Toniolo and Linder, Equation (10+)

Percentage Accurate: 54.8% → 89.2%

Time: 1.0min

Alternatives: 20

Speedup: 3.8×

• 27.0% 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: 54.8% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 89.2% accurate, 0.7× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ \begin{array}{l} t_2 := \frac{{t\_m}^{1.5}}{\ell}\\ t_3 := \sin k\_m \cdot \tan k\_m\\ t_4 := \mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k\_m}{t\_m}\right)\right)\\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k\_m \leq 1.5 \cdot 10^{-125}:\\ \;\;\;\;\frac{2}{{\left(t\_4 \cdot \left(k\_m \cdot t\_2\right)\right)}^{2}}\\ \mathbf{elif}\;k\_m \leq 7 \cdot 10^{+84}:\\ \;\;\;\;\frac{2}{t\_3 \cdot {\left(t\_4 \cdot t\_2\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{t\_3 \cdot {\left(\frac{\sqrt{t\_m} \cdot \left(-k\_m\right)}{\ell}\right)}^{2}}\\ \end{array} \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (let* ((t_2 (/ (pow t_m 1.5) l))
        (t_3 (* (sin k_m) (tan k_m)))
        (t_4 (hypot 1.0 (hypot 1.0 (/ k_m t_m)))))
   (*
    t_s
    (if (<= k_m 1.5e-125)
      (/ 2.0 (pow (* t_4 (* k_m t_2)) 2.0))
      (if (<= k_m 7e+84)
        (/ 2.0 (* t_3 (pow (* t_4 t_2) 2.0)))
        (/ 2.0 (* t_3 (pow (/ (* (sqrt t_m) (- k_m)) l) 2.0))))))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double t_2 = pow(t_m, 1.5) / l;
	double t_3 = sin(k_m) * tan(k_m);
	double t_4 = hypot(1.0, hypot(1.0, (k_m / t_m)));
	double tmp;
	if (k_m <= 1.5e-125) {
		tmp = 2.0 / pow((t_4 * (k_m * t_2)), 2.0);
	} else if (k_m <= 7e+84) {
		tmp = 2.0 / (t_3 * pow((t_4 * t_2), 2.0));
	} else {
		tmp = 2.0 / (t_3 * pow(((sqrt(t_m) * -k_m) / l), 2.0));
	}
	return t_s * tmp;
}

Java

k_m = Math.abs(k);
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_m) {
	double t_2 = Math.pow(t_m, 1.5) / l;
	double t_3 = Math.sin(k_m) * Math.tan(k_m);
	double t_4 = Math.hypot(1.0, Math.hypot(1.0, (k_m / t_m)));
	double tmp;
	if (k_m <= 1.5e-125) {
		tmp = 2.0 / Math.pow((t_4 * (k_m * t_2)), 2.0);
	} else if (k_m <= 7e+84) {
		tmp = 2.0 / (t_3 * Math.pow((t_4 * t_2), 2.0));
	} else {
		tmp = 2.0 / (t_3 * Math.pow(((Math.sqrt(t_m) * -k_m) / l), 2.0));
	}
	return t_s * tmp;
}

Python

k_m = math.fabs(k)
t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k_m):
	t_2 = math.pow(t_m, 1.5) / l
	t_3 = math.sin(k_m) * math.tan(k_m)
	t_4 = math.hypot(1.0, math.hypot(1.0, (k_m / t_m)))
	tmp = 0
	if k_m <= 1.5e-125:
		tmp = 2.0 / math.pow((t_4 * (k_m * t_2)), 2.0)
	elif k_m <= 7e+84:
		tmp = 2.0 / (t_3 * math.pow((t_4 * t_2), 2.0))
	else:
		tmp = 2.0 / (t_3 * math.pow(((math.sqrt(t_m) * -k_m) / l), 2.0))
	return t_s * tmp

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	t_2 = Float64((t_m ^ 1.5) / l)
	t_3 = Float64(sin(k_m) * tan(k_m))
	t_4 = hypot(1.0, hypot(1.0, Float64(k_m / t_m)))
	tmp = 0.0
	if (k_m <= 1.5e-125)
		tmp = Float64(2.0 / (Float64(t_4 * Float64(k_m * t_2)) ^ 2.0));
	elseif (k_m <= 7e+84)
		tmp = Float64(2.0 / Float64(t_3 * (Float64(t_4 * t_2) ^ 2.0)));
	else
		tmp = Float64(2.0 / Float64(t_3 * (Float64(Float64(sqrt(t_m) * Float64(-k_m)) / l) ^ 2.0)));
	end
	return Float64(t_s * tmp)
end

MATLAB

k_m = abs(k);
t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k_m)
	t_2 = (t_m ^ 1.5) / l;
	t_3 = sin(k_m) * tan(k_m);
	t_4 = hypot(1.0, hypot(1.0, (k_m / t_m)));
	tmp = 0.0;
	if (k_m <= 1.5e-125)
		tmp = 2.0 / ((t_4 * (k_m * t_2)) ^ 2.0);
	elseif (k_m <= 7e+84)
		tmp = 2.0 / (t_3 * ((t_4 * t_2) ^ 2.0));
	else
		tmp = 2.0 / (t_3 * (((sqrt(t_m) * -k_m) / l) ^ 2.0));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := Block[{t$95$2 = N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision]}, Block[{t$95$3 = N[(N[Sin[k$95$m], $MachinePrecision] * N[Tan[k$95$m], $MachinePrecision]), $MachinePrecision]}, Block[{t$95$4 = N[Sqrt[1.0 ^ 2 + N[Sqrt[1.0 ^ 2 + N[(k$95$m / t$95$m), $MachinePrecision] ^ 2], $MachinePrecision] ^ 2], $MachinePrecision]}, N[(t$95$s * If[LessEqual[k$95$m, 1.5e-125], N[(2.0 / N[Power[N[(t$95$4 * N[(k$95$m * t$95$2), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[k$95$m, 7e+84], N[(2.0 / N[(t$95$3 * N[Power[N[(t$95$4 * t$95$2), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(t$95$3 * N[Power[N[(N[(N[Sqrt[t$95$m], $MachinePrecision] * (-k$95$m)), $MachinePrecision] / l), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]]

Derivation

1. Split input into 3 regimes

2. if k < 1.49999999999999995e-125

   1. Initial program 59.2%

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

      1. pow1 59.2%

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

   4. Applied egg-rr 28.4%

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

      1. unpow1 28.4%

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

   6. Simplified 28.4%

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

   7. Taylor expanded in k around 0 39.4%

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

   if 1.49999999999999995e-125 < k < 6.9999999999999998e84

   1. Initial program 70.7%

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

      1. pow1 70.7%

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

   4. Applied egg-rr 35.0%

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

      1. unpow1 35.0%

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

   6. Simplified 35.0%

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

      1. associate-*r* 35.0%

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

      2. unpow-prod-down 35.1%

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

      3. pow2 35.1%

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

      4. add-sqr-sqrt 43.5%

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

   8. Applied egg-rr 43.5%

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

   if 6.9999999999999998e84 < k

   1. Initial program 48.5%

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

      1. pow1 48.5%

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

   4. Applied egg-rr 23.1%

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

      1. unpow1 23.1%

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

   6. Simplified 23.1%

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

      1. associate-*r* 23.1%

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

      2. unpow-prod-down 23.1%

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

      3. pow2 23.1%

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

      4. add-sqr-sqrt 39.5%

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

   8. Applied egg-rr 39.5%

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

   9. Taylor expanded in k around -inf 44.1%

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

       1. mul-1-neg 44.1%

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

       2. associate-*l/ 44.0%

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

       3. distribute-neg-frac2 44.0%

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

   11. Simplified 44.0%

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

4. Final simplification 40.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 1.5 \cdot 10^{-125}:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(k \cdot \frac{{t}^{1.5}}{\ell}\right)\right)}^{2}}\\ \mathbf{elif}\;k \leq 7 \cdot 10^{+84}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot \tan k\right) \cdot {\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \frac{{t}^{1.5}}{\ell}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot \tan k\right) \cdot {\left(\frac{\sqrt{t} \cdot \left(-k\right)}{\ell}\right)}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 89.5% accurate, 0.7× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 3.2 \cdot 10^{-86}:\\ \;\;\;\;\frac{2}{{\left(\frac{k\_m}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k\_m \cdot \tan k\_m\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{{\left(\frac{t\_m}{\sqrt[3]{\ell}} \cdot \sqrt[3]{\sin k\_m}\right)}^{3} \cdot \left(\tan k\_m \cdot \left(2 + {\left(\frac{k\_m}{t\_m}\right)}^{2}\right)\right)}{\ell}}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= t_m 3.2e-86)
    (/ 2.0 (* (pow (* (/ k_m l) (sqrt t_m)) 2.0) (* (sin k_m) (tan k_m))))
    (/
     2.0
     (/
      (*
       (pow (* (/ t_m (cbrt l)) (cbrt (sin k_m))) 3.0)
       (* (tan k_m) (+ 2.0 (pow (/ k_m t_m) 2.0))))
      l)))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (t_m <= 3.2e-86) {
		tmp = 2.0 / (pow(((k_m / l) * sqrt(t_m)), 2.0) * (sin(k_m) * tan(k_m)));
	} else {
		tmp = 2.0 / ((pow(((t_m / cbrt(l)) * cbrt(sin(k_m))), 3.0) * (tan(k_m) * (2.0 + pow((k_m / t_m), 2.0)))) / l);
	}
	return t_s * tmp;
}

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (t_m <= 3.2e-86) {
		tmp = 2.0 / (Math.pow(((k_m / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k_m) * Math.tan(k_m)));
	} else {
		tmp = 2.0 / ((Math.pow(((t_m / Math.cbrt(l)) * Math.cbrt(Math.sin(k_m))), 3.0) * (Math.tan(k_m) * (2.0 + Math.pow((k_m / t_m), 2.0)))) / l);
	}
	return t_s * tmp;
}

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (t_m <= 3.2e-86)
		tmp = Float64(2.0 / Float64((Float64(Float64(k_m / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k_m) * tan(k_m))));
	else
		tmp = Float64(2.0 / Float64(Float64((Float64(Float64(t_m / cbrt(l)) * cbrt(sin(k_m))) ^ 3.0) * Float64(tan(k_m) * Float64(2.0 + (Float64(k_m / t_m) ^ 2.0)))) / l));
	end
	return Float64(t_s * tmp)
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[t$95$m, 3.2e-86], N[(2.0 / N[(N[Power[N[(N[(k$95$m / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k$95$m], $MachinePrecision] * N[Tan[k$95$m], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[Power[N[(N[(t$95$m / N[Power[l, 1/3], $MachinePrecision]), $MachinePrecision] * N[Power[N[Sin[k$95$m], $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision] * N[(N[Tan[k$95$m], $MachinePrecision] * N[(2.0 + N[Power[N[(k$95$m / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if t < 3.20000000000000006e-86

   1. Initial program 50.9%

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

      1. pow1 50.9%

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

   4. Applied egg-rr 13.4%

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

      1. unpow1 13.4%

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

   6. Simplified 13.4%

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

      1. associate-*r* 13.4%

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

      2. unpow-prod-down 13.4%

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

      3. pow2 13.4%

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

      4. add-sqr-sqrt 17.5%

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

   8. Applied egg-rr 17.5%

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

   9. Taylor expanded in k around inf 25.4%

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

   if 3.20000000000000006e-86 < t

   1. Initial program 75.7%

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

   3. Simplified 73.1%

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

      1. associate-*l/ 73.7%

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

      2. associate-*l* 73.7%

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

   6. Applied egg-rr 73.7%

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

      1. associate-*r* 78.9%

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

   8. Simplified 78.9%

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

      1. add-cube-cbrt 78.7%

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

      2. pow3 78.7%

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

      3. cbrt-prod 78.6%

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

      4. cbrt-div 78.6%

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

      5. unpow3 78.6%

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

      6. add-cbrt-cube 89.8%

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

   10. Applied egg-rr 89.8%

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

Alternative 3: 83.4% accurate, 1.0× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k\_m \leq 1.9 \cdot 10^{-159}:\\ \;\;\;\;\frac{2}{{\left(\sqrt{{t\_m}^{3}} \cdot \left(k\_m \cdot \frac{\sqrt{2}}{\ell}\right)\right)}^{2}}\\ \mathbf{elif}\;k\_m \leq 10:\\ \;\;\;\;\frac{2}{{\left(\frac{t\_m}{{\left(\sqrt[3]{\ell}\right)}^{2}}\right)}^{3} \cdot \left(2 \cdot {k\_m}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\sin k\_m \cdot \tan k\_m\right) \cdot {\left(\frac{\sqrt{t\_m} \cdot \left(-k\_m\right)}{\ell}\right)}^{2}}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= k_m 1.9e-159)
    (/ 2.0 (pow (* (sqrt (pow t_m 3.0)) (* k_m (/ (sqrt 2.0) l))) 2.0))
    (if (<= k_m 10.0)
      (/ 2.0 (* (pow (/ t_m (pow (cbrt l) 2.0)) 3.0) (* 2.0 (pow k_m 2.0))))
      (/
       2.0
       (* (* (sin k_m) (tan k_m)) (pow (/ (* (sqrt t_m) (- k_m)) l) 2.0)))))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (k_m <= 1.9e-159) {
		tmp = 2.0 / pow((sqrt(pow(t_m, 3.0)) * (k_m * (sqrt(2.0) / l))), 2.0);
	} else if (k_m <= 10.0) {
		tmp = 2.0 / (pow((t_m / pow(cbrt(l), 2.0)), 3.0) * (2.0 * pow(k_m, 2.0)));
	} else {
		tmp = 2.0 / ((sin(k_m) * tan(k_m)) * pow(((sqrt(t_m) * -k_m) / l), 2.0));
	}
	return t_s * tmp;
}

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (k_m <= 1.9e-159) {
		tmp = 2.0 / Math.pow((Math.sqrt(Math.pow(t_m, 3.0)) * (k_m * (Math.sqrt(2.0) / l))), 2.0);
	} else if (k_m <= 10.0) {
		tmp = 2.0 / (Math.pow((t_m / Math.pow(Math.cbrt(l), 2.0)), 3.0) * (2.0 * Math.pow(k_m, 2.0)));
	} else {
		tmp = 2.0 / ((Math.sin(k_m) * Math.tan(k_m)) * Math.pow(((Math.sqrt(t_m) * -k_m) / l), 2.0));
	}
	return t_s * tmp;
}

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (k_m <= 1.9e-159)
		tmp = Float64(2.0 / (Float64(sqrt((t_m ^ 3.0)) * Float64(k_m * Float64(sqrt(2.0) / l))) ^ 2.0));
	elseif (k_m <= 10.0)
		tmp = Float64(2.0 / Float64((Float64(t_m / (cbrt(l) ^ 2.0)) ^ 3.0) * Float64(2.0 * (k_m ^ 2.0))));
	else
		tmp = Float64(2.0 / Float64(Float64(sin(k_m) * tan(k_m)) * (Float64(Float64(sqrt(t_m) * Float64(-k_m)) / l) ^ 2.0)));
	end
	return Float64(t_s * tmp)
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[k$95$m, 1.9e-159], N[(2.0 / N[Power[N[(N[Sqrt[N[Power[t$95$m, 3.0], $MachinePrecision]], $MachinePrecision] * N[(k$95$m * N[(N[Sqrt[2.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[k$95$m, 10.0], N[(2.0 / N[(N[Power[N[(t$95$m / N[Power[N[Power[l, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision] * N[(2.0 * N[Power[k$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[Sin[k$95$m], $MachinePrecision] * N[Tan[k$95$m], $MachinePrecision]), $MachinePrecision] * N[Power[N[(N[(N[Sqrt[t$95$m], $MachinePrecision] * (-k$95$m)), $MachinePrecision] / l), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if k < 1.9e-159

   1. Initial program 57.8%

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

      1. pow1 57.8%

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

   4. Applied egg-rr 26.3%

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

      1. unpow1 26.3%

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

   6. Simplified 26.3%

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

   7. Taylor expanded in k around 0 30.4%

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

      1. *-commutative 30.4%

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

      2. associate-/l* 30.4%

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

   9. Simplified 30.4%

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

   if 1.9e-159 < k < 10

   1. Initial program 83.9%

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

   3. Simplified 84.2%

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

   5. Taylor expanded in k around 0 84.8%

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

      1. add-cube-cbrt 84.6%

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

      2. pow3 84.6%

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

      3. cbrt-div 84.4%

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

      4. unpow3 84.4%

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

      5. add-cbrt-cube 92.4%

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

      6. pow3 92.3%

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

      7. unpow2 92.3%

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

      8. *-un-lft-identity 92.3%

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

      9. frac-times 95.5%

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

      10. add-cube-cbrt 95.6%

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

   7. Applied egg-rr 87.8%

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

      1. unpow2 87.8%

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

      2. pow3 87.7%

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

      3. unpow2 87.7%

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

      4. cbrt-prod 95.4%

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

      5. pow2 95.4%

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

   9. Applied egg-rr 95.4%

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

   if 10 < k

   1. Initial program 52.5%

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

      1. pow1 52.5%

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

   4. Applied egg-rr 18.9%

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

      1. unpow1 18.9%

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

   6. Simplified 18.9%

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

      1. associate-*r* 18.9%

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

      2. unpow-prod-down 19.0%

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

      3. pow2 19.0%

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

      4. add-sqr-sqrt 35.0%

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

   8. Applied egg-rr 35.0%

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

   9. Taylor expanded in k around -inf 39.6%

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

       1. mul-1-neg 39.6%

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

       2. associate-*l/ 39.5%

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

       3. distribute-neg-frac2 39.5%

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

   11. Simplified 39.5%

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

4. Final simplification 39.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 1.9 \cdot 10^{-159}:\\ \;\;\;\;\frac{2}{{\left(\sqrt{{t}^{3}} \cdot \left(k \cdot \frac{\sqrt{2}}{\ell}\right)\right)}^{2}}\\ \mathbf{elif}\;k \leq 10:\\ \;\;\;\;\frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}}\right)}^{3} \cdot \left(2 \cdot {k}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot \tan k\right) \cdot {\left(\frac{\sqrt{t} \cdot \left(-k\right)}{\ell}\right)}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 83.7% accurate, 1.0× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k\_m \leq 2.8 \cdot 10^{-160}:\\ \;\;\;\;\frac{2}{{\left(\sqrt{{t\_m}^{3}} \cdot \left(k\_m \cdot \frac{\sqrt{2}}{\ell}\right)\right)}^{2}}\\ \mathbf{elif}\;k\_m \leq 7.5:\\ \;\;\;\;\frac{2}{{\left(\frac{t\_m}{{\left(\sqrt[3]{\ell}\right)}^{2}}\right)}^{3} \cdot \left(2 \cdot {k\_m}^{2}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{k\_m}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k\_m \cdot \tan k\_m\right)}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= k_m 2.8e-160)
    (/ 2.0 (pow (* (sqrt (pow t_m 3.0)) (* k_m (/ (sqrt 2.0) l))) 2.0))
    (if (<= k_m 7.5)
      (/ 2.0 (* (pow (/ t_m (pow (cbrt l) 2.0)) 3.0) (* 2.0 (pow k_m 2.0))))
      (/
       2.0
       (* (pow (* (/ k_m l) (sqrt t_m)) 2.0) (* (sin k_m) (tan k_m))))))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (k_m <= 2.8e-160) {
		tmp = 2.0 / pow((sqrt(pow(t_m, 3.0)) * (k_m * (sqrt(2.0) / l))), 2.0);
	} else if (k_m <= 7.5) {
		tmp = 2.0 / (pow((t_m / pow(cbrt(l), 2.0)), 3.0) * (2.0 * pow(k_m, 2.0)));
	} else {
		tmp = 2.0 / (pow(((k_m / l) * sqrt(t_m)), 2.0) * (sin(k_m) * tan(k_m)));
	}
	return t_s * tmp;
}

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (k_m <= 2.8e-160) {
		tmp = 2.0 / Math.pow((Math.sqrt(Math.pow(t_m, 3.0)) * (k_m * (Math.sqrt(2.0) / l))), 2.0);
	} else if (k_m <= 7.5) {
		tmp = 2.0 / (Math.pow((t_m / Math.pow(Math.cbrt(l), 2.0)), 3.0) * (2.0 * Math.pow(k_m, 2.0)));
	} else {
		tmp = 2.0 / (Math.pow(((k_m / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k_m) * Math.tan(k_m)));
	}
	return t_s * tmp;
}

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (k_m <= 2.8e-160)
		tmp = Float64(2.0 / (Float64(sqrt((t_m ^ 3.0)) * Float64(k_m * Float64(sqrt(2.0) / l))) ^ 2.0));
	elseif (k_m <= 7.5)
		tmp = Float64(2.0 / Float64((Float64(t_m / (cbrt(l) ^ 2.0)) ^ 3.0) * Float64(2.0 * (k_m ^ 2.0))));
	else
		tmp = Float64(2.0 / Float64((Float64(Float64(k_m / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k_m) * tan(k_m))));
	end
	return Float64(t_s * tmp)
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[k$95$m, 2.8e-160], N[(2.0 / N[Power[N[(N[Sqrt[N[Power[t$95$m, 3.0], $MachinePrecision]], $MachinePrecision] * N[(k$95$m * N[(N[Sqrt[2.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[k$95$m, 7.5], N[(2.0 / N[(N[Power[N[(t$95$m / N[Power[N[Power[l, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision] * N[(2.0 * N[Power[k$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[Power[N[(N[(k$95$m / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k$95$m], $MachinePrecision] * N[Tan[k$95$m], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if k < 2.80000000000000016e-160

   1. Initial program 57.8%

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

      1. pow1 57.8%

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

   4. Applied egg-rr 26.3%

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

      1. unpow1 26.3%

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

   6. Simplified 26.3%

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

   7. Taylor expanded in k around 0 30.4%

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

      1. *-commutative 30.4%

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

      2. associate-/l* 30.4%

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

   9. Simplified 30.4%

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

   if 2.80000000000000016e-160 < k < 7.5

   1. Initial program 83.9%

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

   3. Simplified 84.2%

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

   5. Taylor expanded in k around 0 84.8%

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

      1. add-cube-cbrt 84.6%

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

      2. pow3 84.6%

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

      3. cbrt-div 84.4%

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

      4. unpow3 84.4%

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

      5. add-cbrt-cube 92.4%

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

      6. pow3 92.3%

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

      7. unpow2 92.3%

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

      8. *-un-lft-identity 92.3%

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

      9. frac-times 95.5%

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

      10. add-cube-cbrt 95.6%

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

   7. Applied egg-rr 87.8%

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

      1. unpow2 87.8%

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

      2. pow3 87.7%

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

      3. unpow2 87.7%

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

      4. cbrt-prod 95.4%

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

      5. pow2 95.4%

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

   9. Applied egg-rr 95.4%

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

   if 7.5 < k

   1. Initial program 52.5%

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

      1. pow1 52.5%

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

   4. Applied egg-rr 18.9%

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

      1. unpow1 18.9%

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

   6. Simplified 18.9%

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

      1. associate-*r* 18.9%

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

      2. unpow-prod-down 19.0%

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

      3. pow2 19.0%

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

      4. add-sqr-sqrt 35.0%

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

   8. Applied egg-rr 35.0%

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

   9. Taylor expanded in k around inf 39.6%

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

Alternative 5: 82.5% accurate, 1.0× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k\_m \leq 4.8 \cdot 10^{-159}:\\ \;\;\;\;\frac{2}{{\left(\sqrt{{t\_m}^{3}} \cdot \left(k\_m \cdot \frac{\sqrt{2}}{\ell}\right)\right)}^{2}}\\ \mathbf{elif}\;k\_m \leq 52000000000000:\\ \;\;\;\;\frac{2}{\left({\left(\frac{t\_m}{\sqrt[3]{\ell}}\right)}^{2} \cdot \frac{t\_m}{\ell \cdot \sqrt[3]{\ell}}\right) \cdot \left(2 \cdot \left(k\_m \cdot k\_m\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{k\_m}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k\_m \cdot \tan k\_m\right)}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= k_m 4.8e-159)
    (/ 2.0 (pow (* (sqrt (pow t_m 3.0)) (* k_m (/ (sqrt 2.0) l))) 2.0))
    (if (<= k_m 52000000000000.0)
      (/
       2.0
       (*
        (* (pow (/ t_m (cbrt l)) 2.0) (/ t_m (* l (cbrt l))))
        (* 2.0 (* k_m k_m))))
      (/
       2.0
       (* (pow (* (/ k_m l) (sqrt t_m)) 2.0) (* (sin k_m) (tan k_m))))))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (k_m <= 4.8e-159) {
		tmp = 2.0 / pow((sqrt(pow(t_m, 3.0)) * (k_m * (sqrt(2.0) / l))), 2.0);
	} else if (k_m <= 52000000000000.0) {
		tmp = 2.0 / ((pow((t_m / cbrt(l)), 2.0) * (t_m / (l * cbrt(l)))) * (2.0 * (k_m * k_m)));
	} else {
		tmp = 2.0 / (pow(((k_m / l) * sqrt(t_m)), 2.0) * (sin(k_m) * tan(k_m)));
	}
	return t_s * tmp;
}

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (k_m <= 4.8e-159) {
		tmp = 2.0 / Math.pow((Math.sqrt(Math.pow(t_m, 3.0)) * (k_m * (Math.sqrt(2.0) / l))), 2.0);
	} else if (k_m <= 52000000000000.0) {
		tmp = 2.0 / ((Math.pow((t_m / Math.cbrt(l)), 2.0) * (t_m / (l * Math.cbrt(l)))) * (2.0 * (k_m * k_m)));
	} else {
		tmp = 2.0 / (Math.pow(((k_m / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k_m) * Math.tan(k_m)));
	}
	return t_s * tmp;
}

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (k_m <= 4.8e-159)
		tmp = Float64(2.0 / (Float64(sqrt((t_m ^ 3.0)) * Float64(k_m * Float64(sqrt(2.0) / l))) ^ 2.0));
	elseif (k_m <= 52000000000000.0)
		tmp = Float64(2.0 / Float64(Float64((Float64(t_m / cbrt(l)) ^ 2.0) * Float64(t_m / Float64(l * cbrt(l)))) * Float64(2.0 * Float64(k_m * k_m))));
	else
		tmp = Float64(2.0 / Float64((Float64(Float64(k_m / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k_m) * tan(k_m))));
	end
	return Float64(t_s * tmp)
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[k$95$m, 4.8e-159], N[(2.0 / N[Power[N[(N[Sqrt[N[Power[t$95$m, 3.0], $MachinePrecision]], $MachinePrecision] * N[(k$95$m * N[(N[Sqrt[2.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[k$95$m, 52000000000000.0], N[(2.0 / N[(N[(N[Power[N[(t$95$m / N[Power[l, 1/3], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(t$95$m / N[(l * N[Power[l, 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(2.0 * N[(k$95$m * k$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[Power[N[(N[(k$95$m / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k$95$m], $MachinePrecision] * N[Tan[k$95$m], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if k < 4.79999999999999995e-159

   1. Initial program 57.8%

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

      1. pow1 57.8%

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

   4. Applied egg-rr 26.3%

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

      1. unpow1 26.3%

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

   6. Simplified 26.3%

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

   7. Taylor expanded in k around 0 30.4%

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

      1. *-commutative 30.4%

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

      2. associate-/l* 30.4%

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

   9. Simplified 30.4%

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

   if 4.79999999999999995e-159 < k < 5.2e13

   1. Initial program 86.6%

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

   3. Simplified 86.7%

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

   5. Taylor expanded in k around 0 81.3%

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

      1. add-cube-cbrt 81.2%

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

      2. pow3 81.2%

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

      3. cbrt-div 81.0%

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

      4. unpow3 81.0%

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

      5. add-cbrt-cube 87.6%

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

      6. pow3 87.6%

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

      7. unpow2 87.6%

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

      8. associate-*r/ 90.2%

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

      9. associate-/l/ 89.9%

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

   7. Applied egg-rr 89.9%

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

      1. unpow2 89.9%

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

   9. Applied egg-rr 89.9%

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

   if 5.2e13 < k

   1. Initial program 48.4%

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

      1. pow1 48.4%

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

   4. Applied egg-rr 17.1%

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

      1. unpow1 17.1%

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

   6. Simplified 17.1%

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

      1. associate-*r* 17.1%

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

      2. unpow-prod-down 17.2%

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

      3. pow2 17.2%

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

      4. add-sqr-sqrt 34.6%

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

   8. Applied egg-rr 34.6%

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

   9. Taylor expanded in k around inf 39.6%

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

Alternative 6: 88.7% accurate, 1.0× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k\_m \leq 52000000000000:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k\_m}{t\_m}\right)\right) \cdot \left(k\_m \cdot \frac{{t\_m}^{1.5}}{\ell}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\sin k\_m \cdot \tan k\_m\right) \cdot {\left(\frac{\sqrt{t\_m} \cdot \left(-k\_m\right)}{\ell}\right)}^{2}}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= k_m 52000000000000.0)
    (/
     2.0
     (pow
      (* (hypot 1.0 (hypot 1.0 (/ k_m t_m))) (* k_m (/ (pow t_m 1.5) l)))
      2.0))
    (/
     2.0
     (* (* (sin k_m) (tan k_m)) (pow (/ (* (sqrt t_m) (- k_m)) l) 2.0))))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (k_m <= 52000000000000.0) {
		tmp = 2.0 / pow((hypot(1.0, hypot(1.0, (k_m / t_m))) * (k_m * (pow(t_m, 1.5) / l))), 2.0);
	} else {
		tmp = 2.0 / ((sin(k_m) * tan(k_m)) * pow(((sqrt(t_m) * -k_m) / l), 2.0));
	}
	return t_s * tmp;
}

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (k_m <= 52000000000000.0) {
		tmp = 2.0 / Math.pow((Math.hypot(1.0, Math.hypot(1.0, (k_m / t_m))) * (k_m * (Math.pow(t_m, 1.5) / l))), 2.0);
	} else {
		tmp = 2.0 / ((Math.sin(k_m) * Math.tan(k_m)) * Math.pow(((Math.sqrt(t_m) * -k_m) / l), 2.0));
	}
	return t_s * tmp;
}

Python

k_m = math.fabs(k)
t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k_m):
	tmp = 0
	if k_m <= 52000000000000.0:
		tmp = 2.0 / math.pow((math.hypot(1.0, math.hypot(1.0, (k_m / t_m))) * (k_m * (math.pow(t_m, 1.5) / l))), 2.0)
	else:
		tmp = 2.0 / ((math.sin(k_m) * math.tan(k_m)) * math.pow(((math.sqrt(t_m) * -k_m) / l), 2.0))
	return t_s * tmp

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (k_m <= 52000000000000.0)
		tmp = Float64(2.0 / (Float64(hypot(1.0, hypot(1.0, Float64(k_m / t_m))) * Float64(k_m * Float64((t_m ^ 1.5) / l))) ^ 2.0));
	else
		tmp = Float64(2.0 / Float64(Float64(sin(k_m) * tan(k_m)) * (Float64(Float64(sqrt(t_m) * Float64(-k_m)) / l) ^ 2.0)));
	end
	return Float64(t_s * tmp)
end

MATLAB

k_m = abs(k);
t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k_m)
	tmp = 0.0;
	if (k_m <= 52000000000000.0)
		tmp = 2.0 / ((hypot(1.0, hypot(1.0, (k_m / t_m))) * (k_m * ((t_m ^ 1.5) / l))) ^ 2.0);
	else
		tmp = 2.0 / ((sin(k_m) * tan(k_m)) * (((sqrt(t_m) * -k_m) / l) ^ 2.0));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[k$95$m, 52000000000000.0], N[(2.0 / N[Power[N[(N[Sqrt[1.0 ^ 2 + N[Sqrt[1.0 ^ 2 + N[(k$95$m / t$95$m), $MachinePrecision] ^ 2], $MachinePrecision] ^ 2], $MachinePrecision] * N[(k$95$m * N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[Sin[k$95$m], $MachinePrecision] * N[Tan[k$95$m], $MachinePrecision]), $MachinePrecision] * N[Power[N[(N[(N[Sqrt[t$95$m], $MachinePrecision] * (-k$95$m)), $MachinePrecision] / l), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 5.2e13

   1. Initial program 62.2%

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

      1. pow1 62.2%

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

   4. Applied egg-rr 31.8%

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

      1. unpow1 31.8%

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

   6. Simplified 31.8%

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

   7. Taylor expanded in k around 0 41.6%

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

   if 5.2e13 < k

   1. Initial program 48.4%

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

      1. pow1 48.4%

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

   4. Applied egg-rr 17.1%

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

      1. unpow1 17.1%

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

   6. Simplified 17.1%

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

      1. associate-*r* 17.1%

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

      2. unpow-prod-down 17.2%

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

      3. pow2 17.2%

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

      4. add-sqr-sqrt 34.6%

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

   8. Applied egg-rr 34.6%

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

   9. Taylor expanded in k around -inf 39.6%

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

       1. mul-1-neg 39.6%

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

       2. associate-*l/ 39.5%

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

       3. distribute-neg-frac2 39.5%

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

   11. Simplified 39.5%

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

4. Final simplification 41.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 52000000000000:\\ \;\;\;\;\frac{2}{{\left(\mathsf{hypot}\left(1, \mathsf{hypot}\left(1, \frac{k}{t}\right)\right) \cdot \left(k \cdot \frac{{t}^{1.5}}{\ell}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot \tan k\right) \cdot {\left(\frac{\sqrt{t} \cdot \left(-k\right)}{\ell}\right)}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 68.0% accurate, 1.0× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k\_m \leq 2.2 \cdot 10^{-159}:\\ \;\;\;\;\frac{2}{{\left(\sqrt{{t\_m}^{3}} \cdot \left(k\_m \cdot \frac{\sqrt{2}}{\ell}\right)\right)}^{2}}\\ \mathbf{elif}\;k\_m \leq 2.3 \cdot 10^{-25}:\\ \;\;\;\;\frac{2}{\left({\left(\frac{t\_m}{\sqrt[3]{\ell}}\right)}^{2} \cdot \frac{t\_m}{\ell \cdot \sqrt[3]{\ell}}\right) \cdot \left(2 \cdot \left(k\_m \cdot k\_m\right)\right)}\\ \mathbf{elif}\;k\_m \leq 2.8 \cdot 10^{+82}:\\ \;\;\;\;\frac{2}{2 \cdot \left(\tan k\_m \cdot \left(\sin k\_m \cdot \frac{{t\_m}^{3}}{\ell \cdot \ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{{k\_m}^{2} \cdot \left(t\_m \cdot {k\_m}^{2}\right)}{\ell \cdot \cos k\_m}}{\ell}}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= k_m 2.2e-159)
    (/ 2.0 (pow (* (sqrt (pow t_m 3.0)) (* k_m (/ (sqrt 2.0) l))) 2.0))
    (if (<= k_m 2.3e-25)
      (/
       2.0
       (*
        (* (pow (/ t_m (cbrt l)) 2.0) (/ t_m (* l (cbrt l))))
        (* 2.0 (* k_m k_m))))
      (if (<= k_m 2.8e+82)
        (/ 2.0 (* 2.0 (* (tan k_m) (* (sin k_m) (/ (pow t_m 3.0) (* l l))))))
        (/
         2.0
         (/
          (/ (* (pow k_m 2.0) (* t_m (pow k_m 2.0))) (* l (cos k_m)))
          l)))))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (k_m <= 2.2e-159) {
		tmp = 2.0 / pow((sqrt(pow(t_m, 3.0)) * (k_m * (sqrt(2.0) / l))), 2.0);
	} else if (k_m <= 2.3e-25) {
		tmp = 2.0 / ((pow((t_m / cbrt(l)), 2.0) * (t_m / (l * cbrt(l)))) * (2.0 * (k_m * k_m)));
	} else if (k_m <= 2.8e+82) {
		tmp = 2.0 / (2.0 * (tan(k_m) * (sin(k_m) * (pow(t_m, 3.0) / (l * l)))));
	} else {
		tmp = 2.0 / (((pow(k_m, 2.0) * (t_m * pow(k_m, 2.0))) / (l * cos(k_m))) / l);
	}
	return t_s * tmp;
}

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (k_m <= 2.2e-159) {
		tmp = 2.0 / Math.pow((Math.sqrt(Math.pow(t_m, 3.0)) * (k_m * (Math.sqrt(2.0) / l))), 2.0);
	} else if (k_m <= 2.3e-25) {
		tmp = 2.0 / ((Math.pow((t_m / Math.cbrt(l)), 2.0) * (t_m / (l * Math.cbrt(l)))) * (2.0 * (k_m * k_m)));
	} else if (k_m <= 2.8e+82) {
		tmp = 2.0 / (2.0 * (Math.tan(k_m) * (Math.sin(k_m) * (Math.pow(t_m, 3.0) / (l * l)))));
	} else {
		tmp = 2.0 / (((Math.pow(k_m, 2.0) * (t_m * Math.pow(k_m, 2.0))) / (l * Math.cos(k_m))) / l);
	}
	return t_s * tmp;
}

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (k_m <= 2.2e-159)
		tmp = Float64(2.0 / (Float64(sqrt((t_m ^ 3.0)) * Float64(k_m * Float64(sqrt(2.0) / l))) ^ 2.0));
	elseif (k_m <= 2.3e-25)
		tmp = Float64(2.0 / Float64(Float64((Float64(t_m / cbrt(l)) ^ 2.0) * Float64(t_m / Float64(l * cbrt(l)))) * Float64(2.0 * Float64(k_m * k_m))));
	elseif (k_m <= 2.8e+82)
		tmp = Float64(2.0 / Float64(2.0 * Float64(tan(k_m) * Float64(sin(k_m) * Float64((t_m ^ 3.0) / Float64(l * l))))));
	else
		tmp = Float64(2.0 / Float64(Float64(Float64((k_m ^ 2.0) * Float64(t_m * (k_m ^ 2.0))) / Float64(l * cos(k_m))) / l));
	end
	return Float64(t_s * tmp)
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[k$95$m, 2.2e-159], N[(2.0 / N[Power[N[(N[Sqrt[N[Power[t$95$m, 3.0], $MachinePrecision]], $MachinePrecision] * N[(k$95$m * N[(N[Sqrt[2.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[k$95$m, 2.3e-25], N[(2.0 / N[(N[(N[Power[N[(t$95$m / N[Power[l, 1/3], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(t$95$m / N[(l * N[Power[l, 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(2.0 * N[(k$95$m * k$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[k$95$m, 2.8e+82], N[(2.0 / N[(2.0 * N[(N[Tan[k$95$m], $MachinePrecision] * N[(N[Sin[k$95$m], $MachinePrecision] * N[(N[Power[t$95$m, 3.0], $MachinePrecision] / N[(l * l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[(N[Power[k$95$m, 2.0], $MachinePrecision] * N[(t$95$m * N[Power[k$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(l * N[Cos[k$95$m], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]]]]), $MachinePrecision]

Derivation

1. Split input into 4 regimes

2. if k < 2.2e-159

   1. Initial program 57.8%

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

      1. pow1 57.8%

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

   4. Applied egg-rr 26.3%

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

      1. unpow1 26.3%

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

   6. Simplified 26.3%

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

   7. Taylor expanded in k around 0 30.4%

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

      1. *-commutative 30.4%

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

      2. associate-/l* 30.4%

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

   9. Simplified 30.4%

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

   if 2.2e-159 < k < 2.2999999999999999e-25

   1. Initial program 81.7%

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

   3. Simplified 82.1%

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

   5. Taylor expanded in k around 0 86.6%

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

      1. add-cube-cbrt 86.4%

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

      2. pow3 86.4%

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

      3. cbrt-div 86.2%

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

      4. unpow3 86.2%

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

      5. add-cbrt-cube 95.2%

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

      6. pow3 95.1%

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

      7. unpow2 95.1%

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

      8. associate-*r/ 98.7%

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

      9. associate-/l/ 94.5%

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

   7. Applied egg-rr 94.5%

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

      1. unpow2 94.5%

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

   9. Applied egg-rr 94.5%

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

   if 2.2999999999999999e-25 < k < 2.8e82

   1. Initial program 69.1%

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

   3. Taylor expanded in k around 0 75.2%

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

   if 2.8e82 < k

   1. Initial program 47.4%

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

   3. Simplified 52.4%

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

      1. associate-*l/ 52.4%

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

      2. associate-*l* 52.4%

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

   6. Applied egg-rr 52.4%

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

      1. associate-*r* 52.4%

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

   8. Simplified 52.4%

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

   9. Taylor expanded in t around 0 75.4%

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

   10. Taylor expanded in k around 0 64.6%

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

4. Final simplification 45.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 2.2 \cdot 10^{-159}:\\ \;\;\;\;\frac{2}{{\left(\sqrt{{t}^{3}} \cdot \left(k \cdot \frac{\sqrt{2}}{\ell}\right)\right)}^{2}}\\ \mathbf{elif}\;k \leq 2.3 \cdot 10^{-25}:\\ \;\;\;\;\frac{2}{\left({\left(\frac{t}{\sqrt[3]{\ell}}\right)}^{2} \cdot \frac{t}{\ell \cdot \sqrt[3]{\ell}}\right) \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \mathbf{elif}\;k \leq 2.8 \cdot 10^{+82}:\\ \;\;\;\;\frac{2}{2 \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t}^{3}}{\ell \cdot \ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{{k}^{2} \cdot \left(t \cdot {k}^{2}\right)}{\ell \cdot \cos k}}{\ell}}\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 62.9% accurate, 1.3× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k\_m \leq 2.5 \cdot 10^{-25}:\\ \;\;\;\;\frac{2}{\left(2 \cdot {k\_m}^{2}\right) \cdot \left({t\_m}^{1.5} \cdot \frac{\frac{{t\_m}^{1.5}}{\ell}}{\ell}\right)}\\ \mathbf{elif}\;k\_m \leq 2.4 \cdot 10^{+82}:\\ \;\;\;\;\frac{2}{2 \cdot \left(\tan k\_m \cdot \left(\sin k\_m \cdot \frac{{t\_m}^{3}}{\ell \cdot \ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{{k\_m}^{2} \cdot \left(t\_m \cdot {k\_m}^{2}\right)}{\ell \cdot \cos k\_m}}{\ell}}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= k_m 2.5e-25)
    (/
     2.0
     (* (* 2.0 (pow k_m 2.0)) (* (pow t_m 1.5) (/ (/ (pow t_m 1.5) l) l))))
    (if (<= k_m 2.4e+82)
      (/ 2.0 (* 2.0 (* (tan k_m) (* (sin k_m) (/ (pow t_m 3.0) (* l l))))))
      (/
       2.0
       (/ (/ (* (pow k_m 2.0) (* t_m (pow k_m 2.0))) (* l (cos k_m))) l))))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (k_m <= 2.5e-25) {
		tmp = 2.0 / ((2.0 * pow(k_m, 2.0)) * (pow(t_m, 1.5) * ((pow(t_m, 1.5) / l) / l)));
	} else if (k_m <= 2.4e+82) {
		tmp = 2.0 / (2.0 * (tan(k_m) * (sin(k_m) * (pow(t_m, 3.0) / (l * l)))));
	} else {
		tmp = 2.0 / (((pow(k_m, 2.0) * (t_m * pow(k_m, 2.0))) / (l * cos(k_m))) / l);
	}
	return t_s * tmp;
}

Fortran

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k_m)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k_m
    real(8) :: tmp
    if (k_m <= 2.5d-25) then
        tmp = 2.0d0 / ((2.0d0 * (k_m ** 2.0d0)) * ((t_m ** 1.5d0) * (((t_m ** 1.5d0) / l) / l)))
    else if (k_m <= 2.4d+82) then
        tmp = 2.0d0 / (2.0d0 * (tan(k_m) * (sin(k_m) * ((t_m ** 3.0d0) / (l * l)))))
    else
        tmp = 2.0d0 / ((((k_m ** 2.0d0) * (t_m * (k_m ** 2.0d0))) / (l * cos(k_m))) / l)
    end if
    code = t_s * tmp
end function

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (k_m <= 2.5e-25) {
		tmp = 2.0 / ((2.0 * Math.pow(k_m, 2.0)) * (Math.pow(t_m, 1.5) * ((Math.pow(t_m, 1.5) / l) / l)));
	} else if (k_m <= 2.4e+82) {
		tmp = 2.0 / (2.0 * (Math.tan(k_m) * (Math.sin(k_m) * (Math.pow(t_m, 3.0) / (l * l)))));
	} else {
		tmp = 2.0 / (((Math.pow(k_m, 2.0) * (t_m * Math.pow(k_m, 2.0))) / (l * Math.cos(k_m))) / l);
	}
	return t_s * tmp;
}

Python

k_m = math.fabs(k)
t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k_m):
	tmp = 0
	if k_m <= 2.5e-25:
		tmp = 2.0 / ((2.0 * math.pow(k_m, 2.0)) * (math.pow(t_m, 1.5) * ((math.pow(t_m, 1.5) / l) / l)))
	elif k_m <= 2.4e+82:
		tmp = 2.0 / (2.0 * (math.tan(k_m) * (math.sin(k_m) * (math.pow(t_m, 3.0) / (l * l)))))
	else:
		tmp = 2.0 / (((math.pow(k_m, 2.0) * (t_m * math.pow(k_m, 2.0))) / (l * math.cos(k_m))) / l)
	return t_s * tmp

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (k_m <= 2.5e-25)
		tmp = Float64(2.0 / Float64(Float64(2.0 * (k_m ^ 2.0)) * Float64((t_m ^ 1.5) * Float64(Float64((t_m ^ 1.5) / l) / l))));
	elseif (k_m <= 2.4e+82)
		tmp = Float64(2.0 / Float64(2.0 * Float64(tan(k_m) * Float64(sin(k_m) * Float64((t_m ^ 3.0) / Float64(l * l))))));
	else
		tmp = Float64(2.0 / Float64(Float64(Float64((k_m ^ 2.0) * Float64(t_m * (k_m ^ 2.0))) / Float64(l * cos(k_m))) / l));
	end
	return Float64(t_s * tmp)
end

MATLAB

k_m = abs(k);
t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k_m)
	tmp = 0.0;
	if (k_m <= 2.5e-25)
		tmp = 2.0 / ((2.0 * (k_m ^ 2.0)) * ((t_m ^ 1.5) * (((t_m ^ 1.5) / l) / l)));
	elseif (k_m <= 2.4e+82)
		tmp = 2.0 / (2.0 * (tan(k_m) * (sin(k_m) * ((t_m ^ 3.0) / (l * l)))));
	else
		tmp = 2.0 / ((((k_m ^ 2.0) * (t_m * (k_m ^ 2.0))) / (l * cos(k_m))) / l);
	end
	tmp_2 = t_s * tmp;
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[k$95$m, 2.5e-25], N[(2.0 / N[(N[(2.0 * N[Power[k$95$m, 2.0], $MachinePrecision]), $MachinePrecision] * N[(N[Power[t$95$m, 1.5], $MachinePrecision] * N[(N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[k$95$m, 2.4e+82], N[(2.0 / N[(2.0 * N[(N[Tan[k$95$m], $MachinePrecision] * N[(N[Sin[k$95$m], $MachinePrecision] * N[(N[Power[t$95$m, 3.0], $MachinePrecision] / N[(l * l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[(N[Power[k$95$m, 2.0], $MachinePrecision] * N[(t$95$m * N[Power[k$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(l * N[Cos[k$95$m], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if k < 2.49999999999999981e-25

   1. Initial program 60.6%

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

   3. Simplified 63.0%

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

   5. Taylor expanded in k around 0 62.1%

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

      1. sqr-pow 31.3%

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

      2. *-un-lft-identity 31.3%

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

      3. times-frac 33.9%

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

      4. metadata-eval 33.9%

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

      5. metadata-eval 33.9%

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

   7. Applied egg-rr 33.9%

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

      1. /-rgt-identity 33.9%

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

      2. associate-/l* 35.2%

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

   9. Applied egg-rr 35.2%

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

   if 2.49999999999999981e-25 < k < 2.39999999999999998e82

   1. Initial program 69.1%

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

   3. Taylor expanded in k around 0 75.2%

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

   if 2.39999999999999998e82 < k

   1. Initial program 47.4%

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

   3. Simplified 52.4%

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

      1. associate-*l/ 52.4%

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

      2. associate-*l* 52.4%

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

   6. Applied egg-rr 52.4%

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

      1. associate-*r* 52.4%

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

   8. Simplified 52.4%

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

   9. Taylor expanded in t around 0 75.4%

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

   10. Taylor expanded in k around 0 64.6%

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

4. Final simplification 43.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 2.5 \cdot 10^{-25}:\\ \;\;\;\;\frac{2}{\left(2 \cdot {k}^{2}\right) \cdot \left({t}^{1.5} \cdot \frac{\frac{{t}^{1.5}}{\ell}}{\ell}\right)}\\ \mathbf{elif}\;k \leq 2.4 \cdot 10^{+82}:\\ \;\;\;\;\frac{2}{2 \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t}^{3}}{\ell \cdot \ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{{k}^{2} \cdot \left(t \cdot {k}^{2}\right)}{\ell \cdot \cos k}}{\ell}}\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 62.8% accurate, 1.3× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k\_m \leq 7 \cdot 10^{-26}:\\ \;\;\;\;\frac{2}{\left(2 \cdot {k\_m}^{2}\right) \cdot \frac{{\left(\frac{t\_m}{\sqrt[3]{\ell}}\right)}^{3}}{\ell}}\\ \mathbf{elif}\;k\_m \leq 3.4 \cdot 10^{+82}:\\ \;\;\;\;\frac{2}{2 \cdot \left(\tan k\_m \cdot \left(\sin k\_m \cdot \frac{{t\_m}^{3}}{\ell \cdot \ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{{k\_m}^{2} \cdot \left(t\_m \cdot {k\_m}^{2}\right)}{\ell \cdot \cos k\_m}}{\ell}}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= k_m 7e-26)
    (/ 2.0 (* (* 2.0 (pow k_m 2.0)) (/ (pow (/ t_m (cbrt l)) 3.0) l)))
    (if (<= k_m 3.4e+82)
      (/ 2.0 (* 2.0 (* (tan k_m) (* (sin k_m) (/ (pow t_m 3.0) (* l l))))))
      (/
       2.0
       (/ (/ (* (pow k_m 2.0) (* t_m (pow k_m 2.0))) (* l (cos k_m))) l))))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (k_m <= 7e-26) {
		tmp = 2.0 / ((2.0 * pow(k_m, 2.0)) * (pow((t_m / cbrt(l)), 3.0) / l));
	} else if (k_m <= 3.4e+82) {
		tmp = 2.0 / (2.0 * (tan(k_m) * (sin(k_m) * (pow(t_m, 3.0) / (l * l)))));
	} else {
		tmp = 2.0 / (((pow(k_m, 2.0) * (t_m * pow(k_m, 2.0))) / (l * cos(k_m))) / l);
	}
	return t_s * tmp;
}

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (k_m <= 7e-26) {
		tmp = 2.0 / ((2.0 * Math.pow(k_m, 2.0)) * (Math.pow((t_m / Math.cbrt(l)), 3.0) / l));
	} else if (k_m <= 3.4e+82) {
		tmp = 2.0 / (2.0 * (Math.tan(k_m) * (Math.sin(k_m) * (Math.pow(t_m, 3.0) / (l * l)))));
	} else {
		tmp = 2.0 / (((Math.pow(k_m, 2.0) * (t_m * Math.pow(k_m, 2.0))) / (l * Math.cos(k_m))) / l);
	}
	return t_s * tmp;
}

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (k_m <= 7e-26)
		tmp = Float64(2.0 / Float64(Float64(2.0 * (k_m ^ 2.0)) * Float64((Float64(t_m / cbrt(l)) ^ 3.0) / l)));
	elseif (k_m <= 3.4e+82)
		tmp = Float64(2.0 / Float64(2.0 * Float64(tan(k_m) * Float64(sin(k_m) * Float64((t_m ^ 3.0) / Float64(l * l))))));
	else
		tmp = Float64(2.0 / Float64(Float64(Float64((k_m ^ 2.0) * Float64(t_m * (k_m ^ 2.0))) / Float64(l * cos(k_m))) / l));
	end
	return Float64(t_s * tmp)
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[k$95$m, 7e-26], N[(2.0 / N[(N[(2.0 * N[Power[k$95$m, 2.0], $MachinePrecision]), $MachinePrecision] * N[(N[Power[N[(t$95$m / N[Power[l, 1/3], $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[k$95$m, 3.4e+82], N[(2.0 / N[(2.0 * N[(N[Tan[k$95$m], $MachinePrecision] * N[(N[Sin[k$95$m], $MachinePrecision] * N[(N[Power[t$95$m, 3.0], $MachinePrecision] / N[(l * l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[(N[Power[k$95$m, 2.0], $MachinePrecision] * N[(t$95$m * N[Power[k$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(l * N[Cos[k$95$m], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if k < 6.9999999999999997e-26

   1. Initial program 60.6%

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

   3. Simplified 63.0%

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

   5. Taylor expanded in k around 0 62.1%

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

      1. add-cube-cbrt 62.0%

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

      2. pow3 62.0%

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

      3. cbrt-div 61.9%

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

      4. rem-cbrt-cube 66.2%

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

   7. Applied egg-rr 66.2%

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

   if 6.9999999999999997e-26 < k < 3.39999999999999994e82

   1. Initial program 69.1%

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

   3. Taylor expanded in k around 0 75.2%

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

   if 3.39999999999999994e82 < k

   1. Initial program 47.4%

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

   3. Simplified 52.4%

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

      1. associate-*l/ 52.4%

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

      2. associate-*l* 52.4%

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

   6. Applied egg-rr 52.4%

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

      1. associate-*r* 52.4%

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

   8. Simplified 52.4%

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

   9. Taylor expanded in t around 0 75.4%

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

   10. Taylor expanded in k around 0 64.6%

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

4. Final simplification 66.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 7 \cdot 10^{-26}:\\ \;\;\;\;\frac{2}{\left(2 \cdot {k}^{2}\right) \cdot \frac{{\left(\frac{t}{\sqrt[3]{\ell}}\right)}^{3}}{\ell}}\\ \mathbf{elif}\;k \leq 3.4 \cdot 10^{+82}:\\ \;\;\;\;\frac{2}{2 \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t}^{3}}{\ell \cdot \ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{{k}^{2} \cdot \left(t \cdot {k}^{2}\right)}{\ell \cdot \cos k}}{\ell}}\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 62.5% accurate, 1.3× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k\_m \leq 2.5 \cdot 10^{-25}:\\ \;\;\;\;\frac{2}{\left(2 \cdot {k\_m}^{2}\right) \cdot \frac{{\left(\frac{t\_m}{\sqrt[3]{\ell}}\right)}^{3}}{\ell}}\\ \mathbf{elif}\;k\_m \leq 1.05 \cdot 10^{+83}:\\ \;\;\;\;\frac{2}{2 \cdot \left(\tan k\_m \cdot \left(\sin k\_m \cdot \frac{{t\_m}^{3}}{\ell \cdot \ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{t\_m \cdot {k\_m}^{4}}{\ell}}{\ell}}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= k_m 2.5e-25)
    (/ 2.0 (* (* 2.0 (pow k_m 2.0)) (/ (pow (/ t_m (cbrt l)) 3.0) l)))
    (if (<= k_m 1.05e+83)
      (/ 2.0 (* 2.0 (* (tan k_m) (* (sin k_m) (/ (pow t_m 3.0) (* l l))))))
      (/ 2.0 (/ (/ (* t_m (pow k_m 4.0)) l) l))))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (k_m <= 2.5e-25) {
		tmp = 2.0 / ((2.0 * pow(k_m, 2.0)) * (pow((t_m / cbrt(l)), 3.0) / l));
	} else if (k_m <= 1.05e+83) {
		tmp = 2.0 / (2.0 * (tan(k_m) * (sin(k_m) * (pow(t_m, 3.0) / (l * l)))));
	} else {
		tmp = 2.0 / (((t_m * pow(k_m, 4.0)) / l) / l);
	}
	return t_s * tmp;
}

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (k_m <= 2.5e-25) {
		tmp = 2.0 / ((2.0 * Math.pow(k_m, 2.0)) * (Math.pow((t_m / Math.cbrt(l)), 3.0) / l));
	} else if (k_m <= 1.05e+83) {
		tmp = 2.0 / (2.0 * (Math.tan(k_m) * (Math.sin(k_m) * (Math.pow(t_m, 3.0) / (l * l)))));
	} else {
		tmp = 2.0 / (((t_m * Math.pow(k_m, 4.0)) / l) / l);
	}
	return t_s * tmp;
}

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (k_m <= 2.5e-25)
		tmp = Float64(2.0 / Float64(Float64(2.0 * (k_m ^ 2.0)) * Float64((Float64(t_m / cbrt(l)) ^ 3.0) / l)));
	elseif (k_m <= 1.05e+83)
		tmp = Float64(2.0 / Float64(2.0 * Float64(tan(k_m) * Float64(sin(k_m) * Float64((t_m ^ 3.0) / Float64(l * l))))));
	else
		tmp = Float64(2.0 / Float64(Float64(Float64(t_m * (k_m ^ 4.0)) / l) / l));
	end
	return Float64(t_s * tmp)
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[k$95$m, 2.5e-25], N[(2.0 / N[(N[(2.0 * N[Power[k$95$m, 2.0], $MachinePrecision]), $MachinePrecision] * N[(N[Power[N[(t$95$m / N[Power[l, 1/3], $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[k$95$m, 1.05e+83], N[(2.0 / N[(2.0 * N[(N[Tan[k$95$m], $MachinePrecision] * N[(N[Sin[k$95$m], $MachinePrecision] * N[(N[Power[t$95$m, 3.0], $MachinePrecision] / N[(l * l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[(t$95$m * N[Power[k$95$m, 4.0], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if k < 2.49999999999999981e-25

   1. Initial program 60.6%

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

   3. Simplified 63.0%

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

   5. Taylor expanded in k around 0 62.1%

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

      1. add-cube-cbrt 62.0%

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

      2. pow3 62.0%

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

      3. cbrt-div 61.9%

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

      4. rem-cbrt-cube 66.2%

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

   7. Applied egg-rr 66.2%

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

   if 2.49999999999999981e-25 < k < 1.05000000000000001e83

   1. Initial program 69.1%

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

   3. Taylor expanded in k around 0 75.2%

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

   if 1.05000000000000001e83 < k

   1. Initial program 47.4%

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

   3. Simplified 52.4%

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

      1. associate-*l/ 52.4%

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

      2. associate-*l* 52.4%

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

   6. Applied egg-rr 52.4%

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

      1. associate-*r* 52.4%

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

   8. Simplified 52.4%

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

   9. Taylor expanded in t around 0 75.4%

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

   10. Taylor expanded in k around 0 64.5%

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

4. Final simplification 66.7%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 2.5 \cdot 10^{-25}:\\ \;\;\;\;\frac{2}{\left(2 \cdot {k}^{2}\right) \cdot \frac{{\left(\frac{t}{\sqrt[3]{\ell}}\right)}^{3}}{\ell}}\\ \mathbf{elif}\;k \leq 1.05 \cdot 10^{+83}:\\ \;\;\;\;\frac{2}{2 \cdot \left(\tan k \cdot \left(\sin k \cdot \frac{{t}^{3}}{\ell \cdot \ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{t \cdot {k}^{4}}{\ell}}{\ell}}\\ \end{array} \]
5. Add Preprocessing

Alternative 11: 67.5% accurate, 1.3× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 2.25 \cdot 10^{-123}:\\ \;\;\;\;\frac{2}{\frac{\frac{{k\_m}^{2} \cdot \left(t\_m \cdot {k\_m}^{2}\right)}{\ell \cdot \cos k\_m}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\left(\tan k\_m \cdot \left(2 + {\left(\frac{k\_m}{t\_m}\right)}^{2}\right)\right) \cdot \frac{k\_m \cdot {t\_m}^{3}}{\ell}}{\ell}}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= t_m 2.25e-123)
    (/ 2.0 (/ (/ (* (pow k_m 2.0) (* t_m (pow k_m 2.0))) (* l (cos k_m))) l))
    (/
     2.0
     (/
      (*
       (* (tan k_m) (+ 2.0 (pow (/ k_m t_m) 2.0)))
       (/ (* k_m (pow t_m 3.0)) l))
      l)))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (t_m <= 2.25e-123) {
		tmp = 2.0 / (((pow(k_m, 2.0) * (t_m * pow(k_m, 2.0))) / (l * cos(k_m))) / l);
	} else {
		tmp = 2.0 / (((tan(k_m) * (2.0 + pow((k_m / t_m), 2.0))) * ((k_m * pow(t_m, 3.0)) / l)) / l);
	}
	return t_s * tmp;
}

Fortran

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k_m)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k_m
    real(8) :: tmp
    if (t_m <= 2.25d-123) then
        tmp = 2.0d0 / ((((k_m ** 2.0d0) * (t_m * (k_m ** 2.0d0))) / (l * cos(k_m))) / l)
    else
        tmp = 2.0d0 / (((tan(k_m) * (2.0d0 + ((k_m / t_m) ** 2.0d0))) * ((k_m * (t_m ** 3.0d0)) / l)) / l)
    end if
    code = t_s * tmp
end function

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (t_m <= 2.25e-123) {
		tmp = 2.0 / (((Math.pow(k_m, 2.0) * (t_m * Math.pow(k_m, 2.0))) / (l * Math.cos(k_m))) / l);
	} else {
		tmp = 2.0 / (((Math.tan(k_m) * (2.0 + Math.pow((k_m / t_m), 2.0))) * ((k_m * Math.pow(t_m, 3.0)) / l)) / l);
	}
	return t_s * tmp;
}

Python

k_m = math.fabs(k)
t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k_m):
	tmp = 0
	if t_m <= 2.25e-123:
		tmp = 2.0 / (((math.pow(k_m, 2.0) * (t_m * math.pow(k_m, 2.0))) / (l * math.cos(k_m))) / l)
	else:
		tmp = 2.0 / (((math.tan(k_m) * (2.0 + math.pow((k_m / t_m), 2.0))) * ((k_m * math.pow(t_m, 3.0)) / l)) / l)
	return t_s * tmp

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (t_m <= 2.25e-123)
		tmp = Float64(2.0 / Float64(Float64(Float64((k_m ^ 2.0) * Float64(t_m * (k_m ^ 2.0))) / Float64(l * cos(k_m))) / l));
	else
		tmp = Float64(2.0 / Float64(Float64(Float64(tan(k_m) * Float64(2.0 + (Float64(k_m / t_m) ^ 2.0))) * Float64(Float64(k_m * (t_m ^ 3.0)) / l)) / l));
	end
	return Float64(t_s * tmp)
end

MATLAB

k_m = abs(k);
t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k_m)
	tmp = 0.0;
	if (t_m <= 2.25e-123)
		tmp = 2.0 / ((((k_m ^ 2.0) * (t_m * (k_m ^ 2.0))) / (l * cos(k_m))) / l);
	else
		tmp = 2.0 / (((tan(k_m) * (2.0 + ((k_m / t_m) ^ 2.0))) * ((k_m * (t_m ^ 3.0)) / l)) / l);
	end
	tmp_2 = t_s * tmp;
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[t$95$m, 2.25e-123], N[(2.0 / N[(N[(N[(N[Power[k$95$m, 2.0], $MachinePrecision] * N[(t$95$m * N[Power[k$95$m, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[(l * N[Cos[k$95$m], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[(N[Tan[k$95$m], $MachinePrecision] * N[(2.0 + N[Power[N[(k$95$m / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[(k$95$m * N[Power[t$95$m, 3.0], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if t < 2.24999999999999997e-123

   1. Initial program 51.2%

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

   3. Simplified 56.0%

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

      1. associate-*l/ 56.7%

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

      2. associate-*l* 56.7%

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

   6. Applied egg-rr 56.7%

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

      1. associate-*r* 58.1%

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

   8. Simplified 58.1%

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

   9. Taylor expanded in t around 0 69.6%

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

   10. Taylor expanded in k around 0 61.2%

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

   if 2.24999999999999997e-123 < t

   1. Initial program 73.2%

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

   3. Simplified 72.0%

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

      1. associate-*l/ 72.5%

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

      2. associate-*l* 72.5%

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

   6. Applied egg-rr 72.5%

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

      1. associate-*r* 77.3%

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

   8. Simplified 77.3%

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

   9. Taylor expanded in k around 0 73.1%

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

4. Final simplification 65.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 2.25 \cdot 10^{-123}:\\ \;\;\;\;\frac{2}{\frac{\frac{{k}^{2} \cdot \left(t \cdot {k}^{2}\right)}{\ell \cdot \cos k}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\left(\tan k \cdot \left(2 + {\left(\frac{k}{t}\right)}^{2}\right)\right) \cdot \frac{k \cdot {t}^{3}}{\ell}}{\ell}}\\ \end{array} \]
5. Add Preprocessing

Alternative 12: 62.7% accurate, 1.3× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k\_m \leq 65:\\ \;\;\;\;\frac{2}{\left(2 \cdot {k\_m}^{2}\right) \cdot \frac{{\left(\frac{t\_m}{\sqrt[3]{\ell}}\right)}^{3}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{t\_m \cdot {k\_m}^{4}}{\ell \cdot \cos k\_m}}{\ell}}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= k_m 65.0)
    (/ 2.0 (* (* 2.0 (pow k_m 2.0)) (/ (pow (/ t_m (cbrt l)) 3.0) l)))
    (/ 2.0 (/ (/ (* t_m (pow k_m 4.0)) (* l (cos k_m))) l)))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (k_m <= 65.0) {
		tmp = 2.0 / ((2.0 * pow(k_m, 2.0)) * (pow((t_m / cbrt(l)), 3.0) / l));
	} else {
		tmp = 2.0 / (((t_m * pow(k_m, 4.0)) / (l * cos(k_m))) / l);
	}
	return t_s * tmp;
}

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (k_m <= 65.0) {
		tmp = 2.0 / ((2.0 * Math.pow(k_m, 2.0)) * (Math.pow((t_m / Math.cbrt(l)), 3.0) / l));
	} else {
		tmp = 2.0 / (((t_m * Math.pow(k_m, 4.0)) / (l * Math.cos(k_m))) / l);
	}
	return t_s * tmp;
}

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (k_m <= 65.0)
		tmp = Float64(2.0 / Float64(Float64(2.0 * (k_m ^ 2.0)) * Float64((Float64(t_m / cbrt(l)) ^ 3.0) / l)));
	else
		tmp = Float64(2.0 / Float64(Float64(Float64(t_m * (k_m ^ 4.0)) / Float64(l * cos(k_m))) / l));
	end
	return Float64(t_s * tmp)
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[k$95$m, 65.0], N[(2.0 / N[(N[(2.0 * N[Power[k$95$m, 2.0], $MachinePrecision]), $MachinePrecision] * N[(N[Power[N[(t$95$m / N[Power[l, 1/3], $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[(t$95$m * N[Power[k$95$m, 4.0], $MachinePrecision]), $MachinePrecision] / N[(l * N[Cos[k$95$m], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 65

   1. Initial program 61.2%

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

   3. Simplified 63.6%

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

   5. Taylor expanded in k around 0 62.2%

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

      1. add-cube-cbrt 62.1%

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

      2. pow3 62.1%

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

      3. cbrt-div 62.1%

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

      4. rem-cbrt-cube 66.3%

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

   7. Applied egg-rr 66.3%

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

   if 65 < k

   1. Initial program 52.5%

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

   3. Simplified 55.9%

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

      1. associate-*l/ 55.9%

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

      2. associate-*l* 55.9%

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

   6. Applied egg-rr 55.9%

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

      1. associate-*r* 55.9%

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

   8. Simplified 55.9%

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

   9. Taylor expanded in t around 0 77.0%

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

   10. Taylor expanded in k around 0 62.4%

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

4. Final simplification 65.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 65:\\ \;\;\;\;\frac{2}{\left(2 \cdot {k}^{2}\right) \cdot \frac{{\left(\frac{t}{\sqrt[3]{\ell}}\right)}^{3}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{t \cdot {k}^{4}}{\ell \cdot \cos k}}{\ell}}\\ \end{array} \]
5. Add Preprocessing

Alternative 13: 62.6% accurate, 1.9× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k\_m \leq 13.5:\\ \;\;\;\;\frac{2}{\left({t\_m}^{1.5} \cdot \frac{{t\_m}^{1.5}}{\ell}\right) \cdot \frac{2 \cdot \left(k\_m \cdot k\_m\right)}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{t\_m \cdot {k\_m}^{4}}{\ell \cdot \cos k\_m}}{\ell}}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= k_m 13.5)
    (/ 2.0 (* (* (pow t_m 1.5) (/ (pow t_m 1.5) l)) (/ (* 2.0 (* k_m k_m)) l)))
    (/ 2.0 (/ (/ (* t_m (pow k_m 4.0)) (* l (cos k_m))) l)))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (k_m <= 13.5) {
		tmp = 2.0 / ((pow(t_m, 1.5) * (pow(t_m, 1.5) / l)) * ((2.0 * (k_m * k_m)) / l));
	} else {
		tmp = 2.0 / (((t_m * pow(k_m, 4.0)) / (l * cos(k_m))) / l);
	}
	return t_s * tmp;
}

Fortran

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k_m)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k_m
    real(8) :: tmp
    if (k_m <= 13.5d0) then
        tmp = 2.0d0 / (((t_m ** 1.5d0) * ((t_m ** 1.5d0) / l)) * ((2.0d0 * (k_m * k_m)) / l))
    else
        tmp = 2.0d0 / (((t_m * (k_m ** 4.0d0)) / (l * cos(k_m))) / l)
    end if
    code = t_s * tmp
end function

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (k_m <= 13.5) {
		tmp = 2.0 / ((Math.pow(t_m, 1.5) * (Math.pow(t_m, 1.5) / l)) * ((2.0 * (k_m * k_m)) / l));
	} else {
		tmp = 2.0 / (((t_m * Math.pow(k_m, 4.0)) / (l * Math.cos(k_m))) / l);
	}
	return t_s * tmp;
}

Python

k_m = math.fabs(k)
t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k_m):
	tmp = 0
	if k_m <= 13.5:
		tmp = 2.0 / ((math.pow(t_m, 1.5) * (math.pow(t_m, 1.5) / l)) * ((2.0 * (k_m * k_m)) / l))
	else:
		tmp = 2.0 / (((t_m * math.pow(k_m, 4.0)) / (l * math.cos(k_m))) / l)
	return t_s * tmp

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (k_m <= 13.5)
		tmp = Float64(2.0 / Float64(Float64((t_m ^ 1.5) * Float64((t_m ^ 1.5) / l)) * Float64(Float64(2.0 * Float64(k_m * k_m)) / l)));
	else
		tmp = Float64(2.0 / Float64(Float64(Float64(t_m * (k_m ^ 4.0)) / Float64(l * cos(k_m))) / l));
	end
	return Float64(t_s * tmp)
end

MATLAB

k_m = abs(k);
t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k_m)
	tmp = 0.0;
	if (k_m <= 13.5)
		tmp = 2.0 / (((t_m ^ 1.5) * ((t_m ^ 1.5) / l)) * ((2.0 * (k_m * k_m)) / l));
	else
		tmp = 2.0 / (((t_m * (k_m ^ 4.0)) / (l * cos(k_m))) / l);
	end
	tmp_2 = t_s * tmp;
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[k$95$m, 13.5], N[(2.0 / N[(N[(N[Power[t$95$m, 1.5], $MachinePrecision] * N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision] * N[(N[(2.0 * N[(k$95$m * k$95$m), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[(t$95$m * N[Power[k$95$m, 4.0], $MachinePrecision]), $MachinePrecision] / N[(l * N[Cos[k$95$m], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 13.5

   1. Initial program 61.2%

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

   3. Simplified 63.6%

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

   5. Taylor expanded in k around 0 62.2%

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

      1. associate-*l/ 62.1%

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

   7. Applied egg-rr 62.1%

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

      1. associate-/l* 62.2%

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

   9. Simplified 62.2%

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

       1. unpow2 66.1%

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

   11. Applied egg-rr 62.2%

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

       1. sqr-pow 31.3%

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

       2. *-un-lft-identity 31.3%

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

       3. times-frac 33.9%

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

       4. metadata-eval 33.9%

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

       5. metadata-eval 33.9%

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

   13. Applied egg-rr 33.9%

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

   if 13.5 < k

   1. Initial program 52.5%

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

   3. Simplified 55.9%

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

      1. associate-*l/ 55.9%

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

      2. associate-*l* 55.9%

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

   6. Applied egg-rr 55.9%

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

      1. associate-*r* 55.9%

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

   8. Simplified 55.9%

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

   9. Taylor expanded in t around 0 77.0%

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

   10. Taylor expanded in k around 0 62.4%

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

4. Final simplification 40.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 13.5:\\ \;\;\;\;\frac{2}{\left({t}^{1.5} \cdot \frac{{t}^{1.5}}{\ell}\right) \cdot \frac{2 \cdot \left(k \cdot k\right)}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{t \cdot {k}^{4}}{\ell \cdot \cos k}}{\ell}}\\ \end{array} \]
5. Add Preprocessing

Alternative 14: 62.6% accurate, 1.9× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k\_m \leq 13.6:\\ \;\;\;\;\frac{2}{{\left(\frac{t\_m}{\sqrt[3]{\ell}}\right)}^{3} \cdot \frac{2 \cdot \left(k\_m \cdot k\_m\right)}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{t\_m \cdot {k\_m}^{4}}{\ell \cdot \cos k\_m}}{\ell}}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= k_m 13.6)
    (/ 2.0 (* (pow (/ t_m (cbrt l)) 3.0) (/ (* 2.0 (* k_m k_m)) l)))
    (/ 2.0 (/ (/ (* t_m (pow k_m 4.0)) (* l (cos k_m))) l)))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (k_m <= 13.6) {
		tmp = 2.0 / (pow((t_m / cbrt(l)), 3.0) * ((2.0 * (k_m * k_m)) / l));
	} else {
		tmp = 2.0 / (((t_m * pow(k_m, 4.0)) / (l * cos(k_m))) / l);
	}
	return t_s * tmp;
}

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (k_m <= 13.6) {
		tmp = 2.0 / (Math.pow((t_m / Math.cbrt(l)), 3.0) * ((2.0 * (k_m * k_m)) / l));
	} else {
		tmp = 2.0 / (((t_m * Math.pow(k_m, 4.0)) / (l * Math.cos(k_m))) / l);
	}
	return t_s * tmp;
}

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (k_m <= 13.6)
		tmp = Float64(2.0 / Float64((Float64(t_m / cbrt(l)) ^ 3.0) * Float64(Float64(2.0 * Float64(k_m * k_m)) / l)));
	else
		tmp = Float64(2.0 / Float64(Float64(Float64(t_m * (k_m ^ 4.0)) / Float64(l * cos(k_m))) / l));
	end
	return Float64(t_s * tmp)
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[k$95$m, 13.6], N[(2.0 / N[(N[Power[N[(t$95$m / N[Power[l, 1/3], $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision] * N[(N[(2.0 * N[(k$95$m * k$95$m), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[(t$95$m * N[Power[k$95$m, 4.0], $MachinePrecision]), $MachinePrecision] / N[(l * N[Cos[k$95$m], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 13.5999999999999996

   1. Initial program 61.2%

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

   3. Simplified 63.6%

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

   5. Taylor expanded in k around 0 62.2%

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

      1. associate-*l/ 62.1%

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

   7. Applied egg-rr 62.1%

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

      1. associate-/l* 62.2%

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

   9. Simplified 62.2%

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

       1. unpow2 66.1%

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

   11. Applied egg-rr 62.2%

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

       1. add-cube-cbrt 62.1%

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

       2. pow3 62.1%

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

       3. cbrt-div 62.1%

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

       4. rem-cbrt-cube 66.3%

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

   13. Applied egg-rr 66.3%

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

   if 13.5999999999999996 < k

   1. Initial program 52.5%

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

   3. Simplified 55.9%

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

      1. associate-*l/ 55.9%

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

      2. associate-*l* 55.9%

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

   6. Applied egg-rr 55.9%

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

      1. associate-*r* 55.9%

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

   8. Simplified 55.9%

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

   9. Taylor expanded in t around 0 77.0%

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

   10. Taylor expanded in k around 0 62.4%

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

4. Final simplification 65.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 13.6:\\ \;\;\;\;\frac{2}{{\left(\frac{t}{\sqrt[3]{\ell}}\right)}^{3} \cdot \frac{2 \cdot \left(k \cdot k\right)}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{t \cdot {k}^{4}}{\ell \cdot \cos k}}{\ell}}\\ \end{array} \]
5. Add Preprocessing

Alternative 15: 60.7% accurate, 1.9× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k\_m \leq 70:\\ \;\;\;\;\frac{2}{\left(2 \cdot {k\_m}^{2}\right) \cdot \frac{\frac{{t\_m}^{3}}{\ell}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{t\_m \cdot {k\_m}^{4}}{\ell \cdot \cos k\_m}}{\ell}}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= k_m 70.0)
    (/ 2.0 (* (* 2.0 (pow k_m 2.0)) (/ (/ (pow t_m 3.0) l) l)))
    (/ 2.0 (/ (/ (* t_m (pow k_m 4.0)) (* l (cos k_m))) l)))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (k_m <= 70.0) {
		tmp = 2.0 / ((2.0 * pow(k_m, 2.0)) * ((pow(t_m, 3.0) / l) / l));
	} else {
		tmp = 2.0 / (((t_m * pow(k_m, 4.0)) / (l * cos(k_m))) / l);
	}
	return t_s * tmp;
}

Fortran

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k_m)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k_m
    real(8) :: tmp
    if (k_m <= 70.0d0) then
        tmp = 2.0d0 / ((2.0d0 * (k_m ** 2.0d0)) * (((t_m ** 3.0d0) / l) / l))
    else
        tmp = 2.0d0 / (((t_m * (k_m ** 4.0d0)) / (l * cos(k_m))) / l)
    end if
    code = t_s * tmp
end function

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (k_m <= 70.0) {
		tmp = 2.0 / ((2.0 * Math.pow(k_m, 2.0)) * ((Math.pow(t_m, 3.0) / l) / l));
	} else {
		tmp = 2.0 / (((t_m * Math.pow(k_m, 4.0)) / (l * Math.cos(k_m))) / l);
	}
	return t_s * tmp;
}

Python

k_m = math.fabs(k)
t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k_m):
	tmp = 0
	if k_m <= 70.0:
		tmp = 2.0 / ((2.0 * math.pow(k_m, 2.0)) * ((math.pow(t_m, 3.0) / l) / l))
	else:
		tmp = 2.0 / (((t_m * math.pow(k_m, 4.0)) / (l * math.cos(k_m))) / l)
	return t_s * tmp

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (k_m <= 70.0)
		tmp = Float64(2.0 / Float64(Float64(2.0 * (k_m ^ 2.0)) * Float64(Float64((t_m ^ 3.0) / l) / l)));
	else
		tmp = Float64(2.0 / Float64(Float64(Float64(t_m * (k_m ^ 4.0)) / Float64(l * cos(k_m))) / l));
	end
	return Float64(t_s * tmp)
end

MATLAB

k_m = abs(k);
t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k_m)
	tmp = 0.0;
	if (k_m <= 70.0)
		tmp = 2.0 / ((2.0 * (k_m ^ 2.0)) * (((t_m ^ 3.0) / l) / l));
	else
		tmp = 2.0 / (((t_m * (k_m ^ 4.0)) / (l * cos(k_m))) / l);
	end
	tmp_2 = t_s * tmp;
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[k$95$m, 70.0], N[(2.0 / N[(N[(2.0 * N[Power[k$95$m, 2.0], $MachinePrecision]), $MachinePrecision] * N[(N[(N[Power[t$95$m, 3.0], $MachinePrecision] / l), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[(t$95$m * N[Power[k$95$m, 4.0], $MachinePrecision]), $MachinePrecision] / N[(l * N[Cos[k$95$m], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 70

   1. Initial program 61.2%

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

   3. Simplified 63.6%

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

   5. Taylor expanded in k around 0 62.2%

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

   if 70 < k

   1. Initial program 52.5%

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

   3. Simplified 55.9%

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

      1. associate-*l/ 55.9%

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

      2. associate-*l* 55.9%

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

   6. Applied egg-rr 55.9%

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

      1. associate-*r* 55.9%

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

   8. Simplified 55.9%

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

   9. Taylor expanded in t around 0 77.0%

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

   10. Taylor expanded in k around 0 62.4%

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

4. Final simplification 62.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 70:\\ \;\;\;\;\frac{2}{\left(2 \cdot {k}^{2}\right) \cdot \frac{\frac{{t}^{3}}{\ell}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{t \cdot {k}^{4}}{\ell \cdot \cos k}}{\ell}}\\ \end{array} \]
5. Add Preprocessing

Alternative 16: 62.3% accurate, 1.9× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k\_m \leq 13.5:\\ \;\;\;\;\frac{2}{\frac{2 \cdot \left(k\_m \cdot k\_m\right)}{\ell} \cdot \left({t\_m}^{2} \cdot \frac{t\_m}{\ell}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{t\_m \cdot {k\_m}^{4}}{\ell \cdot \cos k\_m}}{\ell}}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= k_m 13.5)
    (/ 2.0 (* (/ (* 2.0 (* k_m k_m)) l) (* (pow t_m 2.0) (/ t_m l))))
    (/ 2.0 (/ (/ (* t_m (pow k_m 4.0)) (* l (cos k_m))) l)))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (k_m <= 13.5) {
		tmp = 2.0 / (((2.0 * (k_m * k_m)) / l) * (pow(t_m, 2.0) * (t_m / l)));
	} else {
		tmp = 2.0 / (((t_m * pow(k_m, 4.0)) / (l * cos(k_m))) / l);
	}
	return t_s * tmp;
}

Fortran

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k_m)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k_m
    real(8) :: tmp
    if (k_m <= 13.5d0) then
        tmp = 2.0d0 / (((2.0d0 * (k_m * k_m)) / l) * ((t_m ** 2.0d0) * (t_m / l)))
    else
        tmp = 2.0d0 / (((t_m * (k_m ** 4.0d0)) / (l * cos(k_m))) / l)
    end if
    code = t_s * tmp
end function

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (k_m <= 13.5) {
		tmp = 2.0 / (((2.0 * (k_m * k_m)) / l) * (Math.pow(t_m, 2.0) * (t_m / l)));
	} else {
		tmp = 2.0 / (((t_m * Math.pow(k_m, 4.0)) / (l * Math.cos(k_m))) / l);
	}
	return t_s * tmp;
}

Python

k_m = math.fabs(k)
t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k_m):
	tmp = 0
	if k_m <= 13.5:
		tmp = 2.0 / (((2.0 * (k_m * k_m)) / l) * (math.pow(t_m, 2.0) * (t_m / l)))
	else:
		tmp = 2.0 / (((t_m * math.pow(k_m, 4.0)) / (l * math.cos(k_m))) / l)
	return t_s * tmp

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (k_m <= 13.5)
		tmp = Float64(2.0 / Float64(Float64(Float64(2.0 * Float64(k_m * k_m)) / l) * Float64((t_m ^ 2.0) * Float64(t_m / l))));
	else
		tmp = Float64(2.0 / Float64(Float64(Float64(t_m * (k_m ^ 4.0)) / Float64(l * cos(k_m))) / l));
	end
	return Float64(t_s * tmp)
end

MATLAB

k_m = abs(k);
t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k_m)
	tmp = 0.0;
	if (k_m <= 13.5)
		tmp = 2.0 / (((2.0 * (k_m * k_m)) / l) * ((t_m ^ 2.0) * (t_m / l)));
	else
		tmp = 2.0 / (((t_m * (k_m ^ 4.0)) / (l * cos(k_m))) / l);
	end
	tmp_2 = t_s * tmp;
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[k$95$m, 13.5], N[(2.0 / N[(N[(N[(2.0 * N[(k$95$m * k$95$m), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[(N[Power[t$95$m, 2.0], $MachinePrecision] * N[(t$95$m / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[(t$95$m * N[Power[k$95$m, 4.0], $MachinePrecision]), $MachinePrecision] / N[(l * N[Cos[k$95$m], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 13.5

   1. Initial program 61.2%

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

   3. Simplified 63.6%

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

   5. Taylor expanded in k around 0 62.2%

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

      1. associate-*l/ 62.1%

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

   7. Applied egg-rr 62.1%

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

      1. associate-/l* 62.2%

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

   9. Simplified 62.2%

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

       1. unpow2 66.1%

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

   11. Applied egg-rr 62.2%

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

       1. unpow3 62.1%

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

       2. *-un-lft-identity 62.1%

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

       3. times-frac 64.3%

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

       4. pow2 64.3%

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

   13. Applied egg-rr 64.3%

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

   if 13.5 < k

   1. Initial program 52.5%

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

   3. Simplified 55.9%

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

      1. associate-*l/ 55.9%

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

      2. associate-*l* 55.9%

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

   6. Applied egg-rr 55.9%

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

      1. associate-*r* 55.9%

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

   8. Simplified 55.9%

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

   9. Taylor expanded in t around 0 77.0%

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

   10. Taylor expanded in k around 0 62.4%

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

4. Final simplification 63.8%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 13.5:\\ \;\;\;\;\frac{2}{\frac{2 \cdot \left(k \cdot k\right)}{\ell} \cdot \left({t}^{2} \cdot \frac{t}{\ell}\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\frac{t \cdot {k}^{4}}{\ell \cdot \cos k}}{\ell}}\\ \end{array} \]
5. Add Preprocessing

Alternative 17: 63.3% accurate, 3.5× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 5.5 \cdot 10^{-78}:\\ \;\;\;\;\frac{2}{\frac{{k\_m}^{4} \cdot \frac{t\_m}{\ell}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{2 \cdot \left(k\_m \cdot k\_m\right)}{\ell} \cdot \left({t\_m}^{2} \cdot \frac{t\_m}{\ell}\right)}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= t_m 5.5e-78)
    (/ 2.0 (/ (* (pow k_m 4.0) (/ t_m l)) l))
    (/ 2.0 (* (/ (* 2.0 (* k_m k_m)) l) (* (pow t_m 2.0) (/ t_m l)))))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (t_m <= 5.5e-78) {
		tmp = 2.0 / ((pow(k_m, 4.0) * (t_m / l)) / l);
	} else {
		tmp = 2.0 / (((2.0 * (k_m * k_m)) / l) * (pow(t_m, 2.0) * (t_m / l)));
	}
	return t_s * tmp;
}

Fortran

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k_m)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k_m
    real(8) :: tmp
    if (t_m <= 5.5d-78) then
        tmp = 2.0d0 / (((k_m ** 4.0d0) * (t_m / l)) / l)
    else
        tmp = 2.0d0 / (((2.0d0 * (k_m * k_m)) / l) * ((t_m ** 2.0d0) * (t_m / l)))
    end if
    code = t_s * tmp
end function

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (t_m <= 5.5e-78) {
		tmp = 2.0 / ((Math.pow(k_m, 4.0) * (t_m / l)) / l);
	} else {
		tmp = 2.0 / (((2.0 * (k_m * k_m)) / l) * (Math.pow(t_m, 2.0) * (t_m / l)));
	}
	return t_s * tmp;
}

Python

k_m = math.fabs(k)
t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k_m):
	tmp = 0
	if t_m <= 5.5e-78:
		tmp = 2.0 / ((math.pow(k_m, 4.0) * (t_m / l)) / l)
	else:
		tmp = 2.0 / (((2.0 * (k_m * k_m)) / l) * (math.pow(t_m, 2.0) * (t_m / l)))
	return t_s * tmp

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (t_m <= 5.5e-78)
		tmp = Float64(2.0 / Float64(Float64((k_m ^ 4.0) * Float64(t_m / l)) / l));
	else
		tmp = Float64(2.0 / Float64(Float64(Float64(2.0 * Float64(k_m * k_m)) / l) * Float64((t_m ^ 2.0) * Float64(t_m / l))));
	end
	return Float64(t_s * tmp)
end

MATLAB

k_m = abs(k);
t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k_m)
	tmp = 0.0;
	if (t_m <= 5.5e-78)
		tmp = 2.0 / (((k_m ^ 4.0) * (t_m / l)) / l);
	else
		tmp = 2.0 / (((2.0 * (k_m * k_m)) / l) * ((t_m ^ 2.0) * (t_m / l)));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[t$95$m, 5.5e-78], N[(2.0 / N[(N[(N[Power[k$95$m, 4.0], $MachinePrecision] * N[(t$95$m / l), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[(2.0 * N[(k$95$m * k$95$m), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[(N[Power[t$95$m, 2.0], $MachinePrecision] * N[(t$95$m / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if t < 5.50000000000000017e-78

   1. Initial program 51.5%

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

   3. Simplified 56.6%

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

      1. associate-*l/ 57.3%

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

      2. associate-*l* 57.3%

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

   6. Applied egg-rr 57.3%

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

      1. associate-*r* 58.6%

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

   8. Simplified 58.6%

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

   9. Taylor expanded in t around 0 70.6%

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

   10. Taylor expanded in k around 0 59.4%

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

       1. associate-/l* 59.7%

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

   12. Simplified 59.7%

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

   if 5.50000000000000017e-78 < t

   1. Initial program 75.1%

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

   3. Simplified 72.5%

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

   5. Taylor expanded in k around 0 67.6%

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

      1. associate-*l/ 67.1%

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

   7. Applied egg-rr 67.1%

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

      1. associate-/l* 68.7%

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

   9. Simplified 68.7%

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

       1. unpow2 73.2%

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

   11. Applied egg-rr 68.7%

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

       1. unpow3 68.7%

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

       2. *-un-lft-identity 68.7%

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

       3. times-frac 68.8%

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

       4. pow2 68.8%

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

   13. Applied egg-rr 68.8%

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

4. Final simplification 62.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 5.5 \cdot 10^{-78}:\\ \;\;\;\;\frac{2}{\frac{{k}^{4} \cdot \frac{t}{\ell}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{2 \cdot \left(k \cdot k\right)}{\ell} \cdot \left({t}^{2} \cdot \frac{t}{\ell}\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 18: 62.1% accurate, 3.5× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 1.06 \cdot 10^{-77}:\\ \;\;\;\;\frac{2}{\frac{{k\_m}^{4} \cdot \frac{t\_m}{\ell}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{2 \cdot \left(k\_m \cdot k\_m\right)}{\ell} \cdot \frac{1}{\frac{\ell}{{t\_m}^{3}}}}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= t_m 1.06e-77)
    (/ 2.0 (/ (* (pow k_m 4.0) (/ t_m l)) l))
    (/ 2.0 (* (/ (* 2.0 (* k_m k_m)) l) (/ 1.0 (/ l (pow t_m 3.0))))))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (t_m <= 1.06e-77) {
		tmp = 2.0 / ((pow(k_m, 4.0) * (t_m / l)) / l);
	} else {
		tmp = 2.0 / (((2.0 * (k_m * k_m)) / l) * (1.0 / (l / pow(t_m, 3.0))));
	}
	return t_s * tmp;
}

Fortran

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k_m)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k_m
    real(8) :: tmp
    if (t_m <= 1.06d-77) then
        tmp = 2.0d0 / (((k_m ** 4.0d0) * (t_m / l)) / l)
    else
        tmp = 2.0d0 / (((2.0d0 * (k_m * k_m)) / l) * (1.0d0 / (l / (t_m ** 3.0d0))))
    end if
    code = t_s * tmp
end function

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (t_m <= 1.06e-77) {
		tmp = 2.0 / ((Math.pow(k_m, 4.0) * (t_m / l)) / l);
	} else {
		tmp = 2.0 / (((2.0 * (k_m * k_m)) / l) * (1.0 / (l / Math.pow(t_m, 3.0))));
	}
	return t_s * tmp;
}

Python

k_m = math.fabs(k)
t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k_m):
	tmp = 0
	if t_m <= 1.06e-77:
		tmp = 2.0 / ((math.pow(k_m, 4.0) * (t_m / l)) / l)
	else:
		tmp = 2.0 / (((2.0 * (k_m * k_m)) / l) * (1.0 / (l / math.pow(t_m, 3.0))))
	return t_s * tmp

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (t_m <= 1.06e-77)
		tmp = Float64(2.0 / Float64(Float64((k_m ^ 4.0) * Float64(t_m / l)) / l));
	else
		tmp = Float64(2.0 / Float64(Float64(Float64(2.0 * Float64(k_m * k_m)) / l) * Float64(1.0 / Float64(l / (t_m ^ 3.0)))));
	end
	return Float64(t_s * tmp)
end

MATLAB

k_m = abs(k);
t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k_m)
	tmp = 0.0;
	if (t_m <= 1.06e-77)
		tmp = 2.0 / (((k_m ^ 4.0) * (t_m / l)) / l);
	else
		tmp = 2.0 / (((2.0 * (k_m * k_m)) / l) * (1.0 / (l / (t_m ^ 3.0))));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[t$95$m, 1.06e-77], N[(2.0 / N[(N[(N[Power[k$95$m, 4.0], $MachinePrecision] * N[(t$95$m / l), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[(2.0 * N[(k$95$m * k$95$m), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[(1.0 / N[(l / N[Power[t$95$m, 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if t < 1.05999999999999991e-77

   1. Initial program 51.5%

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

   3. Simplified 56.6%

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

      1. associate-*l/ 57.3%

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

      2. associate-*l* 57.3%

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

   6. Applied egg-rr 57.3%

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

      1. associate-*r* 58.6%

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

   8. Simplified 58.6%

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

   9. Taylor expanded in t around 0 70.6%

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

   10. Taylor expanded in k around 0 59.4%

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

       1. associate-/l* 59.7%

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

   12. Simplified 59.7%

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

   if 1.05999999999999991e-77 < t

   1. Initial program 75.1%

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

   3. Simplified 72.5%

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

   5. Taylor expanded in k around 0 67.6%

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

      1. associate-*l/ 67.1%

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

   7. Applied egg-rr 67.1%

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

      1. associate-/l* 68.7%

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

   9. Simplified 68.7%

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

       1. unpow2 73.2%

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

   11. Applied egg-rr 68.7%

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

       1. clear-num 68.7%

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

       2. inv-pow 68.7%

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

   13. Applied egg-rr 68.7%

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

       1. unpow-1 68.7%

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

   15. Simplified 68.7%

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

4. Final simplification 62.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 1.06 \cdot 10^{-77}:\\ \;\;\;\;\frac{2}{\frac{{k}^{4} \cdot \frac{t}{\ell}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{2 \cdot \left(k \cdot k\right)}{\ell} \cdot \frac{1}{\frac{\ell}{{t}^{3}}}}\\ \end{array} \]
5. Add Preprocessing

Alternative 19: 62.1% accurate, 3.5× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 9.6 \cdot 10^{-79}:\\ \;\;\;\;\frac{2}{\frac{{k\_m}^{4} \cdot \frac{t\_m}{\ell}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{2 \cdot \left(k\_m \cdot k\_m\right)}{\ell} \cdot \frac{{t\_m}^{3}}{\ell}}\\ \end{array} \end{array} \]

FPCore

k_m = (fabs.f64 k)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k_m)
 :precision binary64
 (*
  t_s
  (if (<= t_m 9.6e-79)
    (/ 2.0 (/ (* (pow k_m 4.0) (/ t_m l)) l))
    (/ 2.0 (* (/ (* 2.0 (* k_m k_m)) l) (/ (pow t_m 3.0) l))))))

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	double tmp;
	if (t_m <= 9.6e-79) {
		tmp = 2.0 / ((pow(k_m, 4.0) * (t_m / l)) / l);
	} else {
		tmp = 2.0 / (((2.0 * (k_m * k_m)) / l) * (pow(t_m, 3.0) / l));
	}
	return t_s * tmp;
}

Fortran

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k_m)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k_m
    real(8) :: tmp
    if (t_m <= 9.6d-79) then
        tmp = 2.0d0 / (((k_m ** 4.0d0) * (t_m / l)) / l)
    else
        tmp = 2.0d0 / (((2.0d0 * (k_m * k_m)) / l) * ((t_m ** 3.0d0) / l))
    end if
    code = t_s * tmp
end function

Java

k_m = Math.abs(k);
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_m) {
	double tmp;
	if (t_m <= 9.6e-79) {
		tmp = 2.0 / ((Math.pow(k_m, 4.0) * (t_m / l)) / l);
	} else {
		tmp = 2.0 / (((2.0 * (k_m * k_m)) / l) * (Math.pow(t_m, 3.0) / l));
	}
	return t_s * tmp;
}

Python

k_m = math.fabs(k)
t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k_m):
	tmp = 0
	if t_m <= 9.6e-79:
		tmp = 2.0 / ((math.pow(k_m, 4.0) * (t_m / l)) / l)
	else:
		tmp = 2.0 / (((2.0 * (k_m * k_m)) / l) * (math.pow(t_m, 3.0) / l))
	return t_s * tmp

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	tmp = 0.0
	if (t_m <= 9.6e-79)
		tmp = Float64(2.0 / Float64(Float64((k_m ^ 4.0) * Float64(t_m / l)) / l));
	else
		tmp = Float64(2.0 / Float64(Float64(Float64(2.0 * Float64(k_m * k_m)) / l) * Float64((t_m ^ 3.0) / l)));
	end
	return Float64(t_s * tmp)
end

MATLAB

k_m = abs(k);
t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k_m)
	tmp = 0.0;
	if (t_m <= 9.6e-79)
		tmp = 2.0 / (((k_m ^ 4.0) * (t_m / l)) / l);
	else
		tmp = 2.0 / (((2.0 * (k_m * k_m)) / l) * ((t_m ^ 3.0) / l));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * If[LessEqual[t$95$m, 9.6e-79], N[(2.0 / N[(N[(N[Power[k$95$m, 4.0], $MachinePrecision] * N[(t$95$m / l), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[(2.0 * N[(k$95$m * k$95$m), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision] * N[(N[Power[t$95$m, 3.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if t < 9.60000000000000023e-79

   1. Initial program 51.5%

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

   3. Simplified 56.6%

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

      1. associate-*l/ 57.3%

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

      2. associate-*l* 57.3%

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

   6. Applied egg-rr 57.3%

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

      1. associate-*r* 58.6%

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

   8. Simplified 58.6%

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

   9. Taylor expanded in t around 0 70.6%

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

   10. Taylor expanded in k around 0 59.4%

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

       1. associate-/l* 59.7%

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

   12. Simplified 59.7%

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

   if 9.60000000000000023e-79 < t

   1. Initial program 75.1%

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

   3. Simplified 72.5%

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

   5. Taylor expanded in k around 0 67.6%

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

      1. associate-*l/ 67.1%

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

   7. Applied egg-rr 67.1%

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

      1. associate-/l* 68.7%

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

   9. Simplified 68.7%

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

       1. unpow2 73.2%

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

   11. Applied egg-rr 68.7%

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

4. Final simplification 62.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 9.6 \cdot 10^{-79}:\\ \;\;\;\;\frac{2}{\frac{{k}^{4} \cdot \frac{t}{\ell}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{2 \cdot \left(k \cdot k\right)}{\ell} \cdot \frac{{t}^{3}}{\ell}}\\ \end{array} \]
5. Add Preprocessing

Alternative 20: 56.3% accurate, 3.8× speedup

Math

\[\begin{array}{l} k_m = \left|k\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \frac{2}{\frac{{k\_m}^{4} \cdot \frac{t\_m}{\ell}}{\ell}} \end{array} \]

FPCore

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

C

k_m = fabs(k);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k_m) {
	return t_s * (2.0 / ((pow(k_m, 4.0) * (t_m / l)) / l));
}

Fortran

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k_m)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k_m
    code = t_s * (2.0d0 / (((k_m ** 4.0d0) * (t_m / l)) / l))
end function

Java

k_m = Math.abs(k);
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_m) {
	return t_s * (2.0 / ((Math.pow(k_m, 4.0) * (t_m / l)) / l));
}

Python

k_m = math.fabs(k)
t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k_m):
	return t_s * (2.0 / ((math.pow(k_m, 4.0) * (t_m / l)) / l))

Julia

k_m = abs(k)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k_m)
	return Float64(t_s * Float64(2.0 / Float64(Float64((k_m ^ 4.0) * Float64(t_m / l)) / l)))
end

MATLAB

k_m = abs(k);
t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp = code(t_s, t_m, l, k_m)
	tmp = t_s * (2.0 / (((k_m ^ 4.0) * (t_m / l)) / l));
end

Wolfram

k_m = N[Abs[k], $MachinePrecision]
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$95$m_] := N[(t$95$s * N[(2.0 / N[(N[(N[Power[k$95$m, 4.0], $MachinePrecision] * N[(t$95$m / l), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 59.0%

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

3. Simplified 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. Step-by-step derivation

   1. associate-*l/ 62.3%

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

   2. associate-*l* 62.3%

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

6. Applied egg-rr 62.3%

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

   1. associate-*r* 64.9%

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

8. Simplified 64.9%

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

9. Taylor expanded in t around 0 66.0%

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

10. Taylor expanded in k around 0 57.4%

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

    1. associate-/l* 57.6%

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

12. Simplified 57.6%

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

Reproduce

herbie shell --seed 2024116 
(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))))