Optimal. Leaf size=61 \[ \frac{1}{5} \left (5-\sqrt{5}\right ) \log \left (-2 \sqrt{x+1}-\sqrt{5}+1\right )+\frac{1}{5} \left (5+\sqrt{5}\right ) \log \left (-2 \sqrt{x+1}+\sqrt{5}+1\right ) \]
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Rubi [A] time = 0.0311208, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {632, 31} \[ \frac{1}{5} \left (5-\sqrt{5}\right ) \log \left (-2 \sqrt{x+1}-\sqrt{5}+1\right )+\frac{1}{5} \left (5+\sqrt{5}\right ) \log \left (-2 \sqrt{x+1}+\sqrt{5}+1\right ) \]
Antiderivative was successfully verified.
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Rule 632
Rule 31
Rubi steps
\begin{align*} \int \frac{1}{x-\sqrt{1+x}} \, dx &=2 \operatorname{Subst}\left (\int \frac{x}{-1-x+x^2} \, dx,x,\sqrt{1+x}\right )\\ &=\frac{1}{5} \left (5-\sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{2}+\frac{\sqrt{5}}{2}+x} \, dx,x,\sqrt{1+x}\right )+\frac{1}{5} \left (5+\sqrt{5}\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{1}{2}-\frac{\sqrt{5}}{2}+x} \, dx,x,\sqrt{1+x}\right )\\ &=\frac{1}{5} \left (5-\sqrt{5}\right ) \log \left (1-\sqrt{5}-2 \sqrt{1+x}\right )+\frac{1}{5} \left (5+\sqrt{5}\right ) \log \left (1+\sqrt{5}-2 \sqrt{1+x}\right )\\ \end{align*}
Mathematica [A] time = 0.0248369, size = 58, normalized size = 0.95 \[ \frac{1}{5} \left (\left (5+\sqrt{5}\right ) \log \left (-2 \sqrt{x+1}+\sqrt{5}+1\right )-\left (\sqrt{5}-5\right ) \log \left (-2 \sqrt{x+1}-\sqrt{5}+1\right )\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.006, size = 91, normalized size = 1.5 \begin{align*}{\frac{\ln \left ({x}^{2}-x-1 \right ) }{2}}-{\frac{\sqrt{5}}{5}{\it Artanh} \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{5}}{5}} \right ) }-{\frac{1}{2}\ln \left ( x+\sqrt{1+x} \right ) }-{\frac{\sqrt{5}}{5}{\it Artanh} \left ({\frac{\sqrt{5}}{5} \left ( 1+2\,\sqrt{1+x} \right ) } \right ) }+{\frac{1}{2}\ln \left ( x-\sqrt{1+x} \right ) }-{\frac{\sqrt{5}}{5}{\it Artanh} \left ({\frac{\sqrt{5}}{5} \left ( 2\,\sqrt{1+x}-1 \right ) } \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.46453, size = 61, normalized size = 1. \begin{align*} \frac{1}{5} \, \sqrt{5} \log \left (-\frac{\sqrt{5} - 2 \, \sqrt{x + 1} + 1}{\sqrt{5} + 2 \, \sqrt{x + 1} - 1}\right ) + \log \left (x - \sqrt{x + 1}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.48591, size = 174, normalized size = 2.85 \begin{align*} \frac{1}{5} \, \sqrt{5} \log \left (\frac{2 \, x^{2} - \sqrt{5}{\left (x + 2\right )} -{\left (\sqrt{5}{\left (2 \, x - 1\right )} - 5\right )} \sqrt{x + 1} + 3 \, x - 2}{x^{2} - x - 1}\right ) + \log \left (x - \sqrt{x + 1}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 1.09701, size = 92, normalized size = 1.51 \begin{align*} 4 \left (\begin{cases} - \frac{\sqrt{5} \operatorname{acoth}{\left (\frac{2 \sqrt{5} \left (\sqrt{x + 1} - \frac{1}{2}\right )}{5} \right )}}{10} & \text{for}\: \left (\sqrt{x + 1} - \frac{1}{2}\right )^{2} > \frac{5}{4} \\- \frac{\sqrt{5} \operatorname{atanh}{\left (\frac{2 \sqrt{5} \left (\sqrt{x + 1} - \frac{1}{2}\right )}{5} \right )}}{10} & \text{for}\: \left (\sqrt{x + 1} - \frac{1}{2}\right )^{2} < \frac{5}{4} \end{cases}\right ) + \log{\left (x - \sqrt{x + 1} \right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.15241, size = 66, normalized size = 1.08 \begin{align*} \frac{1}{5} \, \sqrt{5} \log \left (\frac{{\left | -\sqrt{5} + 2 \, \sqrt{x + 1} - 1 \right |}}{{\left | \sqrt{5} + 2 \, \sqrt{x + 1} - 1 \right |}}\right ) + \log \left ({\left | x - \sqrt{x + 1} \right |}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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