Toniolo and Linder, Equation (10+)

Percentage Accurate: 54.1% → 92.7%

Time: 17.9s

Alternatives: 20

Speedup: 3.7×

• 27.5% of points produce a very large (infinite) output. You may want to add a precondition.

Specification

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Initial Program: 54.1% accurate, 1.0× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Alternative 1: 92.7% accurate, 0.6× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 4.7 \cdot 10^{+25}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\sqrt[3]{\sin k} \cdot \frac{t\_m}{{\left(\sqrt[3]{\ell}\right)}^{2}}\right)}^{3} \cdot \left(\tan k \cdot \left(1 + \left(1 + {\left(\frac{k}{t\_m}\right)}^{2}\right)\right)\right)}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= t_m 4.7e+25)
    (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 2.0) (* (sin k) (tan k))))
    (/
     2.0
     (*
      (pow (* (cbrt (sin k)) (/ t_m (pow (cbrt l) 2.0))) 3.0)
      (* (tan k) (+ 1.0 (+ 1.0 (pow (/ k t_m) 2.0)))))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 4.7e+25) {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * (sin(k) * tan(k)));
	} else {
		tmp = 2.0 / (pow((cbrt(sin(k)) * (t_m / pow(cbrt(l), 2.0))), 3.0) * (tan(k) * (1.0 + (1.0 + pow((k / t_m), 2.0)))));
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 4.7e+25) {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k) * Math.tan(k)));
	} else {
		tmp = 2.0 / (Math.pow((Math.cbrt(Math.sin(k)) * (t_m / Math.pow(Math.cbrt(l), 2.0))), 3.0) * (Math.tan(k) * (1.0 + (1.0 + Math.pow((k / t_m), 2.0)))));
	}
	return t_s * tmp;
}

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (t_m <= 4.7e+25)
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k) * tan(k))));
	else
		tmp = Float64(2.0 / Float64((Float64(cbrt(sin(k)) * Float64(t_m / (cbrt(l) ^ 2.0))) ^ 3.0) * Float64(tan(k) * Float64(1.0 + Float64(1.0 + (Float64(k / t_m) ^ 2.0))))));
	end
	return Float64(t_s * tmp)
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[t$95$m, 4.7e+25], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[Power[N[(N[Power[N[Sin[k], $MachinePrecision], 1/3], $MachinePrecision] * N[(t$95$m / N[Power[N[Power[l, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision] * N[(N[Tan[k], $MachinePrecision] * N[(1.0 + N[(1.0 + N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if t < 4.6999999999999998e25

   1. Initial program 52.0%

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

   3. Simplified 52.0%

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

   6. Applied egg-rr 22.1%

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

      1. associate-*r* 22.1%

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

      2. unpow-prod-down 21.2%

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

      3. pow2 21.2%

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

      4. add-sqr-sqrt 27.6%

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

   8. Applied egg-rr 27.6%

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

   9. Taylor expanded in t around 0 34.3%

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

   if 4.6999999999999998e25 < t

   1. Initial program 75.5%

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

   3. Simplified 75.5%

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

      1. add-cube-cbrt 75.3%

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

      2. pow3 75.3%

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

      3. *-commutative 75.3%

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

      4. cbrt-prod 75.3%

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

      5. cbrt-div 75.3%

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

      6. rem-cbrt-cube 82.6%

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

      7. cbrt-prod 92.3%

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

      8. pow2 92.3%

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

   6. Applied egg-rr 92.3%

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

Alternative 2: 89.8% accurate, 0.7× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 3 \cdot 10^{+29}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\tan k \cdot \left(1 + \left(1 + {\left(\frac{k}{t\_m}\right)}^{2}\right)\right)\right) \cdot \left(\sin k \cdot {\left(\frac{t\_m}{{\left(\sqrt[3]{\ell}\right)}^{2}}\right)}^{3}\right)}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= t_m 3e+29)
    (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 2.0) (* (sin k) (tan k))))
    (/
     2.0
     (*
      (* (tan k) (+ 1.0 (+ 1.0 (pow (/ k t_m) 2.0))))
      (* (sin k) (pow (/ t_m (pow (cbrt l) 2.0)) 3.0)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 3e+29) {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * (sin(k) * tan(k)));
	} else {
		tmp = 2.0 / ((tan(k) * (1.0 + (1.0 + pow((k / t_m), 2.0)))) * (sin(k) * pow((t_m / pow(cbrt(l), 2.0)), 3.0)));
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 3e+29) {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k) * Math.tan(k)));
	} else {
		tmp = 2.0 / ((Math.tan(k) * (1.0 + (1.0 + Math.pow((k / t_m), 2.0)))) * (Math.sin(k) * Math.pow((t_m / Math.pow(Math.cbrt(l), 2.0)), 3.0)));
	}
	return t_s * tmp;
}

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (t_m <= 3e+29)
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k) * tan(k))));
	else
		tmp = Float64(2.0 / Float64(Float64(tan(k) * Float64(1.0 + Float64(1.0 + (Float64(k / t_m) ^ 2.0)))) * Float64(sin(k) * (Float64(t_m / (cbrt(l) ^ 2.0)) ^ 3.0))));
	end
	return Float64(t_s * tmp)
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[t$95$m, 3e+29], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[Tan[k], $MachinePrecision] * N[(1.0 + N[(1.0 + N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Power[N[(t$95$m / N[Power[N[Power[l, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if t < 2.9999999999999999e29

   1. Initial program 52.0%

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

   3. Simplified 52.0%

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

   6. Applied egg-rr 22.1%

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

      1. associate-*r* 22.1%

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

      2. unpow-prod-down 21.2%

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

      3. pow2 21.2%

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

      4. add-sqr-sqrt 27.6%

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

   8. Applied egg-rr 27.6%

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

   9. Taylor expanded in t around 0 34.3%

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

   if 2.9999999999999999e29 < t

   1. Initial program 75.5%

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

   3. Simplified 75.5%

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

      1. add-cube-cbrt 75.4%

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

      2. pow3 75.4%

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

      3. cbrt-div 75.4%

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

      4. rem-cbrt-cube 79.2%

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

      5. cbrt-prod 89.0%

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

      6. pow2 89.0%

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

   6. Applied egg-rr 89.0%

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

4. Final simplification 45.4%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 3 \cdot 10^{+29}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\tan k \cdot \left(1 + \left(1 + {\left(\frac{k}{t}\right)}^{2}\right)\right)\right) \cdot \left(\sin k \cdot {\left(\frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}}\right)}^{3}\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 3: 88.3% accurate, 0.8× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 1.1 \cdot 10^{-8}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{elif}\;t\_m \leq 4.9 \cdot 10^{+151}:\\ \;\;\;\;\frac{2}{\left(\tan k \cdot \left(1 + \left(1 + {\left(\frac{k}{t\_m}\right)}^{2}\right)\right)\right) \cdot \left(\sin k \cdot \left(\frac{{t\_m}^{2}}{\ell} \cdot \frac{t\_m}{\ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{{\left(\frac{{t\_m}^{1.5}}{\ell} \cdot \left(\sin k \cdot \sqrt{2}\right)\right)}^{2}}{\cos k}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= t_m 1.1e-8)
    (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 2.0) (* (sin k) (tan k))))
    (if (<= t_m 4.9e+151)
      (/
       2.0
       (*
        (* (tan k) (+ 1.0 (+ 1.0 (pow (/ k t_m) 2.0))))
        (* (sin k) (* (/ (pow t_m 2.0) l) (/ t_m l)))))
      (/
       2.0
       (/
        (pow (* (/ (pow t_m 1.5) l) (* (sin k) (sqrt 2.0))) 2.0)
        (cos k)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 1.1e-8) {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * (sin(k) * tan(k)));
	} else if (t_m <= 4.9e+151) {
		tmp = 2.0 / ((tan(k) * (1.0 + (1.0 + pow((k / t_m), 2.0)))) * (sin(k) * ((pow(t_m, 2.0) / l) * (t_m / l))));
	} else {
		tmp = 2.0 / (pow(((pow(t_m, 1.5) / l) * (sin(k) * sqrt(2.0))), 2.0) / cos(k));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (t_m <= 1.1d-8) then
        tmp = 2.0d0 / ((((k / l) * sqrt(t_m)) ** 2.0d0) * (sin(k) * tan(k)))
    else if (t_m <= 4.9d+151) then
        tmp = 2.0d0 / ((tan(k) * (1.0d0 + (1.0d0 + ((k / t_m) ** 2.0d0)))) * (sin(k) * (((t_m ** 2.0d0) / l) * (t_m / l))))
    else
        tmp = 2.0d0 / (((((t_m ** 1.5d0) / l) * (sin(k) * sqrt(2.0d0))) ** 2.0d0) / cos(k))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 1.1e-8) {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k) * Math.tan(k)));
	} else if (t_m <= 4.9e+151) {
		tmp = 2.0 / ((Math.tan(k) * (1.0 + (1.0 + Math.pow((k / t_m), 2.0)))) * (Math.sin(k) * ((Math.pow(t_m, 2.0) / l) * (t_m / l))));
	} else {
		tmp = 2.0 / (Math.pow(((Math.pow(t_m, 1.5) / l) * (Math.sin(k) * Math.sqrt(2.0))), 2.0) / Math.cos(k));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if t_m <= 1.1e-8:
		tmp = 2.0 / (math.pow(((k / l) * math.sqrt(t_m)), 2.0) * (math.sin(k) * math.tan(k)))
	elif t_m <= 4.9e+151:
		tmp = 2.0 / ((math.tan(k) * (1.0 + (1.0 + math.pow((k / t_m), 2.0)))) * (math.sin(k) * ((math.pow(t_m, 2.0) / l) * (t_m / l))))
	else:
		tmp = 2.0 / (math.pow(((math.pow(t_m, 1.5) / l) * (math.sin(k) * math.sqrt(2.0))), 2.0) / math.cos(k))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (t_m <= 1.1e-8)
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k) * tan(k))));
	elseif (t_m <= 4.9e+151)
		tmp = Float64(2.0 / Float64(Float64(tan(k) * Float64(1.0 + Float64(1.0 + (Float64(k / t_m) ^ 2.0)))) * Float64(sin(k) * Float64(Float64((t_m ^ 2.0) / l) * Float64(t_m / l)))));
	else
		tmp = Float64(2.0 / Float64((Float64(Float64((t_m ^ 1.5) / l) * Float64(sin(k) * sqrt(2.0))) ^ 2.0) / cos(k)));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (t_m <= 1.1e-8)
		tmp = 2.0 / ((((k / l) * sqrt(t_m)) ^ 2.0) * (sin(k) * tan(k)));
	elseif (t_m <= 4.9e+151)
		tmp = 2.0 / ((tan(k) * (1.0 + (1.0 + ((k / t_m) ^ 2.0)))) * (sin(k) * (((t_m ^ 2.0) / l) * (t_m / l))));
	else
		tmp = 2.0 / (((((t_m ^ 1.5) / l) * (sin(k) * sqrt(2.0))) ^ 2.0) / cos(k));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[t$95$m, 1.1e-8], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$m, 4.9e+151], N[(2.0 / N[(N[(N[Tan[k], $MachinePrecision] * N[(1.0 + N[(1.0 + N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[(N[(N[Power[t$95$m, 2.0], $MachinePrecision] / l), $MachinePrecision] * N[(t$95$m / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[Power[N[(N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] / N[Cos[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if t < 1.0999999999999999e-8

   1. Initial program 50.7%

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

   3. Simplified 50.7%

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

   6. Applied egg-rr 19.5%

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

      1. associate-*r* 19.5%

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

      2. unpow-prod-down 19.0%

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

      3. pow2 19.0%

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

      4. add-sqr-sqrt 24.7%

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

   8. Applied egg-rr 24.7%

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

   9. Taylor expanded in t around 0 31.8%

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

   if 1.0999999999999999e-8 < t < 4.8999999999999999e151

   1. Initial program 85.3%

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

   3. Simplified 85.2%

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

      1. unpow3 85.3%

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

      2. times-frac 94.1%

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

      3. pow2 94.1%

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

   6. Applied egg-rr 94.1%

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

   if 4.8999999999999999e151 < t

   1. Initial program 66.1%

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

   3. Simplified 66.1%

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

   6. Applied egg-rr 42.1%

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

   7. Taylor expanded in t around inf 59.4%

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

      1. *-commutative 59.4%

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

   9. Simplified 59.4%

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

       1. *-un-lft-identity 59.4%

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

       2. associate-*r* 59.4%

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

       3. unpow-prod-down 59.4%

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

       4. pow2 59.4%

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

       5. add-sqr-sqrt 84.6%

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

   11. Applied egg-rr 84.6%

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

       1. *-lft-identity 84.6%

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

       2. associate-*r/ 84.6%

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

       3. *-rgt-identity 84.6%

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

       4. *-commutative 84.6%

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

   13. Simplified 84.6%

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

4. Final simplification 45.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 1.1 \cdot 10^{-8}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{elif}\;t \leq 4.9 \cdot 10^{+151}:\\ \;\;\;\;\frac{2}{\left(\tan k \cdot \left(1 + \left(1 + {\left(\frac{k}{t}\right)}^{2}\right)\right)\right) \cdot \left(\sin k \cdot \left(\frac{{t}^{2}}{\ell} \cdot \frac{t}{\ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{{\left(\frac{{t}^{1.5}}{\ell} \cdot \left(\sin k \cdot \sqrt{2}\right)\right)}^{2}}{\cos k}}\\ \end{array} \]
5. Add Preprocessing

Alternative 4: 89.7% accurate, 0.8× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 1.16 \cdot 10^{-8}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{elif}\;t\_m \leq 4.9 \cdot 10^{+151}:\\ \;\;\;\;\frac{2}{\left(\tan k \cdot \left(1 + \left(1 + {\left(\frac{k}{t\_m}\right)}^{2}\right)\right)\right) \cdot \left(\sin k \cdot \left(\frac{{t\_m}^{2}}{\ell} \cdot \frac{t\_m}{\ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\sqrt[3]{\sin k} \cdot \frac{t\_m}{{\left(\sqrt[3]{\ell}\right)}^{2}}\right)}^{3} \cdot \left(2 \cdot k\right)}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= t_m 1.16e-8)
    (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 2.0) (* (sin k) (tan k))))
    (if (<= t_m 4.9e+151)
      (/
       2.0
       (*
        (* (tan k) (+ 1.0 (+ 1.0 (pow (/ k t_m) 2.0))))
        (* (sin k) (* (/ (pow t_m 2.0) l) (/ t_m l)))))
      (/
       2.0
       (*
        (pow (* (cbrt (sin k)) (/ t_m (pow (cbrt l) 2.0))) 3.0)
        (* 2.0 k)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 1.16e-8) {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * (sin(k) * tan(k)));
	} else if (t_m <= 4.9e+151) {
		tmp = 2.0 / ((tan(k) * (1.0 + (1.0 + pow((k / t_m), 2.0)))) * (sin(k) * ((pow(t_m, 2.0) / l) * (t_m / l))));
	} else {
		tmp = 2.0 / (pow((cbrt(sin(k)) * (t_m / pow(cbrt(l), 2.0))), 3.0) * (2.0 * k));
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 1.16e-8) {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k) * Math.tan(k)));
	} else if (t_m <= 4.9e+151) {
		tmp = 2.0 / ((Math.tan(k) * (1.0 + (1.0 + Math.pow((k / t_m), 2.0)))) * (Math.sin(k) * ((Math.pow(t_m, 2.0) / l) * (t_m / l))));
	} else {
		tmp = 2.0 / (Math.pow((Math.cbrt(Math.sin(k)) * (t_m / Math.pow(Math.cbrt(l), 2.0))), 3.0) * (2.0 * k));
	}
	return t_s * tmp;
}

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (t_m <= 1.16e-8)
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k) * tan(k))));
	elseif (t_m <= 4.9e+151)
		tmp = Float64(2.0 / Float64(Float64(tan(k) * Float64(1.0 + Float64(1.0 + (Float64(k / t_m) ^ 2.0)))) * Float64(sin(k) * Float64(Float64((t_m ^ 2.0) / l) * Float64(t_m / l)))));
	else
		tmp = Float64(2.0 / Float64((Float64(cbrt(sin(k)) * Float64(t_m / (cbrt(l) ^ 2.0))) ^ 3.0) * Float64(2.0 * k)));
	end
	return Float64(t_s * tmp)
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[t$95$m, 1.16e-8], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$m, 4.9e+151], N[(2.0 / N[(N[(N[Tan[k], $MachinePrecision] * N[(1.0 + N[(1.0 + N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[(N[(N[Power[t$95$m, 2.0], $MachinePrecision] / l), $MachinePrecision] * N[(t$95$m / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[Power[N[(N[Power[N[Sin[k], $MachinePrecision], 1/3], $MachinePrecision] * N[(t$95$m / N[Power[N[Power[l, 1/3], $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 3.0], $MachinePrecision] * N[(2.0 * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if t < 1.15999999999999996e-8

   1. Initial program 50.7%

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

   3. Simplified 50.7%

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

   6. Applied egg-rr 19.5%

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

      1. associate-*r* 19.5%

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

      2. unpow-prod-down 19.0%

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

      3. pow2 19.0%

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

      4. add-sqr-sqrt 24.7%

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

   8. Applied egg-rr 24.7%

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

   9. Taylor expanded in t around 0 31.8%

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

   if 1.15999999999999996e-8 < t < 4.8999999999999999e151

   1. Initial program 85.3%

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

   3. Simplified 85.2%

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

      1. unpow3 85.3%

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

      2. times-frac 94.1%

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

      3. pow2 94.1%

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

   6. Applied egg-rr 94.1%

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

   if 4.8999999999999999e151 < t

   1. Initial program 66.1%

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

   3. Simplified 66.1%

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

      1. add-cube-cbrt 66.1%

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

      2. pow3 66.1%

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

      3. *-commutative 66.1%

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

      4. cbrt-prod 66.1%

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

      5. cbrt-div 66.1%

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

      6. rem-cbrt-cube 76.0%

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

      7. cbrt-prod 90.5%

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

      8. pow2 90.5%

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

   6. Applied egg-rr 90.5%

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

   7. Taylor expanded in k around 0 84.3%

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

      1. *-commutative 84.3%

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

   9. Simplified 84.3%

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

4. Final simplification 45.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 1.16 \cdot 10^{-8}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{elif}\;t \leq 4.9 \cdot 10^{+151}:\\ \;\;\;\;\frac{2}{\left(\tan k \cdot \left(1 + \left(1 + {\left(\frac{k}{t}\right)}^{2}\right)\right)\right) \cdot \left(\sin k \cdot \left(\frac{{t}^{2}}{\ell} \cdot \frac{t}{\ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\sqrt[3]{\sin k} \cdot \frac{t}{{\left(\sqrt[3]{\ell}\right)}^{2}}\right)}^{3} \cdot \left(2 \cdot k\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 5: 87.9% accurate, 0.8× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 5:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot {\left(\frac{{t\_m}^{1.5}}{\ell}\right)}^{2}\right) \cdot \left(\tan k \cdot \left(2 + {\left(\frac{k}{t\_m}\right)}^{2}\right)\right)}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= t_m 5.0)
    (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 2.0) (* (sin k) (tan k))))
    (/
     2.0
     (*
      (* (sin k) (pow (/ (pow t_m 1.5) l) 2.0))
      (* (tan k) (+ 2.0 (pow (/ k t_m) 2.0))))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 5.0) {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * (sin(k) * tan(k)));
	} else {
		tmp = 2.0 / ((sin(k) * pow((pow(t_m, 1.5) / l), 2.0)) * (tan(k) * (2.0 + pow((k / t_m), 2.0))));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (t_m <= 5.0d0) then
        tmp = 2.0d0 / ((((k / l) * sqrt(t_m)) ** 2.0d0) * (sin(k) * tan(k)))
    else
        tmp = 2.0d0 / ((sin(k) * (((t_m ** 1.5d0) / l) ** 2.0d0)) * (tan(k) * (2.0d0 + ((k / t_m) ** 2.0d0))))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 5.0) {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k) * Math.tan(k)));
	} else {
		tmp = 2.0 / ((Math.sin(k) * Math.pow((Math.pow(t_m, 1.5) / l), 2.0)) * (Math.tan(k) * (2.0 + Math.pow((k / t_m), 2.0))));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if t_m <= 5.0:
		tmp = 2.0 / (math.pow(((k / l) * math.sqrt(t_m)), 2.0) * (math.sin(k) * math.tan(k)))
	else:
		tmp = 2.0 / ((math.sin(k) * math.pow((math.pow(t_m, 1.5) / l), 2.0)) * (math.tan(k) * (2.0 + math.pow((k / t_m), 2.0))))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (t_m <= 5.0)
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k) * tan(k))));
	else
		tmp = Float64(2.0 / Float64(Float64(sin(k) * (Float64((t_m ^ 1.5) / l) ^ 2.0)) * Float64(tan(k) * Float64(2.0 + (Float64(k / t_m) ^ 2.0)))));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (t_m <= 5.0)
		tmp = 2.0 / ((((k / l) * sqrt(t_m)) ^ 2.0) * (sin(k) * tan(k)));
	else
		tmp = 2.0 / ((sin(k) * (((t_m ^ 1.5) / l) ^ 2.0)) * (tan(k) * (2.0 + ((k / t_m) ^ 2.0))));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[t$95$m, 5.0], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[Sin[k], $MachinePrecision] * N[Power[N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision] * N[(N[Tan[k], $MachinePrecision] * N[(2.0 + N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if t < 5

   1. Initial program 51.0%

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

   3. Simplified 51.0%

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

   6. Applied egg-rr 20.7%

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

      1. associate-*r* 20.7%

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

      2. unpow-prod-down 19.9%

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

      3. pow2 19.9%

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

      4. add-sqr-sqrt 25.4%

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

   8. Applied egg-rr 25.4%

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

   9. Taylor expanded in t around 0 32.4%

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

   if 5 < t

   1. Initial program 76.4%

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

   3. Simplified 76.4%

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

      1. add-sqr-sqrt 76.4%

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

      2. pow2 76.4%

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

      3. sqrt-div 76.3%

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

      4. sqrt-pow1 79.7%

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

      5. metadata-eval 79.7%

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

      6. sqrt-prod 47.1%

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

      7. add-sqr-sqrt 87.2%

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

   6. Applied egg-rr 87.2%

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

      1. pow1 87.2%

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

      2. associate-+r+ 87.2%

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

      3. metadata-eval 87.2%

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

   8. Applied egg-rr 87.2%

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

      1. unpow1 87.2%

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

   10. Simplified 87.2%

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

4. Final simplification 44.8%

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

Alternative 6: 87.9% accurate, 1.0× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 1.15 \cdot 10^{-8}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{elif}\;t\_m \leq 4.8 \cdot 10^{+151}:\\ \;\;\;\;\frac{2}{\left(\tan k \cdot \left(1 + \left(1 + {\left(\frac{k}{t\_m}\right)}^{2}\right)\right)\right) \cdot \left(\sin k \cdot \left(\frac{{t\_m}^{2}}{\ell} \cdot \frac{t\_m}{\ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t\_m}^{1.5}}{\ell} \cdot \left(k \cdot \sqrt{2}\right)\right)}^{2}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= t_m 1.15e-8)
    (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 2.0) (* (sin k) (tan k))))
    (if (<= t_m 4.8e+151)
      (/
       2.0
       (*
        (* (tan k) (+ 1.0 (+ 1.0 (pow (/ k t_m) 2.0))))
        (* (sin k) (* (/ (pow t_m 2.0) l) (/ t_m l)))))
      (/ 2.0 (pow (* (/ (pow t_m 1.5) l) (* k (sqrt 2.0))) 2.0))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 1.15e-8) {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * (sin(k) * tan(k)));
	} else if (t_m <= 4.8e+151) {
		tmp = 2.0 / ((tan(k) * (1.0 + (1.0 + pow((k / t_m), 2.0)))) * (sin(k) * ((pow(t_m, 2.0) / l) * (t_m / l))));
	} else {
		tmp = 2.0 / pow(((pow(t_m, 1.5) / l) * (k * sqrt(2.0))), 2.0);
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (t_m <= 1.15d-8) then
        tmp = 2.0d0 / ((((k / l) * sqrt(t_m)) ** 2.0d0) * (sin(k) * tan(k)))
    else if (t_m <= 4.8d+151) then
        tmp = 2.0d0 / ((tan(k) * (1.0d0 + (1.0d0 + ((k / t_m) ** 2.0d0)))) * (sin(k) * (((t_m ** 2.0d0) / l) * (t_m / l))))
    else
        tmp = 2.0d0 / ((((t_m ** 1.5d0) / l) * (k * sqrt(2.0d0))) ** 2.0d0)
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 1.15e-8) {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k) * Math.tan(k)));
	} else if (t_m <= 4.8e+151) {
		tmp = 2.0 / ((Math.tan(k) * (1.0 + (1.0 + Math.pow((k / t_m), 2.0)))) * (Math.sin(k) * ((Math.pow(t_m, 2.0) / l) * (t_m / l))));
	} else {
		tmp = 2.0 / Math.pow(((Math.pow(t_m, 1.5) / l) * (k * Math.sqrt(2.0))), 2.0);
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if t_m <= 1.15e-8:
		tmp = 2.0 / (math.pow(((k / l) * math.sqrt(t_m)), 2.0) * (math.sin(k) * math.tan(k)))
	elif t_m <= 4.8e+151:
		tmp = 2.0 / ((math.tan(k) * (1.0 + (1.0 + math.pow((k / t_m), 2.0)))) * (math.sin(k) * ((math.pow(t_m, 2.0) / l) * (t_m / l))))
	else:
		tmp = 2.0 / math.pow(((math.pow(t_m, 1.5) / l) * (k * math.sqrt(2.0))), 2.0)
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (t_m <= 1.15e-8)
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k) * tan(k))));
	elseif (t_m <= 4.8e+151)
		tmp = Float64(2.0 / Float64(Float64(tan(k) * Float64(1.0 + Float64(1.0 + (Float64(k / t_m) ^ 2.0)))) * Float64(sin(k) * Float64(Float64((t_m ^ 2.0) / l) * Float64(t_m / l)))));
	else
		tmp = Float64(2.0 / (Float64(Float64((t_m ^ 1.5) / l) * Float64(k * sqrt(2.0))) ^ 2.0));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (t_m <= 1.15e-8)
		tmp = 2.0 / ((((k / l) * sqrt(t_m)) ^ 2.0) * (sin(k) * tan(k)));
	elseif (t_m <= 4.8e+151)
		tmp = 2.0 / ((tan(k) * (1.0 + (1.0 + ((k / t_m) ^ 2.0)))) * (sin(k) * (((t_m ^ 2.0) / l) * (t_m / l))));
	else
		tmp = 2.0 / ((((t_m ^ 1.5) / l) * (k * sqrt(2.0))) ^ 2.0);
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[t$95$m, 1.15e-8], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], If[LessEqual[t$95$m, 4.8e+151], N[(2.0 / N[(N[(N[Tan[k], $MachinePrecision] * N[(1.0 + N[(1.0 + N[Power[N[(k / t$95$m), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[(N[(N[Power[t$95$m, 2.0], $MachinePrecision] / l), $MachinePrecision] * N[(t$95$m / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[Power[N[(N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision] * N[(k * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if t < 1.15e-8

   1. Initial program 50.7%

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

   3. Simplified 50.7%

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

   6. Applied egg-rr 19.5%

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

      1. associate-*r* 19.5%

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

      2. unpow-prod-down 19.0%

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

      3. pow2 19.0%

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

      4. add-sqr-sqrt 24.7%

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

   8. Applied egg-rr 24.7%

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

   9. Taylor expanded in t around 0 31.8%

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

   if 1.15e-8 < t < 4.8000000000000002e151

   1. Initial program 85.3%

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

   3. Simplified 85.2%

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

      1. unpow3 85.3%

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

      2. times-frac 94.1%

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

      3. pow2 94.1%

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

   6. Applied egg-rr 94.1%

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

   if 4.8000000000000002e151 < t

   1. Initial program 66.1%

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

   3. Simplified 66.1%

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

   6. Applied egg-rr 42.1%

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

   7. Taylor expanded in k around 0 81.2%

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

4. Final simplification 45.2%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 1.15 \cdot 10^{-8}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \mathbf{elif}\;t \leq 4.8 \cdot 10^{+151}:\\ \;\;\;\;\frac{2}{\left(\tan k \cdot \left(1 + \left(1 + {\left(\frac{k}{t}\right)}^{2}\right)\right)\right) \cdot \left(\sin k \cdot \left(\frac{{t}^{2}}{\ell} \cdot \frac{t}{\ell}\right)\right)}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t}^{1.5}}{\ell} \cdot \left(k \cdot \sqrt{2}\right)\right)}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 7: 77.2% accurate, 1.0× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 1.2 \cdot 10^{-21}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t\_m}^{1.5}}{\ell} \cdot \left(k \cdot \sqrt{2}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot \tan k\right) \cdot {\left(\frac{k \cdot \sqrt{t\_m}}{-\ell}\right)}^{2}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 1.2e-21)
    (/ 2.0 (pow (* (/ (pow t_m 1.5) l) (* k (sqrt 2.0))) 2.0))
    (/ 2.0 (* (* (sin k) (tan k)) (pow (/ (* k (sqrt t_m)) (- l)) 2.0))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 1.2e-21) {
		tmp = 2.0 / pow(((pow(t_m, 1.5) / l) * (k * sqrt(2.0))), 2.0);
	} else {
		tmp = 2.0 / ((sin(k) * tan(k)) * pow(((k * sqrt(t_m)) / -l), 2.0));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 1.2d-21) then
        tmp = 2.0d0 / ((((t_m ** 1.5d0) / l) * (k * sqrt(2.0d0))) ** 2.0d0)
    else
        tmp = 2.0d0 / ((sin(k) * tan(k)) * (((k * sqrt(t_m)) / -l) ** 2.0d0))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 1.2e-21) {
		tmp = 2.0 / Math.pow(((Math.pow(t_m, 1.5) / l) * (k * Math.sqrt(2.0))), 2.0);
	} else {
		tmp = 2.0 / ((Math.sin(k) * Math.tan(k)) * Math.pow(((k * Math.sqrt(t_m)) / -l), 2.0));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 1.2e-21:
		tmp = 2.0 / math.pow(((math.pow(t_m, 1.5) / l) * (k * math.sqrt(2.0))), 2.0)
	else:
		tmp = 2.0 / ((math.sin(k) * math.tan(k)) * math.pow(((k * math.sqrt(t_m)) / -l), 2.0))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 1.2e-21)
		tmp = Float64(2.0 / (Float64(Float64((t_m ^ 1.5) / l) * Float64(k * sqrt(2.0))) ^ 2.0));
	else
		tmp = Float64(2.0 / Float64(Float64(sin(k) * tan(k)) * (Float64(Float64(k * sqrt(t_m)) / Float64(-l)) ^ 2.0)));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 1.2e-21)
		tmp = 2.0 / ((((t_m ^ 1.5) / l) * (k * sqrt(2.0))) ^ 2.0);
	else
		tmp = 2.0 / ((sin(k) * tan(k)) * (((k * sqrt(t_m)) / -l) ^ 2.0));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 1.2e-21], N[(2.0 / N[Power[N[(N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision] * N[(k * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision] * N[Power[N[(N[(k * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision] / (-l)), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 1.2e-21

   1. Initial program 62.3%

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

   3. Simplified 62.3%

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

   6. Applied egg-rr 31.6%

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

   7. Taylor expanded in k around 0 35.4%

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

   if 1.2e-21 < k

   1. Initial program 44.0%

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

   3. Simplified 44.0%

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

   6. Applied egg-rr 20.6%

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

      1. associate-*r* 20.6%

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

      2. unpow-prod-down 20.6%

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

      3. pow2 20.6%

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

      4. add-sqr-sqrt 41.5%

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

   8. Applied egg-rr 41.5%

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

   9. Taylor expanded in k around -inf 42.6%

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

       1. mul-1-neg 42.6%

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

       2. associate-*l/ 42.6%

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

       3. distribute-neg-frac2 42.6%

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

   11. Simplified 42.6%

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

4. Final simplification 37.6%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 1.2 \cdot 10^{-21}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t}^{1.5}}{\ell} \cdot \left(k \cdot \sqrt{2}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(\sin k \cdot \tan k\right) \cdot {\left(\frac{k \cdot \sqrt{t}}{-\ell}\right)}^{2}}\\ \end{array} \]
5. Add Preprocessing

Alternative 8: 77.5% accurate, 1.0× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 2.2 \cdot 10^{-21}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t\_m}^{1.5}}{\ell} \cdot \left(k \cdot \sqrt{2}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{\left(\frac{k}{\ell} \cdot \sqrt{t\_m}\right)}^{2} \cdot \left(\sin k \cdot \tan k\right)}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 2.2e-21)
    (/ 2.0 (pow (* (/ (pow t_m 1.5) l) (* k (sqrt 2.0))) 2.0))
    (/ 2.0 (* (pow (* (/ k l) (sqrt t_m)) 2.0) (* (sin k) (tan k)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 2.2e-21) {
		tmp = 2.0 / pow(((pow(t_m, 1.5) / l) * (k * sqrt(2.0))), 2.0);
	} else {
		tmp = 2.0 / (pow(((k / l) * sqrt(t_m)), 2.0) * (sin(k) * tan(k)));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 2.2d-21) then
        tmp = 2.0d0 / ((((t_m ** 1.5d0) / l) * (k * sqrt(2.0d0))) ** 2.0d0)
    else
        tmp = 2.0d0 / ((((k / l) * sqrt(t_m)) ** 2.0d0) * (sin(k) * tan(k)))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 2.2e-21) {
		tmp = 2.0 / Math.pow(((Math.pow(t_m, 1.5) / l) * (k * Math.sqrt(2.0))), 2.0);
	} else {
		tmp = 2.0 / (Math.pow(((k / l) * Math.sqrt(t_m)), 2.0) * (Math.sin(k) * Math.tan(k)));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 2.2e-21:
		tmp = 2.0 / math.pow(((math.pow(t_m, 1.5) / l) * (k * math.sqrt(2.0))), 2.0)
	else:
		tmp = 2.0 / (math.pow(((k / l) * math.sqrt(t_m)), 2.0) * (math.sin(k) * math.tan(k)))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 2.2e-21)
		tmp = Float64(2.0 / (Float64(Float64((t_m ^ 1.5) / l) * Float64(k * sqrt(2.0))) ^ 2.0));
	else
		tmp = Float64(2.0 / Float64((Float64(Float64(k / l) * sqrt(t_m)) ^ 2.0) * Float64(sin(k) * tan(k))));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 2.2e-21)
		tmp = 2.0 / ((((t_m ^ 1.5) / l) * (k * sqrt(2.0))) ^ 2.0);
	else
		tmp = 2.0 / ((((k / l) * sqrt(t_m)) ^ 2.0) * (sin(k) * tan(k)));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 2.2e-21], N[(2.0 / N[Power[N[(N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision] * N[(k * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[Power[N[(N[(k / l), $MachinePrecision] * N[Sqrt[t$95$m], $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[Tan[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 2.2000000000000001e-21

   1. Initial program 62.3%

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

   3. Simplified 62.3%

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

   6. Applied egg-rr 31.6%

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

   7. Taylor expanded in k around 0 35.4%

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

   if 2.2000000000000001e-21 < k

   1. Initial program 44.0%

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

   3. Simplified 44.0%

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

   6. Applied egg-rr 20.6%

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

      1. associate-*r* 20.6%

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

      2. unpow-prod-down 20.6%

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

      3. pow2 20.6%

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

      4. add-sqr-sqrt 41.5%

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

   8. Applied egg-rr 41.5%

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

   9. Taylor expanded in t around 0 42.6%

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

Alternative 9: 65.4% accurate, 1.3× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 5.9 \cdot 10^{-5}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t\_m}^{1.5}}{\ell} \cdot \left(k \cdot \sqrt{2}\right)\right)}^{2}}\\ \mathbf{elif}\;k \leq 5.9 \cdot 10^{+161}:\\ \;\;\;\;\left(\ell \cdot \frac{\frac{2}{\tan k \cdot {t\_m}^{3}}}{\sin k}\right) \cdot \left(\ell \cdot 0.5\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\mathsf{expm1}\left(\log t\_m \cdot \left(--3\right) - \log \ell\right)}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 5.9e-5)
    (/ 2.0 (pow (* (/ (pow t_m 1.5) l) (* k (sqrt 2.0))) 2.0))
    (if (<= k 5.9e+161)
      (* (* l (/ (/ 2.0 (* (tan k) (pow t_m 3.0))) (sin k))) (* l 0.5))
      (/
       2.0
       (*
        (/ (expm1 (- (* (log t_m) (- -3.0)) (log l))) l)
        (* 2.0 (* k k))))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 5.9e-5) {
		tmp = 2.0 / pow(((pow(t_m, 1.5) / l) * (k * sqrt(2.0))), 2.0);
	} else if (k <= 5.9e+161) {
		tmp = (l * ((2.0 / (tan(k) * pow(t_m, 3.0))) / sin(k))) * (l * 0.5);
	} else {
		tmp = 2.0 / ((expm1(((log(t_m) * -(-3.0)) - log(l))) / l) * (2.0 * (k * k)));
	}
	return t_s * tmp;
}

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 5.9e-5) {
		tmp = 2.0 / Math.pow(((Math.pow(t_m, 1.5) / l) * (k * Math.sqrt(2.0))), 2.0);
	} else if (k <= 5.9e+161) {
		tmp = (l * ((2.0 / (Math.tan(k) * Math.pow(t_m, 3.0))) / Math.sin(k))) * (l * 0.5);
	} else {
		tmp = 2.0 / ((Math.expm1(((Math.log(t_m) * -(-3.0)) - Math.log(l))) / l) * (2.0 * (k * k)));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 5.9e-5:
		tmp = 2.0 / math.pow(((math.pow(t_m, 1.5) / l) * (k * math.sqrt(2.0))), 2.0)
	elif k <= 5.9e+161:
		tmp = (l * ((2.0 / (math.tan(k) * math.pow(t_m, 3.0))) / math.sin(k))) * (l * 0.5)
	else:
		tmp = 2.0 / ((math.expm1(((math.log(t_m) * -(-3.0)) - math.log(l))) / l) * (2.0 * (k * k)))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 5.9e-5)
		tmp = Float64(2.0 / (Float64(Float64((t_m ^ 1.5) / l) * Float64(k * sqrt(2.0))) ^ 2.0));
	elseif (k <= 5.9e+161)
		tmp = Float64(Float64(l * Float64(Float64(2.0 / Float64(tan(k) * (t_m ^ 3.0))) / sin(k))) * Float64(l * 0.5));
	else
		tmp = Float64(2.0 / Float64(Float64(expm1(Float64(Float64(log(t_m) * Float64(-(-3.0))) - log(l))) / l) * Float64(2.0 * Float64(k * k))));
	end
	return Float64(t_s * tmp)
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 5.9e-5], N[(2.0 / N[Power[N[(N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision] * N[(k * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[k, 5.9e+161], N[(N[(l * N[(N[(2.0 / N[(N[Tan[k], $MachinePrecision] * N[Power[t$95$m, 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[Sin[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(l * 0.5), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(N[(Exp[N[(N[(N[Log[t$95$m], $MachinePrecision] * (--3.0)), $MachinePrecision] - N[Log[l], $MachinePrecision]), $MachinePrecision]] - 1), $MachinePrecision] / l), $MachinePrecision] * N[(2.0 * N[(k * k), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if k < 5.8999999999999998e-5

   1. Initial program 62.7%

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

   3. Simplified 62.7%

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

   6. Applied egg-rr 31.5%

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

   7. Taylor expanded in k around 0 35.1%

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

   if 5.8999999999999998e-5 < k < 5.9000000000000003e161

   1. Initial program 30.5%

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

   3. Simplified 30.5%

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

      1. associate-*r* 41.4%

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

      2. *-un-lft-identity 41.4%

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

      3. times-frac 41.3%

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

      4. associate-/l/ 41.3%

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

   6. Applied egg-rr 41.3%

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

      1. /-rgt-identity 41.3%

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

      2. *-commutative 41.3%

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

      3. metadata-eval 41.3%

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

      4. associate-*r/ 41.3%

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

      5. associate-/l/ 41.3%

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

      6. associate-*r/ 41.3%

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

      7. associate-*r/ 41.3%

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

      8. metadata-eval 41.3%

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

      9. *-lft-identity 41.3%

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

      10. *-commutative 41.3%

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

      11. times-frac 41.2%

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

      12. associate-*l/ 41.3%

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

      13. *-lft-identity 41.3%

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

      14. associate-/l/ 41.3%

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

   8. Simplified 41.3%

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

   9. Taylor expanded in k around 0 46.4%

      \[\leadsto \left(\ell \cdot \frac{\frac{2}{{t}^{3} \cdot \tan k}}{\sin k}\right) \cdot \color{blue}{\left(0.5 \cdot \ell\right)} \]
   10. Step-by-step derivation

       1. *-commutative 46.4%

          \[\leadsto \left(\ell \cdot \frac{\frac{2}{{t}^{3} \cdot \tan k}}{\sin k}\right) \cdot \color{blue}{\left(\ell \cdot 0.5\right)} \]

   11. Simplified 46.4%

       \[\leadsto \left(\ell \cdot \frac{\frac{2}{{t}^{3} \cdot \tan k}}{\sin k}\right) \cdot \color{blue}{\left(\ell \cdot 0.5\right)} \]

   if 5.9000000000000003e161 < k

   1. Initial program 49.0%

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

   3. Simplified 49.4%

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

   5. Taylor expanded in k around 0 49.4%

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

      1. unpow2 49.4%

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

   7. Applied egg-rr 49.4%

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

      1. expm1-log1p-u 34.6%

         \[\leadsto \frac{2}{\frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{{t}^{3}}{\ell}\right)\right)}}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

      2. expm1-undefine 22.2%

         \[\leadsto \frac{2}{\frac{\color{blue}{e^{\mathsf{log1p}\left(\frac{{t}^{3}}{\ell}\right)} - 1}}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

   9. Applied egg-rr 22.2%

      \[\leadsto \frac{2}{\frac{\color{blue}{e^{\mathsf{log1p}\left(\frac{{t}^{3}}{\ell}\right)} - 1}}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]
   10. Step-by-step derivation

       1. expm1-define 34.6%

          \[\leadsto \frac{2}{\frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{{t}^{3}}{\ell}\right)\right)}}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

   11. Simplified 34.6%

       \[\leadsto \frac{2}{\frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\frac{{t}^{3}}{\ell}\right)\right)}}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

   12. Taylor expanded in t around inf 27.1%

       \[\leadsto \frac{2}{\frac{\mathsf{expm1}\left(\color{blue}{\log \left(\frac{1}{\ell}\right) + -3 \cdot \log \left(\frac{1}{t}\right)}\right)}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]
   13. Step-by-step derivation

       1. log-rec 27.1%

          \[\leadsto \frac{2}{\frac{\mathsf{expm1}\left(\color{blue}{\left(-\log \ell\right)} + -3 \cdot \log \left(\frac{1}{t}\right)\right)}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

       2. mul-1-neg 27.1%

          \[\leadsto \frac{2}{\frac{\mathsf{expm1}\left(\color{blue}{-1 \cdot \log \ell} + -3 \cdot \log \left(\frac{1}{t}\right)\right)}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

       3. +-commutative 27.1%

          \[\leadsto \frac{2}{\frac{\mathsf{expm1}\left(\color{blue}{-3 \cdot \log \left(\frac{1}{t}\right) + -1 \cdot \log \ell}\right)}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

       4. log-rec 27.1%

          \[\leadsto \frac{2}{\frac{\mathsf{expm1}\left(-3 \cdot \color{blue}{\left(-\log t\right)} + -1 \cdot \log \ell\right)}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

       5. mul-1-neg 27.1%

          \[\leadsto \frac{2}{\frac{\mathsf{expm1}\left(-3 \cdot \left(-\log t\right) + \color{blue}{\left(-\log \ell\right)}\right)}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]

   14. Simplified 27.1%

       \[\leadsto \frac{2}{\frac{\mathsf{expm1}\left(\color{blue}{-3 \cdot \left(-\log t\right) + \left(-\log \ell\right)}\right)}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)} \]
3. Recombined 3 regimes into one program.

4. Final simplification 35.1%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 5.9 \cdot 10^{-5}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t}^{1.5}}{\ell} \cdot \left(k \cdot \sqrt{2}\right)\right)}^{2}}\\ \mathbf{elif}\;k \leq 5.9 \cdot 10^{+161}:\\ \;\;\;\;\left(\ell \cdot \frac{\frac{2}{\tan k \cdot {t}^{3}}}{\sin k}\right) \cdot \left(\ell \cdot 0.5\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{\mathsf{expm1}\left(\log t \cdot \left(--3\right) - \log \ell\right)}{\ell} \cdot \left(2 \cdot \left(k \cdot k\right)\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 10: 68.9% accurate, 1.3× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 4.6 \cdot 10^{-5}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t\_m}^{1.5}}{\ell} \cdot \left(k \cdot \sqrt{2}\right)\right)}^{2}}\\ \mathbf{elif}\;k \leq 5.9 \cdot 10^{+161}:\\ \;\;\;\;\left(\ell \cdot \frac{\frac{2}{\tan k \cdot {t\_m}^{3}}}{\sin k}\right) \cdot \left(\ell \cdot 0.5\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{t\_m \cdot {k}^{4}}{{\ell}^{2}}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 4.6e-5)
    (/ 2.0 (pow (* (/ (pow t_m 1.5) l) (* k (sqrt 2.0))) 2.0))
    (if (<= k 5.9e+161)
      (* (* l (/ (/ 2.0 (* (tan k) (pow t_m 3.0))) (sin k))) (* l 0.5))
      (/ 2.0 (/ (* t_m (pow k 4.0)) (pow l 2.0)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 4.6e-5) {
		tmp = 2.0 / pow(((pow(t_m, 1.5) / l) * (k * sqrt(2.0))), 2.0);
	} else if (k <= 5.9e+161) {
		tmp = (l * ((2.0 / (tan(k) * pow(t_m, 3.0))) / sin(k))) * (l * 0.5);
	} else {
		tmp = 2.0 / ((t_m * pow(k, 4.0)) / pow(l, 2.0));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 4.6d-5) then
        tmp = 2.0d0 / ((((t_m ** 1.5d0) / l) * (k * sqrt(2.0d0))) ** 2.0d0)
    else if (k <= 5.9d+161) then
        tmp = (l * ((2.0d0 / (tan(k) * (t_m ** 3.0d0))) / sin(k))) * (l * 0.5d0)
    else
        tmp = 2.0d0 / ((t_m * (k ** 4.0d0)) / (l ** 2.0d0))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 4.6e-5) {
		tmp = 2.0 / Math.pow(((Math.pow(t_m, 1.5) / l) * (k * Math.sqrt(2.0))), 2.0);
	} else if (k <= 5.9e+161) {
		tmp = (l * ((2.0 / (Math.tan(k) * Math.pow(t_m, 3.0))) / Math.sin(k))) * (l * 0.5);
	} else {
		tmp = 2.0 / ((t_m * Math.pow(k, 4.0)) / Math.pow(l, 2.0));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 4.6e-5:
		tmp = 2.0 / math.pow(((math.pow(t_m, 1.5) / l) * (k * math.sqrt(2.0))), 2.0)
	elif k <= 5.9e+161:
		tmp = (l * ((2.0 / (math.tan(k) * math.pow(t_m, 3.0))) / math.sin(k))) * (l * 0.5)
	else:
		tmp = 2.0 / ((t_m * math.pow(k, 4.0)) / math.pow(l, 2.0))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 4.6e-5)
		tmp = Float64(2.0 / (Float64(Float64((t_m ^ 1.5) / l) * Float64(k * sqrt(2.0))) ^ 2.0));
	elseif (k <= 5.9e+161)
		tmp = Float64(Float64(l * Float64(Float64(2.0 / Float64(tan(k) * (t_m ^ 3.0))) / sin(k))) * Float64(l * 0.5));
	else
		tmp = Float64(2.0 / Float64(Float64(t_m * (k ^ 4.0)) / (l ^ 2.0)));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 4.6e-5)
		tmp = 2.0 / ((((t_m ^ 1.5) / l) * (k * sqrt(2.0))) ^ 2.0);
	elseif (k <= 5.9e+161)
		tmp = (l * ((2.0 / (tan(k) * (t_m ^ 3.0))) / sin(k))) * (l * 0.5);
	else
		tmp = 2.0 / ((t_m * (k ^ 4.0)) / (l ^ 2.0));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 4.6e-5], N[(2.0 / N[Power[N[(N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision] * N[(k * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], If[LessEqual[k, 5.9e+161], N[(N[(l * N[(N[(2.0 / N[(N[Tan[k], $MachinePrecision] * N[Power[t$95$m, 3.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / N[Sin[k], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] * N[(l * 0.5), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(t$95$m * N[Power[k, 4.0], $MachinePrecision]), $MachinePrecision] / N[Power[l, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]]), $MachinePrecision]

Derivation

1. Split input into 3 regimes

2. if k < 4.6e-5

   1. Initial program 62.7%

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

   3. Simplified 62.7%

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

   6. Applied egg-rr 31.5%

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

   7. Taylor expanded in k around 0 35.1%

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

   if 4.6e-5 < k < 5.9000000000000003e161

   1. Initial program 30.5%

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

   3. Simplified 30.5%

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

      1. associate-*r* 41.4%

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

      2. *-un-lft-identity 41.4%

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

      3. times-frac 41.3%

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

      4. associate-/l/ 41.3%

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

   6. Applied egg-rr 41.3%

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

      1. /-rgt-identity 41.3%

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

      2. *-commutative 41.3%

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

      3. metadata-eval 41.3%

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

      4. associate-*r/ 41.3%

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

      5. associate-/l/ 41.3%

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

      6. associate-*r/ 41.3%

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

      7. associate-*r/ 41.3%

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

      8. metadata-eval 41.3%

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

      9. *-lft-identity 41.3%

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

      10. *-commutative 41.3%

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

      11. times-frac 41.2%

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

      12. associate-*l/ 41.3%

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

      13. *-lft-identity 41.3%

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

      14. associate-/l/ 41.3%

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

   8. Simplified 41.3%

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

   9. Taylor expanded in k around 0 46.4%

      \[\leadsto \left(\ell \cdot \frac{\frac{2}{{t}^{3} \cdot \tan k}}{\sin k}\right) \cdot \color{blue}{\left(0.5 \cdot \ell\right)} \]
   10. Step-by-step derivation

       1. *-commutative 46.4%

          \[\leadsto \left(\ell \cdot \frac{\frac{2}{{t}^{3} \cdot \tan k}}{\sin k}\right) \cdot \color{blue}{\left(\ell \cdot 0.5\right)} \]

   11. Simplified 46.4%

       \[\leadsto \left(\ell \cdot \frac{\frac{2}{{t}^{3} \cdot \tan k}}{\sin k}\right) \cdot \color{blue}{\left(\ell \cdot 0.5\right)} \]

   if 5.9000000000000003e161 < k

   1. Initial program 49.0%

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

   3. Simplified 49.0%

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

   5. Taylor expanded in t around 0 66.7%

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

      1. associate-*r* 66.7%

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

      2. times-frac 66.7%

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

   7. Simplified 66.7%

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

   8. Taylor expanded in k around 0 66.7%

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

4. Final simplification 41.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 4.6 \cdot 10^{-5}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t}^{1.5}}{\ell} \cdot \left(k \cdot \sqrt{2}\right)\right)}^{2}}\\ \mathbf{elif}\;k \leq 5.9 \cdot 10^{+161}:\\ \;\;\;\;\left(\ell \cdot \frac{\frac{2}{\tan k \cdot {t}^{3}}}{\sin k}\right) \cdot \left(\ell \cdot 0.5\right)\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{t \cdot {k}^{4}}{{\ell}^{2}}}\\ \end{array} \]
5. Add Preprocessing

Alternative 11: 68.6% accurate, 1.3× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 2.1 \cdot 10^{-9}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t\_m}^{1.5}}{\ell} \cdot \left(k \cdot \sqrt{2}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{t\_m \cdot {k}^{4}}{{\ell}^{2}}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 2.1e-9)
    (/ 2.0 (pow (* (/ (pow t_m 1.5) l) (* k (sqrt 2.0))) 2.0))
    (/ 2.0 (/ (* t_m (pow k 4.0)) (pow l 2.0))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 2.1e-9) {
		tmp = 2.0 / pow(((pow(t_m, 1.5) / l) * (k * sqrt(2.0))), 2.0);
	} else {
		tmp = 2.0 / ((t_m * pow(k, 4.0)) / pow(l, 2.0));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 2.1d-9) then
        tmp = 2.0d0 / ((((t_m ** 1.5d0) / l) * (k * sqrt(2.0d0))) ** 2.0d0)
    else
        tmp = 2.0d0 / ((t_m * (k ** 4.0d0)) / (l ** 2.0d0))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 2.1e-9) {
		tmp = 2.0 / Math.pow(((Math.pow(t_m, 1.5) / l) * (k * Math.sqrt(2.0))), 2.0);
	} else {
		tmp = 2.0 / ((t_m * Math.pow(k, 4.0)) / Math.pow(l, 2.0));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 2.1e-9:
		tmp = 2.0 / math.pow(((math.pow(t_m, 1.5) / l) * (k * math.sqrt(2.0))), 2.0)
	else:
		tmp = 2.0 / ((t_m * math.pow(k, 4.0)) / math.pow(l, 2.0))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 2.1e-9)
		tmp = Float64(2.0 / (Float64(Float64((t_m ^ 1.5) / l) * Float64(k * sqrt(2.0))) ^ 2.0));
	else
		tmp = Float64(2.0 / Float64(Float64(t_m * (k ^ 4.0)) / (l ^ 2.0)));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 2.1e-9)
		tmp = 2.0 / ((((t_m ^ 1.5) / l) * (k * sqrt(2.0))) ^ 2.0);
	else
		tmp = 2.0 / ((t_m * (k ^ 4.0)) / (l ^ 2.0));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 2.1e-9], N[(2.0 / N[Power[N[(N[(N[Power[t$95$m, 1.5], $MachinePrecision] / l), $MachinePrecision] * N[(k * N[Sqrt[2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], 2.0], $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(t$95$m * N[Power[k, 4.0], $MachinePrecision]), $MachinePrecision] / N[Power[l, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 2.10000000000000019e-9

   1. Initial program 63.1%

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

   3. Simplified 63.1%

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

   6. Applied egg-rr 31.7%

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

   7. Taylor expanded in k around 0 35.3%

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

   if 2.10000000000000019e-9 < k

   1. Initial program 40.6%

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

   3. Simplified 40.6%

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

   5. Taylor expanded in t around 0 66.3%

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

      1. associate-*r* 66.3%

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

      2. times-frac 66.4%

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

   7. Simplified 66.4%

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

   8. Taylor expanded in k around 0 53.9%

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

4. Final simplification 40.5%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 2.1 \cdot 10^{-9}:\\ \;\;\;\;\frac{2}{{\left(\frac{{t}^{1.5}}{\ell} \cdot \left(k \cdot \sqrt{2}\right)\right)}^{2}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{t \cdot {k}^{4}}{{\ell}^{2}}}\\ \end{array} \]
5. Add Preprocessing

Alternative 12: 59.1% accurate, 1.9× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;t\_m \leq 1.42 \cdot 10^{-33}:\\ \;\;\;\;\frac{2}{\frac{t\_m \cdot {k}^{4}}{{\ell}^{2}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(2 \cdot k\right) \cdot \left(\sin k \cdot \frac{{t\_m}^{3}}{\ell \cdot \ell}\right)}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= t_m 1.42e-33)
    (/ 2.0 (/ (* t_m (pow k 4.0)) (pow l 2.0)))
    (/ 2.0 (* (* 2.0 k) (* (sin k) (/ (pow t_m 3.0) (* l l))))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 1.42e-33) {
		tmp = 2.0 / ((t_m * pow(k, 4.0)) / pow(l, 2.0));
	} else {
		tmp = 2.0 / ((2.0 * k) * (sin(k) * (pow(t_m, 3.0) / (l * l))));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (t_m <= 1.42d-33) then
        tmp = 2.0d0 / ((t_m * (k ** 4.0d0)) / (l ** 2.0d0))
    else
        tmp = 2.0d0 / ((2.0d0 * k) * (sin(k) * ((t_m ** 3.0d0) / (l * l))))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (t_m <= 1.42e-33) {
		tmp = 2.0 / ((t_m * Math.pow(k, 4.0)) / Math.pow(l, 2.0));
	} else {
		tmp = 2.0 / ((2.0 * k) * (Math.sin(k) * (Math.pow(t_m, 3.0) / (l * l))));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if t_m <= 1.42e-33:
		tmp = 2.0 / ((t_m * math.pow(k, 4.0)) / math.pow(l, 2.0))
	else:
		tmp = 2.0 / ((2.0 * k) * (math.sin(k) * (math.pow(t_m, 3.0) / (l * l))))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (t_m <= 1.42e-33)
		tmp = Float64(2.0 / Float64(Float64(t_m * (k ^ 4.0)) / (l ^ 2.0)));
	else
		tmp = Float64(2.0 / Float64(Float64(2.0 * k) * Float64(sin(k) * Float64((t_m ^ 3.0) / Float64(l * l)))));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (t_m <= 1.42e-33)
		tmp = 2.0 / ((t_m * (k ^ 4.0)) / (l ^ 2.0));
	else
		tmp = 2.0 / ((2.0 * k) * (sin(k) * ((t_m ^ 3.0) / (l * l))));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[t$95$m, 1.42e-33], N[(2.0 / N[(N[(t$95$m * N[Power[k, 4.0], $MachinePrecision]), $MachinePrecision] / N[Power[l, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(2.0 * k), $MachinePrecision] * N[(N[Sin[k], $MachinePrecision] * N[(N[Power[t$95$m, 3.0], $MachinePrecision] / N[(l * l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if t < 1.42000000000000007e-33

   1. Initial program 50.1%

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

   3. Simplified 50.1%

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

   5. Taylor expanded in t around 0 64.4%

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

      1. associate-*r* 64.5%

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

      2. times-frac 65.5%

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

   7. Simplified 65.5%

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

   8. Taylor expanded in k around 0 52.4%

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

   if 1.42000000000000007e-33 < t

   1. Initial program 75.4%

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

   3. Simplified 75.4%

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

   5. Taylor expanded in k around 0 67.3%

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

      1. *-commutative 77.7%

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

   7. Simplified 67.3%

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

4. Final simplification 56.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;t \leq 1.42 \cdot 10^{-33}:\\ \;\;\;\;\frac{2}{\frac{t \cdot {k}^{4}}{{\ell}^{2}}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\left(2 \cdot k\right) \cdot \left(\sin k \cdot \frac{{t}^{3}}{\ell \cdot \ell}\right)}\\ \end{array} \]
5. Add Preprocessing

Alternative 13: 56.8% accurate, 1.9× speedup

Math

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

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 5.8e-10)
    (/ 2.0 (/ (* (/ (pow t_m 3.0) l) (* 2.0 (pow k 2.0))) l))
    (/ 2.0 (/ (* t_m (pow k 4.0)) (pow l 2.0))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 5.8e-10) {
		tmp = 2.0 / (((pow(t_m, 3.0) / l) * (2.0 * pow(k, 2.0))) / l);
	} else {
		tmp = 2.0 / ((t_m * pow(k, 4.0)) / pow(l, 2.0));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 5.8d-10) then
        tmp = 2.0d0 / ((((t_m ** 3.0d0) / l) * (2.0d0 * (k ** 2.0d0))) / l)
    else
        tmp = 2.0d0 / ((t_m * (k ** 4.0d0)) / (l ** 2.0d0))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 5.8e-10) {
		tmp = 2.0 / (((Math.pow(t_m, 3.0) / l) * (2.0 * Math.pow(k, 2.0))) / l);
	} else {
		tmp = 2.0 / ((t_m * Math.pow(k, 4.0)) / Math.pow(l, 2.0));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 5.8e-10:
		tmp = 2.0 / (((math.pow(t_m, 3.0) / l) * (2.0 * math.pow(k, 2.0))) / l)
	else:
		tmp = 2.0 / ((t_m * math.pow(k, 4.0)) / math.pow(l, 2.0))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 5.8e-10)
		tmp = Float64(2.0 / Float64(Float64(Float64((t_m ^ 3.0) / l) * Float64(2.0 * (k ^ 2.0))) / l));
	else
		tmp = Float64(2.0 / Float64(Float64(t_m * (k ^ 4.0)) / (l ^ 2.0)));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 5.8e-10)
		tmp = 2.0 / ((((t_m ^ 3.0) / l) * (2.0 * (k ^ 2.0))) / l);
	else
		tmp = 2.0 / ((t_m * (k ^ 4.0)) / (l ^ 2.0));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 5.8e-10], N[(2.0 / N[(N[(N[(N[Power[t$95$m, 3.0], $MachinePrecision] / l), $MachinePrecision] * N[(2.0 * N[Power[k, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(t$95$m * N[Power[k, 4.0], $MachinePrecision]), $MachinePrecision] / N[Power[l, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 5.79999999999999962e-10

   1. Initial program 63.1%

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

   3. Simplified 60.9%

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

   5. Taylor expanded in k around 0 57.6%

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

      1. associate-*l/ 58.6%

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

   7. Applied egg-rr 58.6%

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

   if 5.79999999999999962e-10 < k

   1. Initial program 40.6%

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

   3. Simplified 40.6%

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

   5. Taylor expanded in t around 0 66.3%

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

      1. associate-*r* 66.3%

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

      2. times-frac 66.4%

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

   7. Simplified 66.4%

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

   8. Taylor expanded in k around 0 53.9%

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

4. Final simplification 57.3%

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

Alternative 14: 56.6% accurate, 1.9× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 1.82 \cdot 10^{-9}:\\ \;\;\;\;\frac{2}{\frac{{t\_m}^{3}}{\ell} \cdot \frac{2 \cdot {k}^{2}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{t\_m \cdot {k}^{4}}{{\ell}^{2}}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 1.82e-9)
    (/ 2.0 (* (/ (pow t_m 3.0) l) (/ (* 2.0 (pow k 2.0)) l)))
    (/ 2.0 (/ (* t_m (pow k 4.0)) (pow l 2.0))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 1.82e-9) {
		tmp = 2.0 / ((pow(t_m, 3.0) / l) * ((2.0 * pow(k, 2.0)) / l));
	} else {
		tmp = 2.0 / ((t_m * pow(k, 4.0)) / pow(l, 2.0));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 1.82d-9) then
        tmp = 2.0d0 / (((t_m ** 3.0d0) / l) * ((2.0d0 * (k ** 2.0d0)) / l))
    else
        tmp = 2.0d0 / ((t_m * (k ** 4.0d0)) / (l ** 2.0d0))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 1.82e-9) {
		tmp = 2.0 / ((Math.pow(t_m, 3.0) / l) * ((2.0 * Math.pow(k, 2.0)) / l));
	} else {
		tmp = 2.0 / ((t_m * Math.pow(k, 4.0)) / Math.pow(l, 2.0));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 1.82e-9:
		tmp = 2.0 / ((math.pow(t_m, 3.0) / l) * ((2.0 * math.pow(k, 2.0)) / l))
	else:
		tmp = 2.0 / ((t_m * math.pow(k, 4.0)) / math.pow(l, 2.0))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 1.82e-9)
		tmp = Float64(2.0 / Float64(Float64((t_m ^ 3.0) / l) * Float64(Float64(2.0 * (k ^ 2.0)) / l)));
	else
		tmp = Float64(2.0 / Float64(Float64(t_m * (k ^ 4.0)) / (l ^ 2.0)));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 1.82e-9)
		tmp = 2.0 / (((t_m ^ 3.0) / l) * ((2.0 * (k ^ 2.0)) / l));
	else
		tmp = 2.0 / ((t_m * (k ^ 4.0)) / (l ^ 2.0));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 1.82e-9], N[(2.0 / N[(N[(N[Power[t$95$m, 3.0], $MachinePrecision] / l), $MachinePrecision] * N[(N[(2.0 * N[Power[k, 2.0], $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(t$95$m * N[Power[k, 4.0], $MachinePrecision]), $MachinePrecision] / N[Power[l, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 1.8199999999999999e-9

   1. Initial program 63.1%

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

   3. Simplified 60.9%

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

   5. Taylor expanded in k around 0 57.6%

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

      1. associate-*l/ 58.6%

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

   7. Applied egg-rr 58.6%

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

      1. associate-/l* 58.6%

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

   9. Simplified 58.6%

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

   if 1.8199999999999999e-9 < k

   1. Initial program 40.6%

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

   3. Simplified 40.6%

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

   5. Taylor expanded in t around 0 66.3%

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

      1. associate-*r* 66.3%

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

      2. times-frac 66.4%

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

   7. Simplified 66.4%

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

   8. Taylor expanded in k around 0 53.9%

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

4. Final simplification 57.3%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 1.82 \cdot 10^{-9}:\\ \;\;\;\;\frac{2}{\frac{{t}^{3}}{\ell} \cdot \frac{2 \cdot {k}^{2}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{t \cdot {k}^{4}}{{\ell}^{2}}}\\ \end{array} \]
5. Add Preprocessing

Alternative 15: 58.4% accurate, 2.0× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 2.8 \cdot 10^{-9}:\\ \;\;\;\;\frac{2}{\left(2 \cdot \left(k \cdot k\right)\right) \cdot \frac{t\_m \cdot \frac{{t\_m}^{2}}{\ell}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{t\_m \cdot {k}^{4}}{{\ell}^{2}}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 2.8e-9)
    (/ 2.0 (* (* 2.0 (* k k)) (/ (* t_m (/ (pow t_m 2.0) l)) l)))
    (/ 2.0 (/ (* t_m (pow k 4.0)) (pow l 2.0))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 2.8e-9) {
		tmp = 2.0 / ((2.0 * (k * k)) * ((t_m * (pow(t_m, 2.0) / l)) / l));
	} else {
		tmp = 2.0 / ((t_m * pow(k, 4.0)) / pow(l, 2.0));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 2.8d-9) then
        tmp = 2.0d0 / ((2.0d0 * (k * k)) * ((t_m * ((t_m ** 2.0d0) / l)) / l))
    else
        tmp = 2.0d0 / ((t_m * (k ** 4.0d0)) / (l ** 2.0d0))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 2.8e-9) {
		tmp = 2.0 / ((2.0 * (k * k)) * ((t_m * (Math.pow(t_m, 2.0) / l)) / l));
	} else {
		tmp = 2.0 / ((t_m * Math.pow(k, 4.0)) / Math.pow(l, 2.0));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 2.8e-9:
		tmp = 2.0 / ((2.0 * (k * k)) * ((t_m * (math.pow(t_m, 2.0) / l)) / l))
	else:
		tmp = 2.0 / ((t_m * math.pow(k, 4.0)) / math.pow(l, 2.0))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 2.8e-9)
		tmp = Float64(2.0 / Float64(Float64(2.0 * Float64(k * k)) * Float64(Float64(t_m * Float64((t_m ^ 2.0) / l)) / l)));
	else
		tmp = Float64(2.0 / Float64(Float64(t_m * (k ^ 4.0)) / (l ^ 2.0)));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 2.8e-9)
		tmp = 2.0 / ((2.0 * (k * k)) * ((t_m * ((t_m ^ 2.0) / l)) / l));
	else
		tmp = 2.0 / ((t_m * (k ^ 4.0)) / (l ^ 2.0));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 2.8e-9], N[(2.0 / N[(N[(2.0 * N[(k * k), $MachinePrecision]), $MachinePrecision] * N[(N[(t$95$m * N[(N[Power[t$95$m, 2.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[(t$95$m * N[Power[k, 4.0], $MachinePrecision]), $MachinePrecision] / N[Power[l, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 2.79999999999999984e-9

   1. Initial program 63.1%

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

   3. Simplified 60.9%

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

   5. Taylor expanded in k around 0 57.6%

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

      1. unpow2 57.6%

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

   7. Applied egg-rr 57.6%

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

      1. cube-mult 57.6%

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

      2. *-un-lft-identity 57.6%

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

      3. times-frac 58.2%

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

      4. pow2 58.2%

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

   9. Applied egg-rr 58.2%

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

   if 2.79999999999999984e-9 < k

   1. Initial program 40.6%

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

   3. Simplified 40.6%

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

   5. Taylor expanded in t around 0 66.3%

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

      1. associate-*r* 66.3%

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

      2. times-frac 66.4%

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

   7. Simplified 66.4%

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

   8. Taylor expanded in k around 0 53.9%

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

4. Final simplification 57.0%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 2.8 \cdot 10^{-9}:\\ \;\;\;\;\frac{2}{\left(2 \cdot \left(k \cdot k\right)\right) \cdot \frac{t \cdot \frac{{t}^{2}}{\ell}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{\frac{t \cdot {k}^{4}}{{\ell}^{2}}}\\ \end{array} \]
5. Add Preprocessing

Alternative 16: 58.4% accurate, 2.0× speedup

Math

\[\begin{array}{l} t\_m = \left|t\right| \\ t\_s = \mathsf{copysign}\left(1, t\right) \\ t\_s \cdot \begin{array}{l} \mathbf{if}\;k \leq 2.7 \cdot 10^{-9}:\\ \;\;\;\;\frac{2}{\left(2 \cdot \left(k \cdot k\right)\right) \cdot \frac{t\_m \cdot \frac{{t\_m}^{2}}{\ell}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{k}^{4} \cdot \frac{t\_m}{{\ell}^{2}}}\\ \end{array} \end{array} \]

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (*
  t_s
  (if (<= k 2.7e-9)
    (/ 2.0 (* (* 2.0 (* k k)) (/ (* t_m (/ (pow t_m 2.0) l)) l)))
    (/ 2.0 (* (pow k 4.0) (/ t_m (pow l 2.0)))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 2.7e-9) {
		tmp = 2.0 / ((2.0 * (k * k)) * ((t_m * (pow(t_m, 2.0) / l)) / l));
	} else {
		tmp = 2.0 / (pow(k, 4.0) * (t_m / pow(l, 2.0)));
	}
	return t_s * tmp;
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    real(8) :: tmp
    if (k <= 2.7d-9) then
        tmp = 2.0d0 / ((2.0d0 * (k * k)) * ((t_m * ((t_m ** 2.0d0) / l)) / l))
    else
        tmp = 2.0d0 / ((k ** 4.0d0) * (t_m / (l ** 2.0d0)))
    end if
    code = t_s * tmp
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	double tmp;
	if (k <= 2.7e-9) {
		tmp = 2.0 / ((2.0 * (k * k)) * ((t_m * (Math.pow(t_m, 2.0) / l)) / l));
	} else {
		tmp = 2.0 / (Math.pow(k, 4.0) * (t_m / Math.pow(l, 2.0)));
	}
	return t_s * tmp;
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	tmp = 0
	if k <= 2.7e-9:
		tmp = 2.0 / ((2.0 * (k * k)) * ((t_m * (math.pow(t_m, 2.0) / l)) / l))
	else:
		tmp = 2.0 / (math.pow(k, 4.0) * (t_m / math.pow(l, 2.0)))
	return t_s * tmp

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	tmp = 0.0
	if (k <= 2.7e-9)
		tmp = Float64(2.0 / Float64(Float64(2.0 * Float64(k * k)) * Float64(Float64(t_m * Float64((t_m ^ 2.0) / l)) / l)));
	else
		tmp = Float64(2.0 / Float64((k ^ 4.0) * Float64(t_m / (l ^ 2.0))));
	end
	return Float64(t_s * tmp)
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp_2 = code(t_s, t_m, l, k)
	tmp = 0.0;
	if (k <= 2.7e-9)
		tmp = 2.0 / ((2.0 * (k * k)) * ((t_m * ((t_m ^ 2.0) / l)) / l));
	else
		tmp = 2.0 / ((k ^ 4.0) * (t_m / (l ^ 2.0)));
	end
	tmp_2 = t_s * tmp;
end

Wolfram

t\_m = N[Abs[t], $MachinePrecision]
t\_s = N[With[{TMP1 = Abs[1.0], TMP2 = Sign[t]}, TMP1 * If[TMP2 == 0, 1, TMP2]], $MachinePrecision]
code[t$95$s_, t$95$m_, l_, k_] := N[(t$95$s * If[LessEqual[k, 2.7e-9], N[(2.0 / N[(N[(2.0 * N[(k * k), $MachinePrecision]), $MachinePrecision] * N[(N[(t$95$m * N[(N[Power[t$95$m, 2.0], $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision] / l), $MachinePrecision]), $MachinePrecision]), $MachinePrecision], N[(2.0 / N[(N[Power[k, 4.0], $MachinePrecision] * N[(t$95$m / N[Power[l, 2.0], $MachinePrecision]), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]]), $MachinePrecision]

Derivation

1. Split input into 2 regimes

2. if k < 2.7000000000000002e-9

   1. Initial program 63.1%

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

   3. Simplified 60.9%

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

   5. Taylor expanded in k around 0 57.6%

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

      1. unpow2 57.6%

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

   7. Applied egg-rr 57.6%

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

      1. cube-mult 57.6%

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

      2. *-un-lft-identity 57.6%

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

      3. times-frac 58.2%

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

      4. pow2 58.2%

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

   9. Applied egg-rr 58.2%

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

   if 2.7000000000000002e-9 < k

   1. Initial program 40.6%

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

   3. Simplified 40.6%

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

   5. Taylor expanded in t around 0 66.3%

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

      1. associate-*r* 66.3%

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

      2. times-frac 66.4%

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

   7. Simplified 66.4%

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

   8. Taylor expanded in k around 0 53.9%

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

      1. associate-/l* 53.8%

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

   10. Simplified 53.8%

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

4. Final simplification 56.9%

   \[\leadsto \begin{array}{l} \mathbf{if}\;k \leq 2.7 \cdot 10^{-9}:\\ \;\;\;\;\frac{2}{\left(2 \cdot \left(k \cdot k\right)\right) \cdot \frac{t \cdot \frac{{t}^{2}}{\ell}}{\ell}}\\ \mathbf{else}:\\ \;\;\;\;\frac{2}{{k}^{4} \cdot \frac{t}{{\ell}^{2}}}\\ \end{array} \]
5. Add Preprocessing

Alternative 17: 56.8% accurate, 3.6× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 56.7%

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

3. Simplified 56.5%

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

5. Taylor expanded in k around 0 53.1%

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

   1. unpow2 53.1%

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

7. Applied egg-rr 53.1%

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

   1. cube-mult 53.1%

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

   2. *-un-lft-identity 53.1%

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

   3. times-frac 54.3%

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

   4. pow2 54.3%

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

9. Applied egg-rr 54.3%

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

10. Final simplification 54.3%

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

Alternative 18: 54.2% accurate, 3.6× speedup

Math

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

FPCore

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

C

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

Fortran

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

Java

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

Python

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

Julia

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

MATLAB

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

Wolfram

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

Derivation

1. Initial program 56.7%

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

3. Simplified 56.5%

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

5. Taylor expanded in k around 0 53.1%

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

   1. unpow2 53.1%

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

7. Applied egg-rr 53.1%

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

   1. unpow3 53.1%

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

   2. pow2 53.1%

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

9. Applied egg-rr 53.1%

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

10. Final simplification 53.1%

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

Alternative 19: 54.2% accurate, 3.6× speedup

Math

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

FPCore

t\_m = (fabs.f64 t)
t\_s = (copysign.f64 #s(literal 1 binary64) t)
(FPCore (t_s t_m l k)
 :precision binary64
 (* t_s (/ 2.0 (* (* 2.0 (* k k)) (* (/ (pow t_m 3.0) l) (/ 1.0 l))))))

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	return t_s * (2.0 / ((2.0 * (k * k)) * ((pow(t_m, 3.0) / l) * (1.0 / l))));
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    code = t_s * (2.0d0 / ((2.0d0 * (k * k)) * (((t_m ** 3.0d0) / l) * (1.0d0 / l))))
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	return t_s * (2.0 / ((2.0 * (k * k)) * ((Math.pow(t_m, 3.0) / l) * (1.0 / l))));
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	return t_s * (2.0 / ((2.0 * (k * k)) * ((math.pow(t_m, 3.0) / l) * (1.0 / l))))

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	return Float64(t_s * Float64(2.0 / Float64(Float64(2.0 * Float64(k * k)) * Float64(Float64((t_m ^ 3.0) / l) * Float64(1.0 / l)))))
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp = code(t_s, t_m, l, k)
	tmp = t_s * (2.0 / ((2.0 * (k * k)) * (((t_m ^ 3.0) / l) * (1.0 / l))));
end

Wolfram

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

Derivation

1. Initial program 56.7%

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

3. Simplified 56.5%

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

5. Taylor expanded in k around 0 53.1%

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

   1. unpow2 53.1%

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

7. Applied egg-rr 53.1%

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

   1. div-inv 53.1%

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

9. Applied egg-rr 53.1%

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

10. Final simplification 53.1%

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

Alternative 20: 54.2% accurate, 3.7× speedup

Math

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

FPCore

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

C

t\_m = fabs(t);
t\_s = copysign(1.0, t);
double code(double t_s, double t_m, double l, double k) {
	return t_s * (2.0 / ((2.0 * (k * k)) * ((pow(t_m, 3.0) / l) / l)));
}

Fortran

t\_m = abs(t)
t\_s = copysign(1.0d0, t)
real(8) function code(t_s, t_m, l, k)
    real(8), intent (in) :: t_s
    real(8), intent (in) :: t_m
    real(8), intent (in) :: l
    real(8), intent (in) :: k
    code = t_s * (2.0d0 / ((2.0d0 * (k * k)) * (((t_m ** 3.0d0) / l) / l)))
end function

Java

t\_m = Math.abs(t);
t\_s = Math.copySign(1.0, t);
public static double code(double t_s, double t_m, double l, double k) {
	return t_s * (2.0 / ((2.0 * (k * k)) * ((Math.pow(t_m, 3.0) / l) / l)));
}

Python

t\_m = math.fabs(t)
t\_s = math.copysign(1.0, t)
def code(t_s, t_m, l, k):
	return t_s * (2.0 / ((2.0 * (k * k)) * ((math.pow(t_m, 3.0) / l) / l)))

Julia

t\_m = abs(t)
t\_s = copysign(1.0, t)
function code(t_s, t_m, l, k)
	return Float64(t_s * Float64(2.0 / Float64(Float64(2.0 * Float64(k * k)) * Float64(Float64((t_m ^ 3.0) / l) / l))))
end

MATLAB

t\_m = abs(t);
t\_s = sign(t) * abs(1.0);
function tmp = code(t_s, t_m, l, k)
	tmp = t_s * (2.0 / ((2.0 * (k * k)) * (((t_m ^ 3.0) / l) / l)));
end

Wolfram

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

Derivation

1. Initial program 56.7%

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

3. Simplified 56.5%

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

5. Taylor expanded in k around 0 53.1%

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

   1. unpow2 53.1%

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

7. Applied egg-rr 53.1%

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

8. Final simplification 53.1%

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

Reproduce

herbie shell --seed 2024131 
(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))))