Optimal. Leaf size=128 \[ -\frac{i x \text{PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}+\frac{\text{PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{x \tan (a+b x)}{b^2}-\frac{\log (\cos (a+b x))}{b^3}+\frac{x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{x^2 \tan ^2(a+b x)}{2 b}+\frac{x^2}{2 b}-\frac{i x^3}{3} \]
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Rubi [A] time = 0.180095, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {3720, 3475, 30, 3719, 2190, 2531, 2282, 6589} \[ -\frac{i x \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}+\frac{\text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{x \tan (a+b x)}{b^2}-\frac{\log (\cos (a+b x))}{b^3}+\frac{x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac{x^2 \tan ^2(a+b x)}{2 b}+\frac{x^2}{2 b}-\frac{i x^3}{3} \]
Antiderivative was successfully verified.
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Rule 3720
Rule 3475
Rule 30
Rule 3719
Rule 2190
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x^2 \tan ^3(a+b x) \, dx &=\frac{x^2 \tan ^2(a+b x)}{2 b}-\frac{\int x \tan ^2(a+b x) \, dx}{b}-\int x^2 \tan (a+b x) \, dx\\ &=-\frac{i x^3}{3}-\frac{x \tan (a+b x)}{b^2}+\frac{x^2 \tan ^2(a+b x)}{2 b}+2 i \int \frac{e^{2 i (a+b x)} x^2}{1+e^{2 i (a+b x)}} \, dx+\frac{\int \tan (a+b x) \, dx}{b^2}+\frac{\int x \, dx}{b}\\ &=\frac{x^2}{2 b}-\frac{i x^3}{3}+\frac{x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{\log (\cos (a+b x))}{b^3}-\frac{x \tan (a+b x)}{b^2}+\frac{x^2 \tan ^2(a+b x)}{2 b}-\frac{2 \int x \log \left (1+e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=\frac{x^2}{2 b}-\frac{i x^3}{3}+\frac{x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{\log (\cos (a+b x))}{b^3}-\frac{i x \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac{x \tan (a+b x)}{b^2}+\frac{x^2 \tan ^2(a+b x)}{2 b}+\frac{i \int \text{Li}_2\left (-e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=\frac{x^2}{2 b}-\frac{i x^3}{3}+\frac{x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{\log (\cos (a+b x))}{b^3}-\frac{i x \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}-\frac{x \tan (a+b x)}{b^2}+\frac{x^2 \tan ^2(a+b x)}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^3}\\ &=\frac{x^2}{2 b}-\frac{i x^3}{3}+\frac{x^2 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac{\log (\cos (a+b x))}{b^3}-\frac{i x \text{Li}_2\left (-e^{2 i (a+b x)}\right )}{b^2}+\frac{\text{Li}_3\left (-e^{2 i (a+b x)}\right )}{2 b^3}-\frac{x \tan (a+b x)}{b^2}+\frac{x^2 \tan ^2(a+b x)}{2 b}\\ \end{align*}
Mathematica [A] time = 2.4844, size = 172, normalized size = 1.34 \[ \frac{e^{-i a} \sec (a) \left (6 i \left (1+e^{2 i a}\right ) b x \text{PolyLog}\left (2,-e^{-2 i (a+b x)}\right )+3 \left (1+e^{2 i a}\right ) \text{PolyLog}\left (3,-e^{-2 i (a+b x)}\right )+2 b^2 x^2 \left (3 \left (1+e^{2 i a}\right ) \log \left (1+e^{-2 i (a+b x)}\right )+2 i b x\right )\right )-4 b^3 x^3 \tan (a)+6 b^2 x^2 \sec ^2(a+b x)-12 b x \sec (a) \sin (b x) \sec (a+b x)-12 (b x \tan (a)+\log (\cos (a+b x)))}{12 b^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.058, size = 180, normalized size = 1.4 \begin{align*} -{\frac{i}{3}}{x}^{3}+2\,{\frac{x \left ( bx{{\rm e}^{2\,i \left ( bx+a \right ) }}-i{{\rm e}^{2\,i \left ( bx+a \right ) }}-i \right ) }{{b}^{2} \left ( 1+{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) ^{2}}}-2\,{\frac{{a}^{2}\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+{\frac{2\,i{a}^{2}x}{{b}^{2}}}+{\frac{{x}^{2}\ln \left ( 1+{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{b}}-{\frac{ix{\it polylog} \left ( 2,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{2}}}+{\frac{{\it polylog} \left ( 3,-{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{2\,{b}^{3}}}-{\frac{\ln \left ( 1+{{\rm e}^{2\,i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+2\,{\frac{\ln \left ({{\rm e}^{i \left ( bx+a \right ) }} \right ) }{{b}^{3}}}+{\frac{{\frac{4\,i}{3}}{a}^{3}}{{b}^{3}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.98648, size = 999, normalized size = 7.8 \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 = 1.75364, size = 657, normalized size = 5.13 \begin{align*} \frac{2 \, b^{2} x^{2} \tan \left (b x + a\right )^{2} + 2 \, b^{2} x^{2} + 2 i \, b x{\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 i \, b x{\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 x \tan \left (b x + a\right ) + 2 \,{\left (b^{2} x^{2} - 1\right )} \log \left (-\frac{2 \,{\left (i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \,{\left (b^{2} x^{2} - 1\right )} \log \left (-\frac{2 \,{\left (-i \, \tan \left (b x + a\right ) - 1\right )}}{\tan \left (b x + a\right )^{2} + 1}\right ) +{\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 ) +{\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 )}{4 \, b^{3}} \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^{2} \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^{2} \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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