Optimal. Leaf size=106 \[ \frac{3 i x^2 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{3 x \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 i \text{PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}-\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i x^4}{4} \]
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Rubi [A] time = 0.155191, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {3719, 2190, 2531, 6609, 2282, 6589} \[ \frac{3 i x^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{3 x \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 i \text{Li}_4\left (-e^{2 i (a+b x)}\right )}{4 b^4}-\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i x^4}{4} \]
Antiderivative was successfully verified.
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Rule 3719
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x^3 \tan (a+b x) \, dx &=\frac{i x^4}{4}-2 i \int \frac{e^{2 i (a+b x)} x^3}{1+e^{2 i (a+b x)}} \, dx\\ &=\frac{i x^4}{4}-\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{3 \int x^2 \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=\frac{i x^4}{4}-\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{3 i x^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{(3 i) \int x \text{Li}_2\left (-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=\frac{i x^4}{4}-\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{3 i x^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{3 x \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac{3 \int \text{Li}_3\left (-e^{2 i (a+b x)}\right ) \, dx}{2 b^3}\\ &=\frac{i x^4}{4}-\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{3 i x^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{3 x \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{4 b^4}\\ &=\frac{i x^4}{4}-\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{3 i x^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{3 x \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 i \text{Li}_4\left (-e^{2 i (a+b x)}\right )}{4 b^4}\\ \end{align*}
Mathematica [A] time = 0.0639529, size = 106, normalized size = 1. \[ \frac{3 i x^2 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}-\frac{3 x \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{3 i \text{PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}-\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{i x^4}{4} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.087, size = 125, normalized size = 1.2 \begin{align*}{\frac{i}{4}}{x}^{4}+{\frac{2\,i{a}^{3}x}{{b}^{3}}}+{\frac{{\frac{3\,i}{2}}{a}^{4}}{{b}^{4}}}-{\frac{{x}^{3}\ln \left ( 1+{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{b}}+{\frac{{\frac{3\,i}{2}}{x}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}-{\frac{3\,x{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{2\,{b}^{3}}}-{\frac{{\frac{3\,i}{4}}{\it polylog} \left ( 4,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{4}}}-2\,{\frac{{a}^{3}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{4}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.69412, size = 325, normalized size = 3.07 \begin{align*} -\frac{-3 i \,{\left (b x + a\right )}^{4} + 12 i \,{\left (b x + a\right )}^{3} a - 18 i \,{\left (b x + a\right )}^{2} a^{2} + 12 \, a^{3} \log \left (\sec \left (b x + a\right )\right ) +{\left (16 i \,{\left (b x + a\right )}^{3} - 36 i \,{\left (b x + a\right )}^{2} a + 36 i \,{\left (b x + a\right )} a^{2}\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) +{\left (-24 i \,{\left (b x + a\right )}^{2} + 36 i \,{\left (b x + a\right )} a - 18 i \, a^{2}\right )}{\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 2 \,{\left (4 \,{\left (b x + a\right )}^{3} - 9 \,{\left (b x + a\right )}^{2} a + 9 \,{\left (b x + a\right )} a^{2}\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 ) + 6 \,{\left (4 \, b x + a\right )}{\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )}) + 12 i \,{\rm Li}_{4}(-e^{\left (2 i \, b x + 2 i \, a\right )})}{12 \, b^{4}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 1.61102, size = 802, normalized size = 7.57 \begin{align*} -\frac{4 \, b^{3} x^{3} \log \left (-\frac{2 \,{\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 4 \, b^{3} x^{3} \log \left (-\frac{2 \,{\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 6 i \, b^{2} x^{2}{\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 ) - 6 i \, b^{2} x^{2}{\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 ) + 6 \, b x{\rm polylog}\left (3, \frac{\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 6 \, b x{\rm polylog}\left (3, \frac{\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 3 i \,{\rm polylog}\left (4, \frac{\tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 3 i \,{\rm polylog}\left (4, \frac{\tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) - 1}{\tan \left (b x + a\right )^{2} + 1}\right )}{8 \, b^{4}} \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^{3} \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^{3} \tan \left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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