Optimal. Leaf size=142 \[ -\frac{i x \text{PolyLog}\left (3,-i c e^{2 a+2 b x}\right )}{4 b^2}+\frac{i \text{PolyLog}\left (4,-i c e^{2 a+2 b x}\right )}{8 b^3}+\frac{i x^2 \text{PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{4 b}+\frac{1}{6} i x^3 \log \left (1+i c e^{2 a+2 b x}\right )+\frac{1}{3} x^3 \tan ^{-1}(c+(c+i) \tanh (a+b x))-\frac{1}{12} i b x^4 \]
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Rubi [A] time = 0.228086, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {5195, 2184, 2190, 2531, 6609, 2282, 6589} \[ -\frac{i x \text{PolyLog}\left (3,-i c e^{2 a+2 b x}\right )}{4 b^2}+\frac{i \text{PolyLog}\left (4,-i c e^{2 a+2 b x}\right )}{8 b^3}+\frac{i x^2 \text{PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{4 b}+\frac{1}{6} i x^3 \log \left (1+i c e^{2 a+2 b x}\right )+\frac{1}{3} x^3 \tan ^{-1}(c+(c+i) \tanh (a+b x))-\frac{1}{12} i b x^4 \]
Antiderivative was successfully verified.
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Rule 5195
Rule 2184
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x^2 \tan ^{-1}(c+(i+c) \tanh (a+b x)) \, dx &=\frac{1}{3} x^3 \tan ^{-1}(c+(i+c) \tanh (a+b x))-\frac{1}{3} b \int \frac{x^3}{-i+c e^{2 a+2 b x}} \, dx\\ &=-\frac{1}{12} i b x^4+\frac{1}{3} x^3 \tan ^{-1}(c+(i+c) \tanh (a+b x))+\frac{1}{3} (i b c) \int \frac{e^{2 a+2 b x} x^3}{-i+c e^{2 a+2 b x}} \, dx\\ &=-\frac{1}{12} i b x^4+\frac{1}{3} x^3 \tan ^{-1}(c+(i+c) \tanh (a+b x))+\frac{1}{6} i x^3 \log \left (1+i c e^{2 a+2 b x}\right )-\frac{1}{2} i \int x^2 \log \left (1+i c e^{2 a+2 b x}\right ) \, dx\\ &=-\frac{1}{12} i b x^4+\frac{1}{3} x^3 \tan ^{-1}(c+(i+c) \tanh (a+b x))+\frac{1}{6} i x^3 \log \left (1+i c e^{2 a+2 b x}\right )+\frac{i x^2 \text{Li}_2\left (-i c e^{2 a+2 b x}\right )}{4 b}-\frac{i \int x \text{Li}_2\left (-i c e^{2 a+2 b x}\right ) \, dx}{2 b}\\ &=-\frac{1}{12} i b x^4+\frac{1}{3} x^3 \tan ^{-1}(c+(i+c) \tanh (a+b x))+\frac{1}{6} i x^3 \log \left (1+i c e^{2 a+2 b x}\right )+\frac{i x^2 \text{Li}_2\left (-i c e^{2 a+2 b x}\right )}{4 b}-\frac{i x \text{Li}_3\left (-i c e^{2 a+2 b x}\right )}{4 b^2}+\frac{i \int \text{Li}_3\left (-i c e^{2 a+2 b x}\right ) \, dx}{4 b^2}\\ &=-\frac{1}{12} i b x^4+\frac{1}{3} x^3 \tan ^{-1}(c+(i+c) \tanh (a+b x))+\frac{1}{6} i x^3 \log \left (1+i c e^{2 a+2 b x}\right )+\frac{i x^2 \text{Li}_2\left (-i c e^{2 a+2 b x}\right )}{4 b}-\frac{i x \text{Li}_3\left (-i c e^{2 a+2 b x}\right )}{4 b^2}+\frac{i \operatorname{Subst}\left (\int \frac{\text{Li}_3(-i c x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3}\\ &=-\frac{1}{12} i b x^4+\frac{1}{3} x^3 \tan ^{-1}(c+(i+c) \tanh (a+b x))+\frac{1}{6} i x^3 \log \left (1+i c e^{2 a+2 b x}\right )+\frac{i x^2 \text{Li}_2\left (-i c e^{2 a+2 b x}\right )}{4 b}-\frac{i x \text{Li}_3\left (-i c e^{2 a+2 b x}\right )}{4 b^2}+\frac{i \text{Li}_4\left (-i c e^{2 a+2 b x}\right )}{8 b^3}\\ \end{align*}
Mathematica [A] time = 5.12782, size = 128, normalized size = 0.9 \[ \frac{i \left (-6 b^2 x^2 \text{PolyLog}\left (2,\frac{i e^{-2 (a+b x)}}{c}\right )-6 b x \text{PolyLog}\left (3,\frac{i e^{-2 (a+b x)}}{c}\right )-3 \text{PolyLog}\left (4,\frac{i e^{-2 (a+b x)}}{c}\right )+4 b^3 x^3 \log \left (1-\frac{i e^{-2 (a+b x)}}{c}\right )\right )}{24 b^3}+\frac{1}{3} x^3 \tan ^{-1}(c+(c+i) \tanh (a+b x)) \]
Antiderivative was successfully verified.
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Maple [C] time = 11.773, size = 1555, normalized size = 11. \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 [A] time = 5.93211, size = 174, normalized size = 1.23 \begin{align*} \frac{1}{3} \, x^{3} \arctan \left ({\left (c + i\right )} \tanh \left (b x + a\right ) + c\right ) + \frac{4}{9} \,{\left (\frac{3 \, x^{4}}{4 i \, c - 4} - \frac{4 \, b^{3} x^{3} \log \left (i \, c e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2}{\rm Li}_2\left (-i \, c e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x{\rm Li}_{3}(-i \, c e^{\left (2 \, b x + 2 \, a\right )}) + 3 \,{\rm Li}_{4}(-i \, c e^{\left (2 \, b x + 2 \, a\right )})}{-2 \, b^{4}{\left (-i \, c + 1\right )}}\right )} b{\left (c + i\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 1.98988, size = 868, normalized size = 6.11 \begin{align*} \frac{-i \, b^{4} x^{4} + 2 i \, b^{3} x^{3} \log \left (-\frac{{\left (c + i\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c e^{\left (2 \, b x + 2 \, a\right )} - i}\right ) + 6 i \, b^{2} x^{2}{\rm Li}_2\left (\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )}\right ) + 6 i \, b^{2} x^{2}{\rm Li}_2\left (-\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )}\right ) + i \, a^{4} - 2 i \, a^{3} \log \left (\frac{2 \, c e^{\left (b x + a\right )} + i \, \sqrt{-4 i \, c}}{2 \, c}\right ) - 2 i \, a^{3} \log \left (\frac{2 \, c e^{\left (b x + 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 (b x + a\right )}\right ) - 12 i \, b x{\rm polylog}\left (3, -\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + 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 (b x + 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 (b x + a\right )} + 1\right ) + 12 i \,{\rm polylog}\left (4, \frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + a\right )}\right ) + 12 i \,{\rm polylog}\left (4, -\frac{1}{2} \, \sqrt{-4 i \, c} e^{\left (b x + 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{i b \int \frac{x^{3}}{i c e^{2 a} e^{2 b x} + 1}\, dx}{3} + \frac{i x^{3} \log{\left (- i c + \frac{i c}{e^{2 a} e^{2 b x} + 1} - \frac{i c e^{a} e^{b x}}{e^{a} e^{b x} + e^{- a} e^{- b x}} + 1 - \frac{1}{e^{2 a} e^{2 b x} + 1} + \frac{e^{a} e^{b x}}{e^{a} e^{b x} + e^{- a} e^{- b x}} \right )}}{6} - \frac{i x^{3} \log{\left (i c - \frac{i c}{e^{2 a} e^{2 b x} + 1} + \frac{i c e^{a} e^{b x}}{e^{a} e^{b x} + e^{- a} e^{- b x}} + 1 + \frac{1}{e^{2 a} e^{2 b x} + 1} - \frac{e^{a} e^{b x}}{e^{a} e^{b x} + e^{- a} e^{- 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 (c + i\right )} \tanh \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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