Average Error: 53.1 → 0.4
Time: 6.8s
Precision: 64
\[\log \left(x + \sqrt{x \cdot x + 1}\right)\]
↓
\[\begin{array}{l} \mathbf{if}\;x \le -1.0051038436432536:\\ \;\;\;\;\log \left(\frac{0.125}{{x}^{3}} - \left(\frac{0.5}{x} - \frac{-0.0625}{{x}^{5}}\right)\right)\\ \mathbf{elif}\;x \le 0.886907278544093791:\\ \;\;\;\;\left(\log \left(\sqrt{1}\right) + \frac{x}{\sqrt{1}}\right) - \frac{1}{6} \cdot \frac{{x}^{3}}{{\left(\sqrt{1}\right)}^{3}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + \left(\left(x + \frac{0.5}{x}\right) - \frac{0.125}{{x}^{3}}\right)\right)\\ \end{array}\]
\log \left(x + \sqrt{x \cdot x + 1}\right)
↓
\begin{array}{l}
\mathbf{if}\;x \le -1.0051038436432536:\\
\;\;\;\;\log \left(\frac{0.125}{{x}^{3}} - \left(\frac{0.5}{x} - \frac{-0.0625}{{x}^{5}}\right)\right)\\

\mathbf{elif}\;x \le 0.886907278544093791:\\
\;\;\;\;\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 r211187 = x;
        double r211188 = r211187 * r211187;
        double r211189 = 1.0;
        double r211190 = r211188 + r211189;
        double r211191 = sqrt(r211190);
        double r211192 = r211187 + r211191;
        double r211193 = log(r211192);
        return r211193;
}
↓
double f(double x) {
        double r211194 = x;
        double r211195 = -1.0051038436432536;
        bool r211196 = r211194 <= r211195;
        double r211197 = 0.125;
        double r211198 = 3.0;
        double r211199 = pow(r211194, r211198);
        double r211200 = r211197 / r211199;
        double r211201 = 0.5;
        double r211202 = r211201 / r211194;
        double r211203 = 0.0625;
        double r211204 = -r211203;
        double r211205 = 5.0;
        double r211206 = pow(r211194, r211205);
        double r211207 = r211204 / r211206;
        double r211208 = r211202 - r211207;
        double r211209 = r211200 - r211208;
        double r211210 = log(r211209);
        double r211211 = 0.8869072785440938;
        bool r211212 = r211194 <= r211211;
        double r211213 = 1.0;
        double r211214 = sqrt(r211213);
        double r211215 = log(r211214);
        double r211216 = r211194 / r211214;
        double r211217 = r211215 + r211216;
        double r211218 = 0.16666666666666666;
        double r211219 = pow(r211214, r211198);
        double r211220 = r211199 / r211219;
        double r211221 = r211218 * r211220;
        double r211222 = r211217 - r211221;
        double r211223 = r211194 + r211202;
        double r211224 = r211223 - r211200;
        double r211225 = r211194 + r211224;
        double r211226 = log(r211225);
        double r211227 = r211212 ? r211222 : r211226;
        double r211228 = r211196 ? r211210 : r211227;
        return r211228;
}

Error

Bits error versus x

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original53.1
Target45.2
Herbie0.4
\[\begin{array}{l} \mathbf{if}\;x \lt 0.0:\\ \;\;\;\;\log \left(\frac{-1}{x - \sqrt{x \cdot x + 1}}\right)\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + \sqrt{x \cdot x + 1}\right)\\ \end{array}\]

Derivation

  1. Split input into 3 regimes

  2. if x < -1.0051038436432536

    1. Initial program 62.9

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

    2. Taylor expanded around -inf 0.1

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

    3. Simplified0.1

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

    if -1.0051038436432536 < x < 0.8869072785440938

    1. Initial program 58.2

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

    2. Taylor expanded around 0 0.5

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

    if 0.8869072785440938 < x

    1. Initial program 32.7

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

    2. Taylor expanded around inf 0.4

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

    3. Simplified0.4

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

  4. Final simplification0.4

    \[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.0051038436432536:\\ \;\;\;\;\log \left(\frac{0.125}{{x}^{3}} - \left(\frac{0.5}{x} - \frac{-0.0625}{{x}^{5}}\right)\right)\\ \mathbf{elif}\;x \le 0.886907278544093791:\\ \;\;\;\;\left(\log \left(\sqrt{1}\right) + \frac{x}{\sqrt{1}}\right) - \frac{1}{6} \cdot \frac{{x}^{3}}{{\left(\sqrt{1}\right)}^{3}}\\ \mathbf{else}:\\ \;\;\;\;\log \left(x + \left(\left(x + \frac{0.5}{x}\right) - \frac{0.125}{{x}^{3}}\right)\right)\\ \end{array}\]

Reproduce

herbie shell --seed 2020034 
(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)))))