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

↓

\[\begin{array}{l} \mathbf{if}\;x \le -4.54501292903996873 \cdot 10^{-10}:\\ \;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x + x} - 1 \cdot 1}} \cdot \sqrt{e^{x} + 1}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{2 + x \cdot \left(0.5 \cdot x + 1\right)}\\ \end{array}\]

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

↓

\begin{array}{l}
\mathbf{if}\;x \le -4.54501292903996873 \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 r19587 = 2.0;
        double r19588 = x;
        double r19589 = r19587 * r19588;
        double r19590 = exp(r19589);
        double r19591 = 1.0;
        double r19592 = r19590 - r19591;
        double r19593 = exp(r19588);
        double r19594 = r19593 - r19591;
        double r19595 = r19592 / r19594;
        double r19596 = sqrt(r19595);
        return r19596;
}

↓

double f(double x) {
        double r19597 = x;
        double r19598 = -4.5450129290399687e-10;
        bool r19599 = r19597 <= r19598;
        double r19600 = 2.0;
        double r19601 = r19600 * r19597;
        double r19602 = exp(r19601);
        double r19603 = 1.0;
        double r19604 = r19602 - r19603;
        double r19605 = r19597 + r19597;
        double r19606 = exp(r19605);
        double r19607 = r19603 * r19603;
        double r19608 = r19606 - r19607;
        double r19609 = r19604 / r19608;
        double r19610 = sqrt(r19609);
        double r19611 = exp(r19597);
        double r19612 = r19611 + r19603;
        double r19613 = sqrt(r19612);
        double r19614 = r19610 * r19613;
        double r19615 = 0.5;
        double r19616 = r19615 * r19597;
        double r19617 = r19616 + r19603;
        double r19618 = r19597 * r19617;
        double r19619 = r19600 + r19618;
        double r19620 = sqrt(r19619);
        double r19621 = r19599 ? r19614 : r19620;
        return r19621;
}

Derivation

Split input into 2 regimes
if x < -4.5450129290399687e-10
1. Initial program 0.3
  \[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
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 -4.5450129290399687e-10 < x
1. Initial program 38.3
  \[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
2. Taylor expanded around 0 7.0
  \[\leadsto \sqrt{\color{blue}{0.5 \cdot {x}^{2} + \left(1 \cdot x + 2\right)}}\]
3. Simplified7.0
  \[\leadsto \sqrt{\color{blue}{2 + x \cdot \left(0.5 \cdot x + 1\right)}}\]
Recombined 2 regimes into one program.
Final simplification0.8
\[\leadsto \begin{array}{l} \mathbf{if}\;x \le -4.54501292903996873 \cdot 10^{-10}:\\ \;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x + x} - 1 \cdot 1}} \cdot \sqrt{e^{x} + 1}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{2 + x \cdot \left(0.5 \cdot x + 1\right)}\\ \end{array}\]

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

Error

Try it out

Derivation

`if x < -4.5450129290399687e-10`

`if -4.5450129290399687e-10 < x`

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