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

↓

\[\begin{array}{l} \mathbf{if}\;x \le -5.69778124812825276 \cdot 10^{-17}:\\ \;\;\;\;\sqrt{\frac{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{fma}\left(\sqrt{e^{2 \cdot x}}, \sqrt{e^{2 \cdot x}}, -1\right)\right)\right)}{e^{x} - 1}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}\\ \end{array}\]

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

↓

\begin{array}{l}
\mathbf{if}\;x \le -5.69778124812825276 \cdot 10^{-17}:\\
\;\;\;\;\sqrt{\frac{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{fma}\left(\sqrt{e^{2 \cdot x}}, \sqrt{e^{2 \cdot x}}, -1\right)\right)\right)}{e^{x} - 1}}\\

\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 r25474 = 2.0;
        double r25475 = x;
        double r25476 = r25474 * r25475;
        double r25477 = exp(r25476);
        double r25478 = 1.0;
        double r25479 = r25477 - r25478;
        double r25480 = exp(r25475);
        double r25481 = r25480 - r25478;
        double r25482 = r25479 / r25481;
        double r25483 = sqrt(r25482);
        return r25483;
}

↓

double f(double x) {
        double r25484 = x;
        double r25485 = -5.697781248128253e-17;
        bool r25486 = r25484 <= r25485;
        double r25487 = 2.0;
        double r25488 = r25487 * r25484;
        double r25489 = exp(r25488);
        double r25490 = sqrt(r25489);
        double r25491 = 1.0;
        double r25492 = -r25491;
        double r25493 = fma(r25490, r25490, r25492);
        double r25494 = log1p(r25493);
        double r25495 = expm1(r25494);
        double r25496 = exp(r25484);
        double r25497 = r25496 - r25491;
        double r25498 = r25495 / r25497;
        double r25499 = sqrt(r25498);
        double r25500 = 0.5;
        double r25501 = 2.0;
        double r25502 = pow(r25484, r25501);
        double r25503 = fma(r25491, r25484, r25487);
        double r25504 = fma(r25500, r25502, r25503);
        double r25505 = sqrt(r25504);
        double r25506 = r25486 ? r25499 : r25505;
        return r25506;
}

Derivation

Split input into 2 regimes
if x < -5.697781248128253e-17
1. Initial program 1.0
  \[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
2. Using strategy rm
3. Applied add-sqr-sqrt0.8
  \[\leadsto \sqrt{\frac{\color{blue}{\sqrt{e^{2 \cdot x}} \cdot \sqrt{e^{2 \cdot x}}} - 1}{e^{x} - 1}}\]
4. Applied fma-neg0.3
  \[\leadsto \sqrt{\frac{\color{blue}{\mathsf{fma}\left(\sqrt{e^{2 \cdot x}}, \sqrt{e^{2 \cdot x}}, -1\right)}}{e^{x} - 1}}\]
5. Using strategy rm
6. Applied expm1-log1p-u0.3
  \[\leadsto \sqrt{\frac{\color{blue}{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{fma}\left(\sqrt{e^{2 \cdot x}}, \sqrt{e^{2 \cdot x}}, -1\right)\right)\right)}}{e^{x} - 1}}\]
if -5.697781248128253e-17 < x
1. Initial program 37.1
  \[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
2. Taylor expanded around 0 9.3
  \[\leadsto \sqrt{\color{blue}{0.5 \cdot {x}^{2} + \left(1 \cdot x + 2\right)}}\]
3. Simplified9.3
  \[\leadsto \sqrt{\color{blue}{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}}\]
Recombined 2 regimes into one program.
Final simplification1.3
\[\leadsto \begin{array}{l} \mathbf{if}\;x \le -5.69778124812825276 \cdot 10^{-17}:\\ \;\;\;\;\sqrt{\frac{\mathsf{expm1}\left(\mathsf{log1p}\left(\mathsf{fma}\left(\sqrt{e^{2 \cdot x}}, \sqrt{e^{2 \cdot x}}, -1\right)\right)\right)}{e^{x} - 1}}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}\\ \end{array}\]

Error

Derivation

`if x < -5.697781248128253e-17`

`if -5.697781248128253e-17 < x`

Reproduce