Toniolo and Linder, Equation (10-)

Percentage Accurate: 35.5% → 88.8%

Time: 33.8s

Alternatives: 14

Speedup: 3.8×

• 38.7% 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: 35.5% 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: 88.8% accurate, 0.2× speedup

Math

\[\begin{array}{l} l_m = \left|\ell\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ \begin{array}{l} t_2 := {\left(\sqrt[3]{l\_m}\right)}^{2}\\ t_3 := \sqrt[3]{\sin k \cdot \tan k}\\ t\_s \cdot \begin{array}{l} \mathbf{if}\;l\_m \leq 4.5 \cdot 10^{-153}:\\ \;\;\;\;\frac{2}{{\left({\left(\frac{k}{\sqrt{l\_m}}\right)}^{2} \cdot \sqrt{t\_m}\right)}^{2}}\\ \mathbf{elif}\;l\_m \leq 2.6 \cdot 10^{+163}:\\ \;\;\;\;\frac{2}{t\_m \cdot {k}^{2}} \cdot \frac{\cos k \cdot {l\_m}^{2}}{{\sin k}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{{\left(\frac{t\_m}{t\_2} \cdot \left({\left({t\_3}^{2}\right)}^{0.3333333333333333} \cdot \sqrt[3]{t\_3}\right)\right)}^{2} \cdot \frac{k}{t\_m}} \cdot \frac{t\_m \cdot \frac{\sqrt{2}}{k}}{\frac{t\_m \cdot t\_3}{t\_2}}\\ \end{array} \end{array} \end{array} \]

FPCore

l_m = (fabs.f64 l)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l_m k)
 :precision binary64
 (let* ((t_2 (pow (cbrt l_m) 2.0)) (t_3 (cbrt (* (sin k) (tan k)))))
   (*
    t_s
    (if (<= l_m 4.5e-153)
      (/ 2.0 (pow (* (pow (/ k (sqrt l_m)) 2.0) (sqrt t_m)) 2.0))
      (if (<= l_m 2.6e+163)
        (*
         (/ 2.0 (* t_m (pow k 2.0)))
         (/ (* (cos k) (pow l_m 2.0)) (pow (sin k) 2.0)))
        (*
         (/
          (sqrt 2.0)
          (*
           (pow
            (*
             (/ t_m t_2)
             (* (pow (pow t_3 2.0) 0.3333333333333333) (cbrt t_3)))
            2.0)
           (/ k t_m)))
         (/ (* t_m (/ (sqrt 2.0) k)) (/ (* t_m t_3) t_2))))))))

C

l_m = fabs(l);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l_m, double k) {
	double t_2 = pow(cbrt(l_m), 2.0);
	double t_3 = cbrt((sin(k) * tan(k)));
	double tmp;
	if (l_m <= 4.5e-153) {
		tmp = 2.0 / pow((pow((k / sqrt(l_m)), 2.0) * sqrt(t_m)), 2.0);
	} else if (l_m <= 2.6e+163) {
		tmp = (2.0 / (t_m * pow(k, 2.0))) * ((cos(k) * pow(l_m, 2.0)) / pow(sin(k), 2.0));
	} else {
		tmp = (sqrt(2.0) / (pow(((t_m / t_2) * (pow(pow(t_3, 2.0), 0.3333333333333333) * cbrt(t_3))), 2.0) * (k / t_m))) * ((t_m * (sqrt(2.0) / k)) / ((t_m * t_3) / t_2));
	}
	return t_s * tmp;
}

Java

l_m = Math.abs(l);
t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l_m, double k) {
	double t_2 = Math.pow(Math.cbrt(l_m), 2.0);
	double t_3 = Math.cbrt((Math.sin(k) * Math.tan(k)));
	double tmp;
	if (l_m <= 4.5e-153) {
		tmp = 2.0 / Math.pow((Math.pow((k / Math.sqrt(l_m)), 2.0) * Math.sqrt(t_m)), 2.0);
	} else if (l_m <= 2.6e+163) {
		tmp = (2.0 / (t_m * Math.pow(k, 2.0))) * ((Math.cos(k) * Math.pow(l_m, 2.0)) / Math.pow(Math.sin(k), 2.0));
	} else {
		tmp = (Math.sqrt(2.0) / (Math.pow(((t_m / t_2) * (Math.pow(Math.pow(t_3, 2.0), 0.3333333333333333) * Math.cbrt(t_3))), 2.0) * (k / t_m))) * ((t_m * (Math.sqrt(2.0) / k)) / ((t_m * t_3) / t_2));
	}
	return t_s * tmp;
}

Julia

l_m = abs(l)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l_m, k)
	t_2 = cbrt(l_m) ^ 2.0
	t_3 = cbrt(Float64(sin(k) * tan(k)))
	tmp = 0.0
	if (l_m <= 4.5e-153)
		tmp = Float64(2.0 / (Float64((Float64(k / sqrt(l_m)) ^ 2.0) * sqrt(t_m)) ^ 2.0));
	elseif (l_m <= 2.6e+163)
		tmp = Float64(Float64(2.0 / Float64(t_m * (k ^ 2.0))) * Float64(Float64(cos(k) * (l_m ^ 2.0)) / (sin(k) ^ 2.0)));
	else
		tmp = Float64(Float64(sqrt(2.0) / Float64((Float64(Float64(t_m / t_2) * Float64(((t_3 ^ 2.0) ^ 0.3333333333333333) * cbrt(t_3))) ^ 2.0) * Float64(k / t_m))) * Float64(Float64(t_m * Float64(sqrt(2.0) / k)) / Float64(Float64(t_m * t_3) / t_2)));
	end
	return Float64(t_s * tmp)
end

Wolfram

l_m = N[Abs[l], $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$95$m_, k_] := Block[{t$95$2 = N[Power[N[Power[l$95$m, 1/3], $MachinePrecision], 2.0], $MachinePrecision]}, Block[{t$95$3 = N[Power[N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision]}, N[(t$95$s * If[LessEqual[l$95$m, 4.5e-153], N[(2.0 / N[Power[N[(N[Power[N[(k / N[Sqrt[l$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[l$95$m, 2.6e+163], N[(N[(2.0 / N[(t$95$m * N[Power[k, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[(N[Cos[k], $MachinePrecision] * N[Power[l$95$m, 2.0], $MachinePrecision]), $MachinePrecision] / N[Power[N[Sin[k], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / N[(N[Power[N[(N[(t$95$m / t$95$2), $MachinePrecision] * N[(N[Power[N[Power[t$95$3, 2.0], $MachinePrecision], 0.3333333333333333], $MachinePrecision] * N[Power[t$95$3, 1/3], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(k / t$95$m), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[(t$95$m * N[(N[Sqrt[2.0], $MachinePrecision] / k), $MachinePrecision]), $MachinePrecision] / N[(N[(t$95$m * t$95$3), $MachinePrecision] / t$95$2), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if l < 4.5e-153

   1. Initial program 33.1%

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

      1. add-sqr-sqrt 17.1%

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

      2. pow2 17.1%

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

   4. Applied egg-rr 26.6%

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

   5. Taylor expanded in k around 0 41.3%

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

      1. add-sqr-sqrt 20.0%

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

      2. sqrt-div 9.2%

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

      3. sqrt-pow1 9.2%

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

      4. metadata-eval 9.2%

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

      5. pow1 9.2%

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

      6. sqrt-div 9.2%

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

      7. sqrt-pow1 10.2%

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

      8. metadata-eval 10.2%

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

      9. pow1 10.2%

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

   7. Applied egg-rr 10.2%

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

      1. unpow2 10.2%

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

   9. Simplified 10.2%

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

   if 4.5e-153 < l < 2.6000000000000002e163

   1. Initial program 41.6%

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

   3. Simplified 54.8%

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

   5. Taylor expanded in t around 0 90.8%

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

      1. associate-*r/ 90.8%

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

      2. associate-*r* 90.7%

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

      3. times-frac 90.7%

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

      4. *-commutative 90.7%

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

   7. Simplified 90.7%

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

   if 2.6000000000000002e163 < l

   1. Initial program 32.3%

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

      1. *-commutative 32.3%

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

      2. associate-/r* 32.3%

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

   3. Simplified 32.3%

      \[\leadsto \color{blue}{\frac{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}}{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. add-sqr-sqrt 32.3%

         \[\leadsto \frac{\color{blue}{\sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}} \cdot \sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}}}}{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)} \]

      2. add-cube-cbrt 32.3%

         \[\leadsto \frac{\sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}} \cdot \sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}}}{\color{blue}{\left(\sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)} \cdot \sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}\right) \cdot \sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}}} \]

      3. times-frac 32.3%

         \[\leadsto \color{blue}{\frac{\sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}}}{\sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)} \cdot \sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}} \cdot \frac{\sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}}}{\sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}}} \]

   6. Applied egg-rr 75.0%

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

      1. associate-/l/ 75.0%

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

      2. associate-/r/ 75.0%

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

   8. Simplified 75.0%

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

      1. associate-*l/ 75.0%

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

   10. Applied egg-rr 75.0%

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

       1. pow1/3 53.6%

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

       2. add-cube-cbrt 53.6%

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

       3. unpow-prod-down 53.6%

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

       4. pow2 53.6%

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

       5. pow1/3 75.1%

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

   12. Applied egg-rr 75.1%

       \[\leadsto \frac{\sqrt{2}}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \color{blue}{\left({\left({\left(\sqrt[3]{\sin k \cdot \tan k}\right)}^{2}\right)}^{0.3333333333333333} \cdot \sqrt[3]{\sqrt[3]{\sin k \cdot \tan k}}\right)}\right)}^{2} \cdot \frac{k}{t}} \cdot \frac{\frac{\sqrt{2}}{k} \cdot t}{\frac{t \cdot \sqrt[3]{\sin k \cdot \tan k}}{{\left(\sqrt[3]{\ell}\right)}^{2}}} \]
3. Recombined 3 regimes into one program.

4. Final simplification 34.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\ell \leq 4.5 \cdot 10^{-153}:\\ \;\;\;\;\frac{2}{{\left({\left(\frac{k}{\sqrt{\ell}}\right)}^{2} \cdot \sqrt{t}\right)}^{2}}\\ \mathbf{elif}\;\ell \leq 2.6 \cdot 10^{+163}:\\ \;\;\;\;\frac{2}{t \cdot {k}^{2}} \cdot \frac{\cos k \cdot {\ell}^{2}}{{\sin k}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \left({\left({\left(\sqrt[3]{\sin k \cdot \tan k}\right)}^{2}\right)}^{0.3333333333333333} \cdot \sqrt[3]{\sqrt[3]{\sin k \cdot \tan k}}\right)\right)}^{2} \cdot \frac{k}{t}} \cdot \frac{t \cdot \frac{\sqrt{2}}{k}}{\frac{t \cdot \sqrt[3]{\sin k \cdot \tan k}}{{\left(\sqrt[3]{\ell}\right)}^{2}}}\\ \end{array} \]
5. Add Preprocessing

Alternative 2: 88.8% accurate, 0.3× speedup

Math

\[\begin{array}{l} l_m = \left|\ell\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ \begin{array}{l} t_2 := \frac{t\_m}{{\left(\sqrt[3]{l\_m}\right)}^{2}} \cdot \sqrt[3]{\sin k \cdot \tan k}\\ t\_s \cdot \begin{array}{l} \mathbf{if}\;l\_m \leq 3.8 \cdot 10^{-153}:\\ \;\;\;\;\frac{2}{{\left({\left(\frac{k}{\sqrt{l\_m}}\right)}^{2} \cdot \sqrt{t\_m}\right)}^{2}}\\ \mathbf{elif}\;l\_m \leq 2.6 \cdot 10^{+163}:\\ \;\;\;\;\frac{2}{t\_m \cdot {k}^{2}} \cdot \frac{\cos k \cdot {l\_m}^{2}}{{\sin k}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{\frac{k}{t\_m} \cdot {t\_2}^{2}} \cdot \frac{t\_m \cdot \frac{\sqrt{2}}{k}}{t\_2}\\ \end{array} \end{array} \end{array} \]

FPCore

l_m = (fabs.f64 l)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l_m k)
 :precision binary64
 (let* ((t_2 (* (/ t_m (pow (cbrt l_m) 2.0)) (cbrt (* (sin k) (tan k))))))
   (*
    t_s
    (if (<= l_m 3.8e-153)
      (/ 2.0 (pow (* (pow (/ k (sqrt l_m)) 2.0) (sqrt t_m)) 2.0))
      (if (<= l_m 2.6e+163)
        (*
         (/ 2.0 (* t_m (pow k 2.0)))
         (/ (* (cos k) (pow l_m 2.0)) (pow (sin k) 2.0)))
        (*
         (/ (sqrt 2.0) (* (/ k t_m) (pow t_2 2.0)))
         (/ (* t_m (/ (sqrt 2.0) k)) t_2)))))))

C

l_m = fabs(l);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l_m, double k) {
	double t_2 = (t_m / pow(cbrt(l_m), 2.0)) * cbrt((sin(k) * tan(k)));
	double tmp;
	if (l_m <= 3.8e-153) {
		tmp = 2.0 / pow((pow((k / sqrt(l_m)), 2.0) * sqrt(t_m)), 2.0);
	} else if (l_m <= 2.6e+163) {
		tmp = (2.0 / (t_m * pow(k, 2.0))) * ((cos(k) * pow(l_m, 2.0)) / pow(sin(k), 2.0));
	} else {
		tmp = (sqrt(2.0) / ((k / t_m) * pow(t_2, 2.0))) * ((t_m * (sqrt(2.0) / k)) / t_2);
	}
	return t_s * tmp;
}

Java

l_m = Math.abs(l);
t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l_m, double k) {
	double t_2 = (t_m / Math.pow(Math.cbrt(l_m), 2.0)) * Math.cbrt((Math.sin(k) * Math.tan(k)));
	double tmp;
	if (l_m <= 3.8e-153) {
		tmp = 2.0 / Math.pow((Math.pow((k / Math.sqrt(l_m)), 2.0) * Math.sqrt(t_m)), 2.0);
	} else if (l_m <= 2.6e+163) {
		tmp = (2.0 / (t_m * Math.pow(k, 2.0))) * ((Math.cos(k) * Math.pow(l_m, 2.0)) / Math.pow(Math.sin(k), 2.0));
	} else {
		tmp = (Math.sqrt(2.0) / ((k / t_m) * Math.pow(t_2, 2.0))) * ((t_m * (Math.sqrt(2.0) / k)) / t_2);
	}
	return t_s * tmp;
}

Julia

l_m = abs(l)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l_m, k)
	t_2 = Float64(Float64(t_m / (cbrt(l_m) ^ 2.0)) * cbrt(Float64(sin(k) * tan(k))))
	tmp = 0.0
	if (l_m <= 3.8e-153)
		tmp = Float64(2.0 / (Float64((Float64(k / sqrt(l_m)) ^ 2.0) * sqrt(t_m)) ^ 2.0));
	elseif (l_m <= 2.6e+163)
		tmp = Float64(Float64(2.0 / Float64(t_m * (k ^ 2.0))) * Float64(Float64(cos(k) * (l_m ^ 2.0)) / (sin(k) ^ 2.0)));
	else
		tmp = Float64(Float64(sqrt(2.0) / Float64(Float64(k / t_m) * (t_2 ^ 2.0))) * Float64(Float64(t_m * Float64(sqrt(2.0) / k)) / t_2));
	end
	return Float64(t_s * tmp)
end

Wolfram

l_m = N[Abs[l], $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$95$m_, k_] := Block[{t$95$2 = N[(N[(t$95$m / N[Power[N[Power[l$95$m, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] * N[Power[N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]}, N[(t$95$s * If[LessEqual[l$95$m, 3.8e-153], N[(2.0 / N[Power[N[(N[Power[N[(k / N[Sqrt[l$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[l$95$m, 2.6e+163], N[(N[(2.0 / N[(t$95$m * N[Power[k, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[(N[Cos[k], $MachinePrecision] * N[Power[l$95$m, 2.0], $MachinePrecision]), $MachinePrecision] / N[Power[N[Sin[k], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[Sqrt[2.0], $MachinePrecision] / N[(N[(k / t$95$m), $MachinePrecision] * N[Power[t$95$2, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[(t$95$m * N[(N[Sqrt[2.0], $MachinePrecision] / k), $MachinePrecision]), $MachinePrecision] / t$95$2), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]

Derivation

1. Split input into 3 regimes

2. if l < 3.80000000000000023e-153

   1. Initial program 33.1%

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

      1. add-sqr-sqrt 17.1%

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

      2. pow2 17.1%

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

   4. Applied egg-rr 26.6%

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

   5. Taylor expanded in k around 0 41.3%

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

      1. add-sqr-sqrt 20.0%

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

      2. sqrt-div 9.2%

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

      3. sqrt-pow1 9.2%

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

      4. metadata-eval 9.2%

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

      5. pow1 9.2%

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

      6. sqrt-div 9.2%

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

      7. sqrt-pow1 10.2%

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

      8. metadata-eval 10.2%

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

      9. pow1 10.2%

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

   7. Applied egg-rr 10.2%

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

      1. unpow2 10.2%

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

   9. Simplified 10.2%

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

   if 3.80000000000000023e-153 < l < 2.6000000000000002e163

   1. Initial program 41.6%

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

   3. Simplified 54.8%

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

   5. Taylor expanded in t around 0 90.8%

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

      1. associate-*r/ 90.8%

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

      2. associate-*r* 90.7%

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

      3. times-frac 90.7%

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

      4. *-commutative 90.7%

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

   7. Simplified 90.7%

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

   if 2.6000000000000002e163 < l

   1. Initial program 32.3%

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

      1. *-commutative 32.3%

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

      2. associate-/r* 32.3%

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

   3. Simplified 32.3%

      \[\leadsto \color{blue}{\frac{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}}{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. add-sqr-sqrt 32.3%

         \[\leadsto \frac{\color{blue}{\sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}} \cdot \sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}}}}{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)} \]

      2. add-cube-cbrt 32.3%

         \[\leadsto \frac{\sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}} \cdot \sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}}}{\color{blue}{\left(\sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)} \cdot \sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}\right) \cdot \sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}}} \]

      3. times-frac 32.3%

         \[\leadsto \color{blue}{\frac{\sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}}}{\sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)} \cdot \sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}} \cdot \frac{\sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}}}{\sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}}} \]

   6. Applied egg-rr 75.0%

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

      1. associate-/l/ 75.0%

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

      2. associate-/r/ 75.0%

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

   8. Simplified 75.0%

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

4. Final simplification 34.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\ell \leq 3.8 \cdot 10^{-153}:\\ \;\;\;\;\frac{2}{{\left({\left(\frac{k}{\sqrt{\ell}}\right)}^{2} \cdot \sqrt{t}\right)}^{2}}\\ \mathbf{elif}\;\ell \leq 2.6 \cdot 10^{+163}:\\ \;\;\;\;\frac{2}{t \cdot {k}^{2}} \cdot \frac{\cos k \cdot {\ell}^{2}}{{\sin k}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{\sqrt{2}}{\frac{k}{t} \cdot {\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \sqrt[3]{\sin k \cdot \tan k}\right)}^{2}} \cdot \frac{t \cdot \frac{\sqrt{2}}{k}}{\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \sqrt[3]{\sin k \cdot \tan k}}\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 88.8% accurate, 0.3× speedup

Math

\[\begin{array}{l} l_m = \left|\ell\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ \begin{array}{l} t_2 := \sqrt[3]{\sin k \cdot \tan k}\\ t_3 := {\left(\sqrt[3]{l\_m}\right)}^{2}\\ t\_s \cdot \begin{array}{l} \mathbf{if}\;l\_m \leq 4 \cdot 10^{-153}:\\ \;\;\;\;\frac{2}{{\left({\left(\frac{k}{\sqrt{l\_m}}\right)}^{2} \cdot \sqrt{t\_m}\right)}^{2}}\\ \mathbf{elif}\;l\_m \leq 3 \cdot 10^{+163}:\\ \;\;\;\;\frac{2}{t\_m \cdot {k}^{2}} \cdot \frac{\cos k \cdot {l\_m}^{2}}{{\sin k}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{t\_m \cdot \frac{\sqrt{2}}{k}}{\frac{t\_m \cdot t\_2}{t\_3}} \cdot \frac{\sqrt{2}}{\frac{k}{t\_m} \cdot {\left(\frac{t\_m}{t\_3} \cdot t\_2\right)}^{2}}\\ \end{array} \end{array} \end{array} \]

FPCore

l_m = (fabs.f64 l)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l_m k)
 :precision binary64
 (let* ((t_2 (cbrt (* (sin k) (tan k)))) (t_3 (pow (cbrt l_m) 2.0)))
   (*
    t_s
    (if (<= l_m 4e-153)
      (/ 2.0 (pow (* (pow (/ k (sqrt l_m)) 2.0) (sqrt t_m)) 2.0))
      (if (<= l_m 3e+163)
        (*
         (/ 2.0 (* t_m (pow k 2.0)))
         (/ (* (cos k) (pow l_m 2.0)) (pow (sin k) 2.0)))
        (*
         (/ (* t_m (/ (sqrt 2.0) k)) (/ (* t_m t_2) t_3))
         (/ (sqrt 2.0) (* (/ k t_m) (pow (* (/ t_m t_3) t_2) 2.0)))))))))

C

l_m = fabs(l);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l_m, double k) {
	double t_2 = cbrt((sin(k) * tan(k)));
	double t_3 = pow(cbrt(l_m), 2.0);
	double tmp;
	if (l_m <= 4e-153) {
		tmp = 2.0 / pow((pow((k / sqrt(l_m)), 2.0) * sqrt(t_m)), 2.0);
	} else if (l_m <= 3e+163) {
		tmp = (2.0 / (t_m * pow(k, 2.0))) * ((cos(k) * pow(l_m, 2.0)) / pow(sin(k), 2.0));
	} else {
		tmp = ((t_m * (sqrt(2.0) / k)) / ((t_m * t_2) / t_3)) * (sqrt(2.0) / ((k / t_m) * pow(((t_m / t_3) * t_2), 2.0)));
	}
	return t_s * tmp;
}

Java

l_m = Math.abs(l);
t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l_m, double k) {
	double t_2 = Math.cbrt((Math.sin(k) * Math.tan(k)));
	double t_3 = Math.pow(Math.cbrt(l_m), 2.0);
	double tmp;
	if (l_m <= 4e-153) {
		tmp = 2.0 / Math.pow((Math.pow((k / Math.sqrt(l_m)), 2.0) * Math.sqrt(t_m)), 2.0);
	} else if (l_m <= 3e+163) {
		tmp = (2.0 / (t_m * Math.pow(k, 2.0))) * ((Math.cos(k) * Math.pow(l_m, 2.0)) / Math.pow(Math.sin(k), 2.0));
	} else {
		tmp = ((t_m * (Math.sqrt(2.0) / k)) / ((t_m * t_2) / t_3)) * (Math.sqrt(2.0) / ((k / t_m) * Math.pow(((t_m / t_3) * t_2), 2.0)));
	}
	return t_s * tmp;
}

Julia

l_m = abs(l)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l_m, k)
	t_2 = cbrt(Float64(sin(k) * tan(k)))
	t_3 = cbrt(l_m) ^ 2.0
	tmp = 0.0
	if (l_m <= 4e-153)
		tmp = Float64(2.0 / (Float64((Float64(k / sqrt(l_m)) ^ 2.0) * sqrt(t_m)) ^ 2.0));
	elseif (l_m <= 3e+163)
		tmp = Float64(Float64(2.0 / Float64(t_m * (k ^ 2.0))) * Float64(Float64(cos(k) * (l_m ^ 2.0)) / (sin(k) ^ 2.0)));
	else
		tmp = Float64(Float64(Float64(t_m * Float64(sqrt(2.0) / k)) / Float64(Float64(t_m * t_2) / t_3)) * Float64(sqrt(2.0) / Float64(Float64(k / t_m) * (Float64(Float64(t_m / t_3) * t_2) ^ 2.0))));
	end
	return Float64(t_s * tmp)
end

Wolfram

l_m = N[Abs[l], $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$95$m_, k_] := Block[{t$95$2 = N[Power[N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision]}, Block[{t$95$3 = N[Power[N[Power[l$95$m, 1/3], $MachinePrecision], 2.0], $MachinePrecision]}, N[(t$95$s * If[LessEqual[l$95$m, 4e-153], N[(2.0 / N[Power[N[(N[Power[N[(k / N[Sqrt[l$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[l$95$m, 3e+163], N[(N[(2.0 / N[(t$95$m * N[Power[k, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[(N[Cos[k], $MachinePrecision] * N[Power[l$95$m, 2.0], $MachinePrecision]), $MachinePrecision] / N[Power[N[Sin[k], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(N[(N[(t$95$m * N[(N[Sqrt[2.0], $MachinePrecision] / k), $MachinePrecision]), $MachinePrecision] / N[(N[(t$95$m * t$95$2), $MachinePrecision] / t$95$3), $MachinePrecision]), $MachinePrecision] * N[(N[Sqrt[2.0], $MachinePrecision] / N[(N[(k / t$95$m), $MachinePrecision] * N[Power[N[(N[(t$95$m / t$95$3), $MachinePrecision] * t$95$2), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]]

Derivation

1. Split input into 3 regimes

2. if l < 4.00000000000000016e-153

   1. Initial program 33.1%

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

      1. add-sqr-sqrt 17.1%

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

      2. pow2 17.1%

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

   4. Applied egg-rr 26.6%

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

   5. Taylor expanded in k around 0 41.3%

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

      1. add-sqr-sqrt 20.0%

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

      2. sqrt-div 9.2%

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

      3. sqrt-pow1 9.2%

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

      4. metadata-eval 9.2%

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

      5. pow1 9.2%

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

      6. sqrt-div 9.2%

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

      7. sqrt-pow1 10.2%

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

      8. metadata-eval 10.2%

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

      9. pow1 10.2%

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

   7. Applied egg-rr 10.2%

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

      1. unpow2 10.2%

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

   9. Simplified 10.2%

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

   if 4.00000000000000016e-153 < l < 3.00000000000000013e163

   1. Initial program 41.6%

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

   3. Simplified 54.8%

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

   5. Taylor expanded in t around 0 90.8%

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

      1. associate-*r/ 90.8%

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

      2. associate-*r* 90.7%

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

      3. times-frac 90.7%

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

      4. *-commutative 90.7%

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

   7. Simplified 90.7%

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

   if 3.00000000000000013e163 < l

   1. Initial program 32.3%

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

      1. *-commutative 32.3%

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

      2. associate-/r* 32.3%

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

   3. Simplified 32.3%

      \[\leadsto \color{blue}{\frac{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}}{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. add-sqr-sqrt 32.3%

         \[\leadsto \frac{\color{blue}{\sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}} \cdot \sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}}}}{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)} \]

      2. add-cube-cbrt 32.3%

         \[\leadsto \frac{\sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}} \cdot \sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}}}{\color{blue}{\left(\sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)} \cdot \sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}\right) \cdot \sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}}} \]

      3. times-frac 32.3%

         \[\leadsto \color{blue}{\frac{\sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}}}{\sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)} \cdot \sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}} \cdot \frac{\sqrt{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}}}{\sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}}} \]

   6. Applied egg-rr 75.0%

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

      1. associate-/l/ 75.0%

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

      2. associate-/r/ 75.0%

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

   8. Simplified 75.0%

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

      1. associate-*l/ 75.0%

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

   10. Applied egg-rr 75.0%

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

4. Final simplification 34.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\ell \leq 4 \cdot 10^{-153}:\\ \;\;\;\;\frac{2}{{\left({\left(\frac{k}{\sqrt{\ell}}\right)}^{2} \cdot \sqrt{t}\right)}^{2}}\\ \mathbf{elif}\;\ell \leq 3 \cdot 10^{+163}:\\ \;\;\;\;\frac{2}{t \cdot {k}^{2}} \cdot \frac{\cos k \cdot {\ell}^{2}}{{\sin k}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{t \cdot \frac{\sqrt{2}}{k}}{\frac{t \cdot \sqrt[3]{\sin k \cdot \tan k}}{{\left(\sqrt[3]{\ell}\right)}^{2}}} \cdot \frac{\sqrt{2}}{\frac{k}{t} \cdot {\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \sqrt[3]{\sin k \cdot \tan k}\right)}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 79.8% accurate, 0.3× speedup

Math

\[\begin{array}{l} l_m = \left|\ell\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ \begin{array}{l} t_2 := \frac{t\_m}{{\left(\sqrt[3]{l\_m}\right)}^{2}} \cdot \sqrt[3]{\sin k \cdot \tan k}\\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 2.5 \cdot 10^{-36}:\\ \;\;\;\;\frac{2}{{\left(k \cdot \left(\frac{\sin k}{l\_m} \cdot \frac{\sqrt{t\_m}}{\sqrt{\cos k}}\right)\right)}^{2}}\\ \mathbf{elif}\;k \leq 1.25 \cdot 10^{+173}:\\ \;\;\;\;\frac{2}{{k}^{2} \cdot \frac{t\_m \cdot {\sin k}^{2}}{\cos k}} \cdot \left(l\_m \cdot l\_m\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{t\_2}^{2}} \cdot \frac{{\left(\frac{k}{t\_m}\right)}^{-2}}{t\_2}\\ \end{array} \end{array} \end{array} \]

FPCore

l_m = (fabs.f64 l)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l_m k)
 :precision binary64
 (let* ((t_2 (* (/ t_m (pow (cbrt l_m) 2.0)) (cbrt (* (sin k) (tan k))))))
   (*
    t_s
    (if (<= k 2.5e-36)
      (/ 2.0 (pow (* k (* (/ (sin k) l_m) (/ (sqrt t_m) (sqrt (cos k))))) 2.0))
      (if (<= k 1.25e+173)
        (*
         (/ 2.0 (* (pow k 2.0) (/ (* t_m (pow (sin k) 2.0)) (cos k))))
         (* l_m l_m))
        (* (/ 2.0 (pow t_2 2.0)) (/ (pow (/ k t_m) -2.0) t_2)))))))

C

l_m = fabs(l);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l_m, double k) {
	double t_2 = (t_m / pow(cbrt(l_m), 2.0)) * cbrt((sin(k) * tan(k)));
	double tmp;
	if (k <= 2.5e-36) {
		tmp = 2.0 / pow((k * ((sin(k) / l_m) * (sqrt(t_m) / sqrt(cos(k))))), 2.0);
	} else if (k <= 1.25e+173) {
		tmp = (2.0 / (pow(k, 2.0) * ((t_m * pow(sin(k), 2.0)) / cos(k)))) * (l_m * l_m);
	} else {
		tmp = (2.0 / pow(t_2, 2.0)) * (pow((k / t_m), -2.0) / t_2);
	}
	return t_s * tmp;
}

Java

l_m = Math.abs(l);
t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l_m, double k) {
	double t_2 = (t_m / Math.pow(Math.cbrt(l_m), 2.0)) * Math.cbrt((Math.sin(k) * Math.tan(k)));
	double tmp;
	if (k <= 2.5e-36) {
		tmp = 2.0 / Math.pow((k * ((Math.sin(k) / l_m) * (Math.sqrt(t_m) / Math.sqrt(Math.cos(k))))), 2.0);
	} else if (k <= 1.25e+173) {
		tmp = (2.0 / (Math.pow(k, 2.0) * ((t_m * Math.pow(Math.sin(k), 2.0)) / Math.cos(k)))) * (l_m * l_m);
	} else {
		tmp = (2.0 / Math.pow(t_2, 2.0)) * (Math.pow((k / t_m), -2.0) / t_2);
	}
	return t_s * tmp;
}

Julia

l_m = abs(l)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l_m, k)
	t_2 = Float64(Float64(t_m / (cbrt(l_m) ^ 2.0)) * cbrt(Float64(sin(k) * tan(k))))
	tmp = 0.0
	if (k <= 2.5e-36)
		tmp = Float64(2.0 / (Float64(k * Float64(Float64(sin(k) / l_m) * Float64(sqrt(t_m) / sqrt(cos(k))))) ^ 2.0));
	elseif (k <= 1.25e+173)
		tmp = Float64(Float64(2.0 / Float64((k ^ 2.0) * Float64(Float64(t_m * (sin(k) ^ 2.0)) / cos(k)))) * Float64(l_m * l_m));
	else
		tmp = Float64(Float64(2.0 / (t_2 ^ 2.0)) * Float64((Float64(k / t_m) ^ -2.0) / t_2));
	end
	return Float64(t_s * tmp)
end

Wolfram

l_m = N[Abs[l], $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$95$m_, k_] := Block[{t$95$2 = N[(N[(t$95$m / N[Power[N[Power[l$95$m, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] * N[Power[N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision], 1/3], $MachinePrecision]), $MachinePrecision]}, N[(t$95$s * If[LessEqual[k, 2.5e-36], N[(2.0 / N[Power[N[(k * N[(N[(N[Sin[k], $MachinePrecision] / l$95$m), $MachinePrecision] * N[(N[Sqrt[t$95$m], $MachinePrecision] / N[Sqrt[N[Cos[k], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[k, 1.25e+173], N[(N[(2.0 / N[(N[Power[k, 2.0], $MachinePrecision] * N[(N[(t$95$m * N[Power[N[Sin[k], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] / N[Cos[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(l$95$m * l$95$m), $MachinePrecision]), $MachinePrecision], N[(N[(2.0 / N[Power[t$95$2, 2.0], $MachinePrecision]), $MachinePrecision] * N[(N[Power[N[(k / t$95$m), $MachinePrecision], -2.0], $MachinePrecision] / t$95$2), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]]

Derivation

1. Split input into 3 regimes

2. if k < 2.50000000000000002e-36

   1. Initial program 38.6%

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

   3. Taylor expanded in t around 0 76.1%

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

      1. associate-/l* 76.7%

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

      2. *-commutative 76.7%

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

   5. Simplified 76.7%

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

      1. pow1 76.7%

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

   7. Applied egg-rr 43.3%

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

      1. unpow1 43.3%

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

      2. times-frac 43.3%

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

   9. Simplified 43.3%

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

   if 2.50000000000000002e-36 < k < 1.25000000000000009e173

   1. Initial program 12.9%

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

   3. Simplified 38.6%

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

   5. Taylor expanded in t around 0 92.2%

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

      1. associate-/l* 92.2%

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

   7. Simplified 92.2%

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

   if 1.25000000000000009e173 < k

   1. Initial program 39.3%

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

      1. *-commutative 39.3%

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

      2. associate-/r* 39.3%

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

   3. Simplified 45.4%

      \[\leadsto \color{blue}{\frac{\frac{2}{{\left(\frac{k}{t}\right)}^{2} + 0}}{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}} \]
   4. Add Preprocessing
   5. Step-by-step derivation

      1. div-inv 45.4%

         \[\leadsto \frac{\color{blue}{2 \cdot \frac{1}{{\left(\frac{k}{t}\right)}^{2} + 0}}}{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)} \]

      2. add-cube-cbrt 45.4%

         \[\leadsto \frac{2 \cdot \frac{1}{{\left(\frac{k}{t}\right)}^{2} + 0}}{\color{blue}{\left(\sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)} \cdot \sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}\right) \cdot \sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}}} \]

      3. times-frac 45.4%

         \[\leadsto \color{blue}{\frac{2}{\sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)} \cdot \sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}} \cdot \frac{\frac{1}{{\left(\frac{k}{t}\right)}^{2} + 0}}{\sqrt[3]{\frac{{t}^{3}}{\ell \cdot \ell} \cdot \left(\sin k \cdot \tan k\right)}}} \]

   6. Applied egg-rr 69.4%

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

4. Final simplification 54.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 2.5 \cdot 10^{-36}:\\ \;\;\;\;\frac{2}{{\left(k \cdot \left(\frac{\sin k}{\ell} \cdot \frac{\sqrt{t}}{\sqrt{\cos k}}\right)\right)}^{2}}\\ \mathbf{elif}\;k \leq 1.25 \cdot 10^{+173}:\\ \;\;\;\;\frac{2}{{k}^{2} \cdot \frac{t \cdot {\sin k}^{2}}{\cos k}} \cdot \left(\ell \cdot \ell\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \sqrt[3]{\sin k \cdot \tan k}\right)}^{2}} \cdot \frac{{\left(\frac{k}{t}\right)}^{-2}}{\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}} \cdot \sqrt[3]{\sin k \cdot \tan k}}\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 79.1% accurate, 0.8× speedup

Math

\[\begin{array}{l} l_m = \left|\ell\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 2.05 \cdot 10^{-36}:\\ \;\;\;\;\frac{2}{{\left(k \cdot \left(\frac{\sin k}{l\_m} \cdot \frac{\sqrt{t\_m}}{\sqrt{\cos k}}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{k}^{2} \cdot \frac{t\_m \cdot {\sin k}^{2}}{\cos k}} \cdot \left(l\_m \cdot l\_m\right)\\ \end{array} \end{array} \]

FPCore

l_m = (fabs.f64 l)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l_m k)
 :precision binary64
 (*
  t_s
  (if (<= k 2.05e-36)
    (/ 2.0 (pow (* k (* (/ (sin k) l_m) (/ (sqrt t_m) (sqrt (cos k))))) 2.0))
    (*
     (/ 2.0 (* (pow k 2.0) (/ (* t_m (pow (sin k) 2.0)) (cos k))))
     (* l_m l_m)))))

C

l_m = fabs(l);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l_m, double k) {
	double tmp;
	if (k <= 2.05e-36) {
		tmp = 2.0 / pow((k * ((sin(k) / l_m) * (sqrt(t_m) / sqrt(cos(k))))), 2.0);
	} else {
		tmp = (2.0 / (pow(k, 2.0) * ((t_m * pow(sin(k), 2.0)) / cos(k)))) * (l_m * l_m);
	}
	return t_s * tmp;
}

Fortran

l_m = abs(l)
t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l_m, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l_m
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 2.05d-36) then
        tmp = 2.0d0 / ((k * ((sin(k) / l_m) * (sqrt(t_m) / sqrt(cos(k))))) ** 2.0d0)
    else
        tmp = (2.0d0 / ((k ** 2.0d0) * ((t_m * (sin(k) ** 2.0d0)) / cos(k)))) * (l_m * l_m)
    end if
    code = t_s * tmp
end function

Java

l_m = Math.abs(l);
t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l_m, double k) {
	double tmp;
	if (k <= 2.05e-36) {
		tmp = 2.0 / Math.pow((k * ((Math.sin(k) / l_m) * (Math.sqrt(t_m) / Math.sqrt(Math.cos(k))))), 2.0);
	} else {
		tmp = (2.0 / (Math.pow(k, 2.0) * ((t_m * Math.pow(Math.sin(k), 2.0)) / Math.cos(k)))) * (l_m * l_m);
	}
	return t_s * tmp;
}

Python

l_m = math.fabs(l)
t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l_m, k):
	tmp = 0
	if k <= 2.05e-36:
		tmp = 2.0 / math.pow((k * ((math.sin(k) / l_m) * (math.sqrt(t_m) / math.sqrt(math.cos(k))))), 2.0)
	else:
		tmp = (2.0 / (math.pow(k, 2.0) * ((t_m * math.pow(math.sin(k), 2.0)) / math.cos(k)))) * (l_m * l_m)
	return t_s * tmp

Julia

l_m = abs(l)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l_m, k)
	tmp = 0.0
	if (k <= 2.05e-36)
		tmp = Float64(2.0 / (Float64(k * Float64(Float64(sin(k) / l_m) * Float64(sqrt(t_m) / sqrt(cos(k))))) ^ 2.0));
	else
		tmp = Float64(Float64(2.0 / Float64((k ^ 2.0) * Float64(Float64(t_m * (sin(k) ^ 2.0)) / cos(k)))) * Float64(l_m * l_m));
	end
	return Float64(t_s * tmp)
end

MATLAB

l_m = abs(l);
t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l_m, k)
	tmp = 0.0;
	if (k <= 2.05e-36)
		tmp = 2.0 / ((k * ((sin(k) / l_m) * (sqrt(t_m) / sqrt(cos(k))))) ^ 2.0);
	else
		tmp = (2.0 / ((k ^ 2.0) * ((t_m * (sin(k) ^ 2.0)) / cos(k)))) * (l_m * l_m);
	end
	tmp_2 = t_s * tmp;
end

Wolfram

l_m = N[Abs[l], $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$95$m_, k_] := N[(t$95$s * If[LessEqual[k, 2.05e-36], N[(2.0 / N[Power[N[(k * N[(N[(N[Sin[k], $MachinePrecision] / l$95$m), $MachinePrecision] * N[(N[Sqrt[t$95$m], $MachinePrecision] / N[Sqrt[N[Cos[k], $MachinePrecision]], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(N[(2.0 / N[(N[Power[k, 2.0], $MachinePrecision] * N[(N[(t$95$m * N[Power[N[Sin[k], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] / N[Cos[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(l$95$m * l$95$m), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 2.05000000000000006e-36

   1. Initial program 38.6%

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

   3. Taylor expanded in t around 0 76.1%

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

      1. associate-/l* 76.7%

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

      2. *-commutative 76.7%

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

   5. Simplified 76.7%

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

      1. pow1 76.7%

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

   7. Applied egg-rr 43.3%

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

      1. unpow1 43.3%

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

      2. times-frac 43.3%

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

   9. Simplified 43.3%

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

   if 2.05000000000000006e-36 < k

   1. Initial program 25.0%

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

   3. Simplified 40.5%

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

   5. Taylor expanded in t around 0 76.5%

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

      1. associate-/l* 76.5%

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

   7. Simplified 76.5%

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

4. Final simplification 52.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 2.05 \cdot 10^{-36}:\\ \;\;\;\;\frac{2}{{\left(k \cdot \left(\frac{\sin k}{\ell} \cdot \frac{\sqrt{t}}{\sqrt{\cos k}}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{k}^{2} \cdot \frac{t \cdot {\sin k}^{2}}{\cos k}} \cdot \left(\ell \cdot \ell\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 6: 79.3% accurate, 1.0× speedup

Math

\[\begin{array}{l} l_m = \left|\ell\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 2.7 \cdot 10^{-36}:\\ \;\;\;\;\frac{2}{{\left(\left(k \cdot \frac{\sin k}{l\_m}\right) \cdot \sqrt{\frac{t\_m}{\cos k}}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{k}^{2} \cdot \frac{t\_m \cdot {\sin k}^{2}}{\cos k}} \cdot \left(l\_m \cdot l\_m\right)\\ \end{array} \end{array} \]

FPCore

l_m = (fabs.f64 l)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l_m k)
 :precision binary64
 (*
  t_s
  (if (<= k 2.7e-36)
    (/ 2.0 (pow (* (* k (/ (sin k) l_m)) (sqrt (/ t_m (cos k)))) 2.0))
    (*
     (/ 2.0 (* (pow k 2.0) (/ (* t_m (pow (sin k) 2.0)) (cos k))))
     (* l_m l_m)))))

C

l_m = fabs(l);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l_m, double k) {
	double tmp;
	if (k <= 2.7e-36) {
		tmp = 2.0 / pow(((k * (sin(k) / l_m)) * sqrt((t_m / cos(k)))), 2.0);
	} else {
		tmp = (2.0 / (pow(k, 2.0) * ((t_m * pow(sin(k), 2.0)) / cos(k)))) * (l_m * l_m);
	}
	return t_s * tmp;
}

Fortran

l_m = abs(l)
t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l_m, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l_m
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 2.7d-36) then
        tmp = 2.0d0 / (((k * (sin(k) / l_m)) * sqrt((t_m / cos(k)))) ** 2.0d0)
    else
        tmp = (2.0d0 / ((k ** 2.0d0) * ((t_m * (sin(k) ** 2.0d0)) / cos(k)))) * (l_m * l_m)
    end if
    code = t_s * tmp
end function

Java

l_m = Math.abs(l);
t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l_m, double k) {
	double tmp;
	if (k <= 2.7e-36) {
		tmp = 2.0 / Math.pow(((k * (Math.sin(k) / l_m)) * Math.sqrt((t_m / Math.cos(k)))), 2.0);
	} else {
		tmp = (2.0 / (Math.pow(k, 2.0) * ((t_m * Math.pow(Math.sin(k), 2.0)) / Math.cos(k)))) * (l_m * l_m);
	}
	return t_s * tmp;
}

Python

l_m = math.fabs(l)
t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l_m, k):
	tmp = 0
	if k <= 2.7e-36:
		tmp = 2.0 / math.pow(((k * (math.sin(k) / l_m)) * math.sqrt((t_m / math.cos(k)))), 2.0)
	else:
		tmp = (2.0 / (math.pow(k, 2.0) * ((t_m * math.pow(math.sin(k), 2.0)) / math.cos(k)))) * (l_m * l_m)
	return t_s * tmp

Julia

l_m = abs(l)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l_m, k)
	tmp = 0.0
	if (k <= 2.7e-36)
		tmp = Float64(2.0 / (Float64(Float64(k * Float64(sin(k) / l_m)) * sqrt(Float64(t_m / cos(k)))) ^ 2.0));
	else
		tmp = Float64(Float64(2.0 / Float64((k ^ 2.0) * Float64(Float64(t_m * (sin(k) ^ 2.0)) / cos(k)))) * Float64(l_m * l_m));
	end
	return Float64(t_s * tmp)
end

MATLAB

l_m = abs(l);
t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l_m, k)
	tmp = 0.0;
	if (k <= 2.7e-36)
		tmp = 2.0 / (((k * (sin(k) / l_m)) * sqrt((t_m / cos(k)))) ^ 2.0);
	else
		tmp = (2.0 / ((k ^ 2.0) * ((t_m * (sin(k) ^ 2.0)) / cos(k)))) * (l_m * l_m);
	end
	tmp_2 = t_s * tmp;
end

Wolfram

l_m = N[Abs[l], $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$95$m_, k_] := N[(t$95$s * If[LessEqual[k, 2.7e-36], N[(2.0 / N[Power[N[(N[(k * N[(N[Sin[k], $MachinePrecision] / l$95$m), $MachinePrecision]), $MachinePrecision] * N[Sqrt[N[(t$95$m / N[Cos[k], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(N[(2.0 / N[(N[Power[k, 2.0], $MachinePrecision] * N[(N[(t$95$m * N[Power[N[Sin[k], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] / N[Cos[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(l$95$m * l$95$m), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 2.70000000000000007e-36

   1. Initial program 38.6%

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

      1. add-sqr-sqrt 21.7%

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

      2. pow2 21.7%

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

   4. Applied egg-rr 30.7%

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

   5. Taylor expanded in k around inf 50.5%

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

      1. associate-/l* 52.2%

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

   7. Simplified 52.2%

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

   if 2.70000000000000007e-36 < k

   1. Initial program 25.0%

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

   3. Simplified 40.5%

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

   5. Taylor expanded in t around 0 76.5%

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

      1. associate-/l* 76.5%

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

   7. Simplified 76.5%

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

4. Final simplification 59.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 2.7 \cdot 10^{-36}:\\ \;\;\;\;\frac{2}{{\left(\left(k \cdot \frac{\sin k}{\ell}\right) \cdot \sqrt{\frac{t}{\cos k}}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{k}^{2} \cdot \frac{t \cdot {\sin k}^{2}}{\cos k}} \cdot \left(\ell \cdot \ell\right)\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 82.4% accurate, 1.0× speedup

Math

\[\begin{array}{l} l_m = \left|\ell\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;l\_m \cdot l\_m \leq 5 \cdot 10^{+144}:\\ \;\;\;\;\frac{2}{{\left({\left(\frac{k}{\sqrt{l\_m}}\right)}^{2} \cdot \sqrt{t\_m}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\sqrt{\frac{t\_m}{\cos k}} \cdot \frac{k \cdot \sin k}{l\_m}\right)}^{2}}\\ \end{array} \end{array} \]

FPCore

l_m = (fabs.f64 l)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l_m k)
 :precision binary64
 (*
  t_s
  (if (<= (* l_m l_m) 5e+144)
    (/ 2.0 (pow (* (pow (/ k (sqrt l_m)) 2.0) (sqrt t_m)) 2.0))
    (/ 2.0 (pow (* (sqrt (/ t_m (cos k))) (/ (* k (sin k)) l_m)) 2.0)))))

C

l_m = fabs(l);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l_m, double k) {
	double tmp;
	if ((l_m * l_m) <= 5e+144) {
		tmp = 2.0 / pow((pow((k / sqrt(l_m)), 2.0) * sqrt(t_m)), 2.0);
	} else {
		tmp = 2.0 / pow((sqrt((t_m / cos(k))) * ((k * sin(k)) / l_m)), 2.0);
	}
	return t_s * tmp;
}

Fortran

l_m = abs(l)
t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l_m, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l_m
    real(8), intent (in) :: k
    real(8) :: tmp
    if ((l_m * l_m) <= 5d+144) then
        tmp = 2.0d0 / ((((k / sqrt(l_m)) ** 2.0d0) * sqrt(t_m)) ** 2.0d0)
    else
        tmp = 2.0d0 / ((sqrt((t_m / cos(k))) * ((k * sin(k)) / l_m)) ** 2.0d0)
    end if
    code = t_s * tmp
end function

Java

l_m = Math.abs(l);
t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l_m, double k) {
	double tmp;
	if ((l_m * l_m) <= 5e+144) {
		tmp = 2.0 / Math.pow((Math.pow((k / Math.sqrt(l_m)), 2.0) * Math.sqrt(t_m)), 2.0);
	} else {
		tmp = 2.0 / Math.pow((Math.sqrt((t_m / Math.cos(k))) * ((k * Math.sin(k)) / l_m)), 2.0);
	}
	return t_s * tmp;
}

Python

l_m = math.fabs(l)
t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l_m, k):
	tmp = 0
	if (l_m * l_m) <= 5e+144:
		tmp = 2.0 / math.pow((math.pow((k / math.sqrt(l_m)), 2.0) * math.sqrt(t_m)), 2.0)
	else:
		tmp = 2.0 / math.pow((math.sqrt((t_m / math.cos(k))) * ((k * math.sin(k)) / l_m)), 2.0)
	return t_s * tmp

Julia

l_m = abs(l)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l_m, k)
	tmp = 0.0
	if (Float64(l_m * l_m) <= 5e+144)
		tmp = Float64(2.0 / (Float64((Float64(k / sqrt(l_m)) ^ 2.0) * sqrt(t_m)) ^ 2.0));
	else
		tmp = Float64(2.0 / (Float64(sqrt(Float64(t_m / cos(k))) * Float64(Float64(k * sin(k)) / l_m)) ^ 2.0));
	end
	return Float64(t_s * tmp)
end

MATLAB

l_m = abs(l);
t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l_m, k)
	tmp = 0.0;
	if ((l_m * l_m) <= 5e+144)
		tmp = 2.0 / ((((k / sqrt(l_m)) ^ 2.0) * sqrt(t_m)) ^ 2.0);
	else
		tmp = 2.0 / ((sqrt((t_m / cos(k))) * ((k * sin(k)) / l_m)) ^ 2.0);
	end
	tmp_2 = t_s * tmp;
end

Wolfram

l_m = N[Abs[l], $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$95$m_, k_] := N[(t$95$s * If[LessEqual[N[(l$95$m * l$95$m), $MachinePrecision], 5e+144], N[(2.0 / N[Power[N[(N[Power[N[(k / N[Sqrt[l$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(2.0 / N[Power[N[(N[Sqrt[N[(t$95$m / N[Cos[k], $MachinePrecision]), $MachinePrecision]], $MachinePrecision] * N[(N[(k * N[Sin[k], $MachinePrecision]), $MachinePrecision] / l$95$m), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if (*.f64 l l) < 4.9999999999999999e144

   1. Initial program 32.3%

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

      1. add-sqr-sqrt 16.6%

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

      2. pow2 16.6%

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

   4. Applied egg-rr 27.9%

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

   5. Taylor expanded in k around 0 46.8%

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

      1. add-sqr-sqrt 25.4%

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

      2. sqrt-div 21.2%

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

      3. sqrt-pow1 21.2%

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

      4. metadata-eval 21.2%

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

      5. pow1 21.2%

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

      6. sqrt-div 21.2%

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

      7. sqrt-pow1 22.3%

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

      8. metadata-eval 22.3%

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

      9. pow1 22.3%

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

   7. Applied egg-rr 22.3%

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

      1. unpow2 22.3%

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

   9. Simplified 22.3%

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

   if 4.9999999999999999e144 < (*.f64 l l)

   1. Initial program 39.5%

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

      1. add-sqr-sqrt 22.6%

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

      2. pow2 22.6%

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

   4. Applied egg-rr 32.5%

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

   5. Taylor expanded in k around inf 55.9%

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

4. Final simplification 34.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\ell \cdot \ell \leq 5 \cdot 10^{+144}:\\ \;\;\;\;\frac{2}{{\left({\left(\frac{k}{\sqrt{\ell}}\right)}^{2} \cdot \sqrt{t}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\sqrt{\frac{t}{\cos k}} \cdot \frac{k \cdot \sin k}{\ell}\right)}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 82.4% accurate, 1.0× speedup

Math

\[\begin{array}{l} l_m = \left|\ell\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;l\_m \leq 1.9 \cdot 10^{+73}:\\ \;\;\;\;\frac{2}{{\left({\left(\frac{k}{\sqrt{l\_m}}\right)}^{2} \cdot \sqrt{t\_m}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\left(k \cdot \frac{\sin k}{l\_m}\right) \cdot \sqrt{\frac{t\_m}{\cos k}}\right)}^{2}}\\ \end{array} \end{array} \]

FPCore

l_m = (fabs.f64 l)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l_m k)
 :precision binary64
 (*
  t_s
  (if (<= l_m 1.9e+73)
    (/ 2.0 (pow (* (pow (/ k (sqrt l_m)) 2.0) (sqrt t_m)) 2.0))
    (/ 2.0 (pow (* (* k (/ (sin k) l_m)) (sqrt (/ t_m (cos k)))) 2.0)))))

C

l_m = fabs(l);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l_m, double k) {
	double tmp;
	if (l_m <= 1.9e+73) {
		tmp = 2.0 / pow((pow((k / sqrt(l_m)), 2.0) * sqrt(t_m)), 2.0);
	} else {
		tmp = 2.0 / pow(((k * (sin(k) / l_m)) * sqrt((t_m / cos(k)))), 2.0);
	}
	return t_s * tmp;
}

Fortran

l_m = abs(l)
t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l_m, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l_m
    real(8), intent (in) :: k
    real(8) :: tmp
    if (l_m <= 1.9d+73) then
        tmp = 2.0d0 / ((((k / sqrt(l_m)) ** 2.0d0) * sqrt(t_m)) ** 2.0d0)
    else
        tmp = 2.0d0 / (((k * (sin(k) / l_m)) * sqrt((t_m / cos(k)))) ** 2.0d0)
    end if
    code = t_s * tmp
end function

Java

l_m = Math.abs(l);
t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l_m, double k) {
	double tmp;
	if (l_m <= 1.9e+73) {
		tmp = 2.0 / Math.pow((Math.pow((k / Math.sqrt(l_m)), 2.0) * Math.sqrt(t_m)), 2.0);
	} else {
		tmp = 2.0 / Math.pow(((k * (Math.sin(k) / l_m)) * Math.sqrt((t_m / Math.cos(k)))), 2.0);
	}
	return t_s * tmp;
}

Python

l_m = math.fabs(l)
t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l_m, k):
	tmp = 0
	if l_m <= 1.9e+73:
		tmp = 2.0 / math.pow((math.pow((k / math.sqrt(l_m)), 2.0) * math.sqrt(t_m)), 2.0)
	else:
		tmp = 2.0 / math.pow(((k * (math.sin(k) / l_m)) * math.sqrt((t_m / math.cos(k)))), 2.0)
	return t_s * tmp

Julia

l_m = abs(l)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l_m, k)
	tmp = 0.0
	if (l_m <= 1.9e+73)
		tmp = Float64(2.0 / (Float64((Float64(k / sqrt(l_m)) ^ 2.0) * sqrt(t_m)) ^ 2.0));
	else
		tmp = Float64(2.0 / (Float64(Float64(k * Float64(sin(k) / l_m)) * sqrt(Float64(t_m / cos(k)))) ^ 2.0));
	end
	return Float64(t_s * tmp)
end

MATLAB

l_m = abs(l);
t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l_m, k)
	tmp = 0.0;
	if (l_m <= 1.9e+73)
		tmp = 2.0 / ((((k / sqrt(l_m)) ^ 2.0) * sqrt(t_m)) ^ 2.0);
	else
		tmp = 2.0 / (((k * (sin(k) / l_m)) * sqrt((t_m / cos(k)))) ^ 2.0);
	end
	tmp_2 = t_s * tmp;
end

Wolfram

l_m = N[Abs[l], $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$95$m_, k_] := N[(t$95$s * If[LessEqual[l$95$m, 1.9e+73], N[(2.0 / N[Power[N[(N[Power[N[(k / N[Sqrt[l$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(2.0 / N[Power[N[(N[(k * N[(N[Sin[k], $MachinePrecision] / l$95$m), $MachinePrecision]), $MachinePrecision] * N[Sqrt[N[(t$95$m / N[Cos[k], $MachinePrecision]), $MachinePrecision]], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if l < 1.90000000000000011e73

   1. Initial program 34.6%

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

      1. add-sqr-sqrt 18.6%

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

      2. pow2 18.6%

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

   4. Applied egg-rr 29.2%

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

   5. Taylor expanded in k around 0 42.8%

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

      1. add-sqr-sqrt 25.5%

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

      2. sqrt-div 16.6%

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

      3. sqrt-pow1 16.6%

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

      4. metadata-eval 16.6%

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

      5. pow1 16.6%

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

      6. sqrt-div 16.6%

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

      7. sqrt-pow1 17.4%

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

      8. metadata-eval 17.4%

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

      9. pow1 17.4%

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

   7. Applied egg-rr 17.4%

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

      1. unpow2 17.4%

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

   9. Simplified 17.4%

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

   if 1.90000000000000011e73 < l

   1. Initial program 35.8%

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

      1. add-sqr-sqrt 19.1%

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

      2. pow2 19.1%

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

   4. Applied egg-rr 30.8%

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

   5. Taylor expanded in k around inf 54.5%

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

      1. associate-/l* 54.4%

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

   7. Simplified 54.4%

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

4. Final simplification 23.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;\ell \leq 1.9 \cdot 10^{+73}:\\ \;\;\;\;\frac{2}{{\left({\left(\frac{k}{\sqrt{\ell}}\right)}^{2} \cdot \sqrt{t}\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\left(k \cdot \frac{\sin k}{\ell}\right) \cdot \sqrt{\frac{t}{\cos k}}\right)}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 9: 75.9% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

l_m = Math.abs(l);
t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l_m, double k) {
	return t_s * (2.0 / Math.pow((Math.pow((k / Math.sqrt(l_m)), 2.0) * Math.sqrt(t_m)), 2.0));
}

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 34.8%

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

   1. add-sqr-sqrt 18.7%

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

   2. pow2 18.7%

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

4. Applied egg-rr 29.5%

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

5. Taylor expanded in k around 0 40.9%

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

   1. add-sqr-sqrt 26.4%

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

   2. sqrt-div 19.0%

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

   3. sqrt-pow1 19.0%

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

   4. metadata-eval 19.0%

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

   5. pow1 19.0%

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

   6. sqrt-div 19.0%

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

   7. sqrt-pow1 19.7%

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

   8. metadata-eval 19.7%

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

   9. pow1 19.7%

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

7. Applied egg-rr 19.7%

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

   1. unpow2 19.7%

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

9. Simplified 19.7%

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

10. Final simplification 19.7%

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

Alternative 10: 74.2% accurate, 1.4× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

l_m = Math.abs(l);
t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l_m, double k) {
	return t_s * (2.0 * Math.pow((Math.sqrt(t_m) * (Math.pow(k, 2.0) / l_m)), -2.0));
}

Python

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

Julia

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

MATLAB

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

Wolfram

l_m = N[Abs[l], $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$95$m_, k_] := N[(t$95$s * N[(2.0 * N[Power[N[(N[Sqrt[t$95$m], $MachinePrecision] * N[(N[Power[k, 2.0], $MachinePrecision] / l$95$m), $MachinePrecision]), $MachinePrecision], -2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 34.8%

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

   1. add-sqr-sqrt 18.7%

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

   2. pow2 18.7%

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

4. Applied egg-rr 29.5%

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

5. Taylor expanded in k around 0 40.9%

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

   1. div-inv 40.9%

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

   2. pow-flip 40.9%

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

   3. metadata-eval 40.9%

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

7. Applied egg-rr 40.9%

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

8. Final simplification 40.9%

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

Alternative 11: 74.1% accurate, 1.4× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

l_m = Math.abs(l);
t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l_m, double k) {
	return t_s * (2.0 / Math.pow((Math.sqrt(t_m) * (Math.pow(k, 2.0) / l_m)), 2.0));
}

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 34.8%

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

   1. add-sqr-sqrt 18.7%

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

   2. pow2 18.7%

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

4. Applied egg-rr 29.5%

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

5. Taylor expanded in k around 0 40.9%

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

6. Final simplification 40.9%

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

Alternative 12: 72.6% accurate, 2.0× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

l_m = Math.abs(l);
t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l_m, double k) {
	return t_s * (2.0 / (t_m * Math.pow((Math.pow(k, 2.0) / l_m), 2.0)));
}

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 34.8%

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

   1. add-sqr-sqrt 18.7%

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

   2. pow2 18.7%

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

4. Applied egg-rr 29.5%

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

5. Taylor expanded in k around 0 40.9%

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

   1. unpow-prod-down 40.2%

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

   2. pow2 40.2%

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

   3. add-sqr-sqrt 74.3%

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

7. Applied egg-rr 74.3%

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

8. Final simplification 74.3%

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

Alternative 13: 62.2% accurate, 3.8× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

l_m = Math.abs(l);
t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l_m, double k) {
	return t_s * ((l_m * l_m) * (2.0 / (t_m * Math.pow(k, 4.0))));
}

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 34.8%

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

3. Simplified 43.4%

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

5. Taylor expanded in k around 0 64.6%

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

6. Final simplification 64.6%

   \[\leadsto \left(\ell \cdot \ell\right) \cdot \frac{2}{t \cdot {k}^{4}} \]
7. Add Preprocessing

Alternative 14: 33.3% accurate, 60.1× speedup

Math

\[\begin{array}{l} l_m = \left|\ell\right| \\ t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \left(\left(l\_m \cdot l\_m\right) \cdot \frac{2}{0}\right) \end{array} \]

FPCore

l_m = (fabs.f64 l)
t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l_m k)
 :precision binary64
 (* t_s (* (* l_m l_m) (/ 2.0 0.0))))

C

l_m = fabs(l);
t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l_m, double k) {
	return t_s * ((l_m * l_m) * (2.0 / 0.0));
}

Fortran

l_m = abs(l)
t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l_m, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l_m
    real(8), intent (in) :: k
    code = t_s * ((l_m * l_m) * (2.0d0 / 0.0d0))
end function

Java

l_m = Math.abs(l);
t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l_m, double k) {
	return t_s * ((l_m * l_m) * (2.0 / 0.0));
}

Python

l_m = math.fabs(l)
t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l_m, k):
	return t_s * ((l_m * l_m) * (2.0 / 0.0))

Julia

l_m = abs(l)
t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l_m, k)
	return Float64(t_s * Float64(Float64(l_m * l_m) * Float64(2.0 / 0.0)))
end

MATLAB

l_m = abs(l);
t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp = code(t_s, t_m, l_m, k)
	tmp = t_s * ((l_m * l_m) * (2.0 / 0.0));
end

Wolfram

l_m = N[Abs[l], $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$95$m_, k_] := N[(t$95$s * N[(N[(l$95$m * l$95$m), $MachinePrecision] * N[(2.0 / 0.0), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]

Derivation

1. Initial program 34.8%

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

3. Simplified 43.4%

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

5. Taylor expanded in k around 0 64.6%

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

   1. add-log-exp 46.1%

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

   2. exp-prod 43.4%

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

7. Applied egg-rr 43.4%

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

8. Taylor expanded in k around 0 21.1%

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

9. Final simplification 21.1%

   \[\leadsto \left(\ell \cdot \ell\right) \cdot \frac{2}{0} \]
10. Add Preprocessing

Reproduce

herbie shell --seed 2024081 
(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))))