Optimal. Leaf size=102 \[ -\frac{1}{4} i \text{PolyLog}(2,-i x)+\frac{1}{4} i \text{PolyLog}(2,i x)+\frac{\log \left (x^2+1\right )}{4 x}-\frac{\log \left (x^2+1\right )}{12 x^3}-\frac{\tan ^{-1}(x)}{4 x^2}-\frac{\log \left (x^2+1\right ) \tan ^{-1}(x)}{4 x^4}+\frac{1}{4} \log \left (x^2+1\right ) \tan ^{-1}(x)-\frac{5}{12 x}-\frac{11}{12} \tan ^{-1}(x) \]
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Rubi [A] time = 0.129238, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {4852, 325, 203, 5021, 453, 4980, 4848, 2391} \[ -\frac{1}{4} i \text{PolyLog}(2,-i x)+\frac{1}{4} i \text{PolyLog}(2,i x)+\frac{\log \left (x^2+1\right )}{4 x}-\frac{\log \left (x^2+1\right )}{12 x^3}-\frac{\tan ^{-1}(x)}{4 x^2}-\frac{\log \left (x^2+1\right ) \tan ^{-1}(x)}{4 x^4}+\frac{1}{4} \log \left (x^2+1\right ) \tan ^{-1}(x)-\frac{5}{12 x}-\frac{11}{12} \tan ^{-1}(x) \]
Antiderivative was successfully verified.
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Rule 4852
Rule 325
Rule 203
Rule 5021
Rule 453
Rule 4980
Rule 4848
Rule 2391
Rubi steps
\begin{align*} \int \frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{x^5} \, dx &=-\frac{\log \left (1+x^2\right )}{12 x^3}+\frac{\log \left (1+x^2\right )}{4 x}+\frac{1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{4 x^4}-2 \int \left (\frac{-1+3 x^2}{12 x^2 \left (1+x^2\right )}+\frac{\left (-1+x^2\right ) \tan ^{-1}(x)}{4 x^3}\right ) \, dx\\ &=-\frac{\log \left (1+x^2\right )}{12 x^3}+\frac{\log \left (1+x^2\right )}{4 x}+\frac{1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{4 x^4}-\frac{1}{6} \int \frac{-1+3 x^2}{x^2 \left (1+x^2\right )} \, dx-\frac{1}{2} \int \frac{\left (-1+x^2\right ) \tan ^{-1}(x)}{x^3} \, dx\\ &=-\frac{1}{6 x}-\frac{\log \left (1+x^2\right )}{12 x^3}+\frac{\log \left (1+x^2\right )}{4 x}+\frac{1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{4 x^4}-\frac{1}{2} \int \left (-\frac{\tan ^{-1}(x)}{x^3}+\frac{\tan ^{-1}(x)}{x}\right ) \, dx-\frac{2}{3} \int \frac{1}{1+x^2} \, dx\\ &=-\frac{1}{6 x}-\frac{2}{3} \tan ^{-1}(x)-\frac{\log \left (1+x^2\right )}{12 x^3}+\frac{\log \left (1+x^2\right )}{4 x}+\frac{1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{4 x^4}+\frac{1}{2} \int \frac{\tan ^{-1}(x)}{x^3} \, dx-\frac{1}{2} \int \frac{\tan ^{-1}(x)}{x} \, dx\\ &=-\frac{1}{6 x}-\frac{2}{3} \tan ^{-1}(x)-\frac{\tan ^{-1}(x)}{4 x^2}-\frac{\log \left (1+x^2\right )}{12 x^3}+\frac{\log \left (1+x^2\right )}{4 x}+\frac{1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{4 x^4}-\frac{1}{4} i \int \frac{\log (1-i x)}{x} \, dx+\frac{1}{4} i \int \frac{\log (1+i x)}{x} \, dx+\frac{1}{4} \int \frac{1}{x^2 \left (1+x^2\right )} \, dx\\ &=-\frac{5}{12 x}-\frac{2}{3} \tan ^{-1}(x)-\frac{\tan ^{-1}(x)}{4 x^2}-\frac{\log \left (1+x^2\right )}{12 x^3}+\frac{\log \left (1+x^2\right )}{4 x}+\frac{1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{4 x^4}-\frac{1}{4} i \text{Li}_2(-i x)+\frac{1}{4} i \text{Li}_2(i x)-\frac{1}{4} \int \frac{1}{1+x^2} \, dx\\ &=-\frac{5}{12 x}-\frac{11}{12} \tan ^{-1}(x)-\frac{\tan ^{-1}(x)}{4 x^2}-\frac{\log \left (1+x^2\right )}{12 x^3}+\frac{\log \left (1+x^2\right )}{4 x}+\frac{1}{4} \tan ^{-1}(x) \log \left (1+x^2\right )-\frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{4 x^4}-\frac{1}{4} i \text{Li}_2(-i x)+\frac{1}{4} i \text{Li}_2(i x)\\ \end{align*}
Mathematica [A] time = 0.0350324, size = 98, normalized size = 0.96 \[ -\frac{1}{4} i (\text{PolyLog}(2,-i x)-\text{PolyLog}(2,i x))+\frac{1}{2} \left (\frac{1}{2} \left (-\frac{1}{x}-\tan ^{-1}(x)\right )-\frac{\tan ^{-1}(x)}{2 x^2}\right )+\frac{\log \left (x^2+1\right ) \left (3 x^3+3 x^4 \tan ^{-1}(x)-x-3 \tan ^{-1}(x)\right )}{12 x^4}-\frac{1}{6 x}-\frac{2}{3} \tan ^{-1}(x) \]
Antiderivative was successfully verified.
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Maple [F] time = 3.61, size = 0, normalized size = 0. \begin{align*} \int{\frac{\arctan \left ( x \right ) \ln \left ({x}^{2}+1 \right ) }{{x}^{5}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.68755, size = 120, normalized size = 1.18 \begin{align*} -\frac{12 \, x^{4} \arctan \left (x\right ) \log \left (x\right ) - 6 i \, x^{4}{\rm Li}_2\left (i \, x + 1\right ) + 6 i \, x^{4}{\rm Li}_2\left (-i \, x + 1\right ) + 10 \, x^{3} + 2 \,{\left (11 \, x^{4} + 3 \, x^{2}\right )} \arctan \left (x\right ) -{\left (3 \, \pi x^{4} + 6 \, x^{3} + 6 \,{\left (x^{4} - 1\right )} \arctan \left (x\right ) - 2 \, x\right )} \log \left (x^{2} + 1\right )}{24 \, x^{4}} \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^{5}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\log{\left (x^{2} + 1 \right )} \operatorname{atan}{\left (x \right )}}{x^{5}}\, dx \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^{5}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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