Optimal. Leaf size=155 \[ \frac{i x \text{PolyLog}\left (3,-i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac{\text{PolyLog}\left (4,-i c e^{2 i a+2 i b x}\right )}{8 b^3}+\frac{x^2 \text{PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac{1}{6} i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac{1}{3} x^3 \tan ^{-1}(c-(1-i c) \tan (a+b x))+\frac{b x^4}{12} \]
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Rubi [A] time = 0.244238, antiderivative size = 155, 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 = 0.439728, size = 137, normalized size = 0.88 \[ \frac{1}{24} \left (\frac{6 i x \text{PolyLog}\left (3,\frac{i e^{-2 i (a+b x)}}{c}\right )}{b^2}+\frac{3 \text{PolyLog}\left (4,\frac{i e^{-2 i (a+b x)}}{c}\right )}{b^3}-\frac{6 x^2 \text{PolyLog}\left (2,\frac{i e^{-2 i (a+b x)}}{c}\right )}{b}+4 i x^3 \log \left (1-\frac{i e^{-2 i (a+b x)}}{c}\right )+8 x^3 \tan ^{-1}(c+i (c+i) \tan (a+b x))\right ) \]
Antiderivative was successfully verified.
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Maple [C] time = 19.335, size = 1533, normalized size = 9.9 \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 = 1.13526, size = 419, normalized size = 2.7 \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 = 1.98512, size = 915, normalized size = 5.9 \begin{align*} \frac{b^{4} x^{4} + 2 i \, b^{3} x^{3} \log \left (-\frac{{\left (c + i\right )} e^{\left (2 i \, b x + 2 i \, a\right )}}{c e^{\left (2 i \, b x + 2 i \, a\right )} - 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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