Average Error: 53.2 → 0.3
Time: 5.9s
Precision: 64
\[\log \left(x + \sqrt{x \cdot x + 1}\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.01388042333944495:\\ \;\;\;\;\log \left(\frac{0.125}{{x}^{3}} - \left(\frac{0.5}{x} - \frac{-0.0625}{{x}^{5}}\right)\right)\\ \mathbf{elif}\;x \le 0.8851102844894212:\\ \;\;\;\;\left(\log \left(\sqrt{1}\right) + \frac{x}{\sqrt{1}}\right) - \frac{1}{6} \cdot \frac{{x}^{3}}{{\left(\sqrt{1}\right)}^{3}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + \left(\left(x + \frac{0.5}{x}\right) - \frac{0.125}{{x}^{3}}\right)\right)\\ \end{array}\]
\log \left(x + \sqrt{x \cdot x + 1}\right)
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.01388042333944495:\\
\;\;\;\;\log \left(\frac{0.125}{{x}^{3}} - \left(\frac{0.5}{x} - \frac{-0.0625}{{x}^{5}}\right)\right)\\

\mathbf{elif}\;x \le 0.8851102844894212:\\
\;\;\;\;\left(\log \left(\sqrt{1}\right) + \frac{x}{\sqrt{1}}\right) - \frac{1}{6} \cdot \frac{{x}^{3}}{{\left(\sqrt{1}\right)}^{3}}\\

\mathbf{else}:\\
\;\;\;\;\log \left(x + \left(\left(x + \frac{0.5}{x}\right) - \frac{0.125}{{x}^{3}}\right)\right)\\

\end{array}
double f(double x) {
        double r176010 = x;
        double r176011 = r176010 * r176010;
        double r176012 = 1.0;
        double r176013 = r176011 + r176012;
        double r176014 = sqrt(r176013);
        double r176015 = r176010 + r176014;
        double r176016 = log(r176015);
        return r176016;
}
↓
double f(double x) {
        double r176017 = x;
        double r176018 = -1.013880423339445;
        bool r176019 = r176017 <= r176018;
        double r176020 = 0.125;
        double r176021 = 3.0;
        double r176022 = pow(r176017, r176021);
        double r176023 = r176020 / r176022;
        double r176024 = 0.5;
        double r176025 = r176024 / r176017;
        double r176026 = 0.0625;
        double r176027 = -r176026;
        double r176028 = 5.0;
        double r176029 = pow(r176017, r176028);
        double r176030 = r176027 / r176029;
        double r176031 = r176025 - r176030;
        double r176032 = r176023 - r176031;
        double r176033 = log(r176032);
        double r176034 = 0.8851102844894212;
        bool r176035 = r176017 <= r176034;
        double r176036 = 1.0;
        double r176037 = sqrt(r176036);
        double r176038 = log(r176037);
        double r176039 = r176017 / r176037;
        double r176040 = r176038 + r176039;
        double r176041 = 0.16666666666666666;
        double r176042 = pow(r176037, r176021);
        double r176043 = r176022 / r176042;
        double r176044 = r176041 * r176043;
        double r176045 = r176040 - r176044;
        double r176046 = r176017 + r176025;
        double r176047 = r176046 - r176023;
        double r176048 = r176017 + r176047;
        double r176049 = log(r176048);
        double r176050 = r176035 ? r176045 : r176049;
        double r176051 = r176019 ? r176033 : r176050;
        return r176051;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original53.2
Target45.1
Herbie0.3
\[\begin{array}{l} \mathbf{if}\;x \lt 0.0:\\ \;\;\;\;\log \left(\frac{-1}{x - \sqrt{x \cdot x + 1}}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + \sqrt{x \cdot x + 1}\right)\\ \end{array}\]

Derivation

  1. Split input into 3 regimes

  2. if x < -1.013880423339445

    1. Initial program 63.1

      \[\log \left(x + \sqrt{x \cdot x + 1}\right)\]

    2. Taylor expanded around -inf 0.2

      \[\leadsto \log \color{blue}{\left(0.125 \cdot \frac{1}{{x}^{3}} - \left(0.5 \cdot \frac{1}{x} + 0.0625 \cdot \frac{1}{{x}^{5}}\right)\right)}\]

    3. Simplified0.2

      \[\leadsto \log \color{blue}{\left(\frac{0.125}{{x}^{3}} - \left(\frac{0.5}{x} - \frac{-0.0625}{{x}^{5}}\right)\right)}\]

    if -1.013880423339445 < x < 0.8851102844894212

    1. Initial program 58.6

      \[\log \left(x + \sqrt{x \cdot x + 1}\right)\]

    2. Taylor expanded around 0 0.4

      \[\leadsto \color{blue}{\left(\log \left(\sqrt{1}\right) + \frac{x}{\sqrt{1}}\right) - \frac{1}{6} \cdot \frac{{x}^{3}}{{\left(\sqrt{1}\right)}^{3}}}\]

    if 0.8851102844894212 < x

    1. Initial program 32.9

      \[\log \left(x + \sqrt{x \cdot x + 1}\right)\]

    2. Taylor expanded around inf 0.3

      \[\leadsto \log \left(x + \color{blue}{\left(\left(x + 0.5 \cdot \frac{1}{x}\right) - 0.125 \cdot \frac{1}{{x}^{3}}\right)}\right)\]

    3. Simplified0.3

      \[\leadsto \log \left(x + \color{blue}{\left(\left(x + \frac{0.5}{x}\right) - \frac{0.125}{{x}^{3}}\right)}\right)\]
  3. Recombined 3 regimes into one program.

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.01388042333944495:\\ \;\;\;\;\log \left(\frac{0.125}{{x}^{3}} - \left(\frac{0.5}{x} - \frac{-0.0625}{{x}^{5}}\right)\right)\\ \mathbf{elif}\;x \le 0.8851102844894212:\\ \;\;\;\;\left(\log \left(\sqrt{1}\right) + \frac{x}{\sqrt{1}}\right) - \frac{1}{6} \cdot \frac{{x}^{3}}{{\left(\sqrt{1}\right)}^{3}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + \left(\left(x + \frac{0.5}{x}\right) - \frac{0.125}{{x}^{3}}\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020065 
(FPCore (x)
  :name "Hyperbolic arcsine"
  :precision binary64

  :herbie-target
  (if (< x 0.0) (log (/ -1 (- x (sqrt (+ (* x x) 1))))) (log (+ x (sqrt (+ (* x x) 1)))))

  (log (+ x (sqrt (+ (* x x) 1)))))