\[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]

↓

\[\begin{array}{l} \mathbf{if}\;x \le -1.308379721855004695332726910583787116593 \cdot 10^{-5}:\\ \;\;\;\;\sqrt{\left(\left(e^{2 \cdot x} - 1\right) \cdot \frac{1}{e^{x + x} - 1 \cdot 1}\right) \cdot \left(e^{x} + 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{2 + x \cdot \left(1 + 0.5 \cdot x\right)}\\ \end{array}\]

\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}

↓

\begin{array}{l}
\mathbf{if}\;x \le -1.308379721855004695332726910583787116593 \cdot 10^{-5}:\\
\;\;\;\;\sqrt{\left(\left(e^{2 \cdot x} - 1\right) \cdot \frac{1}{e^{x + x} - 1 \cdot 1}\right) \cdot \left(e^{x} + 1\right)}\\

\mathbf{else}:\\
\;\;\;\;\sqrt{2 + x \cdot \left(1 + 0.5 \cdot x\right)}\\

\end{array}

double f(double x) {
        double r24889 = 2.0;
        double r24890 = x;
        double r24891 = r24889 * r24890;
        double r24892 = exp(r24891);
        double r24893 = 1.0;
        double r24894 = r24892 - r24893;
        double r24895 = exp(r24890);
        double r24896 = r24895 - r24893;
        double r24897 = r24894 / r24896;
        double r24898 = sqrt(r24897);
        return r24898;
}

↓

double f(double x) {
        double r24899 = x;
        double r24900 = -1.3083797218550047e-05;
        bool r24901 = r24899 <= r24900;
        double r24902 = 2.0;
        double r24903 = r24902 * r24899;
        double r24904 = exp(r24903);
        double r24905 = 1.0;
        double r24906 = r24904 - r24905;
        double r24907 = 1.0;
        double r24908 = r24899 + r24899;
        double r24909 = exp(r24908);
        double r24910 = r24905 * r24905;
        double r24911 = r24909 - r24910;
        double r24912 = r24907 / r24911;
        double r24913 = r24906 * r24912;
        double r24914 = exp(r24899);
        double r24915 = r24914 + r24905;
        double r24916 = r24913 * r24915;
        double r24917 = sqrt(r24916);
        double r24918 = 0.5;
        double r24919 = r24918 * r24899;
        double r24920 = r24905 + r24919;
        double r24921 = r24899 * r24920;
        double r24922 = r24902 + r24921;
        double r24923 = sqrt(r24922);
        double r24924 = r24901 ? r24917 : r24923;
        return r24924;
}

Derivation

Split input into 2 regimes
if x < -1.3083797218550047e-05
1. Initial program 0.1
  \[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
2. Using strategy rm
3. Applied flip--0.1
  \[\leadsto \sqrt{\frac{e^{2 \cdot x} - 1}{\color{blue}{\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{e^{x} + 1}}}}\]
4. Applied associate-/r/0.1
  \[\leadsto \sqrt{\color{blue}{\frac{e^{2 \cdot x} - 1}{e^{x} \cdot e^{x} - 1 \cdot 1} \cdot \left(e^{x} + 1\right)}}\]
5. Simplified0.0
  \[\leadsto \sqrt{\color{blue}{\frac{e^{2 \cdot x} - 1}{e^{x + x} - 1 \cdot 1}} \cdot \left(e^{x} + 1\right)}\]
6. Using strategy rm
7. Applied div-inv0.0
  \[\leadsto \sqrt{\color{blue}{\left(\left(e^{2 \cdot x} - 1\right) \cdot \frac{1}{e^{x + x} - 1 \cdot 1}\right)} \cdot \left(e^{x} + 1\right)}\]
if -1.3083797218550047e-05 < x
1. Initial program 34.5
  \[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
2. Taylor expanded around 0 6.4
  \[\leadsto \sqrt{\color{blue}{0.5 \cdot {x}^{2} + \left(1 \cdot x + 2\right)}}\]
3. Simplified6.4
  \[\leadsto \sqrt{\color{blue}{2 + x \cdot \left(1 + 0.5 \cdot x\right)}}\]
Recombined 2 regimes into one program.
Final simplification0.9
\[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.308379721855004695332726910583787116593 \cdot 10^{-5}:\\ \;\;\;\;\sqrt{\left(\left(e^{2 \cdot x} - 1\right) \cdot \frac{1}{e^{x + x} - 1 \cdot 1}\right) \cdot \left(e^{x} + 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{2 + x \cdot \left(1 + 0.5 \cdot x\right)}\\ \end{array}\]

could not determine a ground truth for program body (more)

Error

Try it out

Derivation

`if x < -1.3083797218550047e-05`

`if -1.3083797218550047e-05 < x`

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