Toniolo and Linder, Equation (10+)

Percentage Accurate: 54.1% → 90.5%

Time: 18.2s

Alternatives: 21

Speedup: 3.6×

• 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: 54.1% 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: 90.5% accurate, 0.7× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ \begin{array}{l} t_2 := \frac{t\_m}{\sqrt[3]{\ell}}\\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 4.2 \cdot 10^{-60}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{elif}\;t\_m \leq 9 \cdot 10^{+100}:\\ \;\;\;\;\frac{1}{\sin k \cdot \frac{\tan k}{2 \cdot {\left({t\_m}^{1.5} \cdot \frac{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t\_m}\right)\right)}{\ell}\right)}^{-2}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\tan k \cdot \left(1 + \left(1 + {\left(\frac{k}{t\_m}\right)}^{2}\right)\right)\right) \cdot \left(\sin k \cdot \left({t\_2}^{2} \cdot \frac{t\_2}{\ell}\right)\right)}\\ \end{array} \end{array} \end{array} \]

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 (/ t_m (cbrt l))))
   (*
    t_s
    (if (<= t_m 4.2e-60)
      (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 2.0) (* (sin k) (tan k))))
      (if (<= t_m 9e+100)
        (/
         1.0
         (*
          (sin k)
          (/
           (tan k)
           (*
            2.0
            (pow
             (* (pow t_m 1.5) (/ (hypot 1.0 (hypot 1.0 (/ k t_m))) l))
             -2.0)))))
        (/
         2.0
         (*
          (* (tan k) (+ 1.0 (+ 1.0 (pow (/ k t_m) 2.0))))
          (* (sin k) (* (pow t_2 2.0) (/ t_2 l))))))))))

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 = t_m / cbrt(l);
	double tmp;
	if (t_m <= 4.2e-60) {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * (sin(k) * tan(k)));
	} else if (t_m <= 9e+100) {
		tmp = 1.0 / (sin(k) * (tan(k) / (2.0 * pow((pow(t_m, 1.5) * (hypot(1.0, hypot(1.0, (k / t_m))) / l)), -2.0))));
	} else {
		tmp = 2.0 / ((tan(k) * (1.0 + (1.0 + pow((k / t_m), 2.0)))) * (sin(k) * (pow(t_2, 2.0) * (t_2 / l))));
	}
	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 = t_m / Math.cbrt(l);
	double tmp;
	if (t_m <= 4.2e-60) {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k) * Math.tan(k)));
	} else if (t_m <= 9e+100) {
		tmp = 1.0 / (Math.sin(k) * (Math.tan(k) / (2.0 * Math.pow((Math.pow(t_m, 1.5) * (Math.hypot(1.0, Math.hypot(1.0, (k / t_m))) / l)), -2.0))));
	} else {
		tmp = 2.0 / ((Math.tan(k) * (1.0 + (1.0 + Math.pow((k / t_m), 2.0)))) * (Math.sin(k) * (Math.pow(t_2, 2.0) * (t_2 / l))));
	}
	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(t_m / cbrt(l))
	tmp = 0.0
	if (t_m <= 4.2e-60)
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k) * tan(k))));
	elseif (t_m <= 9e+100)
		tmp = Float64(1.0 / Float64(sin(k) * Float64(tan(k) / Float64(2.0 * (Float64((t_m ^ 1.5) * Float64(hypot(1.0, hypot(1.0, Float64(k / t_m))) / l)) ^ -2.0)))));
	else
		tmp = Float64(2.0 / Float64(Float64(tan(k) * Float64(1.0 + Float64(1.0 + (Float64(k / t_m) ^ 2.0)))) * Float64(sin(k) * Float64((t_2 ^ 2.0) * Float64(t_2 / l)))));
	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[(t$95$m / N[Power[l, 1/3], $MachinePrecision]), $MachinePrecision]}, N[(t$95$s * If[LessEqual[t$95$m, 4.2e-60], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$m, 9e+100], N[(1.0 / N[(N[Sin[k], $MachinePrecision] * N[(N[Tan[k], $MachinePrecision] / N[(2.0 * N[Power[N[(N[Power[t$95$m, 1.5], $MachinePrecision] * N[(N[Sqrt[1.0 ^ 2 + N[Sqrt[1.0 ^ 2 + N[(k / t$95$m), $MachinePrecision] ^ 2], $MachinePrecision] ^ 2], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], -2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[Tan[k], $MachinePrecision] * N[(1.0 + N[(1.0 + N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[(N[Power[t$95$2, 2.0], $MachinePrecision] * N[(t$95$2 / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]

Derivation

1. Split input into 3 regimes

2. if t < 4.19999999999999982e-60

   1. Initial program 50.9%

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

   3. Simplified 50.9%

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

   6. Applied egg-rr 16.0%

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

      1. associate-*r* 16.0%

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

      2. unpow-prod-down 16.0%

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

      3. pow2 16.0%

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

      4. add-sqr-sqrt 20.4%

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

   8. Applied egg-rr 20.4%

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

   9. Taylor expanded in t around 0 25.2%

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

   if 4.19999999999999982e-60 < t < 9.00000000000000073e100

   1. Initial program 78.9%

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

   3. Simplified 78.8%

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

   6. Applied egg-rr 61.9%

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

      1. associate-*r* 61.9%

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

      2. unpow-prod-down 58.7%

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

      3. pow2 58.7%

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

      4. add-sqr-sqrt 88.0%

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

   8. Applied egg-rr 88.0%

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

      1. *-un-lft-identity 88.0%

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

   10. Applied egg-rr 88.0%

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

       1. *-lft-identity 88.0%

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

       2. associate-/r* 88.0%

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

       3. associate-*l/ 90.7%

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

       4. associate-/l* 90.8%

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

   12. Simplified 90.8%

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

       1. clear-num 90.8%

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

       2. inv-pow 90.8%

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

       3. div-inv 90.8%

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

       4. pow-flip 92.6%

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

       5. associate-*r/ 92.6%

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

       6. metadata-eval 92.6%

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

   14. Applied egg-rr 92.6%

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

       1. unpow-1 92.6%

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

       2. associate-/l* 92.7%

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

       3. associate-/l* 92.8%

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

   16. Simplified 92.8%

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

   if 9.00000000000000073e100 < t

   1. Initial program 60.9%

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

   3. Simplified 60.9%

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

      1. associate-/r* 64.0%

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

      2. add-cube-cbrt 64.0%

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

      3. *-un-lft-identity 64.0%

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

      4. times-frac 64.0%

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

      5. pow2 64.0%

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

      6. cbrt-div 64.0%

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

      7. rem-cbrt-cube 64.0%

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

      8. cbrt-div 64.0%

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

      9. rem-cbrt-cube 92.7%

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

   6. Applied egg-rr 92.7%

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

4. Final simplification 44.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 4.2 \cdot 10^{-60}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{elif}\;t \leq 9 \cdot 10^{+100}:\\ \;\;\;\;\frac{1}{\sin k \cdot \frac{\tan k}{2 \cdot {\left({t}^{1.5} \cdot \frac{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}{\ell}\right)}^{-2}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\tan k \cdot \left(1 + \left(1 + {\left(\frac{k}{t}\right)}^{2}\right)\right)\right) \cdot \left(\sin k \cdot \left({\left(\frac{t}{\sqrt[3]{\ell}}\right)}^{2} \cdot \frac{\frac{t}{\sqrt[3]{\ell}}}{\ell}\right)\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 92.3% accurate, 0.6× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 1.25 \cdot 10^{-63}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{t\_m}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \sqrt[3]{\sin k}\right)}^{3} \cdot \left(\tan k \cdot \left(1 + \left(1 + {\left(\frac{k}{t\_m}\right)}^{2}\right)\right)\right)}\\ \end{array} \end{array} \]

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 (<= t_m 1.25e-63)
    (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 2.0) (* (sin k) (tan k))))
    (/
     2.0
     (*
      (pow (* (/ t_m (pow (cbrt l) 2.0)) (cbrt (sin k))) 3.0)
      (* (tan k) (+ 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 tmp;
	if (t_m <= 1.25e-63) {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * (sin(k) * tan(k)));
	} else {
		tmp = 2.0 / (pow(((t_m / pow(cbrt(l), 2.0)) * cbrt(sin(k))), 3.0) * (tan(k) * (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 tmp;
	if (t_m <= 1.25e-63) {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k) * Math.tan(k)));
	} else {
		tmp = 2.0 / (Math.pow(((t_m / Math.pow(Math.cbrt(l), 2.0)) * Math.cbrt(Math.sin(k))), 3.0) * (Math.tan(k) * (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)
	tmp = 0.0
	if (t_m <= 1.25e-63)
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k) * tan(k))));
	else
		tmp = Float64(2.0 / Float64((Float64(Float64(t_m / (cbrt(l) ^ 2.0)) * cbrt(sin(k))) ^ 3.0) * Float64(tan(k) * 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_] := N[(t$95$s * If[LessEqual[t$95$m, 1.25e-63], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[Power[N[(N[(t$95$m / N[Power[N[Power[l, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] * N[Power[N[Sin[k], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision] * N[(N[Tan[k], $MachinePrecision] * N[(1.0 + N[(1.0 + N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if t < 1.25e-63

   1. Initial program 50.3%

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

   3. Simplified 50.3%

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

   6. Applied egg-rr 15.6%

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

      1. associate-*r* 15.6%

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

      2. unpow-prod-down 15.6%

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

      3. pow2 15.6%

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

      4. add-sqr-sqrt 19.5%

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

   8. Applied egg-rr 19.5%

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

   9. Taylor expanded in t around 0 24.4%

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

   if 1.25e-63 < t

   1. Initial program 70.0%

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

   3. Simplified 70.0%

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

      1. associate-/r* 74.1%

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

      2. add-cube-cbrt 73.9%

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

      3. add-sqr-sqrt 40.7%

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

      4. times-frac 40.8%

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

      5. pow2 40.8%

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

      6. cbrt-div 40.7%

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

      7. rem-cbrt-cube 40.7%

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

      8. cbrt-div 40.7%

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

      9. rem-cbrt-cube 46.9%

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

   6. Applied egg-rr 46.9%

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

      1. add-cube-cbrt 46.8%

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

      2. pow3 46.9%

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

   8. Applied egg-rr 92.4%

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

      1. *-commutative 92.4%

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

   10. Simplified 92.4%

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

Alternative 3: 87.6% accurate, 0.7× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ \begin{array}{l} t_2 := \sin k \cdot \tan k\\ t_3 := \frac{t\_m}{\sqrt[3]{\ell}}\\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 10^{-59}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot t\_2}\\ \mathbf{elif}\;t\_m \leq 4.4 \cdot 10^{+105}:\\ \;\;\;\;\left(2 \cdot {\left(\frac{{t\_m}^{1.5} \cdot \mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t\_m}\right)\right)}{\ell}\right)}^{-2}\right) \cdot \frac{1}{t\_2}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot \left({t\_3}^{2} \cdot \frac{t\_3}{\ell}\right)\right) \cdot \left(2 \cdot \tan k\right)}\\ \end{array} \end{array} \end{array} \]

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 (* (sin k) (tan k))) (t_3 (/ t_m (cbrt l))))
   (*
    t_s
    (if (<= t_m 1e-59)
      (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 2.0) t_2))
      (if (<= t_m 4.4e+105)
        (*
         (*
          2.0
          (pow (/ (* (pow t_m 1.5) (hypot 1.0 (hypot 1.0 (/ k t_m)))) l) -2.0))
         (/ 1.0 t_2))
        (/
         2.0
         (* (* (sin k) (* (pow t_3 2.0) (/ t_3 l))) (* 2.0 (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 t_2 = sin(k) * tan(k);
	double t_3 = t_m / cbrt(l);
	double tmp;
	if (t_m <= 1e-59) {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * t_2);
	} else if (t_m <= 4.4e+105) {
		tmp = (2.0 * pow(((pow(t_m, 1.5) * hypot(1.0, hypot(1.0, (k / t_m)))) / l), -2.0)) * (1.0 / t_2);
	} else {
		tmp = 2.0 / ((sin(k) * (pow(t_3, 2.0) * (t_3 / l))) * (2.0 * 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 t_2 = Math.sin(k) * Math.tan(k);
	double t_3 = t_m / Math.cbrt(l);
	double tmp;
	if (t_m <= 1e-59) {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * t_2);
	} else if (t_m <= 4.4e+105) {
		tmp = (2.0 * Math.pow(((Math.pow(t_m, 1.5) * Math.hypot(1.0, Math.hypot(1.0, (k / t_m)))) / l), -2.0)) * (1.0 / t_2);
	} else {
		tmp = 2.0 / ((Math.sin(k) * (Math.pow(t_3, 2.0) * (t_3 / l))) * (2.0 * 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)
	t_2 = Float64(sin(k) * tan(k))
	t_3 = Float64(t_m / cbrt(l))
	tmp = 0.0
	if (t_m <= 1e-59)
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * t_2));
	elseif (t_m <= 4.4e+105)
		tmp = Float64(Float64(2.0 * (Float64(Float64((t_m ^ 1.5) * hypot(1.0, hypot(1.0, Float64(k / t_m)))) / l) ^ -2.0)) * Float64(1.0 / t_2));
	else
		tmp = Float64(2.0 / Float64(Float64(sin(k) * Float64((t_3 ^ 2.0) * Float64(t_3 / l))) * Float64(2.0 * 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_] := Block[{t$95$2 = N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$3 = N[(t$95$m / N[Power[l, 1/3], $MachinePrecision]), $MachinePrecision]}, N[(t$95$s * If[LessEqual[t$95$m, 1e-59], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * t$95$2), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$m, 4.4e+105], N[(N[(2.0 * N[Power[N[(N[(N[Power[t$95$m, 1.5], $MachinePrecision] * N[Sqrt[1.0 ^ 2 + N[Sqrt[1.0 ^ 2 + N[(k / t$95$m), $MachinePrecision] ^ 2], $MachinePrecision] ^ 2], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision], -2.0], $MachinePrecision]), $MachinePrecision] * N[(1.0 / t$95$2), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[Sin[k], $MachinePrecision] * N[(N[Power[t$95$3, 2.0], $MachinePrecision] * N[(t$95$3 / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(2.0 * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if t < 1e-59

   1. Initial program 50.9%

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

   3. Simplified 50.9%

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

   6. Applied egg-rr 16.0%

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

      1. associate-*r* 16.0%

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

      2. unpow-prod-down 16.0%

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

      3. pow2 16.0%

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

      4. add-sqr-sqrt 20.4%

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

   8. Applied egg-rr 20.4%

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

   9. Taylor expanded in t around 0 25.2%

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

   if 1e-59 < t < 4.40000000000000014e105

   1. Initial program 78.9%

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

   3. Simplified 78.8%

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

   6. Applied egg-rr 61.9%

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

      1. associate-*r* 61.9%

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

      2. unpow-prod-down 58.7%

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

      3. pow2 58.7%

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

      4. add-sqr-sqrt 88.0%

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

   8. Applied egg-rr 88.0%

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

      1. *-un-lft-identity 88.0%

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

   10. Applied egg-rr 88.0%

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

       1. *-lft-identity 88.0%

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

       2. associate-/r* 88.0%

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

       3. associate-*l/ 90.7%

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

       4. associate-/l* 90.8%

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

   12. Simplified 90.8%

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

       1. div-inv 90.9%

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

       2. div-inv 90.9%

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

       3. pow-flip 92.7%

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

       4. associate-*r/ 92.7%

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

       5. metadata-eval 92.7%

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

   14. Applied egg-rr 92.7%

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

   if 4.40000000000000014e105 < t

   1. Initial program 60.9%

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

   3. Simplified 60.9%

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

   5. Taylor expanded in k around 0 60.9%

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

      1. associate-/r* 64.0%

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

      2. add-cube-cbrt 64.0%

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

      3. *-un-lft-identity 64.0%

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

      4. times-frac 64.0%

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

      5. pow2 64.0%

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

      6. cbrt-div 64.0%

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

      7. rem-cbrt-cube 64.0%

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

      8. cbrt-div 64.0%

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

      9. rem-cbrt-cube 92.7%

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

   7. Applied egg-rr 92.7%

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

4. Final simplification 44.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 10^{-59}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{elif}\;t \leq 4.4 \cdot 10^{+105}:\\ \;\;\;\;\left(2 \cdot {\left(\frac{{t}^{1.5} \cdot \mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}{\ell}\right)}^{-2}\right) \cdot \frac{1}{\sin k \cdot \tan k}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot \left({\left(\frac{t}{\sqrt[3]{\ell}}\right)}^{2} \cdot \frac{\frac{t}{\sqrt[3]{\ell}}}{\ell}\right)\right) \cdot \left(2 \cdot \tan k\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 87.8% accurate, 0.7× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ \begin{array}{l} t_2 := \sin k \cdot \tan k\\ t_3 := \frac{t\_m}{\sqrt[3]{\ell}}\\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 2.5 \cdot 10^{-60}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot t\_2}\\ \mathbf{elif}\;t\_m \leq 1.05 \cdot 10^{+108}:\\ \;\;\;\;\frac{\frac{2}{{\left({t\_m}^{1.5} \cdot \frac{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t\_m}\right)\right)}{\ell}\right)}^{2}}}{t\_2}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot \left({t\_3}^{2} \cdot \frac{t\_3}{\ell}\right)\right) \cdot \left(2 \cdot \tan k\right)}\\ \end{array} \end{array} \end{array} \]

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 (* (sin k) (tan k))) (t_3 (/ t_m (cbrt l))))
   (*
    t_s
    (if (<= t_m 2.5e-60)
      (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 2.0) t_2))
      (if (<= t_m 1.05e+108)
        (/
         (/
          2.0
          (pow (* (pow t_m 1.5) (/ (hypot 1.0 (hypot 1.0 (/ k t_m))) l)) 2.0))
         t_2)
        (/
         2.0
         (* (* (sin k) (* (pow t_3 2.0) (/ t_3 l))) (* 2.0 (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 t_2 = sin(k) * tan(k);
	double t_3 = t_m / cbrt(l);
	double tmp;
	if (t_m <= 2.5e-60) {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * t_2);
	} else if (t_m <= 1.05e+108) {
		tmp = (2.0 / pow((pow(t_m, 1.5) * (hypot(1.0, hypot(1.0, (k / t_m))) / l)), 2.0)) / t_2;
	} else {
		tmp = 2.0 / ((sin(k) * (pow(t_3, 2.0) * (t_3 / l))) * (2.0 * 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 t_2 = Math.sin(k) * Math.tan(k);
	double t_3 = t_m / Math.cbrt(l);
	double tmp;
	if (t_m <= 2.5e-60) {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * t_2);
	} else if (t_m <= 1.05e+108) {
		tmp = (2.0 / Math.pow((Math.pow(t_m, 1.5) * (Math.hypot(1.0, Math.hypot(1.0, (k / t_m))) / l)), 2.0)) / t_2;
	} else {
		tmp = 2.0 / ((Math.sin(k) * (Math.pow(t_3, 2.0) * (t_3 / l))) * (2.0 * 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)
	t_2 = Float64(sin(k) * tan(k))
	t_3 = Float64(t_m / cbrt(l))
	tmp = 0.0
	if (t_m <= 2.5e-60)
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * t_2));
	elseif (t_m <= 1.05e+108)
		tmp = Float64(Float64(2.0 / (Float64((t_m ^ 1.5) * Float64(hypot(1.0, hypot(1.0, Float64(k / t_m))) / l)) ^ 2.0)) / t_2);
	else
		tmp = Float64(2.0 / Float64(Float64(sin(k) * Float64((t_3 ^ 2.0) * Float64(t_3 / l))) * Float64(2.0 * 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_] := Block[{t$95$2 = N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$3 = N[(t$95$m / N[Power[l, 1/3], $MachinePrecision]), $MachinePrecision]}, N[(t$95$s * If[LessEqual[t$95$m, 2.5e-60], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * t$95$2), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$m, 1.05e+108], N[(N[(2.0 / N[Power[N[(N[Power[t$95$m, 1.5], $MachinePrecision] * N[(N[Sqrt[1.0 ^ 2 + N[Sqrt[1.0 ^ 2 + N[(k / t$95$m), $MachinePrecision] ^ 2], $MachinePrecision] ^ 2], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] / t$95$2), $MachinePrecision], N[(2.0 / N[(N[(N[Sin[k], $MachinePrecision] * N[(N[Power[t$95$3, 2.0], $MachinePrecision] * N[(t$95$3 / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(2.0 * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if t < 2.5000000000000001e-60

   1. Initial program 50.9%

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

   3. Simplified 50.9%

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

   6. Applied egg-rr 16.0%

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

      1. associate-*r* 16.0%

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

      2. unpow-prod-down 16.0%

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

      3. pow2 16.0%

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

      4. add-sqr-sqrt 20.4%

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

   8. Applied egg-rr 20.4%

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

   9. Taylor expanded in t around 0 25.2%

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

   if 2.5000000000000001e-60 < t < 1.05000000000000005e108

   1. Initial program 78.9%

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

   3. Simplified 78.8%

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

   6. Applied egg-rr 61.9%

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

      1. associate-*r* 61.9%

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

      2. unpow-prod-down 58.7%

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

      3. pow2 58.7%

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

      4. add-sqr-sqrt 88.0%

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

   8. Applied egg-rr 88.0%

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

      1. *-un-lft-identity 88.0%

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

   10. Applied egg-rr 88.0%

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

       1. *-lft-identity 88.0%

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

       2. associate-/r* 88.0%

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

       3. associate-*l/ 90.7%

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

       4. associate-/l* 90.8%

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

   12. Simplified 90.8%

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

   if 1.05000000000000005e108 < t

   1. Initial program 60.9%

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

   3. Simplified 60.9%

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

   5. Taylor expanded in k around 0 60.9%

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

      1. associate-/r* 64.0%

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

      2. add-cube-cbrt 64.0%

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

      3. *-un-lft-identity 64.0%

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

      4. times-frac 64.0%

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

      5. pow2 64.0%

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

      6. cbrt-div 64.0%

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

      7. rem-cbrt-cube 64.0%

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

      8. cbrt-div 64.0%

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

      9. rem-cbrt-cube 92.7%

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

   7. Applied egg-rr 92.7%

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

4. Final simplification 44.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 2.5 \cdot 10^{-60}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{elif}\;t \leq 1.05 \cdot 10^{+108}:\\ \;\;\;\;\frac{\frac{2}{{\left({t}^{1.5} \cdot \frac{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right)}{\ell}\right)}^{2}}}{\sin k \cdot \tan k}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot \left({\left(\frac{t}{\sqrt[3]{\ell}}\right)}^{2} \cdot \frac{\frac{t}{\sqrt[3]{\ell}}}{\ell}\right)\right) \cdot \left(2 \cdot \tan k\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 89.6% accurate, 0.7× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 9.5 \cdot 10^{-60}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{1}{\sin k \cdot \frac{\tan k}{2 \cdot {\left({t\_m}^{1.5} \cdot \frac{\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t\_m}\right)\right)}{\ell}\right)}^{-2}}}\\ \end{array} \end{array} \]

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 9.5e-60)
    (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 2.0) (* (sin k) (tan k))))
    (/
     1.0
     (*
      (sin k)
      (/
       (tan k)
       (*
        2.0
        (pow
         (* (pow t_m 1.5) (/ (hypot 1.0 (hypot 1.0 (/ k t_m))) 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 <= 9.5e-60) {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * (sin(k) * tan(k)));
	} else {
		tmp = 1.0 / (sin(k) * (tan(k) / (2.0 * pow((pow(t_m, 1.5) * (hypot(1.0, hypot(1.0, (k / t_m))) / 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 <= 9.5e-60) {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k) * Math.tan(k)));
	} else {
		tmp = 1.0 / (Math.sin(k) * (Math.tan(k) / (2.0 * Math.pow((Math.pow(t_m, 1.5) * (Math.hypot(1.0, Math.hypot(1.0, (k / t_m))) / 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 <= 9.5e-60:
		tmp = 2.0 / (math.pow(((k / l) * math.sqrt(t_m)), 2.0) * (math.sin(k) * math.tan(k)))
	else:
		tmp = 1.0 / (math.sin(k) * (math.tan(k) / (2.0 * math.pow((math.pow(t_m, 1.5) * (math.hypot(1.0, math.hypot(1.0, (k / t_m))) / 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 <= 9.5e-60)
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k) * tan(k))));
	else
		tmp = Float64(1.0 / Float64(sin(k) * Float64(tan(k) / Float64(2.0 * (Float64((t_m ^ 1.5) * Float64(hypot(1.0, hypot(1.0, Float64(k / t_m))) / 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 <= 9.5e-60)
		tmp = 2.0 / ((((k / l) * sqrt(t_m)) ^ 2.0) * (sin(k) * tan(k)));
	else
		tmp = 1.0 / (sin(k) * (tan(k) / (2.0 * (((t_m ^ 1.5) * (hypot(1.0, hypot(1.0, (k / t_m))) / 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, 9.5e-60], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(1.0 / N[(N[Sin[k], $MachinePrecision] * N[(N[Tan[k], $MachinePrecision] / N[(2.0 * N[Power[N[(N[Power[t$95$m, 1.5], $MachinePrecision] * N[(N[Sqrt[1.0 ^ 2 + N[Sqrt[1.0 ^ 2 + N[(k / t$95$m), $MachinePrecision] ^ 2], $MachinePrecision] ^ 2], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], -2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if t < 9.49999999999999958e-60

   1. Initial program 50.9%

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

   3. Simplified 50.9%

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

   6. Applied egg-rr 16.0%

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

      1. associate-*r* 16.0%

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

      2. unpow-prod-down 16.0%

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

      3. pow2 16.0%

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

      4. add-sqr-sqrt 20.4%

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

   8. Applied egg-rr 20.4%

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

   9. Taylor expanded in t around 0 25.2%

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

   if 9.49999999999999958e-60 < t

   1. Initial program 69.2%

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

   3. Simplified 69.2%

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

   6. Applied egg-rr 60.8%

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

      1. associate-*r* 60.8%

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

      2. unpow-prod-down 56.9%

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

      3. pow2 56.9%

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

      4. add-sqr-sqrt 83.8%

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

   8. Applied egg-rr 83.8%

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

      1. *-un-lft-identity 83.8%

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

   10. Applied egg-rr 83.8%

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

       1. *-lft-identity 83.8%

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

       2. associate-/r* 83.8%

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

       3. associate-*l/ 85.0%

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

       4. associate-/l* 85.1%

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

   12. Simplified 85.1%

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

       1. clear-num 85.1%

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

       2. inv-pow 85.1%

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

       3. div-inv 85.1%

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

       4. pow-flip 86.0%

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

       5. associate-*r/ 85.9%

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

       6. metadata-eval 85.9%

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

   14. Applied egg-rr 85.9%

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

       1. unpow-1 85.9%

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

       2. associate-/l* 92.6%

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

       3. associate-/l* 92.7%

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

   16. Simplified 92.7%

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

Alternative 6: 88.1% accurate, 0.8× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ \begin{array}{l} t_2 := \frac{t\_m}{\sqrt[3]{\ell}}\\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 1.85 \cdot 10^{-63}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{elif}\;t\_m \leq 1.32 \cdot 10^{+116}:\\ \;\;\;\;\frac{2}{\frac{{\left(t\_m \cdot \sqrt[3]{\sin k \cdot \left(\tan k \cdot \frac{2 + {\left(\frac{k}{t\_m}\right)}^{2}}{\ell}\right)}\right)}^{3}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot \left({t\_2}^{2} \cdot \frac{t\_2}{\ell}\right)\right) \cdot \left(2 \cdot \tan k\right)}\\ \end{array} \end{array} \end{array} \]

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 (/ t_m (cbrt l))))
   (*
    t_s
    (if (<= t_m 1.85e-63)
      (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 2.0) (* (sin k) (tan k))))
      (if (<= t_m 1.32e+116)
        (/
         2.0
         (/
          (pow
           (*
            t_m
            (cbrt (* (sin k) (* (tan k) (/ (+ 2.0 (pow (/ k t_m) 2.0)) l)))))
           3.0)
          l))
        (/
         2.0
         (* (* (sin k) (* (pow t_2 2.0) (/ t_2 l))) (* 2.0 (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 t_2 = t_m / cbrt(l);
	double tmp;
	if (t_m <= 1.85e-63) {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * (sin(k) * tan(k)));
	} else if (t_m <= 1.32e+116) {
		tmp = 2.0 / (pow((t_m * cbrt((sin(k) * (tan(k) * ((2.0 + pow((k / t_m), 2.0)) / l))))), 3.0) / l);
	} else {
		tmp = 2.0 / ((sin(k) * (pow(t_2, 2.0) * (t_2 / l))) * (2.0 * 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 t_2 = t_m / Math.cbrt(l);
	double tmp;
	if (t_m <= 1.85e-63) {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k) * Math.tan(k)));
	} else if (t_m <= 1.32e+116) {
		tmp = 2.0 / (Math.pow((t_m * Math.cbrt((Math.sin(k) * (Math.tan(k) * ((2.0 + Math.pow((k / t_m), 2.0)) / l))))), 3.0) / l);
	} else {
		tmp = 2.0 / ((Math.sin(k) * (Math.pow(t_2, 2.0) * (t_2 / l))) * (2.0 * 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)
	t_2 = Float64(t_m / cbrt(l))
	tmp = 0.0
	if (t_m <= 1.85e-63)
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k) * tan(k))));
	elseif (t_m <= 1.32e+116)
		tmp = Float64(2.0 / Float64((Float64(t_m * cbrt(Float64(sin(k) * Float64(tan(k) * Float64(Float64(2.0 + (Float64(k / t_m) ^ 2.0)) / l))))) ^ 3.0) / l));
	else
		tmp = Float64(2.0 / Float64(Float64(sin(k) * Float64((t_2 ^ 2.0) * Float64(t_2 / l))) * Float64(2.0 * 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_] := Block[{t$95$2 = N[(t$95$m / N[Power[l, 1/3], $MachinePrecision]), $MachinePrecision]}, N[(t$95$s * If[LessEqual[t$95$m, 1.85e-63], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$m, 1.32e+116], N[(2.0 / N[(N[Power[N[(t$95$m * N[Power[N[(N[Sin[k], $MachinePrecision] * N[(N[Tan[k], $MachinePrecision] * N[(N[(2.0 + N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[Sin[k], $MachinePrecision] * N[(N[Power[t$95$2, 2.0], $MachinePrecision] * N[(t$95$2 / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(2.0 * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]

Derivation

1. Split input into 3 regimes

2. if t < 1.85000000000000006e-63

   1. Initial program 50.3%

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

   3. Simplified 50.3%

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

   6. Applied egg-rr 15.6%

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

      1. associate-*r* 15.6%

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

      2. unpow-prod-down 15.6%

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

      3. pow2 15.6%

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

      4. add-sqr-sqrt 19.5%

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

   8. Applied egg-rr 19.5%

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

   9. Taylor expanded in t around 0 24.4%

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

   if 1.85000000000000006e-63 < t < 1.32000000000000002e116

   1. Initial program 74.7%

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

   3. Simplified 74.6%

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

      1. associate-*l* 74.6%

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

      2. associate-/r* 79.5%

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

      3. associate-+r+ 79.5%

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

      4. metadata-eval 79.5%

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

      5. associate-*l* 79.5%

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

      6. associate-*l/ 80.5%

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

   6. Applied egg-rr 80.5%

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

      1. associate-/l* 82.6%

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

      2. associate-*l* 82.6%

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

   8. Simplified 82.6%

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

      1. associate-*l/ 80.6%

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

      2. associate-/l* 82.9%

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

   10. Applied egg-rr 82.9%

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

       1. add-cube-cbrt 82.5%

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

       2. pow3 82.5%

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

       3. cbrt-prod 82.2%

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

       4. unpow3 82.2%

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

       5. add-cbrt-cube 89.4%

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

       6. associate-/l* 89.4%

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

   12. Applied egg-rr 89.4%

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

   if 1.32000000000000002e116 < t

   1. Initial program 64.8%

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

   3. Simplified 64.8%

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

   5. Taylor expanded in k around 0 64.8%

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

      1. associate-/r* 68.1%

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

      2. add-cube-cbrt 68.1%

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

      3. *-un-lft-identity 68.1%

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

      4. times-frac 68.1%

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

      5. pow2 68.1%

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

      6. cbrt-div 68.1%

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

      7. rem-cbrt-cube 68.1%

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

      8. cbrt-div 68.1%

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

      9. rem-cbrt-cube 94.5%

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

   7. Applied egg-rr 94.5%

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

4. Final simplification 44.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 1.85 \cdot 10^{-63}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{elif}\;t \leq 1.32 \cdot 10^{+116}:\\ \;\;\;\;\frac{2}{\frac{{\left(t \cdot \sqrt[3]{\sin k \cdot \left(\tan k \cdot \frac{2 + {\left(\frac{k}{t}\right)}^{2}}{\ell}\right)}\right)}^{3}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot \left({\left(\frac{t}{\sqrt[3]{\ell}}\right)}^{2} \cdot \frac{\frac{t}{\sqrt[3]{\ell}}}{\ell}\right)\right) \cdot \left(2 \cdot \tan k\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 87.4% accurate, 0.8× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 1.95 \cdot 10^{-63}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{elif}\;t\_m \leq 1.32 \cdot 10^{+116}:\\ \;\;\;\;\frac{2}{\frac{{\left(t\_m \cdot \sqrt[3]{\sin k \cdot \left(\tan k \cdot \frac{2 + {\left(\frac{k}{t\_m}\right)}^{2}}{\ell}\right)}\right)}^{3}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(2 \cdot \tan k\right) \cdot \left(\sin k \cdot {\left(\frac{{t\_m}^{1.5}}{\ell}\right)}^{2}\right)}\\ \end{array} \end{array} \]

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 (<= t_m 1.95e-63)
    (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 2.0) (* (sin k) (tan k))))
    (if (<= t_m 1.32e+116)
      (/
       2.0
       (/
        (pow
         (*
          t_m
          (cbrt (* (sin k) (* (tan k) (/ (+ 2.0 (pow (/ k t_m) 2.0)) l)))))
         3.0)
        l))
      (/ 2.0 (* (* 2.0 (tan k)) (* (sin k) (pow (/ (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 <= 1.95e-63) {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * (sin(k) * tan(k)));
	} else if (t_m <= 1.32e+116) {
		tmp = 2.0 / (pow((t_m * cbrt((sin(k) * (tan(k) * ((2.0 + pow((k / t_m), 2.0)) / l))))), 3.0) / l);
	} else {
		tmp = 2.0 / ((2.0 * tan(k)) * (sin(k) * pow((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 <= 1.95e-63) {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k) * Math.tan(k)));
	} else if (t_m <= 1.32e+116) {
		tmp = 2.0 / (Math.pow((t_m * Math.cbrt((Math.sin(k) * (Math.tan(k) * ((2.0 + Math.pow((k / t_m), 2.0)) / l))))), 3.0) / l);
	} else {
		tmp = 2.0 / ((2.0 * Math.tan(k)) * (Math.sin(k) * Math.pow((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 <= 1.95e-63)
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k) * tan(k))));
	elseif (t_m <= 1.32e+116)
		tmp = Float64(2.0 / Float64((Float64(t_m * cbrt(Float64(sin(k) * Float64(tan(k) * Float64(Float64(2.0 + (Float64(k / t_m) ^ 2.0)) / l))))) ^ 3.0) / l));
	else
		tmp = Float64(2.0 / Float64(Float64(2.0 * tan(k)) * Float64(sin(k) * (Float64((t_m ^ 1.5) / l) ^ 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_] := N[(t$95$s * If[LessEqual[t$95$m, 1.95e-63], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$m, 1.32e+116], N[(2.0 / N[(N[Power[N[(t$95$m * N[Power[N[(N[Sin[k], $MachinePrecision] * N[(N[Tan[k], $MachinePrecision] * N[(N[(2.0 + N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(2.0 * N[Tan[k], $MachinePrecision]), $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Power[N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if t < 1.95000000000000011e-63

   1. Initial program 50.3%

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

   3. Simplified 50.3%

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

   6. Applied egg-rr 15.6%

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

      1. associate-*r* 15.6%

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

      2. unpow-prod-down 15.6%

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

      3. pow2 15.6%

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

      4. add-sqr-sqrt 19.5%

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

   8. Applied egg-rr 19.5%

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

   9. Taylor expanded in t around 0 24.4%

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

   if 1.95000000000000011e-63 < t < 1.32000000000000002e116

   1. Initial program 74.7%

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

   3. Simplified 74.6%

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

      1. associate-*l* 74.6%

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

      2. associate-/r* 79.5%

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

      3. associate-+r+ 79.5%

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

      4. metadata-eval 79.5%

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

      5. associate-*l* 79.5%

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

      6. associate-*l/ 80.5%

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

   6. Applied egg-rr 80.5%

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

      1. associate-/l* 82.6%

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

      2. associate-*l* 82.6%

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

   8. Simplified 82.6%

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

      1. associate-*l/ 80.6%

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

      2. associate-/l* 82.9%

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

   10. Applied egg-rr 82.9%

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

       1. add-cube-cbrt 82.5%

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

       2. pow3 82.5%

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

       3. cbrt-prod 82.2%

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

       4. unpow3 82.2%

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

       5. add-cbrt-cube 89.4%

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

       6. associate-/l* 89.4%

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

   12. Applied egg-rr 89.4%

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

   if 1.32000000000000002e116 < t

   1. Initial program 64.8%

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

   3. Simplified 64.8%

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

   5. Taylor expanded in k around 0 64.8%

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

      1. add-sqr-sqrt 64.8%

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

      2. pow2 64.8%

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

      3. sqrt-div 64.8%

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

      4. sqrt-pow1 83.7%

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

      5. metadata-eval 83.7%

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

      6. sqrt-prod 52.7%

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

      7. add-sqr-sqrt 94.6%

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

   7. Applied egg-rr 94.6%

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

4. Final simplification 44.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 1.95 \cdot 10^{-63}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{elif}\;t \leq 1.32 \cdot 10^{+116}:\\ \;\;\;\;\frac{2}{\frac{{\left(t \cdot \sqrt[3]{\sin k \cdot \left(\tan k \cdot \frac{2 + {\left(\frac{k}{t}\right)}^{2}}{\ell}\right)}\right)}^{3}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(2 \cdot \tan k\right) \cdot \left(\sin k \cdot {\left(\frac{{t}^{1.5}}{\ell}\right)}^{2}\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 86.9% accurate, 1.0× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 1.95 \cdot 10^{-63}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{elif}\;t\_m \leq 1.95 \cdot 10^{+100}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot \left(\tan k \cdot \frac{2 + {\left(\frac{k}{t\_m}\right)}^{2}}{\ell}\right)\right) \cdot \frac{{t\_m}^{3}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(2 \cdot \tan k\right) \cdot \left(\sin k \cdot {\left(\frac{{t\_m}^{1.5}}{\ell}\right)}^{2}\right)}\\ \end{array} \end{array} \]

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 (<= t_m 1.95e-63)
    (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 2.0) (* (sin k) (tan k))))
    (if (<= t_m 1.95e+100)
      (/
       2.0
       (*
        (* (sin k) (* (tan k) (/ (+ 2.0 (pow (/ k t_m) 2.0)) l)))
        (/ (pow t_m 3.0) l)))
      (/ 2.0 (* (* 2.0 (tan k)) (* (sin k) (pow (/ (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 <= 1.95e-63) {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * (sin(k) * tan(k)));
	} else if (t_m <= 1.95e+100) {
		tmp = 2.0 / ((sin(k) * (tan(k) * ((2.0 + pow((k / t_m), 2.0)) / l))) * (pow(t_m, 3.0) / l));
	} else {
		tmp = 2.0 / ((2.0 * tan(k)) * (sin(k) * pow((pow(t_m, 1.5) / l), 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 <= 1.95d-63) then
        tmp = 2.0d0 / ((((k / l) * sqrt(t_m)) ** 2.0d0) * (sin(k) * tan(k)))
    else if (t_m <= 1.95d+100) then
        tmp = 2.0d0 / ((sin(k) * (tan(k) * ((2.0d0 + ((k / t_m) ** 2.0d0)) / l))) * ((t_m ** 3.0d0) / l))
    else
        tmp = 2.0d0 / ((2.0d0 * tan(k)) * (sin(k) * (((t_m ** 1.5d0) / l) ** 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 <= 1.95e-63) {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k) * Math.tan(k)));
	} else if (t_m <= 1.95e+100) {
		tmp = 2.0 / ((Math.sin(k) * (Math.tan(k) * ((2.0 + Math.pow((k / t_m), 2.0)) / l))) * (Math.pow(t_m, 3.0) / l));
	} else {
		tmp = 2.0 / ((2.0 * Math.tan(k)) * (Math.sin(k) * Math.pow((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 <= 1.95e-63:
		tmp = 2.0 / (math.pow(((k / l) * math.sqrt(t_m)), 2.0) * (math.sin(k) * math.tan(k)))
	elif t_m <= 1.95e+100:
		tmp = 2.0 / ((math.sin(k) * (math.tan(k) * ((2.0 + math.pow((k / t_m), 2.0)) / l))) * (math.pow(t_m, 3.0) / l))
	else:
		tmp = 2.0 / ((2.0 * math.tan(k)) * (math.sin(k) * math.pow((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 <= 1.95e-63)
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k) * tan(k))));
	elseif (t_m <= 1.95e+100)
		tmp = Float64(2.0 / Float64(Float64(sin(k) * Float64(tan(k) * Float64(Float64(2.0 + (Float64(k / t_m) ^ 2.0)) / l))) * Float64((t_m ^ 3.0) / l)));
	else
		tmp = Float64(2.0 / Float64(Float64(2.0 * tan(k)) * Float64(sin(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 <= 1.95e-63)
		tmp = 2.0 / ((((k / l) * sqrt(t_m)) ^ 2.0) * (sin(k) * tan(k)));
	elseif (t_m <= 1.95e+100)
		tmp = 2.0 / ((sin(k) * (tan(k) * ((2.0 + ((k / t_m) ^ 2.0)) / l))) * ((t_m ^ 3.0) / l));
	else
		tmp = 2.0 / ((2.0 * tan(k)) * (sin(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, 1.95e-63], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$m, 1.95e+100], N[(2.0 / N[(N[(N[Sin[k], $MachinePrecision] * N[(N[Tan[k], $MachinePrecision] * N[(N[(2.0 + N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[Power[t$95$m, 3.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(2.0 * N[Tan[k], $MachinePrecision]), $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Power[N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if t < 1.95000000000000011e-63

   1. Initial program 50.3%

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

   3. Simplified 50.3%

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

   6. Applied egg-rr 15.6%

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

      1. associate-*r* 15.6%

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

      2. unpow-prod-down 15.6%

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

      3. pow2 15.6%

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

      4. add-sqr-sqrt 19.5%

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

   8. Applied egg-rr 19.5%

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

   9. Taylor expanded in t around 0 24.4%

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

   if 1.95000000000000011e-63 < t < 1.95e100

   1. Initial program 80.1%

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

   3. Simplified 80.0%

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

      1. associate-*l* 79.9%

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

      2. associate-/r* 85.3%

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

      3. associate-+r+ 85.3%

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

      4. metadata-eval 85.3%

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

      5. associate-*l* 85.3%

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

      6. associate-*l/ 86.4%

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

   6. Applied egg-rr 86.4%

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

      1. associate-/l* 88.8%

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

      2. associate-*l* 88.8%

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

   8. Simplified 88.8%

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

      1. associate-/l* 94.2%

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

   10. Applied egg-rr 94.2%

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

       1. associate-/l* 94.2%

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

   12. Simplified 94.2%

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

   if 1.95e100 < t

   1. Initial program 60.9%

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

   3. Simplified 60.9%

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

   5. Taylor expanded in k around 0 60.9%

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

      1. add-sqr-sqrt 60.9%

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

      2. pow2 60.9%

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

      3. sqrt-div 60.9%

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

      4. sqrt-pow1 80.7%

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

      5. metadata-eval 80.7%

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

      6. sqrt-prod 52.5%

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

      7. add-sqr-sqrt 92.7%

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

   7. Applied egg-rr 92.7%

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

4. Final simplification 44.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 1.95 \cdot 10^{-63}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{elif}\;t \leq 1.95 \cdot 10^{+100}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot \left(\tan k \cdot \frac{2 + {\left(\frac{k}{t}\right)}^{2}}{\ell}\right)\right) \cdot \frac{{t}^{3}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(2 \cdot \tan k\right) \cdot \left(\sin k \cdot {\left(\frac{{t}^{1.5}}{\ell}\right)}^{2}\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 84.3% accurate, 1.0× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 1.16 \cdot 10^{+89}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(2 \cdot \tan k\right) \cdot \left(\sin k \cdot {\left(\frac{{t\_m}^{1.5}}{\ell}\right)}^{2}\right)}\\ \end{array} \end{array} \]

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 (<= t_m 1.16e+89)
    (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 2.0) (* (sin k) (tan k))))
    (/ 2.0 (* (* 2.0 (tan k)) (* (sin k) (pow (/ (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 <= 1.16e+89) {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * (sin(k) * tan(k)));
	} else {
		tmp = 2.0 / ((2.0 * tan(k)) * (sin(k) * pow((pow(t_m, 1.5) / l), 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 <= 1.16d+89) then
        tmp = 2.0d0 / ((((k / l) * sqrt(t_m)) ** 2.0d0) * (sin(k) * tan(k)))
    else
        tmp = 2.0d0 / ((2.0d0 * tan(k)) * (sin(k) * (((t_m ** 1.5d0) / l) ** 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 <= 1.16e+89) {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k) * Math.tan(k)));
	} else {
		tmp = 2.0 / ((2.0 * Math.tan(k)) * (Math.sin(k) * Math.pow((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 <= 1.16e+89:
		tmp = 2.0 / (math.pow(((k / l) * math.sqrt(t_m)), 2.0) * (math.sin(k) * math.tan(k)))
	else:
		tmp = 2.0 / ((2.0 * math.tan(k)) * (math.sin(k) * math.pow((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 <= 1.16e+89)
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k) * tan(k))));
	else
		tmp = Float64(2.0 / Float64(Float64(2.0 * tan(k)) * Float64(sin(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 <= 1.16e+89)
		tmp = 2.0 / ((((k / l) * sqrt(t_m)) ^ 2.0) * (sin(k) * tan(k)));
	else
		tmp = 2.0 / ((2.0 * tan(k)) * (sin(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, 1.16e+89], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(2.0 * N[Tan[k], $MachinePrecision]), $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Power[N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if t < 1.16e89

   1. Initial program 54.4%

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

   3. Simplified 54.4%

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

   6. Applied egg-rr 22.2%

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

      1. associate-*r* 22.2%

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

      2. unpow-prod-down 21.7%

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

      3. pow2 21.7%

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

      4. add-sqr-sqrt 29.7%

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

   8. Applied egg-rr 29.7%

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

   9. Taylor expanded in t around 0 32.5%

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

   if 1.16e89 < t

   1. Initial program 64.5%

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

   3. Simplified 64.5%

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

   5. Taylor expanded in k around 0 64.5%

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

      1. add-sqr-sqrt 64.5%

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

      2. pow2 64.5%

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

      3. sqrt-div 64.5%

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

      4. sqrt-pow1 82.4%

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

      5. metadata-eval 82.4%

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

      6. sqrt-prod 54.5%

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

      7. add-sqr-sqrt 93.4%

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

   7. Applied egg-rr 93.4%

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

4. Final simplification 43.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 1.16 \cdot 10^{+89}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(2 \cdot \tan k\right) \cdot \left(\sin k \cdot {\left(\frac{{t}^{1.5}}{\ell}\right)}^{2}\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 77.2% accurate, 1.0× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 1.3 \cdot 10^{-28}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t\_m}^{1.5}}{\ell} \cdot \left(k \cdot \sqrt{2}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \end{array} \end{array} \]

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 1.3e-28)
    (/ 2.0 (pow (* (/ (pow t_m 1.5) l) (* k (sqrt 2.0))) 2.0))
    (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 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 <= 1.3e-28) {
		tmp = 2.0 / pow(((pow(t_m, 1.5) / l) * (k * sqrt(2.0))), 2.0);
	} else {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * (sin(k) * tan(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 <= 1.3d-28) then
        tmp = 2.0d0 / ((((t_m ** 1.5d0) / l) * (k * sqrt(2.0d0))) ** 2.0d0)
    else
        tmp = 2.0d0 / ((((k / l) * sqrt(t_m)) ** 2.0d0) * (sin(k) * tan(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 <= 1.3e-28) {
		tmp = 2.0 / Math.pow(((Math.pow(t_m, 1.5) / l) * (k * Math.sqrt(2.0))), 2.0);
	} else {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k) * Math.tan(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 <= 1.3e-28:
		tmp = 2.0 / math.pow(((math.pow(t_m, 1.5) / l) * (k * math.sqrt(2.0))), 2.0)
	else:
		tmp = 2.0 / (math.pow(((k / l) * math.sqrt(t_m)), 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 <= 1.3e-28)
		tmp = Float64(2.0 / (Float64(Float64((t_m ^ 1.5) / l) * Float64(k * sqrt(2.0))) ^ 2.0));
	else
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k) * tan(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 <= 1.3e-28)
		tmp = 2.0 / ((((t_m ^ 1.5) / l) * (k * sqrt(2.0))) ^ 2.0);
	else
		tmp = 2.0 / ((((k / l) * sqrt(t_m)) ^ 2.0) * (sin(k) * tan(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, 1.3e-28], N[(2.0 / N[Power[N[(N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision] * N[(k * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $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 < 1.3e-28

   1. Initial program 58.4%

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

   3. Simplified 58.4%

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

   6. Applied egg-rr 33.3%

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

   7. Taylor expanded in k around 0 35.4%

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

   if 1.3e-28 < k

   1. Initial program 50.2%

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

   3. Simplified 50.3%

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

   6. Applied egg-rr 17.2%

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

      1. associate-*r* 17.2%

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

      2. unpow-prod-down 17.2%

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

      3. pow2 17.2%

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

      4. add-sqr-sqrt 44.2%

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

   8. Applied egg-rr 44.2%

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

   9. Taylor expanded in t around 0 40.3%

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

Alternative 11: 75.4% accurate, 1.3× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 7.5 \cdot 10^{-28}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t\_m}^{1.5}}{\ell} \cdot \left(k \cdot \sqrt{2}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(k \cdot \frac{\sin k}{\ell}\right)}^{2} \cdot \frac{t\_m}{\cos k}}\\ \end{array} \end{array} \]

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-28)
    (/ 2.0 (pow (* (/ (pow t_m 1.5) l) (* k (sqrt 2.0))) 2.0))
    (/ 2.0 (* (pow (* k (/ (sin k) l)) 2.0) (/ t_m (cos 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 <= 7.5e-28) {
		tmp = 2.0 / pow(((pow(t_m, 1.5) / l) * (k * sqrt(2.0))), 2.0);
	} else {
		tmp = 2.0 / (pow((k * (sin(k) / l)), 2.0) * (t_m / cos(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 <= 7.5d-28) then
        tmp = 2.0d0 / ((((t_m ** 1.5d0) / l) * (k * sqrt(2.0d0))) ** 2.0d0)
    else
        tmp = 2.0d0 / (((k * (sin(k) / l)) ** 2.0d0) * (t_m / cos(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 <= 7.5e-28) {
		tmp = 2.0 / Math.pow(((Math.pow(t_m, 1.5) / l) * (k * Math.sqrt(2.0))), 2.0);
	} else {
		tmp = 2.0 / (Math.pow((k * (Math.sin(k) / l)), 2.0) * (t_m / Math.cos(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 <= 7.5e-28:
		tmp = 2.0 / math.pow(((math.pow(t_m, 1.5) / l) * (k * math.sqrt(2.0))), 2.0)
	else:
		tmp = 2.0 / (math.pow((k * (math.sin(k) / l)), 2.0) * (t_m / math.cos(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 <= 7.5e-28)
		tmp = Float64(2.0 / (Float64(Float64((t_m ^ 1.5) / l) * Float64(k * sqrt(2.0))) ^ 2.0));
	else
		tmp = Float64(2.0 / Float64((Float64(k * Float64(sin(k) / l)) ^ 2.0) * Float64(t_m / cos(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 <= 7.5e-28)
		tmp = 2.0 / ((((t_m ^ 1.5) / l) * (k * sqrt(2.0))) ^ 2.0);
	else
		tmp = 2.0 / (((k * (sin(k) / l)) ^ 2.0) * (t_m / cos(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, 7.5e-28], N[(2.0 / N[Power[N[(N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision] * N[(k * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[Power[N[(k * N[(N[Sin[k], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(t$95$m / N[Cos[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 7.5000000000000003e-28

   1. Initial program 58.4%

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

   3. Simplified 58.4%

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

   6. Applied egg-rr 33.3%

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

   7. Taylor expanded in k around 0 35.4%

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

   if 7.5000000000000003e-28 < k

   1. Initial program 50.2%

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

   3. Simplified 50.3%

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

   6. Applied egg-rr 17.2%

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

   7. Taylor expanded in t around 0 35.7%

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

      1. unpow-prod-down 33.0%

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

      2. associate-/l* 33.0%

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

      3. pow2 33.0%

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

      4. add-sqr-sqrt 79.3%

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

   9. Applied egg-rr 79.3%

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

Alternative 12: 72.8% accurate, 1.3× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 2.45 \cdot 10^{-67}:\\ \;\;\;\;\frac{2}{{\left(\sqrt{t\_m} \cdot \frac{{k}^{2}}{\ell}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t\_m}^{1.5}}{\ell} \cdot \left(k \cdot \sqrt{2}\right)\right)}^{2}}\\ \end{array} \end{array} \]

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 2.45e-67)
    (/ 2.0 (pow (* (sqrt t_m) (/ (pow k 2.0) l)) 2.0))
    (/ 2.0 (pow (* (/ (pow t_m 1.5) l) (* k (sqrt 2.0))) 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 <= 2.45e-67) {
		tmp = 2.0 / pow((sqrt(t_m) * (pow(k, 2.0) / l)), 2.0);
	} else {
		tmp = 2.0 / pow(((pow(t_m, 1.5) / l) * (k * sqrt(2.0))), 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 <= 2.45d-67) then
        tmp = 2.0d0 / ((sqrt(t_m) * ((k ** 2.0d0) / l)) ** 2.0d0)
    else
        tmp = 2.0d0 / ((((t_m ** 1.5d0) / l) * (k * sqrt(2.0d0))) ** 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 <= 2.45e-67) {
		tmp = 2.0 / Math.pow((Math.sqrt(t_m) * (Math.pow(k, 2.0) / l)), 2.0);
	} else {
		tmp = 2.0 / Math.pow(((Math.pow(t_m, 1.5) / l) * (k * Math.sqrt(2.0))), 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 <= 2.45e-67:
		tmp = 2.0 / math.pow((math.sqrt(t_m) * (math.pow(k, 2.0) / l)), 2.0)
	else:
		tmp = 2.0 / math.pow(((math.pow(t_m, 1.5) / l) * (k * math.sqrt(2.0))), 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 <= 2.45e-67)
		tmp = Float64(2.0 / (Float64(sqrt(t_m) * Float64((k ^ 2.0) / l)) ^ 2.0));
	else
		tmp = Float64(2.0 / (Float64(Float64((t_m ^ 1.5) / l) * Float64(k * sqrt(2.0))) ^ 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 <= 2.45e-67)
		tmp = 2.0 / ((sqrt(t_m) * ((k ^ 2.0) / l)) ^ 2.0);
	else
		tmp = 2.0 / ((((t_m ^ 1.5) / l) * (k * sqrt(2.0))) ^ 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, 2.45e-67], N[(2.0 / N[Power[N[(N[Sqrt[t$95$m], $MachinePrecision] * N[(N[Power[k, 2.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(2.0 / N[Power[N[(N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision] * N[(k * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if t < 2.44999999999999997e-67

   1. Initial program 50.3%

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

   3. Simplified 50.3%

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

   6. Applied egg-rr 14.6%

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

   7. Taylor expanded in t around 0 34.0%

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

   8. Taylor expanded in k around 0 19.2%

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

   if 2.44999999999999997e-67 < t

   1. Initial program 69.5%

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

   3. Simplified 69.5%

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

   6. Applied egg-rr 61.5%

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

   7. Taylor expanded in k around 0 80.4%

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

4. Final simplification 37.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 2.45 \cdot 10^{-67}:\\ \;\;\;\;\frac{2}{{\left(\sqrt{t} \cdot \frac{{k}^{2}}{\ell}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t}^{1.5}}{\ell} \cdot \left(k \cdot \sqrt{2}\right)\right)}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 13: 64.8% accurate, 1.3× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 2.12 \cdot 10^{-63}:\\ \;\;\;\;\frac{2}{{\left(\sqrt{t\_m} \cdot \frac{{k}^{2}}{\ell}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{{t\_m}^{3} \cdot \left(\sin k \cdot \frac{2 \cdot k}{\ell}\right)}{\ell}}\\ \end{array} \end{array} \]

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 2.12e-63)
    (/ 2.0 (pow (* (sqrt t_m) (/ (pow k 2.0) l)) 2.0))
    (/ 2.0 (/ (* (pow t_m 3.0) (* (sin k) (/ (* 2.0 k) l))) l)))))

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 <= 2.12e-63) {
		tmp = 2.0 / pow((sqrt(t_m) * (pow(k, 2.0) / l)), 2.0);
	} else {
		tmp = 2.0 / ((pow(t_m, 3.0) * (sin(k) * ((2.0 * k) / l))) / l);
	}
	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 <= 2.12d-63) then
        tmp = 2.0d0 / ((sqrt(t_m) * ((k ** 2.0d0) / l)) ** 2.0d0)
    else
        tmp = 2.0d0 / (((t_m ** 3.0d0) * (sin(k) * ((2.0d0 * k) / l))) / l)
    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 <= 2.12e-63) {
		tmp = 2.0 / Math.pow((Math.sqrt(t_m) * (Math.pow(k, 2.0) / l)), 2.0);
	} else {
		tmp = 2.0 / ((Math.pow(t_m, 3.0) * (Math.sin(k) * ((2.0 * k) / l))) / l);
	}
	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 <= 2.12e-63:
		tmp = 2.0 / math.pow((math.sqrt(t_m) * (math.pow(k, 2.0) / l)), 2.0)
	else:
		tmp = 2.0 / ((math.pow(t_m, 3.0) * (math.sin(k) * ((2.0 * k) / l))) / l)
	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 <= 2.12e-63)
		tmp = Float64(2.0 / (Float64(sqrt(t_m) * Float64((k ^ 2.0) / l)) ^ 2.0));
	else
		tmp = Float64(2.0 / Float64(Float64((t_m ^ 3.0) * Float64(sin(k) * Float64(Float64(2.0 * k) / l))) / l));
	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 <= 2.12e-63)
		tmp = 2.0 / ((sqrt(t_m) * ((k ^ 2.0) / l)) ^ 2.0);
	else
		tmp = 2.0 / (((t_m ^ 3.0) * (sin(k) * ((2.0 * k) / l))) / l);
	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, 2.12e-63], N[(2.0 / N[Power[N[(N[Sqrt[t$95$m], $MachinePrecision] * N[(N[Power[k, 2.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[Power[t$95$m, 3.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[(N[(2.0 * k), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if t < 2.12000000000000008e-63

   1. Initial program 50.6%

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

   3. Simplified 50.6%

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

   6. Applied egg-rr 16.1%

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

   7. Taylor expanded in t around 0 35.1%

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

   8. Taylor expanded in k around 0 20.5%

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

   if 2.12000000000000008e-63 < t

   1. Initial program 69.6%

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

   3. Simplified 69.6%

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

      1. associate-*l* 65.4%

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

      2. associate-/r* 69.6%

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

      3. associate-+r+ 69.6%

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

      4. metadata-eval 69.6%

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

      5. associate-*l* 69.6%

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

      6. associate-*l/ 70.2%

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

   6. Applied egg-rr 70.2%

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

      1. associate-/l* 71.2%

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

      2. associate-*l* 71.2%

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

   8. Simplified 71.2%

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

      1. associate-*l/ 70.1%

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

      2. associate-/l* 75.5%

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

   10. Applied egg-rr 75.5%

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

   11. Taylor expanded in k around 0 66.8%

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

       1. associate-*r/ 66.8%

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

       2. *-commutative 66.8%

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

   13. Simplified 66.8%

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

4. Final simplification 34.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 2.12 \cdot 10^{-63}:\\ \;\;\;\;\frac{2}{{\left(\sqrt{t} \cdot \frac{{k}^{2}}{\ell}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{{t}^{3} \cdot \left(\sin k \cdot \frac{2 \cdot k}{\ell}\right)}{\ell}}\\ \end{array} \]
5. Add Preprocessing

Alternative 14: 59.6% accurate, 1.9× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 5 \cdot 10^{-156}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot \frac{\frac{{t\_m}^{3}}{\ell}}{\ell}\right) \cdot \left(2 \cdot k\right)}\\ \mathbf{elif}\;k \leq 6.2 \cdot 10^{-28}:\\ \;\;\;\;\frac{2}{\frac{{t\_m}^{2} \cdot \frac{t\_m}{\ell}}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{t\_m \cdot {k}^{4}}{\cos k \cdot \left(\ell \cdot \ell\right)}}\\ \end{array} \end{array} \]

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 5e-156)
    (/ 2.0 (* (* (sin k) (/ (/ (pow t_m 3.0) l) l)) (* 2.0 k)))
    (if (<= k 6.2e-28)
      (/ 2.0 (* (/ (* (pow t_m 2.0) (/ t_m l)) l) (* 2.0 (* k k))))
      (/ 2.0 (/ (* t_m (pow k 4.0)) (* (cos k) (* l l))))))))

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 <= 5e-156) {
		tmp = 2.0 / ((sin(k) * ((pow(t_m, 3.0) / l) / l)) * (2.0 * k));
	} else if (k <= 6.2e-28) {
		tmp = 2.0 / (((pow(t_m, 2.0) * (t_m / l)) / l) * (2.0 * (k * k)));
	} else {
		tmp = 2.0 / ((t_m * pow(k, 4.0)) / (cos(k) * (l * l)));
	}
	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 <= 5d-156) then
        tmp = 2.0d0 / ((sin(k) * (((t_m ** 3.0d0) / l) / l)) * (2.0d0 * k))
    else if (k <= 6.2d-28) then
        tmp = 2.0d0 / ((((t_m ** 2.0d0) * (t_m / l)) / l) * (2.0d0 * (k * k)))
    else
        tmp = 2.0d0 / ((t_m * (k ** 4.0d0)) / (cos(k) * (l * l)))
    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 <= 5e-156) {
		tmp = 2.0 / ((Math.sin(k) * ((Math.pow(t_m, 3.0) / l) / l)) * (2.0 * k));
	} else if (k <= 6.2e-28) {
		tmp = 2.0 / (((Math.pow(t_m, 2.0) * (t_m / l)) / l) * (2.0 * (k * k)));
	} else {
		tmp = 2.0 / ((t_m * Math.pow(k, 4.0)) / (Math.cos(k) * (l * l)));
	}
	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 <= 5e-156:
		tmp = 2.0 / ((math.sin(k) * ((math.pow(t_m, 3.0) / l) / l)) * (2.0 * k))
	elif k <= 6.2e-28:
		tmp = 2.0 / (((math.pow(t_m, 2.0) * (t_m / l)) / l) * (2.0 * (k * k)))
	else:
		tmp = 2.0 / ((t_m * math.pow(k, 4.0)) / (math.cos(k) * (l * l)))
	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 <= 5e-156)
		tmp = Float64(2.0 / Float64(Float64(sin(k) * Float64(Float64((t_m ^ 3.0) / l) / l)) * Float64(2.0 * k)));
	elseif (k <= 6.2e-28)
		tmp = Float64(2.0 / Float64(Float64(Float64((t_m ^ 2.0) * Float64(t_m / l)) / l) * Float64(2.0 * Float64(k * k))));
	else
		tmp = Float64(2.0 / Float64(Float64(t_m * (k ^ 4.0)) / Float64(cos(k) * Float64(l * l))));
	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 <= 5e-156)
		tmp = 2.0 / ((sin(k) * (((t_m ^ 3.0) / l) / l)) * (2.0 * k));
	elseif (k <= 6.2e-28)
		tmp = 2.0 / ((((t_m ^ 2.0) * (t_m / l)) / l) * (2.0 * (k * k)));
	else
		tmp = 2.0 / ((t_m * (k ^ 4.0)) / (cos(k) * (l * l)));
	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, 5e-156], N[(2.0 / N[(N[(N[Sin[k], $MachinePrecision] * N[(N[(N[Power[t$95$m, 3.0], $MachinePrecision] / l), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision] * N[(2.0 * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[k, 6.2e-28], N[(2.0 / N[(N[(N[(N[Power[t$95$m, 2.0], $MachinePrecision] * N[(t$95$m / l), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[(2.0 * N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(t$95$m * N[Power[k, 4.0], $MachinePrecision]), $MachinePrecision] / N[(N[Cos[k], $MachinePrecision] * N[(l * l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if k < 5.00000000000000007e-156

   1. Initial program 57.7%

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

   3. Simplified 57.7%

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

      1. associate-/r* 64.8%

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

      2. add-cube-cbrt 64.7%

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

      3. add-sqr-sqrt 29.1%

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

      4. times-frac 29.1%

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

      5. pow2 29.1%

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

      6. cbrt-div 29.1%

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

      7. rem-cbrt-cube 29.1%

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

      8. cbrt-div 29.1%

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

      9. rem-cbrt-cube 36.3%

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

   6. Applied egg-rr 36.3%

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

   7. Taylor expanded in k around 0 30.1%

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

      1. frac-times 28.3%

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

      2. unpow2 28.3%

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

      3. add-sqr-sqrt 64.1%

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

      4. pow3 64.1%

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

      5. add-cbrt-cube 59.9%

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

      6. unpow3 59.9%

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

      7. cbrt-div 59.9%

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

      8. pow3 59.9%

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

      9. add-cube-cbrt 59.9%

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

   9. Applied egg-rr 59.9%

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

   if 5.00000000000000007e-156 < k < 6.19999999999999984e-28

   1. Initial program 63.0%

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

   3. Simplified 71.5%

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

   5. Taylor expanded in k around 0 76.2%

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

      1. unpow2 76.2%

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

   7. Applied egg-rr 76.2%

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

      1. unpow3 76.2%

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

      2. *-un-lft-identity 76.2%

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

      3. times-frac 80.3%

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

      4. pow2 80.3%

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

   9. Applied egg-rr 80.3%

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

   if 6.19999999999999984e-28 < k

   1. Initial program 50.2%

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

   3. Simplified 50.3%

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

   5. Taylor expanded in t around 0 68.9%

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

      1. associate-*r* 68.9%

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

   7. Simplified 68.9%

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

   8. Taylor expanded in k around 0 59.3%

      \[\leadsto \frac{2}{\frac{\color{blue}{{k}^{4} \cdot t}}{{\ell}^{2} \cdot \cos k}} \]
   9. Step-by-step derivation

      1. unpow2 59.3%

         \[\leadsto \frac{2}{\frac{{k}^{4} \cdot t}{\color{blue}{\left(\ell \cdot \ell\right)} \cdot \cos k}} \]

   10. Applied egg-rr 59.3%

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

4. Final simplification 61.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 5 \cdot 10^{-156}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot \frac{\frac{{t}^{3}}{\ell}}{\ell}\right) \cdot \left(2 \cdot k\right)}\\ \mathbf{elif}\;k \leq 6.2 \cdot 10^{-28}:\\ \;\;\;\;\frac{2}{\frac{{t}^{2} \cdot \frac{t}{\ell}}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{t \cdot {k}^{4}}{\cos k \cdot \left(\ell \cdot \ell\right)}}\\ \end{array} \]
5. Add Preprocessing

Alternative 15: 59.1% accurate, 1.9× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 7.5 \cdot 10^{-28}:\\ \;\;\;\;\frac{2}{\frac{{t\_m}^{3} \cdot \left(\sin k \cdot \frac{2 \cdot k}{\ell}\right)}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{t\_m \cdot {k}^{4}}{\cos k \cdot \left(\ell \cdot \ell\right)}}\\ \end{array} \end{array} \]

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 7.5e-28)
    (/ 2.0 (/ (* (pow t_m 3.0) (* (sin k) (/ (* 2.0 k) l))) l))
    (/ 2.0 (/ (* t_m (pow k 4.0)) (* (cos k) (* l l)))))))

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-28) {
		tmp = 2.0 / ((pow(t_m, 3.0) * (sin(k) * ((2.0 * k) / l))) / l);
	} else {
		tmp = 2.0 / ((t_m * pow(k, 4.0)) / (cos(k) * (l * l)));
	}
	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 <= 7.5d-28) then
        tmp = 2.0d0 / (((t_m ** 3.0d0) * (sin(k) * ((2.0d0 * k) / l))) / l)
    else
        tmp = 2.0d0 / ((t_m * (k ** 4.0d0)) / (cos(k) * (l * l)))
    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 <= 7.5e-28) {
		tmp = 2.0 / ((Math.pow(t_m, 3.0) * (Math.sin(k) * ((2.0 * k) / l))) / l);
	} else {
		tmp = 2.0 / ((t_m * Math.pow(k, 4.0)) / (Math.cos(k) * (l * l)));
	}
	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 <= 7.5e-28:
		tmp = 2.0 / ((math.pow(t_m, 3.0) * (math.sin(k) * ((2.0 * k) / l))) / l)
	else:
		tmp = 2.0 / ((t_m * math.pow(k, 4.0)) / (math.cos(k) * (l * l)))
	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-28)
		tmp = Float64(2.0 / Float64(Float64((t_m ^ 3.0) * Float64(sin(k) * Float64(Float64(2.0 * k) / l))) / l));
	else
		tmp = Float64(2.0 / Float64(Float64(t_m * (k ^ 4.0)) / Float64(cos(k) * Float64(l * l))));
	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 <= 7.5e-28)
		tmp = 2.0 / (((t_m ^ 3.0) * (sin(k) * ((2.0 * k) / l))) / l);
	else
		tmp = 2.0 / ((t_m * (k ^ 4.0)) / (cos(k) * (l * l)));
	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, 7.5e-28], N[(2.0 / N[(N[(N[Power[t$95$m, 3.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[(N[(2.0 * k), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(t$95$m * N[Power[k, 4.0], $MachinePrecision]), $MachinePrecision] / N[(N[Cos[k], $MachinePrecision] * N[(l * l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 7.5000000000000003e-28

   1. Initial program 58.4%

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

   3. Simplified 58.4%

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

      1. associate-*l* 54.4%

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

      2. associate-/r* 61.7%

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

      3. associate-+r+ 61.7%

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

      4. metadata-eval 61.7%

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

      5. associate-*l* 61.7%

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

      6. associate-*l/ 62.7%

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

   6. Applied egg-rr 62.7%

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

      1. associate-/l* 63.6%

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

      2. associate-*l* 63.6%

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

   8. Simplified 63.6%

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

      1. associate-*l/ 63.7%

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

      2. associate-/l* 68.5%

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

   10. Applied egg-rr 68.5%

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

   11. Taylor expanded in k around 0 64.9%

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

       1. associate-*r/ 64.9%

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

       2. *-commutative 64.9%

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

   13. Simplified 64.9%

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

   if 7.5000000000000003e-28 < k

   1. Initial program 50.2%

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

   3. Simplified 50.3%

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

   5. Taylor expanded in t around 0 68.9%

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

      1. associate-*r* 68.9%

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

   7. Simplified 68.9%

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

   8. Taylor expanded in k around 0 59.3%

      \[\leadsto \frac{2}{\frac{\color{blue}{{k}^{4} \cdot t}}{{\ell}^{2} \cdot \cos k}} \]
   9. Step-by-step derivation

      1. unpow2 59.3%

         \[\leadsto \frac{2}{\frac{{k}^{4} \cdot t}{\color{blue}{\left(\ell \cdot \ell\right)} \cdot \cos k}} \]

   10. Applied egg-rr 59.3%

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

4. Final simplification 63.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 7.5 \cdot 10^{-28}:\\ \;\;\;\;\frac{2}{\frac{{t}^{3} \cdot \left(\sin k \cdot \frac{2 \cdot k}{\ell}\right)}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{t \cdot {k}^{4}}{\cos k \cdot \left(\ell \cdot \ell\right)}}\\ \end{array} \]
5. Add Preprocessing

Alternative 16: 58.0% accurate, 1.9× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 6.2 \cdot 10^{-28}:\\ \;\;\;\;\frac{2}{\frac{{t\_m}^{2} \cdot \frac{t\_m}{\ell}}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{t\_m \cdot {k}^{4}}{\cos k \cdot \left(\ell \cdot \ell\right)}}\\ \end{array} \end{array} \]

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 6.2e-28)
    (/ 2.0 (* (/ (* (pow t_m 2.0) (/ t_m l)) l) (* 2.0 (* k k))))
    (/ 2.0 (/ (* t_m (pow k 4.0)) (* (cos k) (* l l)))))))

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 <= 6.2e-28) {
		tmp = 2.0 / (((pow(t_m, 2.0) * (t_m / l)) / l) * (2.0 * (k * k)));
	} else {
		tmp = 2.0 / ((t_m * pow(k, 4.0)) / (cos(k) * (l * l)));
	}
	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 <= 6.2d-28) then
        tmp = 2.0d0 / ((((t_m ** 2.0d0) * (t_m / l)) / l) * (2.0d0 * (k * k)))
    else
        tmp = 2.0d0 / ((t_m * (k ** 4.0d0)) / (cos(k) * (l * l)))
    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 <= 6.2e-28) {
		tmp = 2.0 / (((Math.pow(t_m, 2.0) * (t_m / l)) / l) * (2.0 * (k * k)));
	} else {
		tmp = 2.0 / ((t_m * Math.pow(k, 4.0)) / (Math.cos(k) * (l * l)));
	}
	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 <= 6.2e-28:
		tmp = 2.0 / (((math.pow(t_m, 2.0) * (t_m / l)) / l) * (2.0 * (k * k)))
	else:
		tmp = 2.0 / ((t_m * math.pow(k, 4.0)) / (math.cos(k) * (l * l)))
	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 <= 6.2e-28)
		tmp = Float64(2.0 / Float64(Float64(Float64((t_m ^ 2.0) * Float64(t_m / l)) / l) * Float64(2.0 * Float64(k * k))));
	else
		tmp = Float64(2.0 / Float64(Float64(t_m * (k ^ 4.0)) / Float64(cos(k) * Float64(l * l))));
	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 <= 6.2e-28)
		tmp = 2.0 / ((((t_m ^ 2.0) * (t_m / l)) / l) * (2.0 * (k * k)));
	else
		tmp = 2.0 / ((t_m * (k ^ 4.0)) / (cos(k) * (l * l)));
	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, 6.2e-28], N[(2.0 / N[(N[(N[(N[Power[t$95$m, 2.0], $MachinePrecision] * N[(t$95$m / l), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[(2.0 * N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(t$95$m * N[Power[k, 4.0], $MachinePrecision]), $MachinePrecision] / N[(N[Cos[k], $MachinePrecision] * N[(l * l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 6.19999999999999984e-28

   1. Initial program 58.4%

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

   3. Simplified 61.7%

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

   5. Taylor expanded in k around 0 59.1%

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

      1. unpow2 59.1%

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

   7. Applied egg-rr 59.1%

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

      1. unpow3 59.1%

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

      2. *-un-lft-identity 59.1%

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

      3. times-frac 62.4%

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

      4. pow2 62.4%

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

   9. Applied egg-rr 62.4%

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

   if 6.19999999999999984e-28 < k

   1. Initial program 50.2%

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

   3. Simplified 50.3%

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

   5. Taylor expanded in t around 0 68.9%

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

      1. associate-*r* 68.9%

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

   7. Simplified 68.9%

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

   8. Taylor expanded in k around 0 59.3%

      \[\leadsto \frac{2}{\frac{\color{blue}{{k}^{4} \cdot t}}{{\ell}^{2} \cdot \cos k}} \]
   9. Step-by-step derivation

      1. unpow2 59.3%

         \[\leadsto \frac{2}{\frac{{k}^{4} \cdot t}{\color{blue}{\left(\ell \cdot \ell\right)} \cdot \cos k}} \]

   10. Applied egg-rr 59.3%

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

4. Final simplification 61.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 6.2 \cdot 10^{-28}:\\ \;\;\;\;\frac{2}{\frac{{t}^{2} \cdot \frac{t}{\ell}}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{t \cdot {k}^{4}}{\cos k \cdot \left(\ell \cdot \ell\right)}}\\ \end{array} \]
5. Add Preprocessing

Alternative 17: 57.5% accurate, 2.0× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 7.5 \cdot 10^{-28}:\\ \;\;\;\;\frac{2}{\frac{{t\_m}^{2} \cdot \frac{t\_m}{\ell}}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{t\_m \cdot {k}^{4}}{{\ell}^{2}}}\\ \end{array} \end{array} \]

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 7.5e-28)
    (/ 2.0 (* (/ (* (pow t_m 2.0) (/ t_m l)) l) (* 2.0 (* k k))))
    (/ 2.0 (/ (* t_m (pow k 4.0)) (pow 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 (k <= 7.5e-28) {
		tmp = 2.0 / (((pow(t_m, 2.0) * (t_m / l)) / l) * (2.0 * (k * k)));
	} else {
		tmp = 2.0 / ((t_m * pow(k, 4.0)) / pow(l, 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 <= 7.5d-28) then
        tmp = 2.0d0 / ((((t_m ** 2.0d0) * (t_m / l)) / l) * (2.0d0 * (k * k)))
    else
        tmp = 2.0d0 / ((t_m * (k ** 4.0d0)) / (l ** 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 <= 7.5e-28) {
		tmp = 2.0 / (((Math.pow(t_m, 2.0) * (t_m / l)) / l) * (2.0 * (k * k)));
	} else {
		tmp = 2.0 / ((t_m * Math.pow(k, 4.0)) / Math.pow(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 k <= 7.5e-28:
		tmp = 2.0 / (((math.pow(t_m, 2.0) * (t_m / l)) / l) * (2.0 * (k * k)))
	else:
		tmp = 2.0 / ((t_m * math.pow(k, 4.0)) / math.pow(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 (k <= 7.5e-28)
		tmp = Float64(2.0 / Float64(Float64(Float64((t_m ^ 2.0) * Float64(t_m / l)) / l) * Float64(2.0 * Float64(k * k))));
	else
		tmp = Float64(2.0 / Float64(Float64(t_m * (k ^ 4.0)) / (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 (k <= 7.5e-28)
		tmp = 2.0 / ((((t_m ^ 2.0) * (t_m / l)) / l) * (2.0 * (k * k)));
	else
		tmp = 2.0 / ((t_m * (k ^ 4.0)) / (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[k, 7.5e-28], N[(2.0 / N[(N[(N[(N[Power[t$95$m, 2.0], $MachinePrecision] * N[(t$95$m / l), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[(2.0 * N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(t$95$m * N[Power[k, 4.0], $MachinePrecision]), $MachinePrecision] / N[Power[l, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 7.5000000000000003e-28

   1. Initial program 58.4%

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

   3. Simplified 61.7%

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

   5. Taylor expanded in k around 0 59.1%

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

      1. unpow2 59.1%

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

   7. Applied egg-rr 59.1%

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

      1. unpow3 59.1%

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

      2. *-un-lft-identity 59.1%

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

      3. times-frac 62.4%

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

      4. pow2 62.4%

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

   9. Applied egg-rr 62.4%

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

   if 7.5000000000000003e-28 < k

   1. Initial program 50.2%

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

   3. Simplified 50.3%

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

   5. Taylor expanded in t around 0 68.9%

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

      1. associate-*r* 68.9%

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

   7. Simplified 68.9%

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

   8. Taylor expanded in k around 0 53.6%

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

4. Final simplification 60.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 7.5 \cdot 10^{-28}:\\ \;\;\;\;\frac{2}{\frac{{t}^{2} \cdot \frac{t}{\ell}}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{t \cdot {k}^{4}}{{\ell}^{2}}}\\ \end{array} \]
5. Add Preprocessing

Alternative 18: 57.9% accurate, 2.0× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 1.8 \cdot 10^{+106}:\\ \;\;\;\;\frac{2}{\frac{{t\_m}^{2} \cdot \frac{t\_m}{\ell}}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \frac{\frac{{\ell}^{2}}{{k}^{4}}}{t\_m}\\ \end{array} \end{array} \]

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 1.8e+106)
    (/ 2.0 (* (/ (* (pow t_m 2.0) (/ t_m l)) l) (* 2.0 (* k k))))
    (* 2.0 (/ (/ (pow l 2.0) (pow k 4.0)) t_m)))))

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 <= 1.8e+106) {
		tmp = 2.0 / (((pow(t_m, 2.0) * (t_m / l)) / l) * (2.0 * (k * k)));
	} else {
		tmp = 2.0 * ((pow(l, 2.0) / pow(k, 4.0)) / t_m);
	}
	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 <= 1.8d+106) then
        tmp = 2.0d0 / ((((t_m ** 2.0d0) * (t_m / l)) / l) * (2.0d0 * (k * k)))
    else
        tmp = 2.0d0 * (((l ** 2.0d0) / (k ** 4.0d0)) / t_m)
    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 <= 1.8e+106) {
		tmp = 2.0 / (((Math.pow(t_m, 2.0) * (t_m / l)) / l) * (2.0 * (k * k)));
	} else {
		tmp = 2.0 * ((Math.pow(l, 2.0) / Math.pow(k, 4.0)) / t_m);
	}
	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 <= 1.8e+106:
		tmp = 2.0 / (((math.pow(t_m, 2.0) * (t_m / l)) / l) * (2.0 * (k * k)))
	else:
		tmp = 2.0 * ((math.pow(l, 2.0) / math.pow(k, 4.0)) / t_m)
	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 <= 1.8e+106)
		tmp = Float64(2.0 / Float64(Float64(Float64((t_m ^ 2.0) * Float64(t_m / l)) / l) * Float64(2.0 * Float64(k * k))));
	else
		tmp = Float64(2.0 * Float64(Float64((l ^ 2.0) / (k ^ 4.0)) / t_m));
	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 <= 1.8e+106)
		tmp = 2.0 / ((((t_m ^ 2.0) * (t_m / l)) / l) * (2.0 * (k * k)));
	else
		tmp = 2.0 * (((l ^ 2.0) / (k ^ 4.0)) / t_m);
	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, 1.8e+106], N[(2.0 / N[(N[(N[(N[Power[t$95$m, 2.0], $MachinePrecision] * N[(t$95$m / l), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[(2.0 * N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 * N[(N[(N[Power[l, 2.0], $MachinePrecision] / N[Power[k, 4.0], $MachinePrecision]), $MachinePrecision] / t$95$m), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 1.8e106

   1. Initial program 56.8%

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

   3. Simplified 60.5%

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

   5. Taylor expanded in k around 0 58.0%

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

      1. unpow2 58.0%

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

   7. Applied egg-rr 58.0%

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

      1. unpow3 58.0%

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

      2. *-un-lft-identity 58.0%

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

      3. times-frac 60.9%

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

      4. pow2 60.9%

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

   9. Applied egg-rr 60.9%

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

   if 1.8e106 < k

   1. Initial program 52.7%

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

   3. Simplified 52.7%

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

   5. Taylor expanded in t around 0 63.3%

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

      1. associate-*r* 63.3%

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

   7. Simplified 63.3%

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

   8. Taylor expanded in k around 0 55.8%

      \[\leadsto \color{blue}{2 \cdot \frac{{\ell}^{2}}{{k}^{4} \cdot t}} \]
   9. Step-by-step derivation

      1. associate-/r* 55.8%

         \[\leadsto 2 \cdot \color{blue}{\frac{\frac{{\ell}^{2}}{{k}^{4}}}{t}} \]

   10. Simplified 55.8%

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

4. Final simplification 60.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 1.8 \cdot 10^{+106}:\\ \;\;\;\;\frac{2}{\frac{{t}^{2} \cdot \frac{t}{\ell}}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;2 \cdot \frac{\frac{{\ell}^{2}}{{k}^{4}}}{t}\\ \end{array} \]
5. Add Preprocessing

Alternative 19: 56.6% accurate, 3.6× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \frac{2}{\frac{{t\_m}^{2} \cdot \frac{t\_m}{\ell}}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \end{array} \]

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 (* (/ (* (pow t_m 2.0) (/ t_m l)) l) (* 2.0 (* k k))))))

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 / (((pow(t_m, 2.0) * (t_m / l)) / l) * (2.0 * (k * k))));
}

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 / ((((t_m ** 2.0d0) * (t_m / l)) / l) * (2.0d0 * (k * k))))
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 / (((Math.pow(t_m, 2.0) * (t_m / l)) / l) * (2.0 * (k * k))));
}

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 / (((math.pow(t_m, 2.0) * (t_m / l)) / l) * (2.0 * (k * k))))

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(Float64((t_m ^ 2.0) * Float64(t_m / l)) / l) * Float64(2.0 * Float64(k * k)))))
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 / ((((t_m ^ 2.0) * (t_m / l)) / l) * (2.0 * (k * k))));
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[(N[(N[Power[t$95$m, 2.0], $MachinePrecision] * N[(t$95$m / l), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[(2.0 * N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 56.2%

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

3. Simplified 59.9%

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

5. Taylor expanded in k around 0 56.3%

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

   1. unpow2 56.3%

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

7. Applied egg-rr 56.3%

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

   1. unpow3 56.3%

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

   2. *-un-lft-identity 56.3%

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

   3. times-frac 58.7%

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

   4. pow2 58.7%

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

9. Applied egg-rr 58.7%

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

10. Final simplification 58.7%

    \[\leadsto \frac{2}{\frac{{t}^{2} \cdot \frac{t}{\ell}}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]
11. Add Preprocessing

Alternative 20: 53.8% accurate, 3.6× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \frac{2}{\left(2 \cdot \left(k \cdot k\right)\right) \cdot \frac{\frac{1}{\ell \cdot {t\_m}^{-3}}}{\ell}} \end{array} \]

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)) (/ (/ 1.0 (* l (pow t_m -3.0))) 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)) * ((1.0 / (l * pow(t_m, -3.0))) / 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)) * ((1.0d0 / (l * (t_m ** (-3.0d0)))) / 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)) * ((1.0 / (l * Math.pow(t_m, -3.0))) / 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)) * ((1.0 / (l * math.pow(t_m, -3.0))) / 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(1.0 / Float64(l * (t_m ^ -3.0))) / 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)) * ((1.0 / (l * (t_m ^ -3.0))) / 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[(1.0 / N[(l * N[Power[t$95$m, -3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 56.2%

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

3. Simplified 59.9%

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

5. Taylor expanded in k around 0 56.3%

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

   1. unpow2 56.3%

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

7. Applied egg-rr 56.3%

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

   1. clear-num 56.3%

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

   2. inv-pow 56.3%

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

9. Applied egg-rr 56.3%

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

    1. unpow-1 56.3%

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

11. Simplified 56.3%

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

    1. *-un-lft-identity 56.3%

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

    2. div-inv 56.3%

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

    3. pow-flip 56.7%

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

    4. metadata-eval 56.7%

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

13. Applied egg-rr 56.7%

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

    1. *-lft-identity 56.7%

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

15. Simplified 56.7%

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

16. Final simplification 56.7%

    \[\leadsto \frac{2}{\left(2 \cdot \left(k \cdot k\right)\right) \cdot \frac{\frac{1}{\ell \cdot {t}^{-3}}}{\ell}} \]
17. Add Preprocessing

Alternative 21: 53.8% accurate, 3.7× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \frac{2}{\frac{\frac{{t\_m}^{3}}{\ell}}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \end{array} \]

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 (* (/ (/ (pow t_m 3.0) l) l) (* 2.0 (* k k))))))

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 / (((pow(t_m, 3.0) / l) / l) * (2.0 * (k * k))));
}

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 / ((((t_m ** 3.0d0) / l) / l) * (2.0d0 * (k * k))))
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 / (((Math.pow(t_m, 3.0) / l) / l) * (2.0 * (k * k))));
}

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 / (((math.pow(t_m, 3.0) / l) / l) * (2.0 * (k * k))))

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(Float64((t_m ^ 3.0) / l) / l) * Float64(2.0 * Float64(k * k)))))
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 / ((((t_m ^ 3.0) / l) / l) * (2.0 * (k * k))));
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[(N[(N[Power[t$95$m, 3.0], $MachinePrecision] / l), $MachinePrecision] / l), $MachinePrecision] * N[(2.0 * N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 56.2%

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

3. Simplified 59.9%

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

5. Taylor expanded in k around 0 56.3%

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

   1. unpow2 56.3%

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

7. Applied egg-rr 56.3%

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

Reproduce

herbie shell --seed 2024135 
(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))))