Optimal. Leaf size=277 \[ \frac{3 b^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2 d}-\frac{3 b^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d}-\frac{3 i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}-\frac{3 i b^3 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 c^2 d}+\frac{3 i b^3 \text{PolyLog}\left (4,1-\frac{2}{1+i c x}\right )}{4 c^2 d}+\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac{3 i b \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d}-\frac{\log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac{i x \left (a+b \tan ^{-1}(c x)\right )^3}{c d} \]
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Rubi [A] time = 0.503495, antiderivative size = 277, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {4866, 4846, 4920, 4854, 4884, 4994, 6610, 4998} \[ \frac{3 b^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^2 d}-\frac{3 b^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 c^2 d}-\frac{3 i b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^2 d}-\frac{3 i b^3 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 c^2 d}+\frac{3 i b^3 \text{PolyLog}\left (4,1-\frac{2}{1+i c x}\right )}{4 c^2 d}+\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac{3 i b \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d}-\frac{\log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac{i x \left (a+b \tan ^{-1}(c x)\right )^3}{c d} \]
Antiderivative was successfully verified.
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Rule 4866
Rule 4846
Rule 4920
Rule 4854
Rule 4884
Rule 4994
Rule 6610
Rule 4998
Rubi steps
\begin{align*} \int \frac{x \left (a+b \tan ^{-1}(c x)\right )^3}{d+i c d x} \, dx &=\frac{i \int \frac{\left (a+b \tan ^{-1}(c x)\right )^3}{d+i c d x} \, dx}{c}-\frac{i \int \left (a+b \tan ^{-1}(c x)\right )^3 \, dx}{c d}\\ &=-\frac{i x \left (a+b \tan ^{-1}(c x)\right )^3}{c d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac{2}{1+i c x}\right )}{c^2 d}+\frac{(3 i b) \int \frac{x \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{d}+\frac{(3 b) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c d}\\ &=\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac{i x \left (a+b \tan ^{-1}(c x)\right )^3}{c d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac{2}{1+i c x}\right )}{c^2 d}-\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 c^2 d}-\frac{(3 i b) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{i-c x} \, dx}{c d}+\frac{\left (3 i b^2\right ) \int \frac{\left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c d}\\ &=\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac{i x \left (a+b \tan ^{-1}(c x)\right )^3}{c d}-\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac{2}{1+i c x}\right )}{c^2 d}-\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 c^2 d}-\frac{3 b^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^2 d}+\frac{\left (6 i b^2\right ) \int \frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c d}+\frac{\left (3 b^3\right ) \int \frac{\text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{2 c d}\\ &=\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac{i x \left (a+b \tan ^{-1}(c x)\right )^3}{c d}-\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac{2}{1+i c x}\right )}{c^2 d}+\frac{3 b^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^2 d}-\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 c^2 d}-\frac{3 b^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^2 d}+\frac{3 i b^3 \text{Li}_4\left (1-\frac{2}{1+i c x}\right )}{4 c^2 d}-\frac{\left (3 b^3\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c d}\\ &=\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{c^2 d}-\frac{i x \left (a+b \tan ^{-1}(c x)\right )^3}{c d}-\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3 \log \left (\frac{2}{1+i c x}\right )}{c^2 d}+\frac{3 b^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^2 d}-\frac{3 i b \left (a+b \tan ^{-1}(c x)\right )^2 \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{2 c^2 d}-\frac{3 i b^3 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^2 d}-\frac{3 b^2 \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^2 d}+\frac{3 i b^3 \text{Li}_4\left (1-\frac{2}{1+i c x}\right )}{4 c^2 d}\\ \end{align*}
Mathematica [A] time = 0.736524, size = 393, normalized size = 1.42 \[ -\frac{i \left (-6 b \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right ) \left (2 b (a+i b) \tan ^{-1}(c x)+a (a+2 i b)+b^2 \tan ^{-1}(c x)^2\right )+6 b^2 \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(c x)}\right ) \left (-i a-i b \tan ^{-1}(c x)+b\right )+3 b^3 \text{PolyLog}\left (4,-e^{2 i \tan ^{-1}(c x)}\right )-6 a^2 b \log \left (c^2 x^2+1\right )-12 a^2 b \tan ^{-1}(c x)^2+12 a^2 b c x \tan ^{-1}(c x)-12 i a^2 b \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+2 i a^3 \log \left (c^2 x^2+1\right )+4 a^3 c x-4 a^3 \tan ^{-1}(c x)-8 a b^2 \tan ^{-1}(c x)^3-12 i a b^2 \tan ^{-1}(c x)^2+12 a b^2 c x \tan ^{-1}(c x)^2-12 i a b^2 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+24 a b^2 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-2 b^3 \tan ^{-1}(c x)^4-4 i b^3 \tan ^{-1}(c x)^3+4 b^3 c x \tan ^{-1}(c x)^3-4 i b^3 \tan ^{-1}(c x)^3 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+12 b^3 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )}{4 c^2 d} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.645, size = 5478, normalized size = 19.8 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b^{3} x \log \left (-\frac{c x + i}{c x - i}\right )^{3} - 6 i \, a b^{2} x \log \left (-\frac{c x + i}{c x - i}\right )^{2} - 12 \, a^{2} b x \log \left (-\frac{c x + i}{c x - i}\right ) + 8 i \, a^{3} x}{8 \, c d x - 8 i \, d}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \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 \frac{{\left (b \arctan \left (c x\right ) + a\right )}^{3} x}{i \, c d x + d}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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