Optimal. Leaf size=65 \[ \frac{1}{5} \left (5-\sqrt{5}\right ) \log \left (2 \sqrt{1-x}-\sqrt{5}+1\right )+\frac{1}{5} \left (5+\sqrt{5}\right ) \log \left (2 \sqrt{1-x}+\sqrt{5}+1\right ) \]
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Rubi [A] time = 0.0372839, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {632, 31} \[ \frac{1}{5} \left (5-\sqrt{5}\right ) \log \left (2 \sqrt{1-x}-\sqrt{5}+1\right )+\frac{1}{5} \left (5+\sqrt{5}\right ) \log \left (2 \sqrt{1-x}+\sqrt{5}+1\right ) \]
Antiderivative was successfully verified.
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Rule 632
Rule 31
Rubi steps
\begin{align*} \int \frac{1}{-\sqrt{1-x}+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 = 0.0274428, size = 62, normalized size = 0.95 \[ \frac{1}{5} \left (\left (5+\sqrt{5}\right ) \log \left (2 \sqrt{1-x}+\sqrt{5}+1\right )-\left (\sqrt{5}-5\right ) \log \left (2 \sqrt{1-x}-\sqrt{5}+1\right )\right ) \]
Antiderivative was successfully verified.
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Maple [B] time = 0.004, size = 101, normalized size = 1.6 \begin{align*}{\frac{\ln \left ({x}^{2}+x-1 \right ) }{2}}+{\frac{\sqrt{5}}{5}{\it Artanh} \left ({\frac{ \left ( 1+2\,x \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 ( 2\,\sqrt{1-x}+1 \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 = 1.47626, size = 69, normalized size = 1.06 \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 = 1.43954, size = 178, normalized size = 2.74 \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 = 1.15198, size = 92, normalized size = 1.42 \begin{align*} - 4 \left (\begin{cases} - \frac{\sqrt{5} \operatorname{acoth}{\left (\frac{2 \sqrt{5} \left (\sqrt{1 - x} + \frac{1}{2}\right )}{5} \right )}}{10} & \text{for}\: \left (\sqrt{1 - x} + \frac{1}{2}\right )^{2} > \frac{5}{4} \\- \frac{\sqrt{5} \operatorname{atanh}{\left (\frac{2 \sqrt{5} \left (\sqrt{1 - x} + \frac{1}{2}\right )}{5} \right )}}{10} & \text{for}\: \left (\sqrt{1 - x} + \frac{1}{2}\right )^{2} < \frac{5}{4} \end{cases}\right ) + \log{\left (x - \sqrt{1 - x} \right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.13885, size = 73, normalized size = 1.12 \begin{align*} -\frac{1}{5} \, \sqrt{5} \log \left (\frac{{\left | -\sqrt{5} + 2 \, \sqrt{-x + 1} + 1 \right |}}{\sqrt{5} + 2 \, \sqrt{-x + 1} + 1}\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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