Average Error: 6.2 → 1.2
Time: 6.1s
Precision: binary64
\[x + \frac{e^{y \cdot \log \left(\frac{y}{z + y}\right)}}{y}\]
↓
\[\begin{array}{l} \mathbf{if}\;y \le 1.2419084246309027 \cdot 10^{-48}:\\ \;\;\;\;x + \frac{e^{0}}{y}\\ \mathbf{else}:\\ \;\;\;\;x + \frac{e^{\left(\sqrt[3]{y} \cdot \sqrt[3]{y}\right) \cdot \left(\left(\sqrt{\sqrt[3]{y}} \cdot \sqrt{\sqrt[3]{y}}\right) \cdot \log \left(\frac{1}{\frac{z + y}{y}}\right)\right)}}{y}\\ \end{array}\]
x + \frac{e^{y \cdot \log \left(\frac{y}{z + y}\right)}}{y}
↓
\begin{array}{l}
\mathbf{if}\;y \le 1.2419084246309027 \cdot 10^{-48}:\\
\;\;\;\;x + \frac{e^{0}}{y}\\

\mathbf{else}:\\
\;\;\;\;x + \frac{e^{\left(\sqrt[3]{y} \cdot \sqrt[3]{y}\right) \cdot \left(\left(\sqrt{\sqrt[3]{y}} \cdot \sqrt{\sqrt[3]{y}}\right) \cdot \log \left(\frac{1}{\frac{z + y}{y}}\right)\right)}}{y}\\

\end{array}
double code(double x, double y, double z) {
	return ((double) (x + ((double) (((double) exp(((double) (y * ((double) log(((double) (y / ((double) (z + y)))))))))) / y))));
}
↓
double code(double x, double y, double z) {
	double VAR;
	if ((y <= 1.2419084246309027e-48)) {
		VAR = ((double) (x + ((double) (((double) exp(0.0)) / y))));
	} else {
		VAR = ((double) (x + ((double) (((double) exp(((double) (((double) (((double) cbrt(y)) * ((double) cbrt(y)))) * ((double) (((double) (((double) sqrt(((double) cbrt(y)))) * ((double) sqrt(((double) cbrt(y)))))) * ((double) log(((double) (1.0 / ((double) (((double) (z + y)) / y)))))))))))) / y))));
	}
	return VAR;
}

Error

Bits error versus x

Bits error versus y

Bits error versus z

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original6.2
Target1.1
Herbie1.2
\[\begin{array}{l} \mathbf{if}\;\frac{y}{z + y} \lt 7.1154157598 \cdot 10^{-315}:\\ \;\;\;\;x + \frac{e^{\frac{-1}{z}}}{y}\\ \mathbf{else}:\\ \;\;\;\;x + \frac{e^{\log \left({\left(\frac{y}{y + z}\right)}^{y}\right)}}{y}\\ \end{array}\]

Derivation

  1. Split input into 2 regimes

  2. if y < 1.2419084246309027e-48

    1. Initial program 8.6

      \[x + \frac{e^{y \cdot \log \left(\frac{y}{z + y}\right)}}{y}\]

    2. Taylor expanded around inf 1.0

      \[\leadsto x + \frac{e^{\color{blue}{\log 1 \cdot y}}}{y}\]

    3. Simplified1.0

      \[\leadsto x + \frac{e^{\color{blue}{0}}}{y}\]

    if 1.2419084246309027e-48 < y

    1. Initial program 1.5

      \[x + \frac{e^{y \cdot \log \left(\frac{y}{z + y}\right)}}{y}\]
    2. Using strategy rm

    3. Applied clear-num1.6

      \[\leadsto x + \frac{e^{y \cdot \log \color{blue}{\left(\frac{1}{\frac{z + y}{y}}\right)}}}{y}\]
    4. Using strategy rm

    5. Applied add-cube-cbrt1.6

      \[\leadsto x + \frac{e^{\color{blue}{\left(\left(\sqrt[3]{y} \cdot \sqrt[3]{y}\right) \cdot \sqrt[3]{y}\right)} \cdot \log \left(\frac{1}{\frac{z + y}{y}}\right)}}{y}\]

    6. Applied associate-*l*1.6

      \[\leadsto x + \frac{e^{\color{blue}{\left(\sqrt[3]{y} \cdot \sqrt[3]{y}\right) \cdot \left(\sqrt[3]{y} \cdot \log \left(\frac{1}{\frac{z + y}{y}}\right)\right)}}}{y}\]
    7. Using strategy rm

    8. Applied add-sqr-sqrt1.6

      \[\leadsto x + \frac{e^{\left(\sqrt[3]{y} \cdot \sqrt[3]{y}\right) \cdot \left(\color{blue}{\left(\sqrt{\sqrt[3]{y}} \cdot \sqrt{\sqrt[3]{y}}\right)} \cdot \log \left(\frac{1}{\frac{z + y}{y}}\right)\right)}}{y}\]
  3. Recombined 2 regimes into one program.

  4. Final simplification1.2

    \[\leadsto \begin{array}{l} \mathbf{if}\;y \le 1.2419084246309027 \cdot 10^{-48}:\\ \;\;\;\;x + \frac{e^{0}}{y}\\ \mathbf{else}:\\ \;\;\;\;x + \frac{e^{\left(\sqrt[3]{y} \cdot \sqrt[3]{y}\right) \cdot \left(\left(\sqrt{\sqrt[3]{y}} \cdot \sqrt{\sqrt[3]{y}}\right) \cdot \log \left(\frac{1}{\frac{z + y}{y}}\right)\right)}}{y}\\ \end{array}\]

Reproduce

herbie shell --seed 2020173 
(FPCore (x y z)
  :name "Numeric.SpecFunctions:invIncompleteBetaWorker from math-functions-0.1.5.2, G"
  :precision binary64

  :herbie-target
  (if (< (/ y (+ z y)) 7.1154157597908e-315) (+ x (/ (exp (/ -1.0 z)) y)) (+ x (/ (exp (log (pow (/ y (+ y z)) y))) y)))

  (+ x (/ (exp (* y (log (/ y (+ z y))))) y)))