Optimal. Leaf size=81 \[ \frac{1}{6} \text{PolyLog}\left (2,-x^2\right )+\frac{1}{12} \log ^2\left (x^2+1\right )-\frac{\log \left (x^2+1\right )}{6 x^2}-\frac{1}{2} \log \left (x^2+1\right )-\frac{\log \left (x^2+1\right ) \tan ^{-1}(x)}{3 x^3}+\log (x)-\frac{1}{3} \tan ^{-1}(x)^2-\frac{2 \tan ^{-1}(x)}{3 x} \]
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Rubi [A] time = 0.209636, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 15, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.25, Rules used = {4852, 266, 44, 5017, 2475, 2410, 2395, 36, 29, 31, 2391, 2390, 2301, 4918, 4884} \[ \frac{1}{6} \text{PolyLog}\left (2,-x^2\right )+\frac{1}{12} \log ^2\left (x^2+1\right )-\frac{\log \left (x^2+1\right )}{6 x^2}-\frac{1}{2} \log \left (x^2+1\right )-\frac{\log \left (x^2+1\right ) \tan ^{-1}(x)}{3 x^3}+\log (x)-\frac{1}{3} \tan ^{-1}(x)^2-\frac{2 \tan ^{-1}(x)}{3 x} \]
Antiderivative was successfully verified.
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Rule 4852
Rule 266
Rule 44
Rule 5017
Rule 2475
Rule 2410
Rule 2395
Rule 36
Rule 29
Rule 31
Rule 2391
Rule 2390
Rule 2301
Rule 4918
Rule 4884
Rubi steps
\begin{align*} \int \frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{x^4} \, dx &=-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{3 x^3}+\frac{1}{3} \int \frac{\log \left (1+x^2\right )}{x^3 \left (1+x^2\right )} \, dx+\frac{2}{3} \int \frac{\tan ^{-1}(x)}{x^2 \left (1+x^2\right )} \, dx\\ &=-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{3 x^3}+\frac{1}{6} \operatorname{Subst}\left (\int \frac{\log (1+x)}{x^2 (1+x)} \, dx,x,x^2\right )+\frac{2}{3} \int \frac{\tan ^{-1}(x)}{x^2} \, dx-\frac{2}{3} \int \frac{\tan ^{-1}(x)}{1+x^2} \, dx\\ &=-\frac{2 \tan ^{-1}(x)}{3 x}-\frac{1}{3} \tan ^{-1}(x)^2-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{3 x^3}+\frac{1}{6} \operatorname{Subst}\left (\int \left (\frac{\log (1+x)}{x^2}-\frac{\log (1+x)}{x}+\frac{\log (1+x)}{1+x}\right ) \, dx,x,x^2\right )+\frac{2}{3} \int \frac{1}{x \left (1+x^2\right )} \, dx\\ &=-\frac{2 \tan ^{-1}(x)}{3 x}-\frac{1}{3} \tan ^{-1}(x)^2-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{3 x^3}+\frac{1}{6} \operatorname{Subst}\left (\int \frac{\log (1+x)}{x^2} \, dx,x,x^2\right )-\frac{1}{6} \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,x^2\right )+\frac{1}{6} \operatorname{Subst}\left (\int \frac{\log (1+x)}{1+x} \, dx,x,x^2\right )+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{x (1+x)} \, dx,x,x^2\right )\\ &=-\frac{2 \tan ^{-1}(x)}{3 x}-\frac{1}{3} \tan ^{-1}(x)^2-\frac{\log \left (1+x^2\right )}{6 x^2}-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{3 x^3}+\frac{\text{Li}_2\left (-x^2\right )}{6}+\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{x (1+x)} \, dx,x,x^2\right )+\frac{1}{6} \operatorname{Subst}\left (\int \frac{\log (x)}{x} \, dx,x,1+x^2\right )+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )-\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,x^2\right )\\ &=-\frac{2 \tan ^{-1}(x)}{3 x}-\frac{1}{3} \tan ^{-1}(x)^2+\frac{2 \log (x)}{3}-\frac{1}{3} \log \left (1+x^2\right )-\frac{\log \left (1+x^2\right )}{6 x^2}-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{3 x^3}+\frac{1}{12} \log ^2\left (1+x^2\right )+\frac{\text{Li}_2\left (-x^2\right )}{6}+\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,x^2\right )\\ &=-\frac{2 \tan ^{-1}(x)}{3 x}-\frac{1}{3} \tan ^{-1}(x)^2+\log (x)-\frac{1}{2} \log \left (1+x^2\right )-\frac{\log \left (1+x^2\right )}{6 x^2}-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{3 x^3}+\frac{1}{12} \log ^2\left (1+x^2\right )+\frac{\text{Li}_2\left (-x^2\right )}{6}\\ \end{align*}
Mathematica [A] time = 0.0127778, size = 81, normalized size = 1. \[ \frac{1}{6} \text{PolyLog}\left (2,-x^2\right )+\frac{1}{12} \log ^2\left (x^2+1\right )-\frac{\log \left (x^2+1\right )}{6 x^2}-\frac{1}{2} \log \left (x^2+1\right )-\frac{\log \left (x^2+1\right ) \tan ^{-1}(x)}{3 x^3}+\log (x)-\frac{1}{3} \tan ^{-1}(x)^2-\frac{2 \tan ^{-1}(x)}{3 x} \]
Antiderivative was successfully verified.
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Maple [F] time = 1.478, size = 0, normalized size = 0. \begin{align*} \int{\frac{\arctan \left ( x \right ) \ln \left ({x}^{2}+1 \right ) }{{x}^{4}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.64977, size = 128, normalized size = 1.58 \begin{align*} -\frac{1}{3} \,{\left (\frac{2}{x} + \frac{\log \left (x^{2} + 1\right )}{x^{3}} + 2 \, \arctan \left (x\right )\right )} \arctan \left (x\right ) + \frac{4 \, x^{2} \arctan \left (x\right )^{2} + x^{2} \log \left (x^{2} + 1\right )^{2} - 2 \, x^{2}{\rm Li}_2\left (x^{2} + 1\right ) + 12 \, x^{2} \log \left (x\right ) - 2 \,{\left (x^{2} \log \left (-x^{2}\right ) + 3 \, x^{2} + 1\right )} \log \left (x^{2} + 1\right )}{12 \, x^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\arctan \left (x\right ) \log \left (x^{2} + 1\right )}{x^{4}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [C] time = 35.9001, size = 97, normalized size = 1.2 \begin{align*} \frac{2 \log{\left (x \right )}}{3} + \frac{\log{\left (2 x^{2} \right )}}{6} + \frac{\log{\left (x^{2} + 1 \right )}^{2}}{12} - \frac{\log{\left (x^{2} + 1 \right )}}{3} - \frac{\log{\left (2 x^{2} + 2 \right )}}{6} - \frac{\operatorname{atan}^{2}{\left (x \right )}}{3} + \frac{\operatorname{Li}_{2}\left (x^{2} e^{i \pi }\right )}{6} - \frac{2 \operatorname{atan}{\left (x \right )}}{3 x} - \frac{\log{\left (x^{2} + 1 \right )}}{6 x^{2}} - \frac{\log{\left (x^{2} + 1 \right )} \operatorname{atan}{\left (x \right )}}{3 x^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (x\right ) \log \left (x^{2} + 1\right )}{x^{4}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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