Optimal. Leaf size=54 \[ \frac{i \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{x \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i x^2}{2} \]
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Rubi [A] time = 0.0809173, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3719, 2190, 2279, 2391} \[ \frac{i \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{x \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i x^2}{2} \]
Antiderivative was successfully verified.
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Rule 3719
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int x \tan (a+b x) \, dx &=\frac{i x^2}{2}-2 i \int \frac{e^{2 i (a+b x)} x}{1+e^{2 i (a+b x)}} \, dx\\ &=\frac{i x^2}{2}-\frac{x \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{\int \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=\frac{i x^2}{2}-\frac{x \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{i \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^2}\\ &=\frac{i x^2}{2}-\frac{x \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}\\ \end{align*}
Mathematica [A] time = 0.0053405, size = 54, normalized size = 1. \[ \frac{i \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{x \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i x^2}{2} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.033, size = 78, normalized size = 1.4 \begin{align*}{\frac{i}{2}}{x}^{2}+{\frac{2\,iax}{b}}+{\frac{i{a}^{2}}{{b}^{2}}}-{\frac{x\ln \left ( 1+{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{b}}+{\frac{{\frac{i}{2}}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-2\,{\frac{a\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.7094, size = 124, normalized size = 2.3 \begin{align*} -\frac{-i \, b^{2} x^{2} + 2 i \, b x \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + b x \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - i \,{\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right )}{2 \, b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.6213, size = 332, normalized size = 6.15 \begin{align*} -\frac{2 \, b x \log \left (-\frac{2 \,{\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \, b x \log \left (-\frac{2 \,{\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + i \,{\rm Li}_2\left (\frac{2 \,{\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) - i \,{\rm Li}_2\left (\frac{2 \,{\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1} + 1\right )}{4 \, b^{2}} \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 x \tan{\left (a + b x \right )}\, 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 x \tan \left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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