_m}{\sqrt[3]{\ell}} \cdot \sqrt[3]{\frac{1}{\ell}}\right) \cdot \left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t\_m}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= t_m 1.6e-213)
    (/ 2.0 (pow (* (/ (* k (sin k)) l) (sqrt (/ t_m (cos k)))) 2.0))
    (if (<= t_m 6.9e-65)
      (/
       2.0
       (*
        (pow (* (hypot 1.0 (hypot 1.0 (/ k t_m))) (/ (pow t_m 1.5) l)) 2.0)
        (* (sin k) (tan k))))
      (/
       2.0
       (pow
        (*
         (* (/ t_m (cbrt l)) (cbrt (/ 1.0 l)))
         (* (cbrt (* (tan k) (+ 2.0 (pow (/ k t_m) 2.0)))) (cbrt (sin k))))
        3.0))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 1.6e-213) {
		tmp = 2.0 / pow((((k * sin(k)) / l) * sqrt((t_m / cos(k)))), 2.0);
	} else if (t_m <= 6.9e-65) {
		tmp = 2.0 / (pow((hypot(1.0, hypot(1.0, (k / t_m))) * (pow(t_m, 1.5) / l)), 2.0) * (sin(k) * tan(k)));
	} else {
		tmp = 2.0 / pow((((t_m / cbrt(l)) * cbrt((1.0 / l))) * (cbrt((tan(k) * (2.0 + pow((k / t_m), 2.0)))) * cbrt(sin(k)))), 3.0);
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 1.6e-213) {
		tmp = 2.0 / Math.pow((((k * Math.sin(k)) / l) * Math.sqrt((t_m / Math.cos(k)))), 2.0);
	} else if (t_m <= 6.9e-65) {
		tmp = 2.0 / (Math.pow((Math.hypot(1.0, Math.hypot(1.0, (k / t_m))) * (Math.pow(t_m, 1.5) / l)), 2.0) * (Math.sin(k) * Math.tan(k)));
	} else {
		tmp = 2.0 / Math.pow((((t_m / Math.cbrt(l)) * Math.cbrt((1.0 / l))) * (Math.cbrt((Math.tan(k) * (2.0 + Math.pow((k / t_m), 2.0)))) * Math.cbrt(Math.sin(k)))), 3.0);
	}
	return t_s * tmp;
}

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (t_m <= 1.6e-213)
		tmp = Float64(2.0 / (Float64(Float64(Float64(k * sin(k)) / l) * sqrt(Float64(t_m / cos(k)))) ^ 2.0));
	elseif (t_m <= 6.9e-65)
		tmp = Float64(2.0 / Float64((Float64(hypot(1.0, hypot(1.0, Float64(k / t_m))) * Float64((t_m ^ 1.5) / l)) ^ 2.0) * Float64(sin(k) * tan(k))));
	else
		tmp = Float64(2.0 / (Float64(Float64(Float64(t_m / cbrt(l)) * cbrt(Float64(1.0 / l))) * Float64(cbrt(Float64(tan(k) * Float64(2.0 + (Float64(k / t_m) ^ 2.0)))) * cbrt(sin(k)))) ^ 3.0));
	end
	return Float64(t_s * tmp)
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[t$95$m, 1.6e-213], N[(2.0 / N[Power[N[(N[(N[(k * N[Sin[k], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[Sqrt[N[(t$95$m / N[Cos[k], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$m, 6.9e-65], N[(2.0 / N[(N[Power[N[(N[Sqrt[1.0 ^ 2 + N[Sqrt[1.0 ^ 2 + N[(k / t$95$m), $MachinePrecision] ^ 2], $MachinePrecision] ^ 2], $MachinePrecision] * N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[Power[N[(N[(N[(t$95$m / N[Power[l, 1/3], $MachinePrecision]), $MachinePrecision] * N[Power[N[(1.0 / l), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision] * N[(N[Power[N[(N[Tan[k], $MachinePrecision] * N[(2.0 + N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision] * N[Power[N[Sin[k], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if t < 1.59999999999999986e-213

   1. Initial program 49.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 49.4%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 10.2%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 10.2%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 4.1%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 4.1%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 4.1%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 4.1%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around inf 23.0%

      \[\leadsto \frac{2}{{\color{blue}{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t}{\cos k}}\right)}}^{2}} \]

   if 1.59999999999999986e-213 < t < 6.89999999999999991e-65

   1. Initial program 38.6%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 38.6%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 33.1%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 33.1%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 67.4%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 67.4%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 67.4%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 67.4%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   7. Step-by-step derivation

      1. associate-*r* 67.3%

         \[\leadsto \frac{2}{{\color{blue}{\left(\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right) \cdot \sqrt{\sin k \cdot \tan k}\right)}}^{2}} \]

      2. unpow-prod-down 67.3%

         \[\leadsto \frac{2}{\color{blue}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot {\left(\sqrt{\sin k \cdot \tan k}\right)}^{2}}} \]

      3. pow2 67.3%

         \[\leadsto \frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \color{blue}{\left(\sqrt{\sin k \cdot \tan k} \cdot \sqrt{\sin k \cdot \tan k}\right)}} \]

      4. add-sqr-sqrt 88.0%

         \[\leadsto \frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \color{blue}{\left(\sin k \cdot \tan k\right)}} \]

   8. Applied egg-rr 88.0%

      \[\leadsto \frac{2}{\color{blue}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}} \]

   if 6.89999999999999991e-65 < t

   1. Initial program 58.6%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 62.0%

      \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. sqr-pow 62.0%

         \[\leadsto \frac{2}{\frac{\frac{\color{blue}{{t}^{\left(\frac{3}{2}\right)} \cdot {t}^{\left(\frac{3}{2}\right)}}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      2. *-un-lft-identity 62.0%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{\left(\frac{3}{2}\right)} \cdot {t}^{\left(\frac{3}{2}\right)}}{\color{blue}{1 \cdot \ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      3. times-frac 70.0%

         \[\leadsto \frac{2}{\frac{\color{blue}{\frac{{t}^{\left(\frac{3}{2}\right)}}{1} \cdot \frac{{t}^{\left(\frac{3}{2}\right)}}{\ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      4. metadata-eval 70.0%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{\color{blue}{1.5}}}{1} \cdot \frac{{t}^{\left(\frac{3}{2}\right)}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      5. metadata-eval 70.0%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{\color{blue}{1.5}}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   6. Applied egg-rr 70.0%

      \[\leadsto \frac{2}{\frac{\color{blue}{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]
   7. Step-by-step derivation

      1. add-cube-cbrt 69.7%

         \[\leadsto \frac{2}{\color{blue}{\left(\sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \cdot \sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right) \cdot \sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}}} \]

      2. pow3 69.7%

         \[\leadsto \frac{2}{\color{blue}{{\left(\sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right)}^{3}}} \]

   8. Applied egg-rr 81.3%

      \[\leadsto \frac{2}{\color{blue}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \sqrt[3]{\sin k \cdot \left(\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right)}^{3}}} \]
   9. Step-by-step derivation

      1. *-commutative 81.3%

         \[\leadsto \frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \sqrt[3]{\color{blue}{\left(\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right) \cdot \sin k}}\right)}^{3}} \]

      2. cbrt-prod 90.5%

         \[\leadsto \frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \color{blue}{\left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)}\right)}^{3}} \]

   10. Applied egg-rr 90.5%

       \[\leadsto \frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \color{blue}{\left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)}\right)}^{3}} \]
   11. Step-by-step derivation

       1. add-cbrt-cube 71.4%

          \[\leadsto \frac{2}{{\left(\frac{\color{blue}{\sqrt[3]{\left(t \cdot t\right) \cdot t}}}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}} \]

       2. unpow3 71.4%

          \[\leadsto \frac{2}{{\left(\frac{\sqrt[3]{\color{blue}{{t}^{3}}}}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}} \]

       3. unpow2 71.4%

          \[\leadsto \frac{2}{{\left(\frac{\sqrt[3]{{t}^{3}}}{\color{blue}{\sqrt[3]{\ell} \cdot \sqrt[3]{\ell}}} \cdot \left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}} \]

       4. cbrt-prod 58.6%

          \[\leadsto \frac{2}{{\left(\frac{\sqrt[3]{{t}^{3}}}{\color{blue}{\sqrt[3]{\ell \cdot \ell}}} \cdot \left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}} \]

       5. unpow2 58.6%

          \[\leadsto \frac{2}{{\left(\frac{\sqrt[3]{{t}^{3}}}{\sqrt[3]{\color{blue}{{\ell}^{2}}}} \cdot \left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}} \]

       6. cbrt-div 58.4%

          \[\leadsto \frac{2}{{\left(\color{blue}{\sqrt[3]{\frac{{t}^{3}}{{\ell}^{2}}}} \cdot \left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}} \]

       7. unpow2 58.4%

          \[\leadsto \frac{2}{{\left(\sqrt[3]{\frac{{t}^{3}}{\color{blue}{\ell \cdot \ell}}} \cdot \left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}} \]

       8. associate-/l/ 64.8%

          \[\leadsto \frac{2}{{\left(\sqrt[3]{\color{blue}{\frac{\frac{{t}^{3}}{\ell}}{\ell}}} \cdot \left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}} \]

       9. div-inv 64.8%

          \[\leadsto \frac{2}{{\left(\sqrt[3]{\color{blue}{\frac{{t}^{3}}{\ell} \cdot \frac{1}{\ell}}} \cdot \left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}} \]

       10. cbrt-prod 68.7%

           \[\leadsto \frac{2}{{\left(\color{blue}{\left(\sqrt[3]{\frac{{t}^{3}}{\ell}} \cdot \sqrt[3]{\frac{1}{\ell}}\right)} \cdot \left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}} \]

       11. cbrt-div 71.3%

           \[\leadsto \frac{2}{{\left(\left(\color{blue}{\frac{\sqrt[3]{{t}^{3}}}{\sqrt[3]{\ell}}} \cdot \sqrt[3]{\frac{1}{\ell}}\right) \cdot \left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}} \]

       12. unpow3 71.3%

           \[\leadsto \frac{2}{{\left(\left(\frac{\sqrt[3]{\color{blue}{\left(t \cdot t\right) \cdot t}}}{\sqrt[3]{\ell}} \cdot \sqrt[3]{\frac{1}{\ell}}\right) \cdot \left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}} \]

       13. add-cbrt-cube 90.5%

           \[\leadsto \frac{2}{{\left(\left(\frac{\color{blue}{t}}{\sqrt[3]{\ell}} \cdot \sqrt[3]{\frac{1}{\ell}}\right) \cdot \left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}} \]

   12. Applied egg-rr 90.5%

       \[\leadsto \frac{2}{{\left(\color{blue}{\left(\frac{t}{\sqrt[3]{\ell}} \cdot \sqrt[3]{\frac{1}{\ell}}\right)} \cdot \left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}} \]
3. Recombined 3 regimes into one program.
4. Add Preprocessing

Alternative 2: 79.2% accurate, 0.3× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ \begin{array}{l} t_2 := {\left(\frac{k}{t\_m}\right)}^{2}\\ t_3 := \left(1 + \left(1 + t\_2\right)\right) \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t\_m}^{3}}{\ell \cdot \ell}\right)\right)\\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_3 \leq 2 \cdot 10^{+299}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t\_m}^{1.5}}{\ell}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + t\_2\right)\right)}\\ \mathbf{elif}\;t\_3 \leq \infty:\\ \;\;\;\;\frac{2}{{\left(\frac{t\_m}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \left(\sqrt[3]{\sin k} \cdot \sqrt[3]{2 \cdot k}\right)\right)}^{3}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t\_m}{\cos k}}\right)}^{2}}\\ \end{array} \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (let* ((t_2 (pow (/ k t_m) 2.0))
        (t_3
         (*
          (+ 1.0 (+ 1.0 t_2))
          (* (tan k) (* (sin k) (/ (pow t_m 3.0) (* l l)))))))
   (*
    t_s
    (if (<= t_3 2e+299)
      (/
       2.0
       (* (pow (/ (pow t_m 1.5) l) 2.0) (* (* (sin k) (tan k)) (+ 2.0 t_2))))
      (if (<= t_3 INFINITY)
        (/
         2.0
         (pow
          (* (/ t_m (pow (cbrt l) 2.0)) (* (cbrt (sin k)) (cbrt (* 2.0 k))))
          3.0))
        (/ 2.0 (pow (* (/ (* k (sin k)) l) (sqrt (/ t_m (cos k)))) 2.0)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double t_2 = pow((k / t_m), 2.0);
	double t_3 = (1.0 + (1.0 + t_2)) * (tan(k) * (sin(k) * (pow(t_m, 3.0) / (l * l))));
	double tmp;
	if (t_3 <= 2e+299) {
		tmp = 2.0 / (pow((pow(t_m, 1.5) / l), 2.0) * ((sin(k) * tan(k)) * (2.0 + t_2)));
	} else if (t_3 <= ((double) INFINITY)) {
		tmp = 2.0 / pow(((t_m / pow(cbrt(l), 2.0)) * (cbrt(sin(k)) * cbrt((2.0 * k)))), 3.0);
	} else {
		tmp = 2.0 / pow((((k * sin(k)) / l) * sqrt((t_m / cos(k)))), 2.0);
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double t_2 = Math.pow((k / t_m), 2.0);
	double t_3 = (1.0 + (1.0 + t_2)) * (Math.tan(k) * (Math.sin(k) * (Math.pow(t_m, 3.0) / (l * l))));
	double tmp;
	if (t_3 <= 2e+299) {
		tmp = 2.0 / (Math.pow((Math.pow(t_m, 1.5) / l), 2.0) * ((Math.sin(k) * Math.tan(k)) * (2.0 + t_2)));
	} else if (t_3 <= Double.POSITIVE_INFINITY) {
		tmp = 2.0 / Math.pow(((t_m / Math.pow(Math.cbrt(l), 2.0)) * (Math.cbrt(Math.sin(k)) * Math.cbrt((2.0 * k)))), 3.0);
	} else {
		tmp = 2.0 / Math.pow((((k * Math.sin(k)) / l) * Math.sqrt((t_m / Math.cos(k)))), 2.0);
	}
	return t_s * tmp;
}

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	t_2 = Float64(k / t_m) ^ 2.0
	t_3 = Float64(Float64(1.0 + Float64(1.0 + t_2)) * Float64(tan(k) * Float64(sin(k) * Float64((t_m ^ 3.0) / Float64(l * l)))))
	tmp = 0.0
	if (t_3 <= 2e+299)
		tmp = Float64(2.0 / Float64((Float64((t_m ^ 1.5) / l) ^ 2.0) * Float64(Float64(sin(k) * tan(k)) * Float64(2.0 + t_2))));
	elseif (t_3 <= Inf)
		tmp = Float64(2.0 / (Float64(Float64(t_m / (cbrt(l) ^ 2.0)) * Float64(cbrt(sin(k)) * cbrt(Float64(2.0 * k)))) ^ 3.0));
	else
		tmp = Float64(2.0 / (Float64(Float64(Float64(k * sin(k)) / l) * sqrt(Float64(t_m / cos(k)))) ^ 2.0));
	end
	return Float64(t_s * tmp)
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := Block[{t$95$2 = N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]}, Block[{t$95$3 = N[(N[(1.0 + N[(1.0 + t$95$2), $MachinePrecision]), $MachinePrecision] * N[(N[Tan[k], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[(N[Power[t$95$m, 3.0], $MachinePrecision] / N[(l * l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(t$95$s * If[LessEqual[t$95$3, 2e+299], N[(2.0 / N[(N[Power[N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision] * N[(2.0 + t$95$2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$3, Infinity], N[(2.0 / N[Power[N[(N[(t$95$m / N[Power[N[Power[l, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] * N[(N[Power[N[Sin[k], $MachinePrecision], 1/3], $MachinePrecision] * N[Power[N[(2.0 * k), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision], N[(2.0 / N[Power[N[(N[(N[(k * N[Sin[k], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[Sqrt[N[(t$95$m / N[Cos[k], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (*.f64 (*.f64 (*.f64 (/.f64 (pow.f64 t #s(literal 3 binary64)) (*.f64 l l)) (sin.f64 k)) (tan.f64 k)) (+.f64 (+.f64 #s(literal 1 binary64) (pow.f64 (/.f64 k t) #s(literal 2 binary64))) #s(literal 1 binary64))) < 2.0000000000000001e299

   1. Initial program 80.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 81.4%

      \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. add-sqr-sqrt 29.5%

         \[\leadsto \frac{2}{\color{blue}{\left(\sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}} \cdot \sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}}\right)} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      2. pow2 29.5%

         \[\leadsto \frac{2}{\color{blue}{{\left(\sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}}\right)}^{2}} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      3. associate-/r* 28.7%

         \[\leadsto \frac{2}{{\left(\sqrt{\color{blue}{\frac{{t}^{3}}{\ell \cdot \ell}}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      4. sqrt-div 28.7%

         \[\leadsto \frac{2}{{\color{blue}{\left(\frac{\sqrt{{t}^{3}}}{\sqrt{\ell \cdot \ell}}\right)}}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      5. sqrt-pow1 30.2%

         \[\leadsto \frac{2}{{\left(\frac{\color{blue}{{t}^{\left(\frac{3}{2}\right)}}}{\sqrt{\ell \cdot \ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      6. metadata-eval 30.2%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{\color{blue}{1.5}}}{\sqrt{\ell \cdot \ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      7. sqrt-prod 17.4%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{1.5}}{\color{blue}{\sqrt{\ell} \cdot \sqrt{\ell}}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      8. add-sqr-sqrt 30.9%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{1.5}}{\color{blue}{\ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   6. Applied egg-rr 30.9%

      \[\leadsto \frac{2}{\color{blue}{{\left(\frac{{t}^{1.5}}{\ell}\right)}^{2}} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   if 2.0000000000000001e299 < (*.f64 (*.f64 (*.f64 (/.f64 (pow.f64 t #s(literal 3 binary64)) (*.f64 l l)) (sin.f64 k)) (tan.f64 k)) (+.f64 (+.f64 #s(literal 1 binary64) (pow.f64 (/.f64 k t) #s(literal 2 binary64))) #s(literal 1 binary64))) < +inf.0

   1. Initial program 70.2%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 69.6%

      \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. sqr-pow 45.8%

         \[\leadsto \frac{2}{\frac{\frac{\color{blue}{{t}^{\left(\frac{3}{2}\right)} \cdot {t}^{\left(\frac{3}{2}\right)}}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      2. *-un-lft-identity 45.8%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{\left(\frac{3}{2}\right)} \cdot {t}^{\left(\frac{3}{2}\right)}}{\color{blue}{1 \cdot \ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      3. times-frac 48.3%

         \[\leadsto \frac{2}{\frac{\color{blue}{\frac{{t}^{\left(\frac{3}{2}\right)}}{1} \cdot \frac{{t}^{\left(\frac{3}{2}\right)}}{\ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      4. metadata-eval 48.3%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{\color{blue}{1.5}}}{1} \cdot \frac{{t}^{\left(\frac{3}{2}\right)}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      5. metadata-eval 48.3%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{\color{blue}{1.5}}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   6. Applied egg-rr 48.3%

      \[\leadsto \frac{2}{\frac{\color{blue}{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]
   7. Step-by-step derivation

      1. add-cube-cbrt 48.1%

         \[\leadsto \frac{2}{\color{blue}{\left(\sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \cdot \sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right) \cdot \sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}}} \]

      2. pow3 48.1%

         \[\leadsto \frac{2}{\color{blue}{{\left(\sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right)}^{3}}} \]

   8. Applied egg-rr 79.0%

      \[\leadsto \frac{2}{\color{blue}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \sqrt[3]{\sin k \cdot \left(\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right)}^{3}}} \]
   9. Step-by-step derivation

      1. *-commutative 79.0%

         \[\leadsto \frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \sqrt[3]{\color{blue}{\left(\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right) \cdot \sin k}}\right)}^{3}} \]

      2. cbrt-prod 93.2%

         \[\leadsto \frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \color{blue}{\left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)}\right)}^{3}} \]

   10. Applied egg-rr 93.2%

       \[\leadsto \frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \color{blue}{\left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)}\right)}^{3}} \]

   11. Taylor expanded in k around 0 88.5%

       \[\leadsto \frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \left(\sqrt[3]{\color{blue}{2 \cdot k}} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}} \]

   if +inf.0 < (*.f64 (*.f64 (*.f64 (/.f64 (pow.f64 t #s(literal 3 binary64)) (*.f64 l l)) (sin.f64 k)) (tan.f64 k)) (+.f64 (+.f64 #s(literal 1 binary64) (pow.f64 (/.f64 k t) #s(literal 2 binary64))) #s(literal 1 binary64)))

   1. Initial program 0.0%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 0.0%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 0.0%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 0.0%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 24.7%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 24.7%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 24.7%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 24.7%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around inf 41.9%

      \[\leadsto \frac{2}{{\color{blue}{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t}{\cos k}}\right)}}^{2}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 43.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\left(1 + \left(1 + {\left(\frac{k}{t}\right)}^{2}\right)\right) \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t}^{3}}{\ell \cdot \ell}\right)\right) \leq 2 \cdot 10^{+299}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\\ \mathbf{elif}\;\left(1 + \left(1 + {\left(\frac{k}{t}\right)}^{2}\right)\right) \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t}^{3}}{\ell \cdot \ell}\right)\right) \leq \infty:\\ \;\;\;\;\frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \left(\sqrt[3]{\sin k} \cdot \sqrt[3]{2 \cdot k}\right)\right)}^{3}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t}{\cos k}}\right)}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 79.9% accurate, 0.3× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ \begin{array}{l} t_2 := {\left(\frac{k}{t\_m}\right)}^{2}\\ t_3 := \left(1 + \left(1 + t\_2\right)\right) \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t\_m}^{3}}{\ell \cdot \ell}\right)\right)\\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_3 \leq 2 \cdot 10^{+299}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t\_m}^{1.5}}{\ell}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + t\_2\right)\right)}\\ \mathbf{elif}\;t\_3 \leq \infty:\\ \;\;\;\;\frac{2}{{\left(\left(t\_m \cdot \sqrt[3]{2}\right) \cdot \frac{{\left(\sqrt[3]{k}\right)}^{2}}{{\left(\sqrt[3]{\ell}\right)}^{2}}\right)}^{3}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t\_m}{\cos k}}\right)}^{2}}\\ \end{array} \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (let* ((t_2 (pow (/ k t_m) 2.0))
        (t_3
         (*
          (+ 1.0 (+ 1.0 t_2))
          (* (tan k) (* (sin k) (/ (pow t_m 3.0) (* l l)))))))
   (*
    t_s
    (if (<= t_3 2e+299)
      (/
       2.0
       (* (pow (/ (pow t_m 1.5) l) 2.0) (* (* (sin k) (tan k)) (+ 2.0 t_2))))
      (if (<= t_3 INFINITY)
        (/
         2.0
         (pow
          (* (* t_m (cbrt 2.0)) (/ (pow (cbrt k) 2.0) (pow (cbrt l) 2.0)))
          3.0))
        (/ 2.0 (pow (* (/ (* k (sin k)) l) (sqrt (/ t_m (cos k)))) 2.0)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double t_2 = pow((k / t_m), 2.0);
	double t_3 = (1.0 + (1.0 + t_2)) * (tan(k) * (sin(k) * (pow(t_m, 3.0) / (l * l))));
	double tmp;
	if (t_3 <= 2e+299) {
		tmp = 2.0 / (pow((pow(t_m, 1.5) / l), 2.0) * ((sin(k) * tan(k)) * (2.0 + t_2)));
	} else if (t_3 <= ((double) INFINITY)) {
		tmp = 2.0 / pow(((t_m * cbrt(2.0)) * (pow(cbrt(k), 2.0) / pow(cbrt(l), 2.0))), 3.0);
	} else {
		tmp = 2.0 / pow((((k * sin(k)) / l) * sqrt((t_m / cos(k)))), 2.0);
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double t_2 = Math.pow((k / t_m), 2.0);
	double t_3 = (1.0 + (1.0 + t_2)) * (Math.tan(k) * (Math.sin(k) * (Math.pow(t_m, 3.0) / (l * l))));
	double tmp;
	if (t_3 <= 2e+299) {
		tmp = 2.0 / (Math.pow((Math.pow(t_m, 1.5) / l), 2.0) * ((Math.sin(k) * Math.tan(k)) * (2.0 + t_2)));
	} else if (t_3 <= Double.POSITIVE_INFINITY) {
		tmp = 2.0 / Math.pow(((t_m * Math.cbrt(2.0)) * (Math.pow(Math.cbrt(k), 2.0) / Math.pow(Math.cbrt(l), 2.0))), 3.0);
	} else {
		tmp = 2.0 / Math.pow((((k * Math.sin(k)) / l) * Math.sqrt((t_m / Math.cos(k)))), 2.0);
	}
	return t_s * tmp;
}

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	t_2 = Float64(k / t_m) ^ 2.0
	t_3 = Float64(Float64(1.0 + Float64(1.0 + t_2)) * Float64(tan(k) * Float64(sin(k) * Float64((t_m ^ 3.0) / Float64(l * l)))))
	tmp = 0.0
	if (t_3 <= 2e+299)
		tmp = Float64(2.0 / Float64((Float64((t_m ^ 1.5) / l) ^ 2.0) * Float64(Float64(sin(k) * tan(k)) * Float64(2.0 + t_2))));
	elseif (t_3 <= Inf)
		tmp = Float64(2.0 / (Float64(Float64(t_m * cbrt(2.0)) * Float64((cbrt(k) ^ 2.0) / (cbrt(l) ^ 2.0))) ^ 3.0));
	else
		tmp = Float64(2.0 / (Float64(Float64(Float64(k * sin(k)) / l) * sqrt(Float64(t_m / cos(k)))) ^ 2.0));
	end
	return Float64(t_s * tmp)
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := Block[{t$95$2 = N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]}, Block[{t$95$3 = N[(N[(1.0 + N[(1.0 + t$95$2), $MachinePrecision]), $MachinePrecision] * N[(N[Tan[k], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[(N[Power[t$95$m, 3.0], $MachinePrecision] / N[(l * l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(t$95$s * If[LessEqual[t$95$3, 2e+299], N[(2.0 / N[(N[Power[N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision] * N[(2.0 + t$95$2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$3, Infinity], N[(2.0 / N[Power[N[(N[(t$95$m * N[Power[2.0, 1/3], $MachinePrecision]), $MachinePrecision] * N[(N[Power[N[Power[k, 1/3], $MachinePrecision], 2.0], $MachinePrecision] / N[Power[N[Power[l, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision], N[(2.0 / N[Power[N[(N[(N[(k * N[Sin[k], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[Sqrt[N[(t$95$m / N[Cos[k], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (*.f64 (*.f64 (*.f64 (/.f64 (pow.f64 t #s(literal 3 binary64)) (*.f64 l l)) (sin.f64 k)) (tan.f64 k)) (+.f64 (+.f64 #s(literal 1 binary64) (pow.f64 (/.f64 k t) #s(literal 2 binary64))) #s(literal 1 binary64))) < 2.0000000000000001e299

   1. Initial program 80.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 81.4%

      \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. add-sqr-sqrt 29.5%

         \[\leadsto \frac{2}{\color{blue}{\left(\sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}} \cdot \sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}}\right)} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      2. pow2 29.5%

         \[\leadsto \frac{2}{\color{blue}{{\left(\sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}}\right)}^{2}} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      3. associate-/r* 28.7%

         \[\leadsto \frac{2}{{\left(\sqrt{\color{blue}{\frac{{t}^{3}}{\ell \cdot \ell}}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      4. sqrt-div 28.7%

         \[\leadsto \frac{2}{{\color{blue}{\left(\frac{\sqrt{{t}^{3}}}{\sqrt{\ell \cdot \ell}}\right)}}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      5. sqrt-pow1 30.2%

         \[\leadsto \frac{2}{{\left(\frac{\color{blue}{{t}^{\left(\frac{3}{2}\right)}}}{\sqrt{\ell \cdot \ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      6. metadata-eval 30.2%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{\color{blue}{1.5}}}{\sqrt{\ell \cdot \ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      7. sqrt-prod 17.4%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{1.5}}{\color{blue}{\sqrt{\ell} \cdot \sqrt{\ell}}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      8. add-sqr-sqrt 30.9%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{1.5}}{\color{blue}{\ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   6. Applied egg-rr 30.9%

      \[\leadsto \frac{2}{\color{blue}{{\left(\frac{{t}^{1.5}}{\ell}\right)}^{2}} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   if 2.0000000000000001e299 < (*.f64 (*.f64 (*.f64 (/.f64 (pow.f64 t #s(literal 3 binary64)) (*.f64 l l)) (sin.f64 k)) (tan.f64 k)) (+.f64 (+.f64 #s(literal 1 binary64) (pow.f64 (/.f64 k t) #s(literal 2 binary64))) #s(literal 1 binary64))) < +inf.0

   1. Initial program 70.2%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 69.6%

      \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. sqr-pow 45.8%

         \[\leadsto \frac{2}{\frac{\frac{\color{blue}{{t}^{\left(\frac{3}{2}\right)} \cdot {t}^{\left(\frac{3}{2}\right)}}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      2. *-un-lft-identity 45.8%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{\left(\frac{3}{2}\right)} \cdot {t}^{\left(\frac{3}{2}\right)}}{\color{blue}{1 \cdot \ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      3. times-frac 48.3%

         \[\leadsto \frac{2}{\frac{\color{blue}{\frac{{t}^{\left(\frac{3}{2}\right)}}{1} \cdot \frac{{t}^{\left(\frac{3}{2}\right)}}{\ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      4. metadata-eval 48.3%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{\color{blue}{1.5}}}{1} \cdot \frac{{t}^{\left(\frac{3}{2}\right)}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      5. metadata-eval 48.3%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{\color{blue}{1.5}}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   6. Applied egg-rr 48.3%

      \[\leadsto \frac{2}{\frac{\color{blue}{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]
   7. Step-by-step derivation

      1. add-cube-cbrt 48.1%

         \[\leadsto \frac{2}{\color{blue}{\left(\sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \cdot \sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right) \cdot \sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}}} \]

      2. pow3 48.1%

         \[\leadsto \frac{2}{\color{blue}{{\left(\sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right)}^{3}}} \]

   8. Applied egg-rr 79.0%

      \[\leadsto \frac{2}{\color{blue}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \sqrt[3]{\sin k \cdot \left(\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right)}^{3}}} \]

   9. Taylor expanded in k around 0 69.6%

      \[\leadsto \frac{2}{{\color{blue}{\left(\sqrt[3]{\frac{{k}^{2}}{{\ell}^{2}}} \cdot \left(t \cdot \sqrt[3]{2}\right)\right)}}^{3}} \]
   10. Step-by-step derivation

       1. *-commutative 69.6%

          \[\leadsto \frac{2}{{\color{blue}{\left(\left(t \cdot \sqrt[3]{2}\right) \cdot \sqrt[3]{\frac{{k}^{2}}{{\ell}^{2}}}\right)}}^{3}} \]

   11. Simplified 69.6%

       \[\leadsto \frac{2}{{\color{blue}{\left(\left(t \cdot \sqrt[3]{2}\right) \cdot \sqrt[3]{\frac{{k}^{2}}{{\ell}^{2}}}\right)}}^{3}} \]
   12. Step-by-step derivation

       1. pow2 69.6%

          \[\leadsto \frac{2}{{\left(\left(t \cdot \sqrt[3]{2}\right) \cdot \sqrt[3]{\frac{\color{blue}{k \cdot k}}{{\ell}^{2}}}\right)}^{3}} \]

       2. pow2 69.6%

          \[\leadsto \frac{2}{{\left(\left(t \cdot \sqrt[3]{2}\right) \cdot \sqrt[3]{\frac{k \cdot k}{\color{blue}{\ell \cdot \ell}}}\right)}^{3}} \]

       3. cbrt-div 74.2%

          \[\leadsto \frac{2}{{\left(\left(t \cdot \sqrt[3]{2}\right) \cdot \color{blue}{\frac{\sqrt[3]{k \cdot k}}{\sqrt[3]{\ell \cdot \ell}}}\right)}^{3}} \]

       4. cbrt-prod 80.0%

          \[\leadsto \frac{2}{{\left(\left(t \cdot \sqrt[3]{2}\right) \cdot \frac{\color{blue}{\sqrt[3]{k} \cdot \sqrt[3]{k}}}{\sqrt[3]{\ell \cdot \ell}}\right)}^{3}} \]

       5. pow2 80.0%

          \[\leadsto \frac{2}{{\left(\left(t \cdot \sqrt[3]{2}\right) \cdot \frac{\color{blue}{{\left(\sqrt[3]{k}\right)}^{2}}}{\sqrt[3]{\ell \cdot \ell}}\right)}^{3}} \]

       6. cbrt-prod 90.7%

          \[\leadsto \frac{2}{{\left(\left(t \cdot \sqrt[3]{2}\right) \cdot \frac{{\left(\sqrt[3]{k}\right)}^{2}}{\color{blue}{\sqrt[3]{\ell} \cdot \sqrt[3]{\ell}}}\right)}^{3}} \]

       7. unpow2 90.7%

          \[\leadsto \frac{2}{{\left(\left(t \cdot \sqrt[3]{2}\right) \cdot \frac{{\left(\sqrt[3]{k}\right)}^{2}}{\color{blue}{{\left(\sqrt[3]{\ell}\right)}^{2}}}\right)}^{3}} \]

   13. Applied egg-rr 90.7%

       \[\leadsto \frac{2}{{\left(\left(t \cdot \sqrt[3]{2}\right) \cdot \color{blue}{\frac{{\left(\sqrt[3]{k}\right)}^{2}}{{\left(\sqrt[3]{\ell}\right)}^{2}}}\right)}^{3}} \]

   if +inf.0 < (*.f64 (*.f64 (*.f64 (/.f64 (pow.f64 t #s(literal 3 binary64)) (*.f64 l l)) (sin.f64 k)) (tan.f64 k)) (+.f64 (+.f64 #s(literal 1 binary64) (pow.f64 (/.f64 k t) #s(literal 2 binary64))) #s(literal 1 binary64)))

   1. Initial program 0.0%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 0.0%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 0.0%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 0.0%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 24.7%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 24.7%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 24.7%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 24.7%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around inf 41.9%

      \[\leadsto \frac{2}{{\color{blue}{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t}{\cos k}}\right)}}^{2}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 44.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\left(1 + \left(1 + {\left(\frac{k}{t}\right)}^{2}\right)\right) \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t}^{3}}{\ell \cdot \ell}\right)\right) \leq 2 \cdot 10^{+299}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\\ \mathbf{elif}\;\left(1 + \left(1 + {\left(\frac{k}{t}\right)}^{2}\right)\right) \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t}^{3}}{\ell \cdot \ell}\right)\right) \leq \infty:\\ \;\;\;\;\frac{2}{{\left(\left(t \cdot \sqrt[3]{2}\right) \cdot \frac{{\left(\sqrt[3]{k}\right)}^{2}}{{\left(\sqrt[3]{\ell}\right)}^{2}}\right)}^{3}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t}{\cos k}}\right)}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 78.4% accurate, 0.3× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ \begin{array}{l} t_2 := {\left(\frac{k}{t\_m}\right)}^{2}\\ t_3 := \left(1 + \left(1 + t\_2\right)\right) \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t\_m}^{3}}{\ell \cdot \ell}\right)\right)\\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_3 \leq 2 \cdot 10^{+299}:\\ \;\;\;\;\frac{2}{\left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + t\_2\right)\right) \cdot \left(\frac{{t\_m}^{2}}{\ell} \cdot \frac{t\_m}{\ell}\right)}\\ \mathbf{elif}\;t\_3 \leq \infty:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t\_m}\right)\right) \cdot \left(k \cdot \frac{{t\_m}^{1.5}}{\ell}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t\_m}{\cos k}}\right)}^{2}}\\ \end{array} \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (let* ((t_2 (pow (/ k t_m) 2.0))
        (t_3
         (*
          (+ 1.0 (+ 1.0 t_2))
          (* (tan k) (* (sin k) (/ (pow t_m 3.0) (* l l)))))))
   (*
    t_s
    (if (<= t_3 2e+299)
      (/
       2.0
       (*
        (* (* (sin k) (tan k)) (+ 2.0 t_2))
        (* (/ (pow t_m 2.0) l) (/ t_m l))))
      (if (<= t_3 INFINITY)
        (/
         2.0
         (pow
          (* (hypot 1.0 (hypot 1.0 (/ k t_m))) (* k (/ (pow t_m 1.5) l)))
          2.0))
        (/ 2.0 (pow (* (/ (* k (sin k)) l) (sqrt (/ t_m (cos k)))) 2.0)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double t_2 = pow((k / t_m), 2.0);
	double t_3 = (1.0 + (1.0 + t_2)) * (tan(k) * (sin(k) * (pow(t_m, 3.0) / (l * l))));
	double tmp;
	if (t_3 <= 2e+299) {
		tmp = 2.0 / (((sin(k) * tan(k)) * (2.0 + t_2)) * ((pow(t_m, 2.0) / l) * (t_m / l)));
	} else if (t_3 <= ((double) INFINITY)) {
		tmp = 2.0 / pow((hypot(1.0, hypot(1.0, (k / t_m))) * (k * (pow(t_m, 1.5) / l))), 2.0);
	} else {
		tmp = 2.0 / pow((((k * sin(k)) / l) * sqrt((t_m / cos(k)))), 2.0);
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double t_2 = Math.pow((k / t_m), 2.0);
	double t_3 = (1.0 + (1.0 + t_2)) * (Math.tan(k) * (Math.sin(k) * (Math.pow(t_m, 3.0) / (l * l))));
	double tmp;
	if (t_3 <= 2e+299) {
		tmp = 2.0 / (((Math.sin(k) * Math.tan(k)) * (2.0 + t_2)) * ((Math.pow(t_m, 2.0) / l) * (t_m / l)));
	} else if (t_3 <= Double.POSITIVE_INFINITY) {
		tmp = 2.0 / Math.pow((Math.hypot(1.0, Math.hypot(1.0, (k / t_m))) * (k * (Math.pow(t_m, 1.5) / l))), 2.0);
	} else {
		tmp = 2.0 / Math.pow((((k * Math.sin(k)) / l) * Math.sqrt((t_m / Math.cos(k)))), 2.0);
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	t_2 = math.pow((k / t_m), 2.0)
	t_3 = (1.0 + (1.0 + t_2)) * (math.tan(k) * (math.sin(k) * (math.pow(t_m, 3.0) / (l * l))))
	tmp = 0
	if t_3 <= 2e+299:
		tmp = 2.0 / (((math.sin(k) * math.tan(k)) * (2.0 + t_2)) * ((math.pow(t_m, 2.0) / l) * (t_m / l)))
	elif t_3 <= math.inf:
		tmp = 2.0 / math.pow((math.hypot(1.0, math.hypot(1.0, (k / t_m))) * (k * (math.pow(t_m, 1.5) / l))), 2.0)
	else:
		tmp = 2.0 / math.pow((((k * math.sin(k)) / l) * math.sqrt((t_m / math.cos(k)))), 2.0)
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	t_2 = Float64(k / t_m) ^ 2.0
	t_3 = Float64(Float64(1.0 + Float64(1.0 + t_2)) * Float64(tan(k) * Float64(sin(k) * Float64((t_m ^ 3.0) / Float64(l * l)))))
	tmp = 0.0
	if (t_3 <= 2e+299)
		tmp = Float64(2.0 / Float64(Float64(Float64(sin(k) * tan(k)) * Float64(2.0 + t_2)) * Float64(Float64((t_m ^ 2.0) / l) * Float64(t_m / l))));
	elseif (t_3 <= Inf)
		tmp = Float64(2.0 / (Float64(hypot(1.0, hypot(1.0, Float64(k / t_m))) * Float64(k * Float64((t_m ^ 1.5) / l))) ^ 2.0));
	else
		tmp = Float64(2.0 / (Float64(Float64(Float64(k * sin(k)) / l) * sqrt(Float64(t_m / cos(k)))) ^ 2.0));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	t_2 = (k / t_m) ^ 2.0;
	t_3 = (1.0 + (1.0 + t_2)) * (tan(k) * (sin(k) * ((t_m ^ 3.0) / (l * l))));
	tmp = 0.0;
	if (t_3 <= 2e+299)
		tmp = 2.0 / (((sin(k) * tan(k)) * (2.0 + t_2)) * (((t_m ^ 2.0) / l) * (t_m / l)));
	elseif (t_3 <= Inf)
		tmp = 2.0 / ((hypot(1.0, hypot(1.0, (k / t_m))) * (k * ((t_m ^ 1.5) / l))) ^ 2.0);
	else
		tmp = 2.0 / ((((k * sin(k)) / l) * sqrt((t_m / cos(k)))) ^ 2.0);
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := Block[{t$95$2 = N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]}, Block[{t$95$3 = N[(N[(1.0 + N[(1.0 + t$95$2), $MachinePrecision]), $MachinePrecision] * N[(N[Tan[k], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[(N[Power[t$95$m, 3.0], $MachinePrecision] / N[(l * l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(t$95$s * If[LessEqual[t$95$3, 2e+299], N[(2.0 / N[(N[(N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision] * N[(2.0 + t$95$2), $MachinePrecision]), $MachinePrecision] * N[(N[(N[Power[t$95$m, 2.0], $MachinePrecision] / l), $MachinePrecision] * N[(t$95$m / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$3, Infinity], N[(2.0 / N[Power[N[(N[Sqrt[1.0 ^ 2 + N[Sqrt[1.0 ^ 2 + N[(k / t$95$m), $MachinePrecision] ^ 2], $MachinePrecision] ^ 2], $MachinePrecision] * N[(k * N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(2.0 / N[Power[N[(N[(N[(k * N[Sin[k], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[Sqrt[N[(t$95$m / N[Cos[k], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (*.f64 (*.f64 (*.f64 (/.f64 (pow.f64 t #s(literal 3 binary64)) (*.f64 l l)) (sin.f64 k)) (tan.f64 k)) (+.f64 (+.f64 #s(literal 1 binary64) (pow.f64 (/.f64 k t) #s(literal 2 binary64))) #s(literal 1 binary64))) < 2.0000000000000001e299

   1. Initial program 80.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 81.4%

      \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. associate-/r* 78.7%

         \[\leadsto \frac{2}{\color{blue}{\frac{{t}^{3}}{\ell \cdot \ell}} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      2. unpow3 78.6%

         \[\leadsto \frac{2}{\frac{\color{blue}{\left(t \cdot t\right) \cdot t}}{\ell \cdot \ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      3. times-frac 83.2%

         \[\leadsto \frac{2}{\color{blue}{\left(\frac{t \cdot t}{\ell} \cdot \frac{t}{\ell}\right)} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      4. pow2 83.2%

         \[\leadsto \frac{2}{\left(\frac{\color{blue}{{t}^{2}}}{\ell} \cdot \frac{t}{\ell}\right) \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   6. Applied egg-rr 83.2%

      \[\leadsto \frac{2}{\color{blue}{\left(\frac{{t}^{2}}{\ell} \cdot \frac{t}{\ell}\right)} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   if 2.0000000000000001e299 < (*.f64 (*.f64 (*.f64 (/.f64 (pow.f64 t #s(literal 3 binary64)) (*.f64 l l)) (sin.f64 k)) (tan.f64 k)) (+.f64 (+.f64 #s(literal 1 binary64) (pow.f64 (/.f64 k t) #s(literal 2 binary64))) #s(literal 1 binary64))) < +inf.0

   1. Initial program 70.2%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 70.2%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 70.2%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 70.2%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 59.9%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 59.9%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 59.9%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 59.9%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around 0 62.9%

      \[\leadsto \frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \color{blue}{k}\right)\right)}^{2}} \]

   if +inf.0 < (*.f64 (*.f64 (*.f64 (/.f64 (pow.f64 t #s(literal 3 binary64)) (*.f64 l l)) (sin.f64 k)) (tan.f64 k)) (+.f64 (+.f64 #s(literal 1 binary64) (pow.f64 (/.f64 k t) #s(literal 2 binary64))) #s(literal 1 binary64)))

   1. Initial program 0.0%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 0.0%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 0.0%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 0.0%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 24.7%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 24.7%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 24.7%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 24.7%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around inf 41.9%

      \[\leadsto \frac{2}{{\color{blue}{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t}{\cos k}}\right)}}^{2}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 65.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\left(1 + \left(1 + {\left(\frac{k}{t}\right)}^{2}\right)\right) \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t}^{3}}{\ell \cdot \ell}\right)\right) \leq 2 \cdot 10^{+299}:\\ \;\;\;\;\frac{2}{\left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right) \cdot \left(\frac{{t}^{2}}{\ell} \cdot \frac{t}{\ell}\right)}\\ \mathbf{elif}\;\left(1 + \left(1 + {\left(\frac{k}{t}\right)}^{2}\right)\right) \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t}^{3}}{\ell \cdot \ell}\right)\right) \leq \infty:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(k \cdot \frac{{t}^{1.5}}{\ell}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t}{\cos k}}\right)}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 77.7% accurate, 0.3× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ \begin{array}{l} t_2 := {\left(\frac{k}{t\_m}\right)}^{2}\\ t_3 := \left(1 + \left(1 + t\_2\right)\right) \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t\_m}^{3}}{\ell \cdot \ell}\right)\right)\\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_3 \leq 2 \cdot 10^{+299}:\\ \;\;\;\;\frac{2}{\left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + t\_2\right)\right) \cdot \frac{\frac{{t\_m}^{3}}{\ell}}{\ell}}\\ \mathbf{elif}\;t\_3 \leq \infty:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t\_m}\right)\right) \cdot \left(k \cdot \frac{{t\_m}^{1.5}}{\ell}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t\_m}{\cos k}}\right)}^{2}}\\ \end{array} \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (let* ((t_2 (pow (/ k t_m) 2.0))
        (t_3
         (*
          (+ 1.0 (+ 1.0 t_2))
          (* (tan k) (* (sin k) (/ (pow t_m 3.0) (* l l)))))))
   (*
    t_s
    (if (<= t_3 2e+299)
      (/ 2.0 (* (* (* (sin k) (tan k)) (+ 2.0 t_2)) (/ (/ (pow t_m 3.0) l) l)))
      (if (<= t_3 INFINITY)
        (/
         2.0
         (pow
          (* (hypot 1.0 (hypot 1.0 (/ k t_m))) (* k (/ (pow t_m 1.5) l)))
          2.0))
        (/ 2.0 (pow (* (/ (* k (sin k)) l) (sqrt (/ t_m (cos k)))) 2.0)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double t_2 = pow((k / t_m), 2.0);
	double t_3 = (1.0 + (1.0 + t_2)) * (tan(k) * (sin(k) * (pow(t_m, 3.0) / (l * l))));
	double tmp;
	if (t_3 <= 2e+299) {
		tmp = 2.0 / (((sin(k) * tan(k)) * (2.0 + t_2)) * ((pow(t_m, 3.0) / l) / l));
	} else if (t_3 <= ((double) INFINITY)) {
		tmp = 2.0 / pow((hypot(1.0, hypot(1.0, (k / t_m))) * (k * (pow(t_m, 1.5) / l))), 2.0);
	} else {
		tmp = 2.0 / pow((((k * sin(k)) / l) * sqrt((t_m / cos(k)))), 2.0);
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double t_2 = Math.pow((k / t_m), 2.0);
	double t_3 = (1.0 + (1.0 + t_2)) * (Math.tan(k) * (Math.sin(k) * (Math.pow(t_m, 3.0) / (l * l))));
	double tmp;
	if (t_3 <= 2e+299) {
		tmp = 2.0 / (((Math.sin(k) * Math.tan(k)) * (2.0 + t_2)) * ((Math.pow(t_m, 3.0) / l) / l));
	} else if (t_3 <= Double.POSITIVE_INFINITY) {
		tmp = 2.0 / Math.pow((Math.hypot(1.0, Math.hypot(1.0, (k / t_m))) * (k * (Math.pow(t_m, 1.5) / l))), 2.0);
	} else {
		tmp = 2.0 / Math.pow((((k * Math.sin(k)) / l) * Math.sqrt((t_m / Math.cos(k)))), 2.0);
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	t_2 = math.pow((k / t_m), 2.0)
	t_3 = (1.0 + (1.0 + t_2)) * (math.tan(k) * (math.sin(k) * (math.pow(t_m, 3.0) / (l * l))))
	tmp = 0
	if t_3 <= 2e+299:
		tmp = 2.0 / (((math.sin(k) * math.tan(k)) * (2.0 + t_2)) * ((math.pow(t_m, 3.0) / l) / l))
	elif t_3 <= math.inf:
		tmp = 2.0 / math.pow((math.hypot(1.0, math.hypot(1.0, (k / t_m))) * (k * (math.pow(t_m, 1.5) / l))), 2.0)
	else:
		tmp = 2.0 / math.pow((((k * math.sin(k)) / l) * math.sqrt((t_m / math.cos(k)))), 2.0)
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	t_2 = Float64(k / t_m) ^ 2.0
	t_3 = Float64(Float64(1.0 + Float64(1.0 + t_2)) * Float64(tan(k) * Float64(sin(k) * Float64((t_m ^ 3.0) / Float64(l * l)))))
	tmp = 0.0
	if (t_3 <= 2e+299)
		tmp = Float64(2.0 / Float64(Float64(Float64(sin(k) * tan(k)) * Float64(2.0 + t_2)) * Float64(Float64((t_m ^ 3.0) / l) / l)));
	elseif (t_3 <= Inf)
		tmp = Float64(2.0 / (Float64(hypot(1.0, hypot(1.0, Float64(k / t_m))) * Float64(k * Float64((t_m ^ 1.5) / l))) ^ 2.0));
	else
		tmp = Float64(2.0 / (Float64(Float64(Float64(k * sin(k)) / l) * sqrt(Float64(t_m / cos(k)))) ^ 2.0));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	t_2 = (k / t_m) ^ 2.0;
	t_3 = (1.0 + (1.0 + t_2)) * (tan(k) * (sin(k) * ((t_m ^ 3.0) / (l * l))));
	tmp = 0.0;
	if (t_3 <= 2e+299)
		tmp = 2.0 / (((sin(k) * tan(k)) * (2.0 + t_2)) * (((t_m ^ 3.0) / l) / l));
	elseif (t_3 <= Inf)
		tmp = 2.0 / ((hypot(1.0, hypot(1.0, (k / t_m))) * (k * ((t_m ^ 1.5) / l))) ^ 2.0);
	else
		tmp = 2.0 / ((((k * sin(k)) / l) * sqrt((t_m / cos(k)))) ^ 2.0);
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := Block[{t$95$2 = N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]}, Block[{t$95$3 = N[(N[(1.0 + N[(1.0 + t$95$2), $MachinePrecision]), $MachinePrecision] * N[(N[Tan[k], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[(N[Power[t$95$m, 3.0], $MachinePrecision] / N[(l * l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]}, N[(t$95$s * If[LessEqual[t$95$3, 2e+299], N[(2.0 / N[(N[(N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision] * N[(2.0 + t$95$2), $MachinePrecision]), $MachinePrecision] * N[(N[(N[Power[t$95$m, 3.0], $MachinePrecision] / l), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$3, Infinity], N[(2.0 / N[Power[N[(N[Sqrt[1.0 ^ 2 + N[Sqrt[1.0 ^ 2 + N[(k / t$95$m), $MachinePrecision] ^ 2], $MachinePrecision] ^ 2], $MachinePrecision] * N[(k * N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(2.0 / N[Power[N[(N[(N[(k * N[Sin[k], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[Sqrt[N[(t$95$m / N[Cos[k], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if (*.f64 (*.f64 (*.f64 (/.f64 (pow.f64 t #s(literal 3 binary64)) (*.f64 l l)) (sin.f64 k)) (tan.f64 k)) (+.f64 (+.f64 #s(literal 1 binary64) (pow.f64 (/.f64 k t) #s(literal 2 binary64))) #s(literal 1 binary64))) < 2.0000000000000001e299

   1. Initial program 80.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 81.4%

      \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
   4. Add Preprocessing

   if 2.0000000000000001e299 < (*.f64 (*.f64 (*.f64 (/.f64 (pow.f64 t #s(literal 3 binary64)) (*.f64 l l)) (sin.f64 k)) (tan.f64 k)) (+.f64 (+.f64 #s(literal 1 binary64) (pow.f64 (/.f64 k t) #s(literal 2 binary64))) #s(literal 1 binary64))) < +inf.0

   1. Initial program 70.2%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 70.2%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 70.2%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 70.2%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 59.9%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 59.9%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 59.9%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 59.9%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around 0 62.9%

      \[\leadsto \frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \color{blue}{k}\right)\right)}^{2}} \]

   if +inf.0 < (*.f64 (*.f64 (*.f64 (/.f64 (pow.f64 t #s(literal 3 binary64)) (*.f64 l l)) (sin.f64 k)) (tan.f64 k)) (+.f64 (+.f64 #s(literal 1 binary64) (pow.f64 (/.f64 k t) #s(literal 2 binary64))) #s(literal 1 binary64)))

   1. Initial program 0.0%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 0.0%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 0.0%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 0.0%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 24.7%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 24.7%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 24.7%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 24.7%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around inf 41.9%

      \[\leadsto \frac{2}{{\color{blue}{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t}{\cos k}}\right)}}^{2}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 64.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\left(1 + \left(1 + {\left(\frac{k}{t}\right)}^{2}\right)\right) \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t}^{3}}{\ell \cdot \ell}\right)\right) \leq 2 \cdot 10^{+299}:\\ \;\;\;\;\frac{2}{\left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right) \cdot \frac{\frac{{t}^{3}}{\ell}}{\ell}}\\ \mathbf{elif}\;\left(1 + \left(1 + {\left(\frac{k}{t}\right)}^{2}\right)\right) \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t}^{3}}{\ell \cdot \ell}\right)\right) \leq \infty:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(k \cdot \frac{{t}^{1.5}}{\ell}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t}{\cos k}}\right)}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 89.2% accurate, 0.5× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 7.2 \cdot 10^{-213}:\\ \;\;\;\;\frac{2}{{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t\_m}{\cos k}}\right)}^{2}}\\ \mathbf{elif}\;t\_m \leq 4.7 \cdot 10^{-64}:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t\_m}\right)\right) \cdot \frac{{t\_m}^{1.5}}{\ell}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t\_m}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right) \cdot \frac{t\_m}{{\left(\sqrt[3]{\ell}\right)}^{2}}\right)}^{3}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= t_m 7.2e-213)
    (/ 2.0 (pow (* (/ (* k (sin k)) l) (sqrt (/ t_m (cos k)))) 2.0))
    (if (<= t_m 4.7e-64)
      (/
       2.0
       (*
        (pow (* (hypot 1.0 (hypot 1.0 (/ k t_m))) (/ (pow t_m 1.5) l)) 2.0)
        (* (sin k) (tan k))))
      (/
       2.0
       (pow
        (*
         (* (cbrt (* (tan k) (+ 2.0 (pow (/ k t_m) 2.0)))) (cbrt (sin k)))
         (/ t_m (pow (cbrt l) 2.0)))
        3.0))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 7.2e-213) {
		tmp = 2.0 / pow((((k * sin(k)) / l) * sqrt((t_m / cos(k)))), 2.0);
	} else if (t_m <= 4.7e-64) {
		tmp = 2.0 / (pow((hypot(1.0, hypot(1.0, (k / t_m))) * (pow(t_m, 1.5) / l)), 2.0) * (sin(k) * tan(k)));
	} else {
		tmp = 2.0 / pow(((cbrt((tan(k) * (2.0 + pow((k / t_m), 2.0)))) * cbrt(sin(k))) * (t_m / pow(cbrt(l), 2.0))), 3.0);
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 7.2e-213) {
		tmp = 2.0 / Math.pow((((k * Math.sin(k)) / l) * Math.sqrt((t_m / Math.cos(k)))), 2.0);
	} else if (t_m <= 4.7e-64) {
		tmp = 2.0 / (Math.pow((Math.hypot(1.0, Math.hypot(1.0, (k / t_m))) * (Math.pow(t_m, 1.5) / l)), 2.0) * (Math.sin(k) * Math.tan(k)));
	} else {
		tmp = 2.0 / Math.pow(((Math.cbrt((Math.tan(k) * (2.0 + Math.pow((k / t_m), 2.0)))) * Math.cbrt(Math.sin(k))) * (t_m / Math.pow(Math.cbrt(l), 2.0))), 3.0);
	}
	return t_s * tmp;
}

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (t_m <= 7.2e-213)
		tmp = Float64(2.0 / (Float64(Float64(Float64(k * sin(k)) / l) * sqrt(Float64(t_m / cos(k)))) ^ 2.0));
	elseif (t_m <= 4.7e-64)
		tmp = Float64(2.0 / Float64((Float64(hypot(1.0, hypot(1.0, Float64(k / t_m))) * Float64((t_m ^ 1.5) / l)) ^ 2.0) * Float64(sin(k) * tan(k))));
	else
		tmp = Float64(2.0 / (Float64(Float64(cbrt(Float64(tan(k) * Float64(2.0 + (Float64(k / t_m) ^ 2.0)))) * cbrt(sin(k))) * Float64(t_m / (cbrt(l) ^ 2.0))) ^ 3.0));
	end
	return Float64(t_s * tmp)
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[t$95$m, 7.2e-213], N[(2.0 / N[Power[N[(N[(N[(k * N[Sin[k], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[Sqrt[N[(t$95$m / N[Cos[k], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$m, 4.7e-64], N[(2.0 / N[(N[Power[N[(N[Sqrt[1.0 ^ 2 + N[Sqrt[1.0 ^ 2 + N[(k / t$95$m), $MachinePrecision] ^ 2], $MachinePrecision] ^ 2], $MachinePrecision] * N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[Power[N[(N[(N[Power[N[(N[Tan[k], $MachinePrecision] * N[(2.0 + N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision] * N[Power[N[Sin[k], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision] * N[(t$95$m / N[Power[N[Power[l, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if t < 7.2000000000000002e-213

   1. Initial program 49.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 49.4%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 10.2%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 10.2%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 4.1%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 4.1%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 4.1%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 4.1%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around inf 23.0%

      \[\leadsto \frac{2}{{\color{blue}{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t}{\cos k}}\right)}}^{2}} \]

   if 7.2000000000000002e-213 < t < 4.6999999999999998e-64

   1. Initial program 38.6%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 38.6%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 33.1%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 33.1%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 67.4%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 67.4%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 67.4%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 67.4%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   7. Step-by-step derivation

      1. associate-*r* 67.3%

         \[\leadsto \frac{2}{{\color{blue}{\left(\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right) \cdot \sqrt{\sin k \cdot \tan k}\right)}}^{2}} \]

      2. unpow-prod-down 67.3%

         \[\leadsto \frac{2}{\color{blue}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot {\left(\sqrt{\sin k \cdot \tan k}\right)}^{2}}} \]

      3. pow2 67.3%

         \[\leadsto \frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \color{blue}{\left(\sqrt{\sin k \cdot \tan k} \cdot \sqrt{\sin k \cdot \tan k}\right)}} \]

      4. add-sqr-sqrt 88.0%

         \[\leadsto \frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \color{blue}{\left(\sin k \cdot \tan k\right)}} \]

   8. Applied egg-rr 88.0%

      \[\leadsto \frac{2}{\color{blue}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}} \]

   if 4.6999999999999998e-64 < t

   1. Initial program 58.6%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 62.0%

      \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. sqr-pow 62.0%

         \[\leadsto \frac{2}{\frac{\frac{\color{blue}{{t}^{\left(\frac{3}{2}\right)} \cdot {t}^{\left(\frac{3}{2}\right)}}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      2. *-un-lft-identity 62.0%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{\left(\frac{3}{2}\right)} \cdot {t}^{\left(\frac{3}{2}\right)}}{\color{blue}{1 \cdot \ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      3. times-frac 70.0%

         \[\leadsto \frac{2}{\frac{\color{blue}{\frac{{t}^{\left(\frac{3}{2}\right)}}{1} \cdot \frac{{t}^{\left(\frac{3}{2}\right)}}{\ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      4. metadata-eval 70.0%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{\color{blue}{1.5}}}{1} \cdot \frac{{t}^{\left(\frac{3}{2}\right)}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      5. metadata-eval 70.0%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{\color{blue}{1.5}}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   6. Applied egg-rr 70.0%

      \[\leadsto \frac{2}{\frac{\color{blue}{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]
   7. Step-by-step derivation

      1. add-cube-cbrt 69.7%

         \[\leadsto \frac{2}{\color{blue}{\left(\sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \cdot \sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right) \cdot \sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}}} \]

      2. pow3 69.7%

         \[\leadsto \frac{2}{\color{blue}{{\left(\sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right)}^{3}}} \]

   8. Applied egg-rr 81.3%

      \[\leadsto \frac{2}{\color{blue}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \sqrt[3]{\sin k \cdot \left(\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right)}^{3}}} \]
   9. Step-by-step derivation

      1. *-commutative 81.3%

         \[\leadsto \frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \sqrt[3]{\color{blue}{\left(\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right) \cdot \sin k}}\right)}^{3}} \]

      2. cbrt-prod 90.5%

         \[\leadsto \frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \color{blue}{\left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)}\right)}^{3}} \]

   10. Applied egg-rr 90.5%

       \[\leadsto \frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \color{blue}{\left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)}\right)}^{3}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 50.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 7.2 \cdot 10^{-213}:\\ \;\;\;\;\frac{2}{{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t}{\cos k}}\right)}^{2}}\\ \mathbf{elif}\;t \leq 4.7 \cdot 10^{-64}:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right) \cdot \frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}}\right)}^{3}}\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 85.7% accurate, 0.6× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ \begin{array}{l} t_2 := \sqrt[3]{\frac{1}{\ell}}\\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 1.12 \cdot 10^{-212}:\\ \;\;\;\;\frac{2}{{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t\_m}{\cos k}}\right)}^{2}}\\ \mathbf{elif}\;t\_m \leq 6.5 \cdot 10^{+37}:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t\_m}\right)\right) \cdot \frac{{t\_m}^{1.5}}{\ell}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\tan k \cdot {\left(\sqrt[3]{\sin k} \cdot \left(t\_m \cdot \left(t\_2 \cdot t\_2\right)\right)\right)}^{3}\right) \cdot \left(1 + \left(1 + {\left(\frac{k}{t\_m}\right)}^{2}\right)\right)}\\ \end{array} \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (let* ((t_2 (cbrt (/ 1.0 l))))
   (*
    t_s
    (if (<= t_m 1.12e-212)
      (/ 2.0 (pow (* (/ (* k (sin k)) l) (sqrt (/ t_m (cos k)))) 2.0))
      (if (<= t_m 6.5e+37)
        (/
         2.0
         (*
          (pow (* (hypot 1.0 (hypot 1.0 (/ k t_m))) (/ (pow t_m 1.5) l)) 2.0)
          (* (sin k) (tan k))))
        (/
         2.0
         (*
          (* (tan k) (pow (* (cbrt (sin k)) (* t_m (* t_2 t_2))) 3.0))
          (+ 1.0 (+ 1.0 (pow (/ k t_m) 2.0))))))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double t_2 = cbrt((1.0 / l));
	double tmp;
	if (t_m <= 1.12e-212) {
		tmp = 2.0 / pow((((k * sin(k)) / l) * sqrt((t_m / cos(k)))), 2.0);
	} else if (t_m <= 6.5e+37) {
		tmp = 2.0 / (pow((hypot(1.0, hypot(1.0, (k / t_m))) * (pow(t_m, 1.5) / l)), 2.0) * (sin(k) * tan(k)));
	} else {
		tmp = 2.0 / ((tan(k) * pow((cbrt(sin(k)) * (t_m * (t_2 * t_2))), 3.0)) * (1.0 + (1.0 + pow((k / t_m), 2.0))));
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double t_2 = Math.cbrt((1.0 / l));
	double tmp;
	if (t_m <= 1.12e-212) {
		tmp = 2.0 / Math.pow((((k * Math.sin(k)) / l) * Math.sqrt((t_m / Math.cos(k)))), 2.0);
	} else if (t_m <= 6.5e+37) {
		tmp = 2.0 / (Math.pow((Math.hypot(1.0, Math.hypot(1.0, (k / t_m))) * (Math.pow(t_m, 1.5) / l)), 2.0) * (Math.sin(k) * Math.tan(k)));
	} else {
		tmp = 2.0 / ((Math.tan(k) * Math.pow((Math.cbrt(Math.sin(k)) * (t_m * (t_2 * t_2))), 3.0)) * (1.0 + (1.0 + Math.pow((k / t_m), 2.0))));
	}
	return t_s * tmp;
}

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	t_2 = cbrt(Float64(1.0 / l))
	tmp = 0.0
	if (t_m <= 1.12e-212)
		tmp = Float64(2.0 / (Float64(Float64(Float64(k * sin(k)) / l) * sqrt(Float64(t_m / cos(k)))) ^ 2.0));
	elseif (t_m <= 6.5e+37)
		tmp = Float64(2.0 / Float64((Float64(hypot(1.0, hypot(1.0, Float64(k / t_m))) * Float64((t_m ^ 1.5) / l)) ^ 2.0) * Float64(sin(k) * tan(k))));
	else
		tmp = Float64(2.0 / Float64(Float64(tan(k) * (Float64(cbrt(sin(k)) * Float64(t_m * Float64(t_2 * t_2))) ^ 3.0)) * Float64(1.0 + Float64(1.0 + (Float64(k / t_m) ^ 2.0)))));
	end
	return Float64(t_s * tmp)
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := Block[{t$95$2 = N[Power[N[(1.0 / l), $MachinePrecision], 1/3], $MachinePrecision]}, N[(t$95$s * If[LessEqual[t$95$m, 1.12e-212], N[(2.0 / N[Power[N[(N[(N[(k * N[Sin[k], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[Sqrt[N[(t$95$m / N[Cos[k], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$m, 6.5e+37], N[(2.0 / N[(N[Power[N[(N[Sqrt[1.0 ^ 2 + N[Sqrt[1.0 ^ 2 + N[(k / t$95$m), $MachinePrecision] ^ 2], $MachinePrecision] ^ 2], $MachinePrecision] * N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[Tan[k], $MachinePrecision] * N[Power[N[(N[Power[N[Sin[k], $MachinePrecision], 1/3], $MachinePrecision] * N[(t$95$m * N[(t$95$2 * t$95$2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision] * N[(1.0 + N[(1.0 + N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]

Derivation

1. Split input into 3 regimes

2. if t < 1.12e-212

   1. Initial program 49.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 49.4%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 10.2%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 10.2%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 4.1%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 4.1%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 4.1%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 4.1%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around inf 23.0%

      \[\leadsto \frac{2}{{\color{blue}{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t}{\cos k}}\right)}}^{2}} \]

   if 1.12e-212 < t < 6.4999999999999998e37

   1. Initial program 48.6%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 48.6%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 38.0%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 38.0%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 62.4%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 62.4%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 62.4%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 62.4%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   7. Step-by-step derivation

      1. associate-*r* 62.3%

         \[\leadsto \frac{2}{{\color{blue}{\left(\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right) \cdot \sqrt{\sin k \cdot \tan k}\right)}}^{2}} \]

      2. unpow-prod-down 60.7%

         \[\leadsto \frac{2}{\color{blue}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot {\left(\sqrt{\sin k \cdot \tan k}\right)}^{2}}} \]

      3. pow2 60.7%

         \[\leadsto \frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \color{blue}{\left(\sqrt{\sin k \cdot \tan k} \cdot \sqrt{\sin k \cdot \tan k}\right)}} \]

      4. add-sqr-sqrt 85.8%

         \[\leadsto \frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \color{blue}{\left(\sin k \cdot \tan k\right)}} \]

   8. Applied egg-rr 85.8%

      \[\leadsto \frac{2}{\color{blue}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}} \]

   if 6.4999999999999998e37 < t

   1. Initial program 56.2%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. add-cube-cbrt 56.2%

         \[\leadsto \frac{2}{\left(\color{blue}{\left(\left(\sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k} \cdot \sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k}\right) \cdot \sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k}\right)} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      2. pow3 56.2%

         \[\leadsto \frac{2}{\left(\color{blue}{{\left(\sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k}\right)}^{3}} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      3. associate-/r* 65.2%

         \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\color{blue}{\frac{\frac{{t}^{3}}{\ell}}{\ell}} \cdot \sin k}\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      4. *-commutative 65.2%

         \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\color{blue}{\sin k \cdot \frac{\frac{{t}^{3}}{\ell}}{\ell}}}\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      5. cbrt-prod 65.2%

         \[\leadsto \frac{2}{\left({\color{blue}{\left(\sqrt[3]{\sin k} \cdot \sqrt[3]{\frac{\frac{{t}^{3}}{\ell}}{\ell}}\right)}}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      6. associate-/r* 56.2%

         \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \sqrt[3]{\color{blue}{\frac{{t}^{3}}{\ell \cdot \ell}}}\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      7. div-inv 56.2%

         \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \sqrt[3]{\color{blue}{{t}^{3} \cdot \frac{1}{\ell \cdot \ell}}}\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      8. cbrt-prod 56.2%

         \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \color{blue}{\left(\sqrt[3]{{t}^{3}} \cdot \sqrt[3]{\frac{1}{\ell \cdot \ell}}\right)}\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      9. rem-cbrt-cube 66.2%

         \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \left(\color{blue}{t} \cdot \sqrt[3]{\frac{1}{\ell \cdot \ell}}\right)\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      10. pow2 66.2%

          \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \left(t \cdot \sqrt[3]{\frac{1}{\color{blue}{{\ell}^{2}}}}\right)\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      11. pow-flip 67.6%

          \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \left(t \cdot \sqrt[3]{\color{blue}{{\ell}^{\left(-2\right)}}}\right)\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      12. metadata-eval 67.6%

          \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \left(t \cdot \sqrt[3]{{\ell}^{\color{blue}{-2}}}\right)\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   4. Applied egg-rr 67.6%

      \[\leadsto \frac{2}{\left(\color{blue}{{\left(\sqrt[3]{\sin k} \cdot \left(t \cdot \sqrt[3]{{\ell}^{-2}}\right)\right)}^{3}} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   5. Step-by-step derivation

      1. pow1/3 66.7%

         \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \left(t \cdot \color{blue}{{\left({\ell}^{-2}\right)}^{0.3333333333333333}}\right)\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      2. sqr-pow 66.7%

         \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \left(t \cdot {\color{blue}{\left({\ell}^{\left(\frac{-2}{2}\right)} \cdot {\ell}^{\left(\frac{-2}{2}\right)}\right)}}^{0.3333333333333333}\right)\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      3. metadata-eval 66.7%

         \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \left(t \cdot {\left({\ell}^{\color{blue}{-1}} \cdot {\ell}^{\left(\frac{-2}{2}\right)}\right)}^{0.3333333333333333}\right)\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      4. inv-pow 66.7%

         \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \left(t \cdot {\left(\color{blue}{\frac{1}{\ell}} \cdot {\ell}^{\left(\frac{-2}{2}\right)}\right)}^{0.3333333333333333}\right)\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      5. metadata-eval 66.7%

         \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \left(t \cdot {\left(\frac{1}{\ell} \cdot {\ell}^{\color{blue}{-1}}\right)}^{0.3333333333333333}\right)\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      6. inv-pow 66.7%

         \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \left(t \cdot {\left(\frac{1}{\ell} \cdot \color{blue}{\frac{1}{\ell}}\right)}^{0.3333333333333333}\right)\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      7. unpow-prod-down 53.1%

         \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \left(t \cdot \color{blue}{\left({\left(\frac{1}{\ell}\right)}^{0.3333333333333333} \cdot {\left(\frac{1}{\ell}\right)}^{0.3333333333333333}\right)}\right)\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   6. Applied egg-rr 53.1%

      \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \left(t \cdot \color{blue}{\left({\left(\frac{1}{\ell}\right)}^{0.3333333333333333} \cdot {\left(\frac{1}{\ell}\right)}^{0.3333333333333333}\right)}\right)\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   7. Step-by-step derivation

      1. unpow1/3 53.4%

         \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \left(t \cdot \left(\color{blue}{\sqrt[3]{\frac{1}{\ell}}} \cdot {\left(\frac{1}{\ell}\right)}^{0.3333333333333333}\right)\right)\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

      2. unpow1/3 91.1%

         \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \left(t \cdot \left(\sqrt[3]{\frac{1}{\ell}} \cdot \color{blue}{\sqrt[3]{\frac{1}{\ell}}}\right)\right)\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   8. Simplified 91.1%

      \[\leadsto \frac{2}{\left({\left(\sqrt[3]{\sin k} \cdot \left(t \cdot \color{blue}{\left(\sqrt[3]{\frac{1}{\ell}} \cdot \sqrt[3]{\frac{1}{\ell}}\right)}\right)\right)}^{3} \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
3. Recombined 3 regimes into one program.

4. Final simplification 49.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 1.12 \cdot 10^{-212}:\\ \;\;\;\;\frac{2}{{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t}{\cos k}}\right)}^{2}}\\ \mathbf{elif}\;t \leq 6.5 \cdot 10^{+37}:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\tan k \cdot {\left(\sqrt[3]{\sin k} \cdot \left(t \cdot \left(\sqrt[3]{\frac{1}{\ell}} \cdot \sqrt[3]{\frac{1}{\ell}}\right)\right)\right)}^{3}\right) \cdot \left(1 + \left(1 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 76.3% accurate, 0.7× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 9.5 \cdot 10^{-99}:\\ \;\;\;\;\frac{2}{{\left(\frac{t\_m}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \left(\sqrt[3]{\sin k} \cdot \sqrt[3]{2 \cdot k}\right)\right)}^{3}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t\_m}\right)\right) \cdot \frac{{t\_m}^{1.5}}{\ell}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 9.5e-99)
    (/
     2.0
     (pow
      (* (/ t_m (pow (cbrt l) 2.0)) (* (cbrt (sin k)) (cbrt (* 2.0 k))))
      3.0))
    (/
     2.0
     (*
      (pow (* (hypot 1.0 (hypot 1.0 (/ k t_m))) (/ (pow t_m 1.5) l)) 2.0)
      (* (sin k) (tan k)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 9.5e-99) {
		tmp = 2.0 / pow(((t_m / pow(cbrt(l), 2.0)) * (cbrt(sin(k)) * cbrt((2.0 * k)))), 3.0);
	} else {
		tmp = 2.0 / (pow((hypot(1.0, hypot(1.0, (k / t_m))) * (pow(t_m, 1.5) / l)), 2.0) * (sin(k) * tan(k)));
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 9.5e-99) {
		tmp = 2.0 / Math.pow(((t_m / Math.pow(Math.cbrt(l), 2.0)) * (Math.cbrt(Math.sin(k)) * Math.cbrt((2.0 * k)))), 3.0);
	} else {
		tmp = 2.0 / (Math.pow((Math.hypot(1.0, Math.hypot(1.0, (k / t_m))) * (Math.pow(t_m, 1.5) / l)), 2.0) * (Math.sin(k) * Math.tan(k)));
	}
	return t_s * tmp;
}

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 9.5e-99)
		tmp = Float64(2.0 / (Float64(Float64(t_m / (cbrt(l) ^ 2.0)) * Float64(cbrt(sin(k)) * cbrt(Float64(2.0 * k)))) ^ 3.0));
	else
		tmp = Float64(2.0 / Float64((Float64(hypot(1.0, hypot(1.0, Float64(k / t_m))) * Float64((t_m ^ 1.5) / l)) ^ 2.0) * Float64(sin(k) * tan(k))));
	end
	return Float64(t_s * tmp)
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 9.5e-99], N[(2.0 / N[Power[N[(N[(t$95$m / N[Power[N[Power[l, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] * N[(N[Power[N[Sin[k], $MachinePrecision], 1/3], $MachinePrecision] * N[Power[N[(2.0 * k), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[Power[N[(N[Sqrt[1.0 ^ 2 + N[Sqrt[1.0 ^ 2 + N[(k / t$95$m), $MachinePrecision] ^ 2], $MachinePrecision] ^ 2], $MachinePrecision] * N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 9.5000000000000008e-99

   1. Initial program 49.5%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 54.4%

      \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. sqr-pow 25.9%

         \[\leadsto \frac{2}{\frac{\frac{\color{blue}{{t}^{\left(\frac{3}{2}\right)} \cdot {t}^{\left(\frac{3}{2}\right)}}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      2. *-un-lft-identity 25.9%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{\left(\frac{3}{2}\right)} \cdot {t}^{\left(\frac{3}{2}\right)}}{\color{blue}{1 \cdot \ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      3. times-frac 29.5%

         \[\leadsto \frac{2}{\frac{\color{blue}{\frac{{t}^{\left(\frac{3}{2}\right)}}{1} \cdot \frac{{t}^{\left(\frac{3}{2}\right)}}{\ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      4. metadata-eval 29.5%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{\color{blue}{1.5}}}{1} \cdot \frac{{t}^{\left(\frac{3}{2}\right)}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      5. metadata-eval 29.5%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{\color{blue}{1.5}}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   6. Applied egg-rr 29.5%

      \[\leadsto \frac{2}{\frac{\color{blue}{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]
   7. Step-by-step derivation

      1. add-cube-cbrt 29.4%

         \[\leadsto \frac{2}{\color{blue}{\left(\sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \cdot \sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right) \cdot \sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}}} \]

      2. pow3 29.4%

         \[\leadsto \frac{2}{\color{blue}{{\left(\sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right)}^{3}}} \]

   8. Applied egg-rr 73.7%

      \[\leadsto \frac{2}{\color{blue}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \sqrt[3]{\sin k \cdot \left(\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right)}^{3}}} \]
   9. Step-by-step derivation

      1. *-commutative 73.7%

         \[\leadsto \frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \sqrt[3]{\color{blue}{\left(\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right) \cdot \sin k}}\right)}^{3}} \]

      2. cbrt-prod 82.4%

         \[\leadsto \frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \color{blue}{\left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)}\right)}^{3}} \]

   10. Applied egg-rr 82.4%

       \[\leadsto \frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \color{blue}{\left(\sqrt[3]{\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)} \cdot \sqrt[3]{\sin k}\right)}\right)}^{3}} \]

   11. Taylor expanded in k around 0 65.1%

       \[\leadsto \frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \left(\sqrt[3]{\color{blue}{2 \cdot k}} \cdot \sqrt[3]{\sin k}\right)\right)}^{3}} \]

   if 9.5000000000000008e-99 < k

   1. Initial program 52.9%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 52.9%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 23.5%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 23.5%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 21.6%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 21.6%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 21.6%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 21.6%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   7. Step-by-step derivation

      1. associate-*r* 21.6%

         \[\leadsto \frac{2}{{\color{blue}{\left(\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right) \cdot \sqrt{\sin k \cdot \tan k}\right)}}^{2}} \]

      2. unpow-prod-down 21.7%

         \[\leadsto \frac{2}{\color{blue}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot {\left(\sqrt{\sin k \cdot \tan k}\right)}^{2}}} \]

      3. pow2 21.7%

         \[\leadsto \frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \color{blue}{\left(\sqrt{\sin k \cdot \tan k} \cdot \sqrt{\sin k \cdot \tan k}\right)}} \]

      4. add-sqr-sqrt 32.1%

         \[\leadsto \frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \color{blue}{\left(\sin k \cdot \tan k\right)}} \]

   8. Applied egg-rr 32.1%

      \[\leadsto \frac{2}{\color{blue}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 54.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 9.5 \cdot 10^{-99}:\\ \;\;\;\;\frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \left(\sqrt[3]{\sin k} \cdot \sqrt[3]{2 \cdot k}\right)\right)}^{3}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 74.8% accurate, 0.8× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 350:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t\_m}\right)\right) \cdot \left(k \cdot \frac{{t\_m}^{1.5}}{\ell}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;{\ell}^{2} \cdot \frac{\frac{\frac{2 \cdot \cos k}{t\_m}}{{\sin k}^{2}}}{{k}^{2}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 350.0)
    (/
     2.0
     (pow (* (hypot 1.0 (hypot 1.0 (/ k t_m))) (* k (/ (pow t_m 1.5) l))) 2.0))
    (*
     (pow l 2.0)
     (/ (/ (/ (* 2.0 (cos k)) t_m) (pow (sin k) 2.0)) (pow k 2.0))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 350.0) {
		tmp = 2.0 / pow((hypot(1.0, hypot(1.0, (k / t_m))) * (k * (pow(t_m, 1.5) / l))), 2.0);
	} else {
		tmp = pow(l, 2.0) * ((((2.0 * cos(k)) / t_m) / pow(sin(k), 2.0)) / pow(k, 2.0));
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 350.0) {
		tmp = 2.0 / Math.pow((Math.hypot(1.0, Math.hypot(1.0, (k / t_m))) * (k * (Math.pow(t_m, 1.5) / l))), 2.0);
	} else {
		tmp = Math.pow(l, 2.0) * ((((2.0 * Math.cos(k)) / t_m) / Math.pow(Math.sin(k), 2.0)) / Math.pow(k, 2.0));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 350.0:
		tmp = 2.0 / math.pow((math.hypot(1.0, math.hypot(1.0, (k / t_m))) * (k * (math.pow(t_m, 1.5) / l))), 2.0)
	else:
		tmp = math.pow(l, 2.0) * ((((2.0 * math.cos(k)) / t_m) / math.pow(math.sin(k), 2.0)) / math.pow(k, 2.0))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 350.0)
		tmp = Float64(2.0 / (Float64(hypot(1.0, hypot(1.0, Float64(k / t_m))) * Float64(k * Float64((t_m ^ 1.5) / l))) ^ 2.0));
	else
		tmp = Float64((l ^ 2.0) * Float64(Float64(Float64(Float64(2.0 * cos(k)) / t_m) / (sin(k) ^ 2.0)) / (k ^ 2.0)));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 350.0)
		tmp = 2.0 / ((hypot(1.0, hypot(1.0, (k / t_m))) * (k * ((t_m ^ 1.5) / l))) ^ 2.0);
	else
		tmp = (l ^ 2.0) * ((((2.0 * cos(k)) / t_m) / (sin(k) ^ 2.0)) / (k ^ 2.0));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 350.0], N[(2.0 / N[Power[N[(N[Sqrt[1.0 ^ 2 + N[Sqrt[1.0 ^ 2 + N[(k / t$95$m), $MachinePrecision] ^ 2], $MachinePrecision] ^ 2], $MachinePrecision] * N[(k * N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(N[Power[l, 2.0], $MachinePrecision] * N[(N[(N[(N[(2.0 * N[Cos[k], $MachinePrecision]), $MachinePrecision] / t$95$m), $MachinePrecision] / N[Power[N[Sin[k], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] / N[Power[k, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 350

   1. Initial program 50.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 50.4%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 20.7%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 20.7%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 30.1%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 30.1%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 30.1%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 30.1%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around 0 33.5%

      \[\leadsto \frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \color{blue}{k}\right)\right)}^{2}} \]

   if 350 < k

   1. Initial program 51.0%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 50.9%

      \[\leadsto \color{blue}{\frac{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}{2 + {\left(\frac{k}{t}\right)}^{2}}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. add-cube-cbrt 50.8%

         \[\leadsto \frac{\color{blue}{\left(\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}\right) \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

      2. add-sqr-sqrt 50.8%

         \[\leadsto \frac{\left(\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}\right) \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\color{blue}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}} \cdot \sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

      3. times-frac 50.8%

         \[\leadsto \color{blue}{\frac{\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}} \cdot \frac{\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

   6. Applied egg-rr 58.8%

      \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}} \cdot \frac{\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]
   7. Step-by-step derivation

      1. associate-*l/ 58.8%

         \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2} \cdot \frac{\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

   8. Simplified 65.9%

      \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{{\ell}^{2} \cdot \frac{2}{\tan k}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2} \cdot \frac{\frac{\sqrt[3]{{\ell}^{2} \cdot \frac{2}{\tan k}}}{t \cdot \sqrt[3]{\sin k}}}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}} \]

   9. Taylor expanded in k around inf 66.6%

      \[\leadsto \color{blue}{\frac{{\ell}^{2} \cdot \left({\left(\sqrt[3]{2}\right)}^{3} \cdot \cos k\right)}{{k}^{2} \cdot \left(t \cdot {\sin k}^{2}\right)}} \]
   10. Step-by-step derivation

       1. times-frac 61.3%

          \[\leadsto \color{blue}{\frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{{\left(\sqrt[3]{2}\right)}^{3} \cdot \cos k}{t \cdot {\sin k}^{2}}} \]

       2. rem-cube-cbrt 61.5%

          \[\leadsto \frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{\color{blue}{2} \cdot \cos k}{t \cdot {\sin k}^{2}} \]

   11. Simplified 61.5%

       \[\leadsto \color{blue}{\frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}}} \]
   12. Step-by-step derivation

       1. associate-*l/ 64.7%

          \[\leadsto \color{blue}{\frac{{\ell}^{2} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}}}{{k}^{2}}} \]

       2. associate-/r* 64.7%

          \[\leadsto \frac{{\ell}^{2} \cdot \color{blue}{\frac{\frac{2 \cdot \cos k}{t}}{{\sin k}^{2}}}}{{k}^{2}} \]

   13. Applied egg-rr 64.7%

       \[\leadsto \color{blue}{\frac{{\ell}^{2} \cdot \frac{\frac{2 \cdot \cos k}{t}}{{\sin k}^{2}}}{{k}^{2}}} \]
   14. Step-by-step derivation

       1. associate-/l* 67.0%

          \[\leadsto \color{blue}{{\ell}^{2} \cdot \frac{\frac{\frac{2 \cdot \cos k}{t}}{{\sin k}^{2}}}{{k}^{2}}} \]

   15. Simplified 67.0%

       \[\leadsto \color{blue}{{\ell}^{2} \cdot \frac{\frac{\frac{2 \cdot \cos k}{t}}{{\sin k}^{2}}}{{k}^{2}}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 41.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 350:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(k \cdot \frac{{t}^{1.5}}{\ell}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;{\ell}^{2} \cdot \frac{\frac{\frac{2 \cdot \cos k}{t}}{{\sin k}^{2}}}{{k}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 74.8% accurate, 0.8× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 1050:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t\_m}\right)\right) \cdot \left(k \cdot \frac{{t\_m}^{1.5}}{\ell}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{t\_m \cdot {k}^{2}} \cdot \frac{\cos k \cdot {\ell}^{2}}{{\sin k}^{2}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 1050.0)
    (/
     2.0
     (pow (* (hypot 1.0 (hypot 1.0 (/ k t_m))) (* k (/ (pow t_m 1.5) l))) 2.0))
    (*
     (/ 2.0 (* t_m (pow k 2.0)))
     (/ (* (cos k) (pow l 2.0)) (pow (sin k) 2.0))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 1050.0) {
		tmp = 2.0 / pow((hypot(1.0, hypot(1.0, (k / t_m))) * (k * (pow(t_m, 1.5) / l))), 2.0);
	} else {
		tmp = (2.0 / (t_m * pow(k, 2.0))) * ((cos(k) * pow(l, 2.0)) / pow(sin(k), 2.0));
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 1050.0) {
		tmp = 2.0 / Math.pow((Math.hypot(1.0, Math.hypot(1.0, (k / t_m))) * (k * (Math.pow(t_m, 1.5) / l))), 2.0);
	} else {
		tmp = (2.0 / (t_m * Math.pow(k, 2.0))) * ((Math.cos(k) * Math.pow(l, 2.0)) / Math.pow(Math.sin(k), 2.0));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 1050.0:
		tmp = 2.0 / math.pow((math.hypot(1.0, math.hypot(1.0, (k / t_m))) * (k * (math.pow(t_m, 1.5) / l))), 2.0)
	else:
		tmp = (2.0 / (t_m * math.pow(k, 2.0))) * ((math.cos(k) * math.pow(l, 2.0)) / math.pow(math.sin(k), 2.0))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 1050.0)
		tmp = Float64(2.0 / (Float64(hypot(1.0, hypot(1.0, Float64(k / t_m))) * Float64(k * Float64((t_m ^ 1.5) / l))) ^ 2.0));
	else
		tmp = Float64(Float64(2.0 / Float64(t_m * (k ^ 2.0))) * Float64(Float64(cos(k) * (l ^ 2.0)) / (sin(k) ^ 2.0)));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 1050.0)
		tmp = 2.0 / ((hypot(1.0, hypot(1.0, (k / t_m))) * (k * ((t_m ^ 1.5) / l))) ^ 2.0);
	else
		tmp = (2.0 / (t_m * (k ^ 2.0))) * ((cos(k) * (l ^ 2.0)) / (sin(k) ^ 2.0));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 1050.0], N[(2.0 / N[Power[N[(N[Sqrt[1.0 ^ 2 + N[Sqrt[1.0 ^ 2 + N[(k / t$95$m), $MachinePrecision] ^ 2], $MachinePrecision] ^ 2], $MachinePrecision] * N[(k * N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(N[(2.0 / N[(t$95$m * N[Power[k, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[(N[Cos[k], $MachinePrecision] * N[Power[l, 2.0], $MachinePrecision]), $MachinePrecision] / N[Power[N[Sin[k], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 1050

   1. Initial program 50.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 50.4%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 20.7%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 20.7%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 30.1%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 30.1%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 30.1%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 30.1%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around 0 33.5%

      \[\leadsto \frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \color{blue}{k}\right)\right)}^{2}} \]

   if 1050 < k

   1. Initial program 51.0%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in t around 0 66.9%

      \[\leadsto \color{blue}{2 \cdot \frac{{\ell}^{2} \cdot \cos k}{{k}^{2} \cdot \left(t \cdot {\sin k}^{2}\right)}} \]
   4. Step-by-step derivation

      1. associate-*r/ 66.9%

         \[\leadsto \color{blue}{\frac{2 \cdot \left({\ell}^{2} \cdot \cos k\right)}{{k}^{2} \cdot \left(t \cdot {\sin k}^{2}\right)}} \]

      2. associate-*r* 66.9%

         \[\leadsto \frac{2 \cdot \left({\ell}^{2} \cdot \cos k\right)}{\color{blue}{\left({k}^{2} \cdot t\right) \cdot {\sin k}^{2}}} \]

      3. times-frac 66.9%

         \[\leadsto \color{blue}{\frac{2}{{k}^{2} \cdot t} \cdot \frac{{\ell}^{2} \cdot \cos k}{{\sin k}^{2}}} \]

   5. Simplified 66.9%

      \[\leadsto \color{blue}{\frac{2}{{k}^{2} \cdot t} \cdot \frac{{\ell}^{2} \cdot \cos k}{{\sin k}^{2}}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 41.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 1050:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(k \cdot \frac{{t}^{1.5}}{\ell}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{t \cdot {k}^{2}} \cdot \frac{\cos k \cdot {\ell}^{2}}{{\sin k}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 11: 76.7% accurate, 1.0× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 1.85 \cdot 10^{-26}:\\ \;\;\;\;\frac{2}{{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t\_m}{\cos k}}\right)}^{2}}\\ \mathbf{elif}\;t\_m \leq 5.5 \cdot 10^{+102}:\\ \;\;\;\;\left(\ell \cdot \frac{\frac{2}{\tan k}}{\sin k \cdot {t\_m}^{3}}\right) \cdot \frac{\ell}{2 + {\left(\frac{k}{t\_m}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t\_m}\right)\right) \cdot \left(k \cdot \frac{{t\_m}^{1.5}}{\ell}\right)\right)}^{2}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= t_m 1.85e-26)
    (/ 2.0 (pow (* (/ (* k (sin k)) l) (sqrt (/ t_m (cos k)))) 2.0))
    (if (<= t_m 5.5e+102)
      (*
       (* l (/ (/ 2.0 (tan k)) (* (sin k) (pow t_m 3.0))))
       (/ l (+ 2.0 (pow (/ k t_m) 2.0))))
      (/
       2.0
       (pow
        (* (hypot 1.0 (hypot 1.0 (/ k t_m))) (* k (/ (pow t_m 1.5) l)))
        2.0))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 1.85e-26) {
		tmp = 2.0 / pow((((k * sin(k)) / l) * sqrt((t_m / cos(k)))), 2.0);
	} else if (t_m <= 5.5e+102) {
		tmp = (l * ((2.0 / tan(k)) / (sin(k) * pow(t_m, 3.0)))) * (l / (2.0 + pow((k / t_m), 2.0)));
	} else {
		tmp = 2.0 / pow((hypot(1.0, hypot(1.0, (k / t_m))) * (k * (pow(t_m, 1.5) / l))), 2.0);
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 1.85e-26) {
		tmp = 2.0 / Math.pow((((k * Math.sin(k)) / l) * Math.sqrt((t_m / Math.cos(k)))), 2.0);
	} else if (t_m <= 5.5e+102) {
		tmp = (l * ((2.0 / Math.tan(k)) / (Math.sin(k) * Math.pow(t_m, 3.0)))) * (l / (2.0 + Math.pow((k / t_m), 2.0)));
	} else {
		tmp = 2.0 / Math.pow((Math.hypot(1.0, Math.hypot(1.0, (k / t_m))) * (k * (Math.pow(t_m, 1.5) / l))), 2.0);
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if t_m <= 1.85e-26:
		tmp = 2.0 / math.pow((((k * math.sin(k)) / l) * math.sqrt((t_m / math.cos(k)))), 2.0)
	elif t_m <= 5.5e+102:
		tmp = (l * ((2.0 / math.tan(k)) / (math.sin(k) * math.pow(t_m, 3.0)))) * (l / (2.0 + math.pow((k / t_m), 2.0)))
	else:
		tmp = 2.0 / math.pow((math.hypot(1.0, math.hypot(1.0, (k / t_m))) * (k * (math.pow(t_m, 1.5) / l))), 2.0)
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (t_m <= 1.85e-26)
		tmp = Float64(2.0 / (Float64(Float64(Float64(k * sin(k)) / l) * sqrt(Float64(t_m / cos(k)))) ^ 2.0));
	elseif (t_m <= 5.5e+102)
		tmp = Float64(Float64(l * Float64(Float64(2.0 / tan(k)) / Float64(sin(k) * (t_m ^ 3.0)))) * Float64(l / Float64(2.0 + (Float64(k / t_m) ^ 2.0))));
	else
		tmp = Float64(2.0 / (Float64(hypot(1.0, hypot(1.0, Float64(k / t_m))) * Float64(k * Float64((t_m ^ 1.5) / l))) ^ 2.0));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (t_m <= 1.85e-26)
		tmp = 2.0 / ((((k * sin(k)) / l) * sqrt((t_m / cos(k)))) ^ 2.0);
	elseif (t_m <= 5.5e+102)
		tmp = (l * ((2.0 / tan(k)) / (sin(k) * (t_m ^ 3.0)))) * (l / (2.0 + ((k / t_m) ^ 2.0)));
	else
		tmp = 2.0 / ((hypot(1.0, hypot(1.0, (k / t_m))) * (k * ((t_m ^ 1.5) / l))) ^ 2.0);
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[t$95$m, 1.85e-26], N[(2.0 / N[Power[N[(N[(N[(k * N[Sin[k], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[Sqrt[N[(t$95$m / N[Cos[k], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$m, 5.5e+102], N[(N[(l * N[(N[(2.0 / N[Tan[k], $MachinePrecision]), $MachinePrecision] / N[(N[Sin[k], $MachinePrecision] * N[Power[t$95$m, 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(l / N[(2.0 + N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[Power[N[(N[Sqrt[1.0 ^ 2 + N[Sqrt[1.0 ^ 2 + N[(k / t$95$m), $MachinePrecision] ^ 2], $MachinePrecision] ^ 2], $MachinePrecision] * N[(k * N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if t < 1.8499999999999999e-26

   1. Initial program 47.7%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 47.7%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 15.6%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 15.6%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 17.4%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 17.4%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 17.4%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 17.4%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around inf 34.0%

      \[\leadsto \frac{2}{{\color{blue}{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t}{\cos k}}\right)}}^{2}} \]

   if 1.8499999999999999e-26 < t < 5.49999999999999981e102

   1. Initial program 64.0%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 60.5%

      \[\leadsto \color{blue}{\frac{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}{2 + {\left(\frac{k}{t}\right)}^{2}}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. associate-*r* 74.4%

         \[\leadsto \frac{\color{blue}{\left(\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \ell\right) \cdot \ell}}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

      2. *-un-lft-identity 74.4%

         \[\leadsto \frac{\left(\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \ell\right) \cdot \ell}{\color{blue}{1 \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)}} \]

      3. times-frac 78.5%

         \[\leadsto \color{blue}{\frac{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \ell}{1} \cdot \frac{\ell}{2 + {\left(\frac{k}{t}\right)}^{2}}} \]

      4. div-inv 78.5%

         \[\leadsto \frac{\color{blue}{\left(\frac{2}{\tan k} \cdot \frac{1}{{t}^{3} \cdot \sin k}\right)} \cdot \ell}{1} \cdot \frac{\ell}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

      5. frac-times 78.3%

         \[\leadsto \frac{\color{blue}{\frac{2 \cdot 1}{\tan k \cdot \left({t}^{3} \cdot \sin k\right)}} \cdot \ell}{1} \cdot \frac{\ell}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

      6. metadata-eval 78.3%

         \[\leadsto \frac{\frac{\color{blue}{2}}{\tan k \cdot \left({t}^{3} \cdot \sin k\right)} \cdot \ell}{1} \cdot \frac{\ell}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

   6. Applied egg-rr 78.3%

      \[\leadsto \color{blue}{\frac{\frac{2}{\tan k \cdot \left({t}^{3} \cdot \sin k\right)} \cdot \ell}{1} \cdot \frac{\ell}{2 + {\left(\frac{k}{t}\right)}^{2}}} \]
   7. Step-by-step derivation

      1. /-rgt-identity 78.3%

         \[\leadsto \color{blue}{\left(\frac{2}{\tan k \cdot \left({t}^{3} \cdot \sin k\right)} \cdot \ell\right)} \cdot \frac{\ell}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

      2. *-commutative 78.3%

         \[\leadsto \color{blue}{\left(\ell \cdot \frac{2}{\tan k \cdot \left({t}^{3} \cdot \sin k\right)}\right)} \cdot \frac{\ell}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

      3. associate-/r* 78.5%

         \[\leadsto \left(\ell \cdot \color{blue}{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k}}\right) \cdot \frac{\ell}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

      4. *-commutative 78.5%

         \[\leadsto \left(\ell \cdot \frac{\frac{2}{\tan k}}{\color{blue}{\sin k \cdot {t}^{3}}}\right) \cdot \frac{\ell}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

   8. Simplified 78.5%

      \[\leadsto \color{blue}{\left(\ell \cdot \frac{\frac{2}{\tan k}}{\sin k \cdot {t}^{3}}\right) \cdot \frac{\ell}{2 + {\left(\frac{k}{t}\right)}^{2}}} \]

   if 5.49999999999999981e102 < t

   1. Initial program 54.8%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 54.8%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 39.1%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 39.1%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 61.6%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 61.6%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 61.6%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 61.6%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around 0 78.3%

      \[\leadsto \frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \color{blue}{k}\right)\right)}^{2}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 45.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 1.85 \cdot 10^{-26}:\\ \;\;\;\;\frac{2}{{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t}{\cos k}}\right)}^{2}}\\ \mathbf{elif}\;t \leq 5.5 \cdot 10^{+102}:\\ \;\;\;\;\left(\ell \cdot \frac{\frac{2}{\tan k}}{\sin k \cdot {t}^{3}}\right) \cdot \frac{\ell}{2 + {\left(\frac{k}{t}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(k \cdot \frac{{t}^{1.5}}{\ell}\right)\right)}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 12: 64.1% accurate, 1.0× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 7.5 \cdot 10^{-8}:\\ \;\;\;\;\frac{2}{{\left(\left(k \cdot \frac{\sqrt{2}}{\ell}\right) \cdot \sqrt{{t\_m}^{3}}\right)}^{2}}\\ \mathbf{elif}\;k \leq 3.9 \cdot 10^{+86}:\\ \;\;\;\;\frac{2}{2 \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t\_m}^{3}}{\ell \cdot \ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;2 \cdot {\left(t\_m \cdot \sqrt[3]{2 \cdot \left({k}^{2} \cdot {\ell}^{-2}\right)}\right)}^{-3}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 7.5e-8)
    (/ 2.0 (pow (* (* k (/ (sqrt 2.0) l)) (sqrt (pow t_m 3.0))) 2.0))
    (if (<= k 3.9e+86)
      (/ 2.0 (* 2.0 (* (tan k) (* (sin k) (/ (pow t_m 3.0) (* l l))))))
      (*
       2.0
       (pow (* t_m (cbrt (* 2.0 (* (pow k 2.0) (pow l -2.0))))) -3.0))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 7.5e-8) {
		tmp = 2.0 / pow(((k * (sqrt(2.0) / l)) * sqrt(pow(t_m, 3.0))), 2.0);
	} else if (k <= 3.9e+86) {
		tmp = 2.0 / (2.0 * (tan(k) * (sin(k) * (pow(t_m, 3.0) / (l * l)))));
	} else {
		tmp = 2.0 * pow((t_m * cbrt((2.0 * (pow(k, 2.0) * pow(l, -2.0))))), -3.0);
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 7.5e-8) {
		tmp = 2.0 / Math.pow(((k * (Math.sqrt(2.0) / l)) * Math.sqrt(Math.pow(t_m, 3.0))), 2.0);
	} else if (k <= 3.9e+86) {
		tmp = 2.0 / (2.0 * (Math.tan(k) * (Math.sin(k) * (Math.pow(t_m, 3.0) / (l * l)))));
	} else {
		tmp = 2.0 * Math.pow((t_m * Math.cbrt((2.0 * (Math.pow(k, 2.0) * Math.pow(l, -2.0))))), -3.0);
	}
	return t_s * tmp;
}

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 7.5e-8)
		tmp = Float64(2.0 / (Float64(Float64(k * Float64(sqrt(2.0) / l)) * sqrt((t_m ^ 3.0))) ^ 2.0));
	elseif (k <= 3.9e+86)
		tmp = Float64(2.0 / Float64(2.0 * Float64(tan(k) * Float64(sin(k) * Float64((t_m ^ 3.0) / Float64(l * l))))));
	else
		tmp = Float64(2.0 * (Float64(t_m * cbrt(Float64(2.0 * Float64((k ^ 2.0) * (l ^ -2.0))))) ^ -3.0));
	end
	return Float64(t_s * tmp)
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 7.5e-8], N[(2.0 / N[Power[N[(N[(k * N[(N[Sqrt[2.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision] * N[Sqrt[N[Power[t$95$m, 3.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[k, 3.9e+86], N[(2.0 / N[(2.0 * N[(N[Tan[k], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[(N[Power[t$95$m, 3.0], $MachinePrecision] / N[(l * l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 * N[Power[N[(t$95$m * N[Power[N[(2.0 * N[(N[Power[k, 2.0], $MachinePrecision] * N[Power[l, -2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision], -3.0], $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if k < 7.4999999999999997e-8

   1. Initial program 50.1%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 50.1%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 20.5%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 20.5%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 30.1%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 30.1%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 30.1%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 30.1%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around 0 29.2%

      \[\leadsto \frac{2}{{\color{blue}{\left(\frac{k \cdot \sqrt{2}}{\ell} \cdot \sqrt{{t}^{3}}\right)}}^{2}} \]
   8. Step-by-step derivation

      1. associate-/l* 29.2%

         \[\leadsto \frac{2}{{\left(\color{blue}{\left(k \cdot \frac{\sqrt{2}}{\ell}\right)} \cdot \sqrt{{t}^{3}}\right)}^{2}} \]

   9. Simplified 29.2%

      \[\leadsto \frac{2}{{\color{blue}{\left(\left(k \cdot \frac{\sqrt{2}}{\ell}\right) \cdot \sqrt{{t}^{3}}\right)}}^{2}} \]

   if 7.4999999999999997e-8 < k < 3.9000000000000002e86

   1. Initial program 61.9%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in k around 0 62.8%

      \[\leadsto \frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \color{blue}{2}} \]

   if 3.9000000000000002e86 < k

   1. Initial program 46.6%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 49.6%

      \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. sqr-pow 24.5%

         \[\leadsto \frac{2}{\frac{\frac{\color{blue}{{t}^{\left(\frac{3}{2}\right)} \cdot {t}^{\left(\frac{3}{2}\right)}}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      2. *-un-lft-identity 24.5%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{\left(\frac{3}{2}\right)} \cdot {t}^{\left(\frac{3}{2}\right)}}{\color{blue}{1 \cdot \ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      3. times-frac 26.8%

         \[\leadsto \frac{2}{\frac{\color{blue}{\frac{{t}^{\left(\frac{3}{2}\right)}}{1} \cdot \frac{{t}^{\left(\frac{3}{2}\right)}}{\ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      4. metadata-eval 26.8%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{\color{blue}{1.5}}}{1} \cdot \frac{{t}^{\left(\frac{3}{2}\right)}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      5. metadata-eval 26.8%

         \[\leadsto \frac{2}{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{\color{blue}{1.5}}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   6. Applied egg-rr 26.8%

      \[\leadsto \frac{2}{\frac{\color{blue}{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]
   7. Step-by-step derivation

      1. add-cube-cbrt 26.7%

         \[\leadsto \frac{2}{\color{blue}{\left(\sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \cdot \sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right) \cdot \sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}}} \]

      2. pow3 26.6%

         \[\leadsto \frac{2}{\color{blue}{{\left(\sqrt[3]{\frac{\frac{{t}^{1.5}}{1} \cdot \frac{{t}^{1.5}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right)}^{3}}} \]

   8. Applied egg-rr 64.5%

      \[\leadsto \frac{2}{\color{blue}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \sqrt[3]{\sin k \cdot \left(\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}\right)}^{3}}} \]

   9. Taylor expanded in k around 0 51.1%

      \[\leadsto \frac{2}{{\color{blue}{\left(\sqrt[3]{\frac{{k}^{2}}{{\ell}^{2}}} \cdot \left(t \cdot \sqrt[3]{2}\right)\right)}}^{3}} \]
   10. Step-by-step derivation

       1. *-commutative 51.1%

          \[\leadsto \frac{2}{{\color{blue}{\left(\left(t \cdot \sqrt[3]{2}\right) \cdot \sqrt[3]{\frac{{k}^{2}}{{\ell}^{2}}}\right)}}^{3}} \]

   11. Simplified 51.1%

       \[\leadsto \frac{2}{{\color{blue}{\left(\left(t \cdot \sqrt[3]{2}\right) \cdot \sqrt[3]{\frac{{k}^{2}}{{\ell}^{2}}}\right)}}^{3}} \]
   12. Step-by-step derivation

       1. div-inv 51.1%

          \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\left(t \cdot \sqrt[3]{2}\right) \cdot \sqrt[3]{\frac{{k}^{2}}{{\ell}^{2}}}\right)}^{3}}} \]

       2. pow-flip 51.1%

          \[\leadsto 2 \cdot \color{blue}{{\left(\left(t \cdot \sqrt[3]{2}\right) \cdot \sqrt[3]{\frac{{k}^{2}}{{\ell}^{2}}}\right)}^{\left(-3\right)}} \]

       3. associate-*l* 51.1%

          \[\leadsto 2 \cdot {\color{blue}{\left(t \cdot \left(\sqrt[3]{2} \cdot \sqrt[3]{\frac{{k}^{2}}{{\ell}^{2}}}\right)\right)}}^{\left(-3\right)} \]

       4. pow2 51.1%

          \[\leadsto 2 \cdot {\left(t \cdot \left(\sqrt[3]{2} \cdot \sqrt[3]{\frac{\color{blue}{k \cdot k}}{{\ell}^{2}}}\right)\right)}^{\left(-3\right)} \]

       5. pow2 51.1%

          \[\leadsto 2 \cdot {\left(t \cdot \left(\sqrt[3]{2} \cdot \sqrt[3]{\frac{k \cdot k}{\color{blue}{\ell \cdot \ell}}}\right)\right)}^{\left(-3\right)} \]

       6. cbrt-unprod 51.1%

          \[\leadsto 2 \cdot {\left(t \cdot \color{blue}{\sqrt[3]{2 \cdot \frac{k \cdot k}{\ell \cdot \ell}}}\right)}^{\left(-3\right)} \]

       7. div-inv 51.1%

          \[\leadsto 2 \cdot {\left(t \cdot \sqrt[3]{2 \cdot \color{blue}{\left(\left(k \cdot k\right) \cdot \frac{1}{\ell \cdot \ell}\right)}}\right)}^{\left(-3\right)} \]

       8. pow2 51.1%

          \[\leadsto 2 \cdot {\left(t \cdot \sqrt[3]{2 \cdot \left(\color{blue}{{k}^{2}} \cdot \frac{1}{\ell \cdot \ell}\right)}\right)}^{\left(-3\right)} \]

       9. pow2 51.1%

          \[\leadsto 2 \cdot {\left(t \cdot \sqrt[3]{2 \cdot \left({k}^{2} \cdot \frac{1}{\color{blue}{{\ell}^{2}}}\right)}\right)}^{\left(-3\right)} \]

       10. pow-flip 51.1%

           \[\leadsto 2 \cdot {\left(t \cdot \sqrt[3]{2 \cdot \left({k}^{2} \cdot \color{blue}{{\ell}^{\left(-2\right)}}\right)}\right)}^{\left(-3\right)} \]

       11. metadata-eval 51.1%

           \[\leadsto 2 \cdot {\left(t \cdot \sqrt[3]{2 \cdot \left({k}^{2} \cdot {\ell}^{\color{blue}{-2}}\right)}\right)}^{\left(-3\right)} \]

       12. metadata-eval 51.1%

           \[\leadsto 2 \cdot {\left(t \cdot \sqrt[3]{2 \cdot \left({k}^{2} \cdot {\ell}^{-2}\right)}\right)}^{\color{blue}{-3}} \]

   13. Applied egg-rr 51.1%

       \[\leadsto \color{blue}{2 \cdot {\left(t \cdot \sqrt[3]{2 \cdot \left({k}^{2} \cdot {\ell}^{-2}\right)}\right)}^{-3}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 35.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 7.5 \cdot 10^{-8}:\\ \;\;\;\;\frac{2}{{\left(\left(k \cdot \frac{\sqrt{2}}{\ell}\right) \cdot \sqrt{{t}^{3}}\right)}^{2}}\\ \mathbf{elif}\;k \leq 3.9 \cdot 10^{+86}:\\ \;\;\;\;\frac{2}{2 \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t}^{3}}{\ell \cdot \ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;2 \cdot {\left(t \cdot \sqrt[3]{2 \cdot \left({k}^{2} \cdot {\ell}^{-2}\right)}\right)}^{-3}\\ \end{array} \]
5. Add Preprocessing

Alternative 13: 74.2% accurate, 1.0× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 680:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t\_m}\right)\right) \cdot \left(k \cdot \frac{{t\_m}^{1.5}}{\ell}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{{\ell}^{2}}{k \cdot k} \cdot \frac{2 \cdot \cos k}{t\_m \cdot {\sin k}^{2}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 680.0)
    (/
     2.0
     (pow (* (hypot 1.0 (hypot 1.0 (/ k t_m))) (* k (/ (pow t_m 1.5) l))) 2.0))
    (*
     (/ (pow l 2.0) (* k k))
     (/ (* 2.0 (cos k)) (* t_m (pow (sin k) 2.0)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 680.0) {
		tmp = 2.0 / pow((hypot(1.0, hypot(1.0, (k / t_m))) * (k * (pow(t_m, 1.5) / l))), 2.0);
	} else {
		tmp = (pow(l, 2.0) / (k * k)) * ((2.0 * cos(k)) / (t_m * pow(sin(k), 2.0)));
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 680.0) {
		tmp = 2.0 / Math.pow((Math.hypot(1.0, Math.hypot(1.0, (k / t_m))) * (k * (Math.pow(t_m, 1.5) / l))), 2.0);
	} else {
		tmp = (Math.pow(l, 2.0) / (k * k)) * ((2.0 * Math.cos(k)) / (t_m * Math.pow(Math.sin(k), 2.0)));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 680.0:
		tmp = 2.0 / math.pow((math.hypot(1.0, math.hypot(1.0, (k / t_m))) * (k * (math.pow(t_m, 1.5) / l))), 2.0)
	else:
		tmp = (math.pow(l, 2.0) / (k * k)) * ((2.0 * math.cos(k)) / (t_m * math.pow(math.sin(k), 2.0)))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 680.0)
		tmp = Float64(2.0 / (Float64(hypot(1.0, hypot(1.0, Float64(k / t_m))) * Float64(k * Float64((t_m ^ 1.5) / l))) ^ 2.0));
	else
		tmp = Float64(Float64((l ^ 2.0) / Float64(k * k)) * Float64(Float64(2.0 * cos(k)) / Float64(t_m * (sin(k) ^ 2.0))));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 680.0)
		tmp = 2.0 / ((hypot(1.0, hypot(1.0, (k / t_m))) * (k * ((t_m ^ 1.5) / l))) ^ 2.0);
	else
		tmp = ((l ^ 2.0) / (k * k)) * ((2.0 * cos(k)) / (t_m * (sin(k) ^ 2.0)));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 680.0], N[(2.0 / N[Power[N[(N[Sqrt[1.0 ^ 2 + N[Sqrt[1.0 ^ 2 + N[(k / t$95$m), $MachinePrecision] ^ 2], $MachinePrecision] ^ 2], $MachinePrecision] * N[(k * N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(N[(N[Power[l, 2.0], $MachinePrecision] / N[(k * k), $MachinePrecision]), $MachinePrecision] * N[(N[(2.0 * N[Cos[k], $MachinePrecision]), $MachinePrecision] / N[(t$95$m * N[Power[N[Sin[k], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 680

   1. Initial program 50.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 50.4%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 20.7%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 20.7%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 30.1%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 30.1%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 30.1%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 30.1%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around 0 33.5%

      \[\leadsto \frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \color{blue}{k}\right)\right)}^{2}} \]

   if 680 < k

   1. Initial program 51.0%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 50.9%

      \[\leadsto \color{blue}{\frac{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}{2 + {\left(\frac{k}{t}\right)}^{2}}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. add-cube-cbrt 50.8%

         \[\leadsto \frac{\color{blue}{\left(\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}\right) \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

      2. add-sqr-sqrt 50.8%

         \[\leadsto \frac{\left(\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}\right) \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\color{blue}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}} \cdot \sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

      3. times-frac 50.8%

         \[\leadsto \color{blue}{\frac{\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}} \cdot \frac{\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

   6. Applied egg-rr 58.8%

      \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}} \cdot \frac{\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]
   7. Step-by-step derivation

      1. associate-*l/ 58.8%

         \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2} \cdot \frac{\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

   8. Simplified 65.9%

      \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{{\ell}^{2} \cdot \frac{2}{\tan k}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2} \cdot \frac{\frac{\sqrt[3]{{\ell}^{2} \cdot \frac{2}{\tan k}}}{t \cdot \sqrt[3]{\sin k}}}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}} \]

   9. Taylor expanded in k around inf 66.6%

      \[\leadsto \color{blue}{\frac{{\ell}^{2} \cdot \left({\left(\sqrt[3]{2}\right)}^{3} \cdot \cos k\right)}{{k}^{2} \cdot \left(t \cdot {\sin k}^{2}\right)}} \]
   10. Step-by-step derivation

       1. times-frac 61.3%

          \[\leadsto \color{blue}{\frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{{\left(\sqrt[3]{2}\right)}^{3} \cdot \cos k}{t \cdot {\sin k}^{2}}} \]

       2. rem-cube-cbrt 61.5%

          \[\leadsto \frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{\color{blue}{2} \cdot \cos k}{t \cdot {\sin k}^{2}} \]

   11. Simplified 61.5%

       \[\leadsto \color{blue}{\frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}}} \]
   12. Step-by-step derivation

       1. unpow2 61.5%

          \[\leadsto \frac{{\ell}^{2}}{\color{blue}{k \cdot k}} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}} \]

   13. Applied egg-rr 61.5%

       \[\leadsto \frac{{\ell}^{2}}{\color{blue}{k \cdot k}} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 40.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 680:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(k \cdot \frac{{t}^{1.5}}{\ell}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{{\ell}^{2}}{k \cdot k} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 14: 70.6% accurate, 1.0× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 3.4 \cdot 10^{-44}:\\ \;\;\;\;\frac{2}{{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t\_m}{\cos k}}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\left(k \cdot \frac{\sqrt{2}}{\ell}\right) \cdot \sqrt{{t\_m}^{3}}\right)}^{2}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= t_m 3.4e-44)
    (/ 2.0 (pow (* (/ (* k (sin k)) l) (sqrt (/ t_m (cos k)))) 2.0))
    (/ 2.0 (pow (* (* k (/ (sqrt 2.0) l)) (sqrt (pow t_m 3.0))) 2.0)))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 3.4e-44) {
		tmp = 2.0 / pow((((k * sin(k)) / l) * sqrt((t_m / cos(k)))), 2.0);
	} else {
		tmp = 2.0 / pow(((k * (sqrt(2.0) / l)) * sqrt(pow(t_m, 3.0))), 2.0);
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (t_m <= 3.4d-44) then
        tmp = 2.0d0 / ((((k * sin(k)) / l) * sqrt((t_m / cos(k)))) ** 2.0d0)
    else
        tmp = 2.0d0 / (((k * (sqrt(2.0d0) / l)) * sqrt((t_m ** 3.0d0))) ** 2.0d0)
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 3.4e-44) {
		tmp = 2.0 / Math.pow((((k * Math.sin(k)) / l) * Math.sqrt((t_m / Math.cos(k)))), 2.0);
	} else {
		tmp = 2.0 / Math.pow(((k * (Math.sqrt(2.0) / l)) * Math.sqrt(Math.pow(t_m, 3.0))), 2.0);
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if t_m <= 3.4e-44:
		tmp = 2.0 / math.pow((((k * math.sin(k)) / l) * math.sqrt((t_m / math.cos(k)))), 2.0)
	else:
		tmp = 2.0 / math.pow(((k * (math.sqrt(2.0) / l)) * math.sqrt(math.pow(t_m, 3.0))), 2.0)
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (t_m <= 3.4e-44)
		tmp = Float64(2.0 / (Float64(Float64(Float64(k * sin(k)) / l) * sqrt(Float64(t_m / cos(k)))) ^ 2.0));
	else
		tmp = Float64(2.0 / (Float64(Float64(k * Float64(sqrt(2.0) / l)) * sqrt((t_m ^ 3.0))) ^ 2.0));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (t_m <= 3.4e-44)
		tmp = 2.0 / ((((k * sin(k)) / l) * sqrt((t_m / cos(k)))) ^ 2.0);
	else
		tmp = 2.0 / (((k * (sqrt(2.0) / l)) * sqrt((t_m ^ 3.0))) ^ 2.0);
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[t$95$m, 3.4e-44], N[(2.0 / N[Power[N[(N[(N[(k * N[Sin[k], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[Sqrt[N[(t$95$m / N[Cos[k], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(2.0 / N[Power[N[(N[(k * N[(N[Sqrt[2.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision] * N[Sqrt[N[Power[t$95$m, 3.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if t < 3.40000000000000016e-44

   1. Initial program 48.0%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 48.0%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 15.7%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 15.7%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 17.0%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 17.0%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 17.0%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 17.0%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around inf 33.6%

      \[\leadsto \frac{2}{{\color{blue}{\left(\frac{k \cdot \sin k}{\ell} \cdot \sqrt{\frac{t}{\cos k}}\right)}}^{2}} \]

   if 3.40000000000000016e-44 < t

   1. Initial program 57.7%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 57.7%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 39.5%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 39.5%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 56.9%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 56.9%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 56.9%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 56.9%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around 0 63.9%

      \[\leadsto \frac{2}{{\color{blue}{\left(\frac{k \cdot \sqrt{2}}{\ell} \cdot \sqrt{{t}^{3}}\right)}}^{2}} \]
   8. Step-by-step derivation

      1. associate-/l* 63.9%

         \[\leadsto \frac{2}{{\left(\color{blue}{\left(k \cdot \frac{\sqrt{2}}{\ell}\right)} \cdot \sqrt{{t}^{3}}\right)}^{2}} \]

   9. Simplified 63.9%

      \[\leadsto \frac{2}{{\color{blue}{\left(\left(k \cdot \frac{\sqrt{2}}{\ell}\right) \cdot \sqrt{{t}^{3}}\right)}}^{2}} \]
3. Recombined 2 regimes into one program.
4. Add Preprocessing

Alternative 15: 63.9% accurate, 1.0× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 6.5 \cdot 10^{-6}:\\ \;\;\;\;\frac{2}{{\left(\left(k \cdot \frac{\sqrt{2}}{\ell}\right) \cdot \sqrt{{t\_m}^{3}}\right)}^{2}}\\ \mathbf{elif}\;k \leq 1.75 \cdot 10^{+107}:\\ \;\;\;\;\frac{2}{2 \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t\_m}^{3}}{\ell \cdot \ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{t\_m \cdot {k}^{2}} \cdot \frac{{\ell}^{2}}{{k}^{2}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 6.5e-6)
    (/ 2.0 (pow (* (* k (/ (sqrt 2.0) l)) (sqrt (pow t_m 3.0))) 2.0))
    (if (<= k 1.75e+107)
      (/ 2.0 (* 2.0 (* (tan k) (* (sin k) (/ (pow t_m 3.0) (* l l))))))
      (* (/ 2.0 (* t_m (pow k 2.0))) (/ (pow l 2.0) (pow k 2.0)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 6.5e-6) {
		tmp = 2.0 / pow(((k * (sqrt(2.0) / l)) * sqrt(pow(t_m, 3.0))), 2.0);
	} else if (k <= 1.75e+107) {
		tmp = 2.0 / (2.0 * (tan(k) * (sin(k) * (pow(t_m, 3.0) / (l * l)))));
	} else {
		tmp = (2.0 / (t_m * pow(k, 2.0))) * (pow(l, 2.0) / pow(k, 2.0));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 6.5d-6) then
        tmp = 2.0d0 / (((k * (sqrt(2.0d0) / l)) * sqrt((t_m ** 3.0d0))) ** 2.0d0)
    else if (k <= 1.75d+107) then
        tmp = 2.0d0 / (2.0d0 * (tan(k) * (sin(k) * ((t_m ** 3.0d0) / (l * l)))))
    else
        tmp = (2.0d0 / (t_m * (k ** 2.0d0))) * ((l ** 2.0d0) / (k ** 2.0d0))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 6.5e-6) {
		tmp = 2.0 / Math.pow(((k * (Math.sqrt(2.0) / l)) * Math.sqrt(Math.pow(t_m, 3.0))), 2.0);
	} else if (k <= 1.75e+107) {
		tmp = 2.0 / (2.0 * (Math.tan(k) * (Math.sin(k) * (Math.pow(t_m, 3.0) / (l * l)))));
	} else {
		tmp = (2.0 / (t_m * Math.pow(k, 2.0))) * (Math.pow(l, 2.0) / Math.pow(k, 2.0));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 6.5e-6:
		tmp = 2.0 / math.pow(((k * (math.sqrt(2.0) / l)) * math.sqrt(math.pow(t_m, 3.0))), 2.0)
	elif k <= 1.75e+107:
		tmp = 2.0 / (2.0 * (math.tan(k) * (math.sin(k) * (math.pow(t_m, 3.0) / (l * l)))))
	else:
		tmp = (2.0 / (t_m * math.pow(k, 2.0))) * (math.pow(l, 2.0) / math.pow(k, 2.0))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 6.5e-6)
		tmp = Float64(2.0 / (Float64(Float64(k * Float64(sqrt(2.0) / l)) * sqrt((t_m ^ 3.0))) ^ 2.0));
	elseif (k <= 1.75e+107)
		tmp = Float64(2.0 / Float64(2.0 * Float64(tan(k) * Float64(sin(k) * Float64((t_m ^ 3.0) / Float64(l * l))))));
	else
		tmp = Float64(Float64(2.0 / Float64(t_m * (k ^ 2.0))) * Float64((l ^ 2.0) / (k ^ 2.0)));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 6.5e-6)
		tmp = 2.0 / (((k * (sqrt(2.0) / l)) * sqrt((t_m ^ 3.0))) ^ 2.0);
	elseif (k <= 1.75e+107)
		tmp = 2.0 / (2.0 * (tan(k) * (sin(k) * ((t_m ^ 3.0) / (l * l)))));
	else
		tmp = (2.0 / (t_m * (k ^ 2.0))) * ((l ^ 2.0) / (k ^ 2.0));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 6.5e-6], N[(2.0 / N[Power[N[(N[(k * N[(N[Sqrt[2.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision] * N[Sqrt[N[Power[t$95$m, 3.0], $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[k, 1.75e+107], N[(2.0 / N[(2.0 * N[(N[Tan[k], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[(N[Power[t$95$m, 3.0], $MachinePrecision] / N[(l * l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(2.0 / N[(t$95$m * N[Power[k, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[Power[l, 2.0], $MachinePrecision] / N[Power[k, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if k < 6.4999999999999996e-6

   1. Initial program 50.1%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing
   3. Step-by-step derivation

      1. div-inv 50.1%

         \[\leadsto \color{blue}{2 \cdot \frac{1}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}} \]

      2. add-sqr-sqrt 20.5%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \cdot \sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}}} \]

      3. pow2 20.5%

         \[\leadsto 2 \cdot \frac{1}{\color{blue}{{\left(\sqrt{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)}\right)}^{2}}} \]

   4. Applied egg-rr 30.1%

      \[\leadsto \color{blue}{2 \cdot \frac{1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]
   5. Step-by-step derivation

      1. associate-*r/ 30.1%

         \[\leadsto \color{blue}{\frac{2 \cdot 1}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

      2. metadata-eval 30.1%

         \[\leadsto \frac{\color{blue}{2}}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}} \]

   6. Simplified 30.1%

      \[\leadsto \color{blue}{\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(\frac{{t}^{1.5}}{\ell} \cdot \sqrt{\sin k \cdot \tan k}\right)\right)}^{2}}} \]

   7. Taylor expanded in k around 0 29.2%

      \[\leadsto \frac{2}{{\color{blue}{\left(\frac{k \cdot \sqrt{2}}{\ell} \cdot \sqrt{{t}^{3}}\right)}}^{2}} \]
   8. Step-by-step derivation

      1. associate-/l* 29.2%

         \[\leadsto \frac{2}{{\left(\color{blue}{\left(k \cdot \frac{\sqrt{2}}{\ell}\right)} \cdot \sqrt{{t}^{3}}\right)}^{2}} \]

   9. Simplified 29.2%

      \[\leadsto \frac{2}{{\color{blue}{\left(\left(k \cdot \frac{\sqrt{2}}{\ell}\right) \cdot \sqrt{{t}^{3}}\right)}}^{2}} \]

   if 6.4999999999999996e-6 < k < 1.7499999999999999e107

   1. Initial program 61.9%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in k around 0 55.2%

      \[\leadsto \frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \color{blue}{2}} \]

   if 1.7499999999999999e107 < k

   1. Initial program 44.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 44.4%

      \[\leadsto \color{blue}{\frac{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}{2 + {\left(\frac{k}{t}\right)}^{2}}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. add-cube-cbrt 44.4%

         \[\leadsto \frac{\color{blue}{\left(\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}\right) \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

      2. add-sqr-sqrt 44.4%

         \[\leadsto \frac{\left(\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}\right) \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\color{blue}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}} \cdot \sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

      3. times-frac 44.4%

         \[\leadsto \color{blue}{\frac{\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}} \cdot \frac{\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

   6. Applied egg-rr 53.0%

      \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}} \cdot \frac{\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]
   7. Step-by-step derivation

      1. associate-*l/ 53.0%

         \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2} \cdot \frac{\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

   8. Simplified 56.0%

      \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{{\ell}^{2} \cdot \frac{2}{\tan k}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2} \cdot \frac{\frac{\sqrt[3]{{\ell}^{2} \cdot \frac{2}{\tan k}}}{t \cdot \sqrt[3]{\sin k}}}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}} \]

   9. Taylor expanded in k around inf 59.6%

      \[\leadsto \color{blue}{\frac{{\ell}^{2} \cdot \left({\left(\sqrt[3]{2}\right)}^{3} \cdot \cos k\right)}{{k}^{2} \cdot \left(t \cdot {\sin k}^{2}\right)}} \]
   10. Step-by-step derivation

       1. times-frac 59.6%

          \[\leadsto \color{blue}{\frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{{\left(\sqrt[3]{2}\right)}^{3} \cdot \cos k}{t \cdot {\sin k}^{2}}} \]

       2. rem-cube-cbrt 59.7%

          \[\leadsto \frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{\color{blue}{2} \cdot \cos k}{t \cdot {\sin k}^{2}} \]

   11. Simplified 59.7%

       \[\leadsto \color{blue}{\frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}}} \]

   12. Taylor expanded in k around 0 54.2%

       \[\leadsto \frac{{\ell}^{2}}{{k}^{2}} \cdot \color{blue}{\frac{2}{{k}^{2} \cdot t}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 35.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 6.5 \cdot 10^{-6}:\\ \;\;\;\;\frac{2}{{\left(\left(k \cdot \frac{\sqrt{2}}{\ell}\right) \cdot \sqrt{{t}^{3}}\right)}^{2}}\\ \mathbf{elif}\;k \leq 1.75 \cdot 10^{+107}:\\ \;\;\;\;\frac{2}{2 \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t}^{3}}{\ell \cdot \ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{t \cdot {k}^{2}} \cdot \frac{{\ell}^{2}}{{k}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 16: 61.0% accurate, 1.3× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 0.000185:\\ \;\;\;\;\frac{2}{{\left(\frac{{t\_m}^{1.5}}{\ell}\right)}^{2} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \mathbf{elif}\;k \leq 3.5 \cdot 10^{+108}:\\ \;\;\;\;\frac{2}{2 \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t\_m}^{3}}{\ell \cdot \ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{t\_m \cdot {k}^{2}} \cdot \frac{{\ell}^{2}}{{k}^{2}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 0.000185)
    (/ 2.0 (* (pow (/ (pow t_m 1.5) l) 2.0) (* 2.0 (* k k))))
    (if (<= k 3.5e+108)
      (/ 2.0 (* 2.0 (* (tan k) (* (sin k) (/ (pow t_m 3.0) (* l l))))))
      (* (/ 2.0 (* t_m (pow k 2.0))) (/ (pow l 2.0) (pow k 2.0)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 0.000185) {
		tmp = 2.0 / (pow((pow(t_m, 1.5) / l), 2.0) * (2.0 * (k * k)));
	} else if (k <= 3.5e+108) {
		tmp = 2.0 / (2.0 * (tan(k) * (sin(k) * (pow(t_m, 3.0) / (l * l)))));
	} else {
		tmp = (2.0 / (t_m * pow(k, 2.0))) * (pow(l, 2.0) / pow(k, 2.0));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 0.000185d0) then
        tmp = 2.0d0 / ((((t_m ** 1.5d0) / l) ** 2.0d0) * (2.0d0 * (k * k)))
    else if (k <= 3.5d+108) then
        tmp = 2.0d0 / (2.0d0 * (tan(k) * (sin(k) * ((t_m ** 3.0d0) / (l * l)))))
    else
        tmp = (2.0d0 / (t_m * (k ** 2.0d0))) * ((l ** 2.0d0) / (k ** 2.0d0))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 0.000185) {
		tmp = 2.0 / (Math.pow((Math.pow(t_m, 1.5) / l), 2.0) * (2.0 * (k * k)));
	} else if (k <= 3.5e+108) {
		tmp = 2.0 / (2.0 * (Math.tan(k) * (Math.sin(k) * (Math.pow(t_m, 3.0) / (l * l)))));
	} else {
		tmp = (2.0 / (t_m * Math.pow(k, 2.0))) * (Math.pow(l, 2.0) / Math.pow(k, 2.0));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 0.000185:
		tmp = 2.0 / (math.pow((math.pow(t_m, 1.5) / l), 2.0) * (2.0 * (k * k)))
	elif k <= 3.5e+108:
		tmp = 2.0 / (2.0 * (math.tan(k) * (math.sin(k) * (math.pow(t_m, 3.0) / (l * l)))))
	else:
		tmp = (2.0 / (t_m * math.pow(k, 2.0))) * (math.pow(l, 2.0) / math.pow(k, 2.0))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 0.000185)
		tmp = Float64(2.0 / Float64((Float64((t_m ^ 1.5) / l) ^ 2.0) * Float64(2.0 * Float64(k * k))));
	elseif (k <= 3.5e+108)
		tmp = Float64(2.0 / Float64(2.0 * Float64(tan(k) * Float64(sin(k) * Float64((t_m ^ 3.0) / Float64(l * l))))));
	else
		tmp = Float64(Float64(2.0 / Float64(t_m * (k ^ 2.0))) * Float64((l ^ 2.0) / (k ^ 2.0)));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 0.000185)
		tmp = 2.0 / ((((t_m ^ 1.5) / l) ^ 2.0) * (2.0 * (k * k)));
	elseif (k <= 3.5e+108)
		tmp = 2.0 / (2.0 * (tan(k) * (sin(k) * ((t_m ^ 3.0) / (l * l)))));
	else
		tmp = (2.0 / (t_m * (k ^ 2.0))) * ((l ^ 2.0) / (k ^ 2.0));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 0.000185], N[(2.0 / N[(N[Power[N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision], 2.0], $MachinePrecision] * N[(2.0 * N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[k, 3.5e+108], N[(2.0 / N[(2.0 * N[(N[Tan[k], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[(N[Power[t$95$m, 3.0], $MachinePrecision] / N[(l * l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(2.0 / N[(t$95$m * N[Power[k, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[Power[l, 2.0], $MachinePrecision] / N[Power[k, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if k < 1.85e-4

   1. Initial program 50.1%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 55.1%

      \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
   4. Add Preprocessing

   5. Taylor expanded in k around 0 52.6%

      \[\leadsto \frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \color{blue}{\left(2 \cdot {k}^{2}\right)}} \]
   6. Step-by-step derivation

      1. unpow2 57.0%

         \[\leadsto \frac{{\ell}^{2}}{\color{blue}{k \cdot k}} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}} \]

   7. Applied egg-rr 52.6%

      \[\leadsto \frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(2 \cdot \color{blue}{\left(k \cdot k\right)}\right)} \]
   8. Step-by-step derivation

      1. add-sqr-sqrt 25.3%

         \[\leadsto \frac{2}{\color{blue}{\left(\sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}} \cdot \sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}}\right)} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      2. pow2 25.3%

         \[\leadsto \frac{2}{\color{blue}{{\left(\sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}}\right)}^{2}} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      3. associate-/r* 21.5%

         \[\leadsto \frac{2}{{\left(\sqrt{\color{blue}{\frac{{t}^{3}}{\ell \cdot \ell}}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      4. sqrt-div 21.5%

         \[\leadsto \frac{2}{{\color{blue}{\left(\frac{\sqrt{{t}^{3}}}{\sqrt{\ell \cdot \ell}}\right)}}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      5. sqrt-pow1 24.6%

         \[\leadsto \frac{2}{{\left(\frac{\color{blue}{{t}^{\left(\frac{3}{2}\right)}}}{\sqrt{\ell \cdot \ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      6. metadata-eval 24.6%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{\color{blue}{1.5}}}{\sqrt{\ell \cdot \ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      7. sqrt-prod 16.1%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{1.5}}{\color{blue}{\sqrt{\ell} \cdot \sqrt{\ell}}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      8. add-sqr-sqrt 30.8%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{1.5}}{\color{blue}{\ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   9. Applied egg-rr 27.9%

      \[\leadsto \frac{2}{\color{blue}{{\left(\frac{{t}^{1.5}}{\ell}\right)}^{2}} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

   if 1.85e-4 < k < 3.5000000000000002e108

   1. Initial program 61.9%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]
   2. Add Preprocessing

   3. Taylor expanded in k around 0 55.2%

      \[\leadsto \frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \color{blue}{2}} \]

   if 3.5000000000000002e108 < k

   1. Initial program 44.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 44.4%

      \[\leadsto \color{blue}{\frac{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}{2 + {\left(\frac{k}{t}\right)}^{2}}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. add-cube-cbrt 44.4%

         \[\leadsto \frac{\color{blue}{\left(\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}\right) \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

      2. add-sqr-sqrt 44.4%

         \[\leadsto \frac{\left(\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}\right) \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\color{blue}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}} \cdot \sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

      3. times-frac 44.4%

         \[\leadsto \color{blue}{\frac{\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}} \cdot \frac{\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

   6. Applied egg-rr 53.0%

      \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}} \cdot \frac{\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]
   7. Step-by-step derivation

      1. associate-*l/ 53.0%

         \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2} \cdot \frac{\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

   8. Simplified 56.0%

      \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{{\ell}^{2} \cdot \frac{2}{\tan k}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2} \cdot \frac{\frac{\sqrt[3]{{\ell}^{2} \cdot \frac{2}{\tan k}}}{t \cdot \sqrt[3]{\sin k}}}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}} \]

   9. Taylor expanded in k around inf 59.6%

      \[\leadsto \color{blue}{\frac{{\ell}^{2} \cdot \left({\left(\sqrt[3]{2}\right)}^{3} \cdot \cos k\right)}{{k}^{2} \cdot \left(t \cdot {\sin k}^{2}\right)}} \]
   10. Step-by-step derivation

       1. times-frac 59.6%

          \[\leadsto \color{blue}{\frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{{\left(\sqrt[3]{2}\right)}^{3} \cdot \cos k}{t \cdot {\sin k}^{2}}} \]

       2. rem-cube-cbrt 59.7%

          \[\leadsto \frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{\color{blue}{2} \cdot \cos k}{t \cdot {\sin k}^{2}} \]

   11. Simplified 59.7%

       \[\leadsto \color{blue}{\frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}}} \]

   12. Taylor expanded in k around 0 54.2%

       \[\leadsto \frac{{\ell}^{2}}{{k}^{2}} \cdot \color{blue}{\frac{2}{{k}^{2} \cdot t}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 34.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 0.000185:\\ \;\;\;\;\frac{2}{{\left(\frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \mathbf{elif}\;k \leq 3.5 \cdot 10^{+108}:\\ \;\;\;\;\frac{2}{2 \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t}^{3}}{\ell \cdot \ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{t \cdot {k}^{2}} \cdot \frac{{\ell}^{2}}{{k}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 17: 60.3% accurate, 1.3× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 1.35 \cdot 10^{+107}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t\_m}^{1.5}}{\ell}\right)}^{2} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{t\_m \cdot {k}^{2}} \cdot \frac{{\ell}^{2}}{{k}^{2}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 1.35e+107)
    (/ 2.0 (* (pow (/ (pow t_m 1.5) l) 2.0) (* 2.0 (* k k))))
    (* (/ 2.0 (* t_m (pow k 2.0))) (/ (pow l 2.0) (pow k 2.0))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 1.35e+107) {
		tmp = 2.0 / (pow((pow(t_m, 1.5) / l), 2.0) * (2.0 * (k * k)));
	} else {
		tmp = (2.0 / (t_m * pow(k, 2.0))) * (pow(l, 2.0) / pow(k, 2.0));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 1.35d+107) then
        tmp = 2.0d0 / ((((t_m ** 1.5d0) / l) ** 2.0d0) * (2.0d0 * (k * k)))
    else
        tmp = (2.0d0 / (t_m * (k ** 2.0d0))) * ((l ** 2.0d0) / (k ** 2.0d0))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 1.35e+107) {
		tmp = 2.0 / (Math.pow((Math.pow(t_m, 1.5) / l), 2.0) * (2.0 * (k * k)));
	} else {
		tmp = (2.0 / (t_m * Math.pow(k, 2.0))) * (Math.pow(l, 2.0) / Math.pow(k, 2.0));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 1.35e+107:
		tmp = 2.0 / (math.pow((math.pow(t_m, 1.5) / l), 2.0) * (2.0 * (k * k)))
	else:
		tmp = (2.0 / (t_m * math.pow(k, 2.0))) * (math.pow(l, 2.0) / math.pow(k, 2.0))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 1.35e+107)
		tmp = Float64(2.0 / Float64((Float64((t_m ^ 1.5) / l) ^ 2.0) * Float64(2.0 * Float64(k * k))));
	else
		tmp = Float64(Float64(2.0 / Float64(t_m * (k ^ 2.0))) * Float64((l ^ 2.0) / (k ^ 2.0)));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 1.35e+107)
		tmp = 2.0 / ((((t_m ^ 1.5) / l) ^ 2.0) * (2.0 * (k * k)));
	else
		tmp = (2.0 / (t_m * (k ^ 2.0))) * ((l ^ 2.0) / (k ^ 2.0));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 1.35e+107], N[(2.0 / N[(N[Power[N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision], 2.0], $MachinePrecision] * N[(2.0 * N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(2.0 / N[(t$95$m * N[Power[k, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[Power[l, 2.0], $MachinePrecision] / N[Power[k, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 1.3500000000000001e107

   1. Initial program 51.5%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 56.5%

      \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
   4. Add Preprocessing

   5. Taylor expanded in k around 0 51.7%

      \[\leadsto \frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \color{blue}{\left(2 \cdot {k}^{2}\right)}} \]
   6. Step-by-step derivation

      1. unpow2 57.9%

         \[\leadsto \frac{{\ell}^{2}}{\color{blue}{k \cdot k}} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}} \]

   7. Applied egg-rr 51.7%

      \[\leadsto \frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(2 \cdot \color{blue}{\left(k \cdot k\right)}\right)} \]
   8. Step-by-step derivation

      1. add-sqr-sqrt 25.1%

         \[\leadsto \frac{2}{\color{blue}{\left(\sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}} \cdot \sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}}\right)} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      2. pow2 25.1%

         \[\leadsto \frac{2}{\color{blue}{{\left(\sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}}\right)}^{2}} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      3. associate-/r* 21.2%

         \[\leadsto \frac{2}{{\left(\sqrt{\color{blue}{\frac{{t}^{3}}{\ell \cdot \ell}}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      4. sqrt-div 21.2%

         \[\leadsto \frac{2}{{\color{blue}{\left(\frac{\sqrt{{t}^{3}}}{\sqrt{\ell \cdot \ell}}\right)}}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      5. sqrt-pow1 24.4%

         \[\leadsto \frac{2}{{\left(\frac{\color{blue}{{t}^{\left(\frac{3}{2}\right)}}}{\sqrt{\ell \cdot \ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      6. metadata-eval 24.4%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{\color{blue}{1.5}}}{\sqrt{\ell \cdot \ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      7. sqrt-prod 17.0%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{1.5}}{\color{blue}{\sqrt{\ell} \cdot \sqrt{\ell}}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      8. add-sqr-sqrt 30.3%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{1.5}}{\color{blue}{\ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   9. Applied egg-rr 26.0%

      \[\leadsto \frac{2}{\color{blue}{{\left(\frac{{t}^{1.5}}{\ell}\right)}^{2}} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

   if 1.3500000000000001e107 < k

   1. Initial program 44.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 44.4%

      \[\leadsto \color{blue}{\frac{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}{2 + {\left(\frac{k}{t}\right)}^{2}}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. add-cube-cbrt 44.4%

         \[\leadsto \frac{\color{blue}{\left(\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}\right) \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

      2. add-sqr-sqrt 44.4%

         \[\leadsto \frac{\left(\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}\right) \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\color{blue}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}} \cdot \sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

      3. times-frac 44.4%

         \[\leadsto \color{blue}{\frac{\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}} \cdot \frac{\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

   6. Applied egg-rr 53.0%

      \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}} \cdot \frac{\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]
   7. Step-by-step derivation

      1. associate-*l/ 53.0%

         \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2} \cdot \frac{\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

   8. Simplified 56.0%

      \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{{\ell}^{2} \cdot \frac{2}{\tan k}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2} \cdot \frac{\frac{\sqrt[3]{{\ell}^{2} \cdot \frac{2}{\tan k}}}{t \cdot \sqrt[3]{\sin k}}}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}} \]

   9. Taylor expanded in k around inf 59.6%

      \[\leadsto \color{blue}{\frac{{\ell}^{2} \cdot \left({\left(\sqrt[3]{2}\right)}^{3} \cdot \cos k\right)}{{k}^{2} \cdot \left(t \cdot {\sin k}^{2}\right)}} \]
   10. Step-by-step derivation

       1. times-frac 59.6%

          \[\leadsto \color{blue}{\frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{{\left(\sqrt[3]{2}\right)}^{3} \cdot \cos k}{t \cdot {\sin k}^{2}}} \]

       2. rem-cube-cbrt 59.7%

          \[\leadsto \frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{\color{blue}{2} \cdot \cos k}{t \cdot {\sin k}^{2}} \]

   11. Simplified 59.7%

       \[\leadsto \color{blue}{\frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}}} \]

   12. Taylor expanded in k around 0 54.2%

       \[\leadsto \frac{{\ell}^{2}}{{k}^{2}} \cdot \color{blue}{\frac{2}{{k}^{2} \cdot t}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 30.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 1.35 \cdot 10^{+107}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{t \cdot {k}^{2}} \cdot \frac{{\ell}^{2}}{{k}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 18: 60.2% accurate, 1.3× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 2.8 \cdot 10^{+105}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t\_m}^{1.5}}{\ell}\right)}^{2} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \frac{{\ell}^{2}}{{k}^{3} \cdot \left(t\_m \cdot \sin k\right)}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 2.8e+105)
    (/ 2.0 (* (pow (/ (pow t_m 1.5) l) 2.0) (* 2.0 (* k k))))
    (* 2.0 (/ (pow l 2.0) (* (pow k 3.0) (* t_m (sin k))))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 2.8e+105) {
		tmp = 2.0 / (pow((pow(t_m, 1.5) / l), 2.0) * (2.0 * (k * k)));
	} else {
		tmp = 2.0 * (pow(l, 2.0) / (pow(k, 3.0) * (t_m * sin(k))));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 2.8d+105) then
        tmp = 2.0d0 / ((((t_m ** 1.5d0) / l) ** 2.0d0) * (2.0d0 * (k * k)))
    else
        tmp = 2.0d0 * ((l ** 2.0d0) / ((k ** 3.0d0) * (t_m * sin(k))))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 2.8e+105) {
		tmp = 2.0 / (Math.pow((Math.pow(t_m, 1.5) / l), 2.0) * (2.0 * (k * k)));
	} else {
		tmp = 2.0 * (Math.pow(l, 2.0) / (Math.pow(k, 3.0) * (t_m * Math.sin(k))));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 2.8e+105:
		tmp = 2.0 / (math.pow((math.pow(t_m, 1.5) / l), 2.0) * (2.0 * (k * k)))
	else:
		tmp = 2.0 * (math.pow(l, 2.0) / (math.pow(k, 3.0) * (t_m * math.sin(k))))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 2.8e+105)
		tmp = Float64(2.0 / Float64((Float64((t_m ^ 1.5) / l) ^ 2.0) * Float64(2.0 * Float64(k * k))));
	else
		tmp = Float64(2.0 * Float64((l ^ 2.0) / Float64((k ^ 3.0) * Float64(t_m * sin(k)))));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 2.8e+105)
		tmp = 2.0 / ((((t_m ^ 1.5) / l) ^ 2.0) * (2.0 * (k * k)));
	else
		tmp = 2.0 * ((l ^ 2.0) / ((k ^ 3.0) * (t_m * sin(k))));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 2.8e+105], N[(2.0 / N[(N[Power[N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision], 2.0], $MachinePrecision] * N[(2.0 * N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 * N[(N[Power[l, 2.0], $MachinePrecision] / N[(N[Power[k, 3.0], $MachinePrecision] * N[(t$95$m * N[Sin[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 2.8000000000000001e105

   1. Initial program 51.5%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 56.5%

      \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
   4. Add Preprocessing

   5. Taylor expanded in k around 0 51.7%

      \[\leadsto \frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \color{blue}{\left(2 \cdot {k}^{2}\right)}} \]
   6. Step-by-step derivation

      1. unpow2 57.9%

         \[\leadsto \frac{{\ell}^{2}}{\color{blue}{k \cdot k}} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}} \]

   7. Applied egg-rr 51.7%

      \[\leadsto \frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(2 \cdot \color{blue}{\left(k \cdot k\right)}\right)} \]
   8. Step-by-step derivation

      1. add-sqr-sqrt 25.1%

         \[\leadsto \frac{2}{\color{blue}{\left(\sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}} \cdot \sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}}\right)} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      2. pow2 25.1%

         \[\leadsto \frac{2}{\color{blue}{{\left(\sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}}\right)}^{2}} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      3. associate-/r* 21.2%

         \[\leadsto \frac{2}{{\left(\sqrt{\color{blue}{\frac{{t}^{3}}{\ell \cdot \ell}}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      4. sqrt-div 21.2%

         \[\leadsto \frac{2}{{\color{blue}{\left(\frac{\sqrt{{t}^{3}}}{\sqrt{\ell \cdot \ell}}\right)}}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      5. sqrt-pow1 24.4%

         \[\leadsto \frac{2}{{\left(\frac{\color{blue}{{t}^{\left(\frac{3}{2}\right)}}}{\sqrt{\ell \cdot \ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      6. metadata-eval 24.4%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{\color{blue}{1.5}}}{\sqrt{\ell \cdot \ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      7. sqrt-prod 17.0%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{1.5}}{\color{blue}{\sqrt{\ell} \cdot \sqrt{\ell}}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      8. add-sqr-sqrt 30.3%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{1.5}}{\color{blue}{\ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   9. Applied egg-rr 26.0%

      \[\leadsto \frac{2}{\color{blue}{{\left(\frac{{t}^{1.5}}{\ell}\right)}^{2}} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

   if 2.8000000000000001e105 < k

   1. Initial program 44.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 44.4%

      \[\leadsto \color{blue}{\frac{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}{2 + {\left(\frac{k}{t}\right)}^{2}}} \]
   4. Add Preprocessing

   5. Taylor expanded in k around 0 41.8%

      \[\leadsto \frac{\frac{\color{blue}{\frac{2}{k}}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

   6. Taylor expanded in k around inf 54.2%

      \[\leadsto \color{blue}{2 \cdot \frac{{\ell}^{2}}{{k}^{3} \cdot \left(t \cdot \sin k\right)}} \]
3. Recombined 2 regimes into one program.
4. Add Preprocessing

Alternative 19: 60.2% accurate, 1.9× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 2.6 \cdot 10^{+108}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t\_m}^{1.5}}{\ell}\right)}^{2} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \frac{{\ell}^{2}}{t\_m \cdot {k}^{4}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 2.6e+108)
    (/ 2.0 (* (pow (/ (pow t_m 1.5) l) 2.0) (* 2.0 (* k k))))
    (* 2.0 (/ (pow l 2.0) (* t_m (pow k 4.0)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 2.6e+108) {
		tmp = 2.0 / (pow((pow(t_m, 1.5) / l), 2.0) * (2.0 * (k * k)));
	} else {
		tmp = 2.0 * (pow(l, 2.0) / (t_m * pow(k, 4.0)));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 2.6d+108) then
        tmp = 2.0d0 / ((((t_m ** 1.5d0) / l) ** 2.0d0) * (2.0d0 * (k * k)))
    else
        tmp = 2.0d0 * ((l ** 2.0d0) / (t_m * (k ** 4.0d0)))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 2.6e+108) {
		tmp = 2.0 / (Math.pow((Math.pow(t_m, 1.5) / l), 2.0) * (2.0 * (k * k)));
	} else {
		tmp = 2.0 * (Math.pow(l, 2.0) / (t_m * Math.pow(k, 4.0)));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 2.6e+108:
		tmp = 2.0 / (math.pow((math.pow(t_m, 1.5) / l), 2.0) * (2.0 * (k * k)))
	else:
		tmp = 2.0 * (math.pow(l, 2.0) / (t_m * math.pow(k, 4.0)))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 2.6e+108)
		tmp = Float64(2.0 / Float64((Float64((t_m ^ 1.5) / l) ^ 2.0) * Float64(2.0 * Float64(k * k))));
	else
		tmp = Float64(2.0 * Float64((l ^ 2.0) / Float64(t_m * (k ^ 4.0))));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 2.6e+108)
		tmp = 2.0 / ((((t_m ^ 1.5) / l) ^ 2.0) * (2.0 * (k * k)));
	else
		tmp = 2.0 * ((l ^ 2.0) / (t_m * (k ^ 4.0)));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 2.6e+108], N[(2.0 / N[(N[Power[N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision], 2.0], $MachinePrecision] * N[(2.0 * N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 * N[(N[Power[l, 2.0], $MachinePrecision] / N[(t$95$m * N[Power[k, 4.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 2.6000000000000002e108

   1. Initial program 51.5%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 56.5%

      \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
   4. Add Preprocessing

   5. Taylor expanded in k around 0 51.7%

      \[\leadsto \frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \color{blue}{\left(2 \cdot {k}^{2}\right)}} \]
   6. Step-by-step derivation

      1. unpow2 57.9%

         \[\leadsto \frac{{\ell}^{2}}{\color{blue}{k \cdot k}} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}} \]

   7. Applied egg-rr 51.7%

      \[\leadsto \frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(2 \cdot \color{blue}{\left(k \cdot k\right)}\right)} \]
   8. Step-by-step derivation

      1. add-sqr-sqrt 25.1%

         \[\leadsto \frac{2}{\color{blue}{\left(\sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}} \cdot \sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}}\right)} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      2. pow2 25.1%

         \[\leadsto \frac{2}{\color{blue}{{\left(\sqrt{\frac{\frac{{t}^{3}}{\ell}}{\ell}}\right)}^{2}} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      3. associate-/r* 21.2%

         \[\leadsto \frac{2}{{\left(\sqrt{\color{blue}{\frac{{t}^{3}}{\ell \cdot \ell}}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      4. sqrt-div 21.2%

         \[\leadsto \frac{2}{{\color{blue}{\left(\frac{\sqrt{{t}^{3}}}{\sqrt{\ell \cdot \ell}}\right)}}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      5. sqrt-pow1 24.4%

         \[\leadsto \frac{2}{{\left(\frac{\color{blue}{{t}^{\left(\frac{3}{2}\right)}}}{\sqrt{\ell \cdot \ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      6. metadata-eval 24.4%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{\color{blue}{1.5}}}{\sqrt{\ell \cdot \ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      7. sqrt-prod 17.0%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{1.5}}{\color{blue}{\sqrt{\ell} \cdot \sqrt{\ell}}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      8. add-sqr-sqrt 30.3%

         \[\leadsto \frac{2}{{\left(\frac{{t}^{1.5}}{\color{blue}{\ell}}\right)}^{2} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   9. Applied egg-rr 26.0%

      \[\leadsto \frac{2}{\color{blue}{{\left(\frac{{t}^{1.5}}{\ell}\right)}^{2}} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

   if 2.6000000000000002e108 < k

   1. Initial program 44.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 44.4%

      \[\leadsto \color{blue}{\frac{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}{2 + {\left(\frac{k}{t}\right)}^{2}}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. add-cube-cbrt 44.4%

         \[\leadsto \frac{\color{blue}{\left(\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}\right) \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

      2. add-sqr-sqrt 44.4%

         \[\leadsto \frac{\left(\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}\right) \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\color{blue}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}} \cdot \sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

      3. times-frac 44.4%

         \[\leadsto \color{blue}{\frac{\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}} \cdot \frac{\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

   6. Applied egg-rr 53.0%

      \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}} \cdot \frac{\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]
   7. Step-by-step derivation

      1. associate-*l/ 53.0%

         \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2} \cdot \frac{\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

   8. Simplified 56.0%

      \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{{\ell}^{2} \cdot \frac{2}{\tan k}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2} \cdot \frac{\frac{\sqrt[3]{{\ell}^{2} \cdot \frac{2}{\tan k}}}{t \cdot \sqrt[3]{\sin k}}}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}} \]

   9. Taylor expanded in k around inf 59.6%

      \[\leadsto \color{blue}{\frac{{\ell}^{2} \cdot \left({\left(\sqrt[3]{2}\right)}^{3} \cdot \cos k\right)}{{k}^{2} \cdot \left(t \cdot {\sin k}^{2}\right)}} \]
   10. Step-by-step derivation

       1. times-frac 59.6%

          \[\leadsto \color{blue}{\frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{{\left(\sqrt[3]{2}\right)}^{3} \cdot \cos k}{t \cdot {\sin k}^{2}}} \]

       2. rem-cube-cbrt 59.7%

          \[\leadsto \frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{\color{blue}{2} \cdot \cos k}{t \cdot {\sin k}^{2}} \]

   11. Simplified 59.7%

       \[\leadsto \color{blue}{\frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}}} \]

   12. Taylor expanded in k around 0 54.2%

       \[\leadsto \color{blue}{2 \cdot \frac{{\ell}^{2}}{{k}^{4} \cdot t}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 30.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 2.6 \cdot 10^{+108}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t}^{1.5}}{\ell}\right)}^{2} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \frac{{\ell}^{2}}{t \cdot {k}^{4}}\\ \end{array} \]
5. Add Preprocessing

Alternative 20: 56.7% accurate, 1.9× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 5.8 \cdot 10^{+106}:\\ \;\;\;\;\frac{2}{\frac{{t\_m}^{3}}{\ell} \cdot \frac{2 \cdot {k}^{2}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \frac{{\ell}^{2}}{t\_m \cdot {k}^{4}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 5.8e+106)
    (/ 2.0 (* (/ (pow t_m 3.0) l) (/ (* 2.0 (pow k 2.0)) l)))
    (* 2.0 (/ (pow l 2.0) (* t_m (pow k 4.0)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 5.8e+106) {
		tmp = 2.0 / ((pow(t_m, 3.0) / l) * ((2.0 * pow(k, 2.0)) / l));
	} else {
		tmp = 2.0 * (pow(l, 2.0) / (t_m * pow(k, 4.0)));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 5.8d+106) then
        tmp = 2.0d0 / (((t_m ** 3.0d0) / l) * ((2.0d0 * (k ** 2.0d0)) / l))
    else
        tmp = 2.0d0 * ((l ** 2.0d0) / (t_m * (k ** 4.0d0)))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 5.8e+106) {
		tmp = 2.0 / ((Math.pow(t_m, 3.0) / l) * ((2.0 * Math.pow(k, 2.0)) / l));
	} else {
		tmp = 2.0 * (Math.pow(l, 2.0) / (t_m * Math.pow(k, 4.0)));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 5.8e+106:
		tmp = 2.0 / ((math.pow(t_m, 3.0) / l) * ((2.0 * math.pow(k, 2.0)) / l))
	else:
		tmp = 2.0 * (math.pow(l, 2.0) / (t_m * math.pow(k, 4.0)))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 5.8e+106)
		tmp = Float64(2.0 / Float64(Float64((t_m ^ 3.0) / l) * Float64(Float64(2.0 * (k ^ 2.0)) / l)));
	else
		tmp = Float64(2.0 * Float64((l ^ 2.0) / Float64(t_m * (k ^ 4.0))));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 5.8e+106)
		tmp = 2.0 / (((t_m ^ 3.0) / l) * ((2.0 * (k ^ 2.0)) / l));
	else
		tmp = 2.0 * ((l ^ 2.0) / (t_m * (k ^ 4.0)));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 5.8e+106], N[(2.0 / N[(N[(N[Power[t$95$m, 3.0], $MachinePrecision] / l), $MachinePrecision] * N[(N[(2.0 * N[Power[k, 2.0], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 * N[(N[Power[l, 2.0], $MachinePrecision] / N[(t$95$m * N[Power[k, 4.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 5.8000000000000004e106

   1. Initial program 51.5%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 56.5%

      \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
   4. Add Preprocessing

   5. Taylor expanded in k around 0 51.7%

      \[\leadsto \frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \color{blue}{\left(2 \cdot {k}^{2}\right)}} \]
   6. Step-by-step derivation

      1. associate-*l/ 53.8%

         \[\leadsto \frac{2}{\color{blue}{\frac{\frac{{t}^{3}}{\ell} \cdot \left(2 \cdot {k}^{2}\right)}{\ell}}} \]

   7. Applied egg-rr 53.8%

      \[\leadsto \frac{2}{\color{blue}{\frac{\frac{{t}^{3}}{\ell} \cdot \left(2 \cdot {k}^{2}\right)}{\ell}}} \]
   8. Step-by-step derivation

      1. associate-/l* 54.1%

         \[\leadsto \frac{2}{\color{blue}{\frac{{t}^{3}}{\ell} \cdot \frac{2 \cdot {k}^{2}}{\ell}}} \]

   9. Simplified 54.1%

      \[\leadsto \frac{2}{\color{blue}{\frac{{t}^{3}}{\ell} \cdot \frac{2 \cdot {k}^{2}}{\ell}}} \]

   if 5.8000000000000004e106 < k

   1. Initial program 44.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 44.4%

      \[\leadsto \color{blue}{\frac{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}{2 + {\left(\frac{k}{t}\right)}^{2}}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. add-cube-cbrt 44.4%

         \[\leadsto \frac{\color{blue}{\left(\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}\right) \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

      2. add-sqr-sqrt 44.4%

         \[\leadsto \frac{\left(\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}\right) \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\color{blue}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}} \cdot \sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

      3. times-frac 44.4%

         \[\leadsto \color{blue}{\frac{\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}} \cdot \frac{\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

   6. Applied egg-rr 53.0%

      \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}} \cdot \frac{\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]
   7. Step-by-step derivation

      1. associate-*l/ 53.0%

         \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2} \cdot \frac{\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

   8. Simplified 56.0%

      \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{{\ell}^{2} \cdot \frac{2}{\tan k}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2} \cdot \frac{\frac{\sqrt[3]{{\ell}^{2} \cdot \frac{2}{\tan k}}}{t \cdot \sqrt[3]{\sin k}}}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}} \]

   9. Taylor expanded in k around inf 59.6%

      \[\leadsto \color{blue}{\frac{{\ell}^{2} \cdot \left({\left(\sqrt[3]{2}\right)}^{3} \cdot \cos k\right)}{{k}^{2} \cdot \left(t \cdot {\sin k}^{2}\right)}} \]
   10. Step-by-step derivation

       1. times-frac 59.6%

          \[\leadsto \color{blue}{\frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{{\left(\sqrt[3]{2}\right)}^{3} \cdot \cos k}{t \cdot {\sin k}^{2}}} \]

       2. rem-cube-cbrt 59.7%

          \[\leadsto \frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{\color{blue}{2} \cdot \cos k}{t \cdot {\sin k}^{2}} \]

   11. Simplified 59.7%

       \[\leadsto \color{blue}{\frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}}} \]

   12. Taylor expanded in k around 0 54.2%

       \[\leadsto \color{blue}{2 \cdot \frac{{\ell}^{2}}{{k}^{4} \cdot t}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 54.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 5.8 \cdot 10^{+106}:\\ \;\;\;\;\frac{2}{\frac{{t}^{3}}{\ell} \cdot \frac{2 \cdot {k}^{2}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \frac{{\ell}^{2}}{t \cdot {k}^{4}}\\ \end{array} \]
5. Add Preprocessing

Alternative 21: 58.6% accurate, 2.0× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 2.8 \cdot 10^{+105}:\\ \;\;\;\;\frac{2}{\left(2 \cdot \left(k \cdot k\right)\right) \cdot \left(\frac{t\_m}{\ell} \cdot \frac{t\_m \cdot t\_m}{\ell}\right)}\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \frac{{\ell}^{2}}{t\_m \cdot {k}^{4}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 2.8e+105)
    (/ 2.0 (* (* 2.0 (* k k)) (* (/ t_m l) (/ (* t_m t_m) l))))
    (* 2.0 (/ (pow l 2.0) (* t_m (pow k 4.0)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 2.8e+105) {
		tmp = 2.0 / ((2.0 * (k * k)) * ((t_m / l) * ((t_m * t_m) / l)));
	} else {
		tmp = 2.0 * (pow(l, 2.0) / (t_m * pow(k, 4.0)));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 2.8d+105) then
        tmp = 2.0d0 / ((2.0d0 * (k * k)) * ((t_m / l) * ((t_m * t_m) / l)))
    else
        tmp = 2.0d0 * ((l ** 2.0d0) / (t_m * (k ** 4.0d0)))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 2.8e+105) {
		tmp = 2.0 / ((2.0 * (k * k)) * ((t_m / l) * ((t_m * t_m) / l)));
	} else {
		tmp = 2.0 * (Math.pow(l, 2.0) / (t_m * Math.pow(k, 4.0)));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 2.8e+105:
		tmp = 2.0 / ((2.0 * (k * k)) * ((t_m / l) * ((t_m * t_m) / l)))
	else:
		tmp = 2.0 * (math.pow(l, 2.0) / (t_m * math.pow(k, 4.0)))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 2.8e+105)
		tmp = Float64(2.0 / Float64(Float64(2.0 * Float64(k * k)) * Float64(Float64(t_m / l) * Float64(Float64(t_m * t_m) / l))));
	else
		tmp = Float64(2.0 * Float64((l ^ 2.0) / Float64(t_m * (k ^ 4.0))));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 2.8e+105)
		tmp = 2.0 / ((2.0 * (k * k)) * ((t_m / l) * ((t_m * t_m) / l)));
	else
		tmp = 2.0 * ((l ^ 2.0) / (t_m * (k ^ 4.0)));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 2.8e+105], N[(2.0 / N[(N[(2.0 * N[(k * k), $MachinePrecision]), $MachinePrecision] * N[(N[(t$95$m / l), $MachinePrecision] * N[(N[(t$95$m * t$95$m), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 * N[(N[Power[l, 2.0], $MachinePrecision] / N[(t$95$m * N[Power[k, 4.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 2.8000000000000001e105

   1. Initial program 51.5%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 56.5%

      \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
   4. Add Preprocessing

   5. Taylor expanded in k around 0 51.7%

      \[\leadsto \frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \color{blue}{\left(2 \cdot {k}^{2}\right)}} \]
   6. Step-by-step derivation

      1. unpow2 57.9%

         \[\leadsto \frac{{\ell}^{2}}{\color{blue}{k \cdot k}} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}} \]

   7. Applied egg-rr 51.7%

      \[\leadsto \frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(2 \cdot \color{blue}{\left(k \cdot k\right)}\right)} \]
   8. Step-by-step derivation

      1. associate-/r* 50.0%

         \[\leadsto \frac{2}{\color{blue}{\frac{{t}^{3}}{\ell \cdot \ell}} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      2. unpow3 50.0%

         \[\leadsto \frac{2}{\frac{\color{blue}{\left(t \cdot t\right) \cdot t}}{\ell \cdot \ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      3. times-frac 60.2%

         \[\leadsto \frac{2}{\color{blue}{\left(\frac{t \cdot t}{\ell} \cdot \frac{t}{\ell}\right)} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

      4. pow2 60.2%

         \[\leadsto \frac{2}{\left(\frac{\color{blue}{{t}^{2}}}{\ell} \cdot \frac{t}{\ell}\right) \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   9. Applied egg-rr 53.5%

      \[\leadsto \frac{2}{\color{blue}{\left(\frac{{t}^{2}}{\ell} \cdot \frac{t}{\ell}\right)} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]
   10. Step-by-step derivation

       1. unpow2 53.5%

          \[\leadsto \frac{2}{\left(\frac{\color{blue}{t \cdot t}}{\ell} \cdot \frac{t}{\ell}\right) \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

   11. Applied egg-rr 53.5%

       \[\leadsto \frac{2}{\left(\frac{\color{blue}{t \cdot t}}{\ell} \cdot \frac{t}{\ell}\right) \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

   if 2.8000000000000001e105 < k

   1. Initial program 44.4%

      \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

   3. Simplified 44.4%

      \[\leadsto \color{blue}{\frac{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}{2 + {\left(\frac{k}{t}\right)}^{2}}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. add-cube-cbrt 44.4%

         \[\leadsto \frac{\color{blue}{\left(\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}\right) \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}}{2 + {\left(\frac{k}{t}\right)}^{2}} \]

      2. add-sqr-sqrt 44.4%

         \[\leadsto \frac{\left(\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}\right) \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\color{blue}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}} \cdot \sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

      3. times-frac 44.4%

         \[\leadsto \color{blue}{\frac{\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)} \cdot \sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}} \cdot \frac{\sqrt[3]{\frac{\frac{2}{\tan k}}{{t}^{3} \cdot \sin k} \cdot \left(\ell \cdot \ell\right)}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

   6. Applied egg-rr 53.0%

      \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}} \cdot \frac{\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]
   7. Step-by-step derivation

      1. associate-*l/ 53.0%

         \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2} \cdot \frac{\frac{\sqrt[3]{\frac{2}{\tan k} \cdot {\ell}^{2}}}{t \cdot \sqrt[3]{\sin k}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}}{\sqrt{2 + {\left(\frac{k}{t}\right)}^{2}}}} \]

   8. Simplified 56.0%

      \[\leadsto \color{blue}{\frac{{\left(\frac{\sqrt[3]{{\ell}^{2} \cdot \frac{2}{\tan k}}}{t \cdot \sqrt[3]{\sin k}}\right)}^{2} \cdot \frac{\frac{\sqrt[3]{{\ell}^{2} \cdot \frac{2}{\tan k}}}{t \cdot \sqrt[3]{\sin k}}}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}}{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}} \]

   9. Taylor expanded in k around inf 59.6%

      \[\leadsto \color{blue}{\frac{{\ell}^{2} \cdot \left({\left(\sqrt[3]{2}\right)}^{3} \cdot \cos k\right)}{{k}^{2} \cdot \left(t \cdot {\sin k}^{2}\right)}} \]
   10. Step-by-step derivation

       1. times-frac 59.6%

          \[\leadsto \color{blue}{\frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{{\left(\sqrt[3]{2}\right)}^{3} \cdot \cos k}{t \cdot {\sin k}^{2}}} \]

       2. rem-cube-cbrt 59.7%

          \[\leadsto \frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{\color{blue}{2} \cdot \cos k}{t \cdot {\sin k}^{2}} \]

   11. Simplified 59.7%

       \[\leadsto \color{blue}{\frac{{\ell}^{2}}{{k}^{2}} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}}} \]

   12. Taylor expanded in k around 0 54.2%

       \[\leadsto \color{blue}{2 \cdot \frac{{\ell}^{2}}{{k}^{4} \cdot t}} \]
3. Recombined 2 regimes into one program.

4. Final simplification 53.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 2.8 \cdot 10^{+105}:\\ \;\;\;\;\frac{2}{\left(2 \cdot \left(k \cdot k\right)\right) \cdot \left(\frac{t}{\ell} \cdot \frac{t \cdot t}{\ell}\right)}\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \frac{{\ell}^{2}}{t \cdot {k}^{4}}\\ \end{array} \]
5. Add Preprocessing

Alternative 22: 57.2% accurate, 24.8× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \frac{2}{\left(2 \cdot \left(k \cdot k\right)\right) \cdot \left(\frac{t\_m}{\ell} \cdot \frac{t\_m \cdot t\_m}{\ell}\right)} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (* t_s (/ 2.0 (* (* 2.0 (* k k)) (* (/ t_m l) (/ (* t_m t_m) l))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	return t_s * (2.0 / ((2.0 * (k * k)) * ((t_m / l) * ((t_m * t_m) / l))));
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    code = t_s * (2.0d0 / ((2.0d0 * (k * k)) * ((t_m / l) * ((t_m * t_m) / l))))
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	return t_s * (2.0 / ((2.0 * (k * k)) * ((t_m / l) * ((t_m * t_m) / l))));
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	return t_s * (2.0 / ((2.0 * (k * k)) * ((t_m / l) * ((t_m * t_m) / l))))

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	return Float64(t_s * Float64(2.0 / Float64(Float64(2.0 * Float64(k * k)) * Float64(Float64(t_m / l) * Float64(Float64(t_m * t_m) / l)))))
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp = code(t_s, t_m, l, k)
	tmp = t_s * (2.0 / ((2.0 * (k * k)) * ((t_m / l) * ((t_m * t_m) / l))));
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * N[(2.0 / N[(N[(2.0 * N[(k * k), $MachinePrecision]), $MachinePrecision] * N[(N[(t$95$m / l), $MachinePrecision] * N[(N[(t$95$m * t$95$m), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 50.5%

   \[\frac{2}{\left(\left(\frac{{t}^{3}}{\ell \cdot \ell} \cdot \sin k\right) \cdot \tan k\right) \cdot \left(\left(1 + {\left(\frac{k}{t}\right)}^{2}\right) + 1\right)} \]

3. Simplified 55.3%

   \[\leadsto \color{blue}{\frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)}} \]
4. Add Preprocessing

5. Taylor expanded in k around 0 50.8%

   \[\leadsto \frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \color{blue}{\left(2 \cdot {k}^{2}\right)}} \]
6. Step-by-step derivation

   1. unpow2 58.2%

      \[\leadsto \frac{{\ell}^{2}}{\color{blue}{k \cdot k}} \cdot \frac{2 \cdot \cos k}{t \cdot {\sin k}^{2}} \]

7. Applied egg-rr 50.8%

   \[\leadsto \frac{2}{\frac{\frac{{t}^{3}}{\ell}}{\ell} \cdot \left(2 \cdot \color{blue}{\left(k \cdot k\right)}\right)} \]
8. Step-by-step derivation

   1. associate-/r* 49.2%

      \[\leadsto \frac{2}{\color{blue}{\frac{{t}^{3}}{\ell \cdot \ell}} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   2. unpow3 49.2%

      \[\leadsto \frac{2}{\frac{\color{blue}{\left(t \cdot t\right) \cdot t}}{\ell \cdot \ell} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   3. times-frac 58.8%

      \[\leadsto \frac{2}{\color{blue}{\left(\frac{t \cdot t}{\ell} \cdot \frac{t}{\ell}\right)} \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

   4. pow2 58.8%

      \[\leadsto \frac{2}{\left(\frac{\color{blue}{{t}^{2}}}{\ell} \cdot \frac{t}{\ell}\right) \cdot \left(\left(\sin k \cdot \tan k\right) \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right)} \]

9. Applied egg-rr 52.7%

   \[\leadsto \frac{2}{\color{blue}{\left(\frac{{t}^{2}}{\ell} \cdot \frac{t}{\ell}\right)} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]
10. Step-by-step derivation

    1. unpow2 52.7%

       \[\leadsto \frac{2}{\left(\frac{\color{blue}{t \cdot t}}{\ell} \cdot \frac{t}{\ell}\right) \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

11. Applied egg-rr 52.7%

    \[\leadsto \frac{2}{\left(\frac{\color{blue}{t \cdot t}}{\ell} \cdot \frac{t}{\ell}\right) \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

12. Final simplification 52.7%

    \[\leadsto \frac{2}{\left(2 \cdot \left(k \cdot k\right)\right) \cdot \left(\frac{t}{\ell} \cdot \frac{t \cdot t}{\ell}\right)} \]
13. Add Preprocessing

Reproduce

herbie shell --seed 2024172 
(FPCore (t l k)
  :name "Toniolo and Linder, Equation (10+)"
  :precision binary64
  (/ 2.0 (* (* (* (/ (pow t 3.0) (* l l)) (sin k)) (tan k)) (+ (+ 1.0 (pow (/ k t) 2.0)) 1.0))))