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

↓

\[\begin{array}{l} \mathbf{if}\;x \le -1.109419166854957547087282512125511104578 \cdot 10^{-10}:\\ \;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x + x} - 1 \cdot 1}} \cdot \sqrt{e^{x} + 1}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{2 + x \cdot \left(0.5 \cdot x + 1\right)}\\ \end{array}\]

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

↓

\begin{array}{l}
\mathbf{if}\;x \le -1.109419166854957547087282512125511104578 \cdot 10^{-10}:\\
\;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x + x} - 1 \cdot 1}} \cdot \sqrt{e^{x} + 1}\\

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

\end{array}

double f(double x) {
        double r19873 = 2.0;
        double r19874 = x;
        double r19875 = r19873 * r19874;
        double r19876 = exp(r19875);
        double r19877 = 1.0;
        double r19878 = r19876 - r19877;
        double r19879 = exp(r19874);
        double r19880 = r19879 - r19877;
        double r19881 = r19878 / r19880;
        double r19882 = sqrt(r19881);
        return r19882;
}

↓

double f(double x) {
        double r19883 = x;
        double r19884 = -1.1094191668549575e-10;
        bool r19885 = r19883 <= r19884;
        double r19886 = 2.0;
        double r19887 = r19886 * r19883;
        double r19888 = exp(r19887);
        double r19889 = 1.0;
        double r19890 = r19888 - r19889;
        double r19891 = r19883 + r19883;
        double r19892 = exp(r19891);
        double r19893 = r19889 * r19889;
        double r19894 = r19892 - r19893;
        double r19895 = r19890 / r19894;
        double r19896 = sqrt(r19895);
        double r19897 = exp(r19883);
        double r19898 = r19897 + r19889;
        double r19899 = sqrt(r19898);
        double r19900 = r19896 * r19899;
        double r19901 = 0.5;
        double r19902 = r19901 * r19883;
        double r19903 = r19902 + r19889;
        double r19904 = r19883 * r19903;
        double r19905 = r19886 + r19904;
        double r19906 = sqrt(r19905);
        double r19907 = r19885 ? r19900 : r19906;
        return r19907;
}

Derivation

Split input into 2 regimes
if x < -1.1094191668549575e-10
1. Initial program 0.4
  \[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
2. Using strategy rm
3. Applied flip--0.2
  \[\leadsto \sqrt{\frac{e^{2 \cdot x} - 1}{\color{blue}{\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{e^{x} + 1}}}}\]
4. Applied associate-/r/0.2
  \[\leadsto \sqrt{\color{blue}{\frac{e^{2 \cdot x} - 1}{e^{x} \cdot e^{x} - 1 \cdot 1} \cdot \left(e^{x} + 1\right)}}\]
5. Applied sqrt-prod0.2
  \[\leadsto \color{blue}{\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} \cdot e^{x} - 1 \cdot 1}} \cdot \sqrt{e^{x} + 1}}\]
6. Simplified0.0
  \[\leadsto \color{blue}{\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x + x} - 1 \cdot 1}}} \cdot \sqrt{e^{x} + 1}\]
if -1.1094191668549575e-10 < x
1. Initial program 36.5
  \[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
2. Taylor expanded around 0 6.3
  \[\leadsto \sqrt{\color{blue}{0.5 \cdot {x}^{2} + \left(1 \cdot x + 2\right)}}\]
3. Simplified6.3
  \[\leadsto \sqrt{\color{blue}{2 + x \cdot \left(0.5 \cdot x + 1\right)}}\]
Recombined 2 regimes into one program.
Final simplification0.7
\[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.109419166854957547087282512125511104578 \cdot 10^{-10}:\\ \;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x + x} - 1 \cdot 1}} \cdot \sqrt{e^{x} + 1}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{2 + x \cdot \left(0.5 \cdot x + 1\right)}\\ \end{array}\]

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

Error

Try it out

Derivation

`if x < -1.1094191668549575e-10`

`if -1.1094191668549575e-10 < x`

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