Optimal. Leaf size=45 \[ \frac{1}{8} \log \left (x^2+1\right )+\frac{1}{4 (1-x)}-\frac{1}{4} \log (1-x)-\frac{\tan ^{-1}(x)}{2 (1-x)^2} \]
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Rubi [A] time = 0.0317409, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4862, 710, 801, 260} \[ \frac{1}{8} \log \left (x^2+1\right )+\frac{1}{4 (1-x)}-\frac{1}{4} \log (1-x)-\frac{\tan ^{-1}(x)}{2 (1-x)^2} \]
Antiderivative was successfully verified.
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Rule 4862
Rule 710
Rule 801
Rule 260
Rubi steps
\begin{align*} \int \frac{\tan ^{-1}(x)}{(-1+x)^3} \, dx &=-\frac{\tan ^{-1}(x)}{2 (1-x)^2}+\frac{1}{2} \int \frac{1}{(-1+x)^2 \left (1+x^2\right )} \, dx\\ &=\frac{1}{4 (1-x)}-\frac{\tan ^{-1}(x)}{2 (1-x)^2}+\frac{1}{4} \int \frac{-1-x}{(-1+x) \left (1+x^2\right )} \, dx\\ &=\frac{1}{4 (1-x)}-\frac{\tan ^{-1}(x)}{2 (1-x)^2}+\frac{1}{4} \int \left (\frac{1}{1-x}+\frac{x}{1+x^2}\right ) \, dx\\ &=\frac{1}{4 (1-x)}-\frac{\tan ^{-1}(x)}{2 (1-x)^2}-\frac{1}{4} \log (1-x)+\frac{1}{4} \int \frac{x}{1+x^2} \, dx\\ &=\frac{1}{4 (1-x)}-\frac{\tan ^{-1}(x)}{2 (1-x)^2}-\frac{1}{4} \log (1-x)+\frac{1}{8} \log \left (1+x^2\right )\\ \end{align*}
Mathematica [A] time = 0.0328314, size = 35, normalized size = 0.78 \[ \frac{1}{8} \left (\log \left (x^2+1\right )-\frac{2}{x-1}-2 \log (1-x)-\frac{4 \tan ^{-1}(x)}{(x-1)^2}\right ) \]
Antiderivative was successfully verified.
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Maple [A] time = 0.043, size = 32, normalized size = 0.7 \begin{align*} -{\frac{\arctan \left ( x \right ) }{2\, \left ( x-1 \right ) ^{2}}}-{\frac{1}{4\,x-4}}-{\frac{\ln \left ( x-1 \right ) }{4}}+{\frac{\ln \left ({x}^{2}+1 \right ) }{8}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.44204, size = 42, normalized size = 0.93 \begin{align*} -\frac{1}{4 \,{\left (x - 1\right )}} - \frac{\arctan \left (x\right )}{2 \,{\left (x - 1\right )}^{2}} + \frac{1}{8} \, \log \left (x^{2} + 1\right ) - \frac{1}{4} \, \log \left (x - 1\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.39151, size = 144, normalized size = 3.2 \begin{align*} \frac{{\left (x^{2} - 2 \, x + 1\right )} \log \left (x^{2} + 1\right ) - 2 \,{\left (x^{2} - 2 \, x + 1\right )} \log \left (x - 1\right ) - 2 \, x - 4 \, \arctan \left (x\right ) + 2}{8 \,{\left (x^{2} - 2 \, x + 1\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [B] time = 0.652303, size = 153, normalized size = 3.4 \begin{align*} - \frac{2 x^{2} \log{\left (x - 1 \right )}}{8 x^{2} - 16 x + 8} + \frac{x^{2} \log{\left (x^{2} + 1 \right )}}{8 x^{2} - 16 x + 8} - \frac{x^{2}}{8 x^{2} - 16 x + 8} + \frac{4 x \log{\left (x - 1 \right )}}{8 x^{2} - 16 x + 8} - \frac{2 x \log{\left (x^{2} + 1 \right )}}{8 x^{2} - 16 x + 8} - \frac{2 \log{\left (x - 1 \right )}}{8 x^{2} - 16 x + 8} + \frac{\log{\left (x^{2} + 1 \right )}}{8 x^{2} - 16 x + 8} - \frac{4 \operatorname{atan}{\left (x \right )}}{8 x^{2} - 16 x + 8} + \frac{1}{8 x^{2} - 16 x + 8} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.10366, size = 43, normalized size = 0.96 \begin{align*} -\frac{1}{4 \,{\left (x - 1\right )}} - \frac{\arctan \left (x\right )}{2 \,{\left (x - 1\right )}^{2}} + \frac{1}{8} \, \log \left (x^{2} + 1\right ) - \frac{1}{4} \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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