Optimal. Leaf size=90 \[ -\frac{i \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{\tan (a+b x)}{2 b^2}+\frac{x \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{x \tan ^2(a+b x)}{2 b}+\frac{x}{2 b}-\frac{i x^2}{2} \]
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Rubi [A] time = 0.102632, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {3720, 3473, 8, 3719, 2190, 2279, 2391} \[ -\frac{i \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{\tan (a+b x)}{2 b^2}+\frac{x \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{x \tan ^2(a+b x)}{2 b}+\frac{x}{2 b}-\frac{i x^2}{2} \]
Antiderivative was successfully verified.
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Rule 3720
Rule 3473
Rule 8
Rule 3719
Rule 2190
Rule 2279
Rule 2391
Rubi steps
\begin{align*} \int x \tan ^3(a+b x) \, dx &=\frac{x \tan ^2(a+b x)}{2 b}-\frac{\int \tan ^2(a+b x) \, dx}{2 b}-\int x \tan (a+b x) \, dx\\ &=-\frac{i x^2}{2}-\frac{\tan (a+b x)}{2 b^2}+\frac{x \tan ^2(a+b x)}{2 b}+2 i \int \frac{e^{2 i (a+b x)} x}{1+e^{2 i (a+b x)}} \, dx+\frac{\int 1 \, dx}{2 b}\\ &=\frac{x}{2 b}-\frac{i x^2}{2}+\frac{x \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{\tan (a+b x)}{2 b^2}+\frac{x \tan ^2(a+b x)}{2 b}-\frac{\int \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=\frac{x}{2 b}-\frac{i x^2}{2}+\frac{x \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{\tan (a+b x)}{2 b^2}+\frac{x \tan ^2(a+b x)}{2 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{x}{2 b}-\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}-\frac{\tan (a+b x)}{2 b^2}+\frac{x \tan ^2(a+b x)}{2 b}\\ \end{align*}
Mathematica [A] time = 4.31034, size = 171, normalized size = 1.9 \[ \frac{-i \text{PolyLog}\left (2,e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )-b^2 x^2 \tan (a)+b^2 x^2 \tan (a) \sqrt{\csc ^2(a)} e^{-i \tan ^{-1}(\cot (a))}+b x \sec ^2(a+b x)+i b x \left (2 \tan ^{-1}(\cot (a))+\pi \right )-\sec (a) \sin (b x) \sec (a+b x)+2 \left (b x-\tan ^{-1}(\cot (a))\right ) \log \left (1-e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )+2 \tan ^{-1}(\cot (a)) \log \left (\sin \left (b x-\tan ^{-1}(\cot (a))\right )\right )+\pi \log \left (1+e^{-2 i b x}\right )-\pi \log (\cos (b x))}{2 b^2} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.051, size = 122, normalized size = 1.4 \begin{align*} -{\frac{i}{2}}{x}^{2}+{\frac{2\,bx{{\rm e}^{2\,i \left ( bx+a \right ) }}-i{{\rm e}^{2\,i \left ( bx+a \right ) }}-i}{{b}^{2} \left ( 1+{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) ^{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 = 1.87564, size = 524, normalized size = 5.82 \begin{align*} -\frac{b^{2} x^{2} \cos \left (4 \, b x + 4 \, a\right ) + i \, b^{2} x^{2} \sin \left (4 \, b x + 4 \, a\right ) + b^{2} x^{2} -{\left (2 \, b x \cos \left (4 \, b x + 4 \, a\right ) + 4 \, b x \cos \left (2 \, b x + 2 \, a\right ) + 2 i \, b x \sin \left (4 \, b x + 4 \, a\right ) + 4 i \, b x \sin \left (2 \, b x + 2 \, a\right ) + 2 \, b x\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 2 \,{\left (b^{2} x^{2} + 2 i \, b x + 1\right )} \cos \left (2 \, b x + 2 \, a\right ) +{\left (\cos \left (4 \, b x + 4 \, a\right ) + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (4 \, b x + 4 \, a\right ) + 2 i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right )}{\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) -{\left (-i \, b x \cos \left (4 \, b x + 4 \, a\right ) - 2 i \, b x \cos \left (2 \, b x + 2 \, a\right ) + b x \sin \left (4 \, b x + 4 \, a\right ) + 2 \, b x \sin \left (2 \, b x + 2 \, a\right ) - i \, b x\right )} \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 ) -{\left (-2 i \, b^{2} x^{2} + 4 \, b x - 2 i\right )} \sin \left (2 \, b x + 2 \, a\right ) + 2}{-2 i \, b^{2} \cos \left (4 \, b x + 4 \, a\right ) - 4 i \, b^{2} \cos \left (2 \, b x + 2 \, a\right ) + 2 \, b^{2} \sin \left (4 \, b x + 4 \, a\right ) + 4 \, b^{2} \sin \left (2 \, b x + 2 \, a\right ) - 2 i \, b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.60697, size = 396, normalized size = 4.4 \begin{align*} \frac{2 \, b x \tan \left (b x + a\right )^{2} + 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 ) + 2 \, b x + 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 ) - 2 \, \tan \left (b x + a\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 ^{3}{\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 )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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