Optimal. Leaf size=205 \[ -\frac{3 i x^2 \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 x \text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac{3 i \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^4}+\frac{3 i \text{PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}-\frac{3 x^2 \tan (a+b x)}{2 b^2}-\frac{3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{x^3 \tan ^2(a+b x)}{2 b}+\frac{3 i x^2}{2 b^2}+\frac{x^3}{2 b}-\frac{i x^4}{4} \]
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Rubi [A] time = 0.292188, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.833, Rules used = {3720, 3719, 2190, 2279, 2391, 30, 2531, 6609, 2282, 6589} \[ -\frac{3 i x^2 \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^2}+\frac{3 x \text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}+\frac{3 i \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^4}+\frac{3 i \text{Li}_4\left (-e^{2 i (a+b x)}\right )}{4 b^4}-\frac{3 x^2 \tan (a+b x)}{2 b^2}-\frac{3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{x^3 \tan ^2(a+b x)}{2 b}+\frac{3 i x^2}{2 b^2}+\frac{x^3}{2 b}-\frac{i x^4}{4} \]
Antiderivative was successfully verified.
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Rule 3720
Rule 3719
Rule 2190
Rule 2279
Rule 2391
Rule 30
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x^3 \tan ^3(a+b x) \, dx &=\frac{x^3 \tan ^2(a+b x)}{2 b}-\frac{3 \int x^2 \tan ^2(a+b x) \, dx}{2 b}-\int x^3 \tan (a+b x) \, dx\\ &=-\frac{i x^4}{4}-\frac{3 x^2 \tan (a+b x)}{2 b^2}+\frac{x^3 \tan ^2(a+b x)}{2 b}+2 i \int \frac{e^{2 i (a+b x)} x^3}{1+e^{2 i (a+b x)}} \, dx+\frac{3 \int x \tan (a+b x) \, dx}{b^2}+\frac{3 \int x^2 \, dx}{2 b}\\ &=\frac{3 i x^2}{2 b^2}+\frac{x^3}{2 b}-\frac{i x^4}{4}+\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{3 x^2 \tan (a+b x)}{2 b^2}+\frac{x^3 \tan ^2(a+b x)}{2 b}-\frac{(6 i) \int \frac{e^{2 i (a+b x)} x}{1+e^{2 i (a+b x)}} \, dx}{b^2}-\frac{3 \int x^2 \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=\frac{3 i x^2}{2 b^2}+\frac{x^3}{2 b}-\frac{i x^4}{4}-\frac{3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\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^2 \tan (a+b x)}{2 b^2}+\frac{x^3 \tan ^2(a+b x)}{2 b}+\frac{3 \int \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b^3}+\frac{(3 i) \int x \text{Li}_2\left (-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=\frac{3 i x^2}{2 b^2}+\frac{x^3}{2 b}-\frac{i x^4}{4}-\frac{3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\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 x^2 \tan (a+b x)}{2 b^2}+\frac{x^3 \tan ^2(a+b x)}{2 b}-\frac{(3 i) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^4}-\frac{3 \int \text{Li}_3\left (-e^{2 i (a+b x)}\right ) \, dx}{2 b^3}\\ &=\frac{3 i x^2}{2 b^2}+\frac{x^3}{2 b}-\frac{i x^4}{4}-\frac{3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{3 i \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^4}-\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 x^2 \tan (a+b x)}{2 b^2}+\frac{x^3 \tan ^2(a+b x)}{2 b}+\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{3 i x^2}{2 b^2}+\frac{x^3}{2 b}-\frac{i x^4}{4}-\frac{3 x \log \left (1+e^{2 i (a+b x)}\right )}{b^3}+\frac{x^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{3 i \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{2 b^4}-\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{3 x^2 \tan (a+b x)}{2 b^2}+\frac{x^3 \tan ^2(a+b x)}{2 b}\\ \end{align*}
Mathematica [A] time = 6.51339, size = 359, normalized size = 1.75 \[ \frac{1}{8} i e^{i a} \sec (a) \left (\frac{3 e^{-2 i a} \left (1+e^{2 i a}\right ) \left (2 b^2 x^2 \text{PolyLog}\left (2,-e^{-2 i (a+b x)}\right )-2 i b x \text{PolyLog}\left (3,-e^{-2 i (a+b x)}\right )-\text{PolyLog}\left (4,-e^{-2 i (a+b x)}\right )\right )}{b^4}-\frac{4 i \left (1+e^{-2 i a}\right ) x^3 \log \left (1+e^{-2 i (a+b x)}\right )}{b}+2 e^{-2 i a} x^4\right )-\frac{3 \csc (a) \sec (a) \left (b^2 x^2 e^{-i \tan ^{-1}(\cot (a))}-\frac{\cot (a) \left (i \text{PolyLog}\left (2,e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )+i b x \left (-2 \tan ^{-1}(\cot (a))-\pi \right )-2 \left (b x-\tan ^{-1}(\cot (a))\right ) \log \left (1-e^{2 i \left (b x-\tan ^{-1}(\cot (a))\right )}\right )-2 \tan ^{-1}(\cot (a)) \log \left (\sin \left (b x-\tan ^{-1}(\cot (a))\right )\right )-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))\right )}{\sqrt{\cot ^2(a)+1}}\right )}{2 b^4 \sqrt{\csc ^2(a) \left (\sin ^2(a)+\cos ^2(a)\right )}}-\frac{3 x^2 \sec (a) \sin (b x) \sec (a+b x)}{2 b^2}+\frac{x^3 \sec ^2(a+b x)}{2 b}-\frac{1}{4} x^4 \tan (a) \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.069, size = 251, normalized size = 1.2 \begin{align*}{\frac{{\frac{3\,i}{2}}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{4}}}+{\frac{{x}^{2} \left ( 2\,bx{{\rm e}^{2\,i \left ( bx+a \right ) }}-3\,i{{\rm e}^{2\,i \left ( bx+a \right ) }}-3\,i \right ) }{{b}^{2} \left ( 1+{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) ^{2}}}+{\frac{3\,i{x}^{2}}{{b}^{2}}}-{\frac{i}{4}}{x}^{4}+{\frac{{\frac{3\,i}{4}}{\it polylog} \left ( 4,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{4}}}-6\,{\frac{a\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{4}}}+2\,{\frac{{a}^{3}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{4}}}-{\frac{2\,i{a}^{3}x}{{b}^{3}}}-{\frac{{\frac{3\,i}{2}}{x}^{2}{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{6\,iax}{{b}^{3}}}-{\frac{{\frac{3\,i}{2}}{a}^{4}}{{b}^{4}}}+{\frac{3\,i{a}^{2}}{{b}^{4}}}+{\frac{{x}^{3}\ln \left ( 1+{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{b}}-3\,{\frac{x\ln \left ( 1+{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+{\frac{3\,x{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{2\,{b}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 2.19241, size = 1624, normalized size = 7.92 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 1.69524, size = 938, normalized size = 4.58 \begin{align*} \frac{4 \, b^{3} x^{3} \tan \left (b x + a\right )^{2} + 4 \, b^{3} x^{3} - 12 \, b^{2} x^{2} \tan \left (b x + a\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 ) +{\left (6 i \, b^{2} x^{2} - 6 i\right )}{\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 ) +{\left (-6 i \, b^{2} x^{2} + 6 i\right )}{\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 \,{\left (b^{3} x^{3} - 3 \, b x\right )} \log \left (-\frac{2 \,{\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 4 \,{\left (b^{3} x^{3} - 3 \, b x\right )} \log \left (-\frac{2 \,{\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\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 ^{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^{3} \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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