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

↓

\[\begin{array}{l} \mathbf{if}\;x \le -5.3888208797845926 \cdot 10^{-12}:\\ \;\;\;\;\sqrt{\frac{\mathsf{expm1}\left(\mathsf{log1p}\left(e^{2 \cdot x} - 1\right)\right)}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)} \cdot \left(e^{x} + 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}\\ \end{array}\]

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

↓

\begin{array}{l}
\mathbf{if}\;x \le -5.3888208797845926 \cdot 10^{-12}:\\
\;\;\;\;\sqrt{\frac{\mathsf{expm1}\left(\mathsf{log1p}\left(e^{2 \cdot x} - 1\right)\right)}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)} \cdot \left(e^{x} + 1\right)}\\

\mathbf{else}:\\
\;\;\;\;\sqrt{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}\\

\end{array}

double f(double x) {
        double r14776 = 2.0;
        double r14777 = x;
        double r14778 = r14776 * r14777;
        double r14779 = exp(r14778);
        double r14780 = 1.0;
        double r14781 = r14779 - r14780;
        double r14782 = exp(r14777);
        double r14783 = r14782 - r14780;
        double r14784 = r14781 / r14783;
        double r14785 = sqrt(r14784);
        return r14785;
}

↓

double f(double x) {
        double r14786 = x;
        double r14787 = -5.3888208797845926e-12;
        bool r14788 = r14786 <= r14787;
        double r14789 = 2.0;
        double r14790 = r14789 * r14786;
        double r14791 = exp(r14790);
        double r14792 = 1.0;
        double r14793 = r14791 - r14792;
        double r14794 = log1p(r14793);
        double r14795 = expm1(r14794);
        double r14796 = -r14792;
        double r14797 = r14786 + r14786;
        double r14798 = exp(r14797);
        double r14799 = fma(r14796, r14792, r14798);
        double r14800 = r14795 / r14799;
        double r14801 = exp(r14786);
        double r14802 = r14801 + r14792;
        double r14803 = r14800 * r14802;
        double r14804 = sqrt(r14803);
        double r14805 = 0.5;
        double r14806 = 2.0;
        double r14807 = pow(r14786, r14806);
        double r14808 = fma(r14792, r14786, r14789);
        double r14809 = fma(r14805, r14807, r14808);
        double r14810 = sqrt(r14809);
        double r14811 = r14788 ? r14804 : r14810;
        return r14811;
}

Derivation

Split input into 2 regimes
if x < -5.3888208797845926e-12
1. Initial program 0.5
  \[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
2. Using strategy rm
3. Applied flip--0.3
  \[\leadsto \sqrt{\frac{e^{2 \cdot x} - 1}{\color{blue}{\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{e^{x} + 1}}}}\]
4. Applied associate-/r/0.3
  \[\leadsto \sqrt{\color{blue}{\frac{e^{2 \cdot x} - 1}{e^{x} \cdot e^{x} - 1 \cdot 1} \cdot \left(e^{x} + 1\right)}}\]
5. Simplified0.0
  \[\leadsto \sqrt{\color{blue}{\frac{e^{2 \cdot x} - 1}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)}} \cdot \left(e^{x} + 1\right)}\]
6. Using strategy rm
7. Applied expm1-log1p-u0.0
  \[\leadsto \sqrt{\frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(e^{2 \cdot x} - 1\right)\right)}}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)} \cdot \left(e^{x} + 1\right)}\]
if -5.3888208797845926e-12 < x
1. Initial program 35.8
  \[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
2. Taylor expanded around 0 7.9
  \[\leadsto \sqrt{\color{blue}{0.5 \cdot {x}^{2} + \left(1 \cdot x + 2\right)}}\]
3. Simplified7.9
  \[\leadsto \sqrt{\color{blue}{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}}\]
Recombined 2 regimes into one program.
Final simplification0.9
\[\leadsto \begin{array}{l} \mathbf{if}\;x \le -5.3888208797845926 \cdot 10^{-12}:\\ \;\;\;\;\sqrt{\frac{\mathsf{expm1}\left(\mathsf{log1p}\left(e^{2 \cdot x} - 1\right)\right)}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)} \cdot \left(e^{x} + 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}\\ \end{array}\]

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

Error

Derivation

`if x < -5.3888208797845926e-12`

`if -5.3888208797845926e-12 < x`

Reproduce