Optimal. Leaf size=154 \[ -\frac{i x \text{PolyLog}\left (3,i c e^{2 i a+2 i b x}\right )}{4 b^2}+\frac{\text{PolyLog}\left (4,i c e^{2 i a+2 i b x}\right )}{8 b^3}-\frac{x^2 \text{PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b}-\frac{1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )+\frac{1}{3} x^3 \tan ^{-1}(c+(1+i c) \tan (a+b x))-\frac{b x^4}{12} \]
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Rubi [A] time = 0.237697, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {5171, 2184, 2190, 2531, 6609, 2282, 6589} \[ -\frac{i x \text{PolyLog}\left (3,i c e^{2 i a+2 i b x}\right )}{4 b^2}+\frac{\text{PolyLog}\left (4,i c e^{2 i a+2 i b x}\right )}{8 b^3}-\frac{x^2 \text{PolyLog}\left (2,i c e^{2 i a+2 i b x}\right )}{4 b}-\frac{1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )+\frac{1}{3} x^3 \tan ^{-1}(c+(1+i c) \tan (a+b x))-\frac{b x^4}{12} \]
Antiderivative was successfully verified.
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Rule 5171
Rule 2184
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x^2 \tan ^{-1}(c+(1+i c) \tan (a+b x)) \, dx &=\frac{1}{3} x^3 \tan ^{-1}(c+(1+i c) \tan (a+b x))-\frac{1}{3} (i b) \int \frac{x^3}{i (1+i c)+c+c e^{2 i a+2 i b x}} \, dx\\ &=-\frac{b x^4}{12}+\frac{1}{3} x^3 \tan ^{-1}(c+(1+i c) \tan (a+b x))+\frac{1}{3} (b c) \int \frac{e^{2 i a+2 i b x} x^3}{i (1+i c)+c+c e^{2 i a+2 i b x}} \, dx\\ &=-\frac{b x^4}{12}+\frac{1}{3} x^3 \tan ^{-1}(c+(1+i c) \tan (a+b x))-\frac{1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )+\frac{1}{2} i \int x^2 \log \left (1+\frac{c e^{2 i a+2 i b x}}{i (1+i c)+c}\right ) \, dx\\ &=-\frac{b x^4}{12}+\frac{1}{3} x^3 \tan ^{-1}(c+(1+i c) \tan (a+b x))-\frac{1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )-\frac{x^2 \text{Li}_2\left (i c e^{2 i a+2 i b x}\right )}{4 b}+\frac{\int x \text{Li}_2\left (-\frac{c e^{2 i a+2 i b x}}{i (1+i c)+c}\right ) \, dx}{2 b}\\ &=-\frac{b x^4}{12}+\frac{1}{3} x^3 \tan ^{-1}(c+(1+i c) \tan (a+b x))-\frac{1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )-\frac{x^2 \text{Li}_2\left (i c e^{2 i a+2 i b x}\right )}{4 b}-\frac{i x \text{Li}_3\left (i c e^{2 i a+2 i b x}\right )}{4 b^2}+\frac{i \int \text{Li}_3\left (-\frac{c e^{2 i a+2 i b x}}{i (1+i c)+c}\right ) \, dx}{4 b^2}\\ &=-\frac{b x^4}{12}+\frac{1}{3} x^3 \tan ^{-1}(c+(1+i c) \tan (a+b x))-\frac{1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )-\frac{x^2 \text{Li}_2\left (i c e^{2 i a+2 i b x}\right )}{4 b}-\frac{i x \text{Li}_3\left (i c e^{2 i a+2 i b x}\right )}{4 b^2}+\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_3(i c x)}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3}\\ &=-\frac{b x^4}{12}+\frac{1}{3} x^3 \tan ^{-1}(c+(1+i c) \tan (a+b x))-\frac{1}{6} i x^3 \log \left (1-i c e^{2 i a+2 i b x}\right )-\frac{x^2 \text{Li}_2\left (i c e^{2 i a+2 i b x}\right )}{4 b}-\frac{i x \text{Li}_3\left (i c e^{2 i a+2 i b x}\right )}{4 b^2}+\frac{\text{Li}_4\left (i c e^{2 i a+2 i b x}\right )}{8 b^3}\\ \end{align*}
Mathematica [A] time = 0.423886, size = 140, normalized size = 0.91 \[ \frac{1}{3} x^3 \tan ^{-1}(c+(1+i c) \tan (a+b x))-\frac{-6 b^2 x^2 \text{PolyLog}\left (2,-\frac{i e^{-2 i (a+b x)}}{c}\right )+6 i b x \text{PolyLog}\left (3,-\frac{i e^{-2 i (a+b x)}}{c}\right )+3 \text{PolyLog}\left (4,-\frac{i e^{-2 i (a+b x)}}{c}\right )+4 i b^3 x^3 \log \left (1+\frac{i e^{-2 i (a+b x)}}{c}\right )}{24 b^3} \]
Antiderivative was successfully verified.
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Maple [C] time = 17.733, size = 1532, normalized size = 10. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.14926, size = 417, normalized size = 2.71 \begin{align*} \frac{\frac{{\left ({\left (b x + a\right )}^{3} - 3 \,{\left (b x + a\right )}^{2} a + 3 \,{\left (b x + a\right )} a^{2}\right )} \arctan \left ({\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c\right )}{b^{2}} - \frac{3 \,{\left (-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} +{\left (-8 i \,{\left (b x + a\right )}^{3} + 18 i \,{\left (b x + a\right )}^{2} a - 18 i \,{\left (b x + a\right )} a^{2}\right )} \arctan \left (c \cos \left (2 \, b x + 2 \, a\right ), c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) +{\left (-12 i \,{\left (b x + a\right )}^{2} + 18 i \,{\left (b x + a\right )} a - 9 i \, a^{2}\right )}{\rm Li}_2\left (i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right ) +{\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 (c^{2} \cos \left (2 \, b x + 2 \, a\right )^{2} + c^{2} \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, c \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 3 \,{\left (4 \, b x + a\right )}{\rm Li}_{3}(i \, c e^{\left (2 i \, b x + 2 i \, a\right )}) + 6 i \,{\rm Li}_{4}(i \, c e^{\left (2 i \, b x + 2 i \, a\right )})\right )}{\left (i \, c + 1\right )}}{b^{2}{\left (12 \, c - 12 i\right )}}}{3 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.19974, size = 907, normalized size = 5.89 \begin{align*} -\frac{b^{4} x^{4} - 2 i \, b^{3} x^{3} \log \left (-\frac{{\left (c e^{\left (2 i \, b x + 2 i \, a\right )} + i\right )} e^{\left (-2 i \, b x - 2 i \, a\right )}}{c - i}\right ) + 6 \, b^{2} x^{2}{\rm Li}_2\left (\frac{1}{2} \, \sqrt{4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 6 \, b^{2} x^{2}{\rm Li}_2\left (-\frac{1}{2} \, \sqrt{4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) - a^{4} - 2 i \, a^{3} \log \left (\frac{2 \, c e^{\left (i \, b x + i \, a\right )} + i \, \sqrt{4 i \, c}}{2 \, c}\right ) - 2 i \, a^{3} \log \left (\frac{2 \, c e^{\left (i \, b x + i \, a\right )} - i \, \sqrt{4 i \, c}}{2 \, c}\right ) + 12 i \, b x{\rm polylog}\left (3, \frac{1}{2} \, \sqrt{4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) + 12 i \, b x{\rm polylog}\left (3, -\frac{1}{2} \, \sqrt{4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) -{\left (-2 i \, b^{3} x^{3} - 2 i \, a^{3}\right )} \log \left (\frac{1}{2} \, \sqrt{4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) -{\left (-2 i \, b^{3} x^{3} - 2 i \, a^{3}\right )} \log \left (-\frac{1}{2} \, \sqrt{4 i \, c} e^{\left (i \, b x + i \, a\right )} + 1\right ) - 12 \,{\rm polylog}\left (4, \frac{1}{2} \, \sqrt{4 i \, c} e^{\left (i \, b x + i \, a\right )}\right ) - 12 \,{\rm polylog}\left (4, -\frac{1}{2} \, \sqrt{4 i \, c} e^{\left (i \, b x + i \, a\right )}\right )}{12 \, 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*} \frac{b \int \frac{x^{3}}{i c e^{2 i a} e^{2 i b x} - 1}\, dx}{3} + \frac{i x^{3} \log{\left (- i c + \frac{i c}{e^{2 i a} e^{2 i b x} + 1} - \frac{i c e^{i a} e^{i b x}}{e^{i a} e^{i b x} + e^{- i a} e^{- i b x}} + 1 + \frac{1}{e^{2 i a} e^{2 i b x} + 1} - \frac{e^{i a} e^{i b x}}{e^{i a} e^{i b x} + e^{- i a} e^{- i b x}} \right )}}{6} - \frac{i x^{3} \log{\left (i c - \frac{i c}{e^{2 i a} e^{2 i b x} + 1} + \frac{i c e^{i a} e^{i b x}}{e^{i a} e^{i b x} + e^{- i a} e^{- i b x}} + 1 - \frac{1}{e^{2 i a} e^{2 i b x} + 1} + \frac{e^{i a} e^{i b x}}{e^{i a} e^{i b x} + e^{- i a} e^{- i b x}} \right )}}{6} \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} \arctan \left ({\left (i \, c + 1\right )} \tan \left (b x + a\right ) + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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