Optimal. Leaf size=81 \[ -\frac {2 \text {ArcTan}(x)}{3 x}-\frac {\text {ArcTan}(x)^2}{3}+\log (x)-\frac {1}{2} \log \left (1+x^2\right )-\frac {\log \left (1+x^2\right )}{6 x^2}-\frac {\text {ArcTan}(x) \log \left (1+x^2\right )}{3 x^3}+\frac {1}{12} \log ^2\left (1+x^2\right )+\frac {1}{6} \text {PolyLog}\left (2,-x^2\right ) \]
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Rubi [A]
time = 0.14, antiderivative size = 81, normalized size of antiderivative = 1.00, number of
steps used = 18, number of rules used = 15, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.250, Rules used
= {4946, 272, 46, 5137, 2525, 2457, 2442, 36, 29, 31, 2438, 2437, 2338, 5038, 5004}
\begin {gather*} -\frac {\text {ArcTan}(x) \log \left (x^2+1\right )}{3 x^3}-\frac {1}{3} \text {ArcTan}(x)^2-\frac {2 \text {ArcTan}(x)}{3 x}+\frac {\text {Li}_2\left (-x^2\right )}{6}+\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 )+\log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 46
Rule 272
Rule 2338
Rule 2437
Rule 2438
Rule 2442
Rule 2457
Rule 2525
Rule 4946
Rule 5004
Rule 5038
Rule 5137
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} \text {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} \text {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} \text {Subst}\left (\int \frac {\log (1+x)}{x^2} \, dx,x,x^2\right )-\frac {1}{6} \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,x^2\right )+\frac {1}{6} \text {Subst}\left (\int \frac {\log (1+x)}{1+x} \, dx,x,x^2\right )+\frac {1}{3} \text {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} \text {Subst}\left (\int \frac {1}{x (1+x)} \, dx,x,x^2\right )+\frac {1}{6} \text {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,1+x^2\right )+\frac {1}{3} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{3} \text {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} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{6} \text {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*}
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Mathematica [A]
time = 0.01, size = 81, normalized size = 1.00 \begin {gather*} -\frac {2 \text {ArcTan}(x)}{3 x}-\frac {\text {ArcTan}(x)^2}{3}+\log (x)-\frac {1}{2} \log \left (1+x^2\right )-\frac {\log \left (1+x^2\right )}{6 x^2}-\frac {\text {ArcTan}(x) \log \left (1+x^2\right )}{3 x^3}+\frac {1}{12} \log ^2\left (1+x^2\right )+\frac {1}{6} \text {PolyLog}\left (2,-x^2\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 6.70, size = 0, normalized size = 0.00 \[\int \frac {\arctan \left (x \right ) \ln \left (x^{2}+1\right )}{x^{4}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.49, size = 95, normalized size = 1.17 \begin {gather*} -\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 {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [C] Result contains complex when optimal does not.
time = 18.52, size = 97, normalized size = 1.20 \begin {gather*} \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 {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\ln \left (x^2+1\right )\,\mathrm {atan}\left (x\right )}{x^4} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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