Optimal. Leaf size=189 \[ -\frac{1}{2} i \left (-\log \left (x^2+1\right )+\log (1-i x)+\log (1+i x)\right ) \text{PolyLog}(2,-i x)+\frac{1}{2} i \left (-\log \left (x^2+1\right )+\log (1-i x)+\log (1+i x)\right ) \text{PolyLog}(2,i x)-i \text{PolyLog}(3,1-i x)+i \text{PolyLog}(3,1+i x)+i \log (1-i x) \text{PolyLog}(2,1-i x)-i \log (1+i x) \text{PolyLog}(2,1+i x)+\frac{1}{2} i \log (i x) \log ^2(1-i x)-\frac{1}{2} i \log ^2(1+i x) \log (-i x) \]
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Rubi [A] time = 0.18104, antiderivative size = 189, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {4848, 2391, 5011, 2396, 2433, 2374, 6589} \[ -\frac{1}{2} i \left (-\log \left (x^2+1\right )+\log (1-i x)+\log (1+i x)\right ) \text{PolyLog}(2,-i x)+\frac{1}{2} i \left (-\log \left (x^2+1\right )+\log (1-i x)+\log (1+i x)\right ) \text{PolyLog}(2,i x)-i \text{PolyLog}(3,1-i x)+i \text{PolyLog}(3,1+i x)+i \log (1-i x) \text{PolyLog}(2,1-i x)-i \log (1+i x) \text{PolyLog}(2,1+i x)+\frac{1}{2} i \log (i x) \log ^2(1-i x)-\frac{1}{2} i \log ^2(1+i x) \log (-i x) \]
Antiderivative was successfully verified.
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Rule 4848
Rule 2391
Rule 5011
Rule 2396
Rule 2433
Rule 2374
Rule 6589
Rubi steps
\begin{align*} \int \frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{x} \, dx &=\frac{1}{2} i \int \frac{\log ^2(1-i x)}{x} \, dx-\frac{1}{2} i \int \frac{\log ^2(1+i x)}{x} \, dx+\left (-\log (1-i x)-\log (1+i x)+\log \left (1+x^2\right )\right ) \int \frac{\tan ^{-1}(x)}{x} \, dx\\ &=-\frac{1}{2} i \log ^2(1+i x) \log (-i x)+\frac{1}{2} i \log ^2(1-i x) \log (i x)+\frac{1}{2} \left (i \left (\log (1-i x)+\log (1+i x)-\log \left (1+x^2\right )\right )\right ) \int \frac{\log (1+i x)}{x} \, dx+\frac{1}{2} \left (i \left (-\log (1-i x)-\log (1+i x)+\log \left (1+x^2\right )\right )\right ) \int \frac{\log (1-i x)}{x} \, dx-\int \frac{\log (1+i x) \log (-i x)}{1+i x} \, dx-\int \frac{\log (1-i x) \log (i x)}{1-i x} \, dx\\ &=-\frac{1}{2} i \log ^2(1+i x) \log (-i x)+\frac{1}{2} i \log ^2(1-i x) \log (i x)-\frac{1}{2} i \left (\log (1-i x)+\log (1+i x)-\log \left (1+x^2\right )\right ) \text{Li}_2(-i x)+\frac{1}{2} i \left (\log (1-i x)+\log (1+i x)-\log \left (1+x^2\right )\right ) \text{Li}_2(i x)+i \operatorname{Subst}\left (\int \frac{\log (-i (i-i x)) \log (x)}{x} \, dx,x,1+i x\right )-i \operatorname{Subst}\left (\int \frac{\log (i (-i+i x)) \log (x)}{x} \, dx,x,1-i x\right )\\ &=-\frac{1}{2} i \log ^2(1+i x) \log (-i x)+\frac{1}{2} i \log ^2(1-i x) \log (i x)+i \log (1-i x) \text{Li}_2(1-i x)-i \log (1+i x) \text{Li}_2(1+i x)-\frac{1}{2} i \left (\log (1-i x)+\log (1+i x)-\log \left (1+x^2\right )\right ) \text{Li}_2(-i x)+\frac{1}{2} i \left (\log (1-i x)+\log (1+i x)-\log \left (1+x^2\right )\right ) \text{Li}_2(i x)-i \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,1-i x\right )+i \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,1+i x\right )\\ &=-\frac{1}{2} i \log ^2(1+i x) \log (-i x)+\frac{1}{2} i \log ^2(1-i x) \log (i x)+i \log (1-i x) \text{Li}_2(1-i x)-i \log (1+i x) \text{Li}_2(1+i x)-\frac{1}{2} i \left (\log (1-i x)+\log (1+i x)-\log \left (1+x^2\right )\right ) \text{Li}_2(-i x)+\frac{1}{2} i \left (\log (1-i x)+\log (1+i x)-\log \left (1+x^2\right )\right ) \text{Li}_2(i x)-i \text{Li}_3(1-i x)+i \text{Li}_3(1+i x)\\ \end{align*}
Mathematica [F] time = 2.64211, size = 0, normalized size = 0. \[ \int \frac{\tan ^{-1}(x) \log \left (1+x^2\right )}{x} \, dx \]
Verification is Not applicable to the result.
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Maple [C] time = 0.974, size = 5237, normalized size = 27.7 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arctan \left (x\right ) \log \left (x^{2} + 1\right )}{x}\,{d x} \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}, 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}\, 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}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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