Average Error: 53.0 → 0.3
Time: 6.4s
Precision: 64
\[\log \left(x + \sqrt{x \cdot x + 1}\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.02933962805667645:\\ \;\;\;\;\log \left(\frac{0.125}{{x}^{3}} - \left(\frac{0.5}{x} - \frac{-0.0625}{{x}^{5}}\right)\right)\\ \mathbf{elif}\;x \le 0.885438759469176051:\\ \;\;\;\;\left(\log \left(\sqrt{1}\right) + \frac{x}{\sqrt{1}}\right) - \frac{1}{6} \cdot \frac{{x}^{3}}{{\left(\sqrt{1}\right)}^{3}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + \left(\left(x + \frac{0.5}{x}\right) - \frac{0.125}{{x}^{3}}\right)\right)\\ \end{array}\]
\log \left(x + \sqrt{x \cdot x + 1}\right)
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.02933962805667645:\\
\;\;\;\;\log \left(\frac{0.125}{{x}^{3}} - \left(\frac{0.5}{x} - \frac{-0.0625}{{x}^{5}}\right)\right)\\

\mathbf{elif}\;x \le 0.885438759469176051:\\
\;\;\;\;\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 r169896 = x;
        double r169897 = r169896 * r169896;
        double r169898 = 1.0;
        double r169899 = r169897 + r169898;
        double r169900 = sqrt(r169899);
        double r169901 = r169896 + r169900;
        double r169902 = log(r169901);
        return r169902;
}
↓
double f(double x) {
        double r169903 = x;
        double r169904 = -1.0293396280566764;
        bool r169905 = r169903 <= r169904;
        double r169906 = 0.125;
        double r169907 = 3.0;
        double r169908 = pow(r169903, r169907);
        double r169909 = r169906 / r169908;
        double r169910 = 0.5;
        double r169911 = r169910 / r169903;
        double r169912 = 0.0625;
        double r169913 = -r169912;
        double r169914 = 5.0;
        double r169915 = pow(r169903, r169914);
        double r169916 = r169913 / r169915;
        double r169917 = r169911 - r169916;
        double r169918 = r169909 - r169917;
        double r169919 = log(r169918);
        double r169920 = 0.885438759469176;
        bool r169921 = r169903 <= r169920;
        double r169922 = 1.0;
        double r169923 = sqrt(r169922);
        double r169924 = log(r169923);
        double r169925 = r169903 / r169923;
        double r169926 = r169924 + r169925;
        double r169927 = 0.16666666666666666;
        double r169928 = pow(r169923, r169907);
        double r169929 = r169908 / r169928;
        double r169930 = r169927 * r169929;
        double r169931 = r169926 - r169930;
        double r169932 = r169903 + r169911;
        double r169933 = r169932 - r169909;
        double r169934 = r169903 + r169933;
        double r169935 = log(r169934);
        double r169936 = r169921 ? r169931 : r169935;
        double r169937 = r169905 ? r169919 : r169936;
        return r169937;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original53.0
Target45.2
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 < -1.0293396280566764

    1. Initial program 62.5

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

    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 -1.0293396280566764 < x < 0.885438759469176

    1. Initial program 58.3

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

    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 0.885438759469176 < x

    1. Initial program 33.0

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

    2. Taylor expanded around inf 0.2

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

    3. Simplified0.2

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

  4. Final simplification0.3

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.02933962805667645:\\ \;\;\;\;\log \left(\frac{0.125}{{x}^{3}} - \left(\frac{0.5}{x} - \frac{-0.0625}{{x}^{5}}\right)\right)\\ \mathbf{elif}\;x \le 0.885438759469176051:\\ \;\;\;\;\left(\log \left(\sqrt{1}\right) + \frac{x}{\sqrt{1}}\right) - \frac{1}{6} \cdot \frac{{x}^{3}}{{\left(\sqrt{1}\right)}^{3}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + \left(\left(x + \frac{0.5}{x}\right) - \frac{0.125}{{x}^{3}}\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020100 
(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)))))