Optimal. Leaf size=263 \[ -\frac{3 i b^2 c \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac{3 i b^2 c \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 d}+\frac{3 b c \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}+\frac{3 b^3 c \text{PolyLog}\left (3,-1+\frac{2}{1-i c x}\right )}{2 d}-\frac{3 b^3 c \text{PolyLog}\left (4,-1+\frac{2}{1+i c x}\right )}{4 d}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{d x}+\frac{3 b c \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac{i c \log \left (2-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{d} \]
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Rubi [A] time = 0.60013, antiderivative size = 263, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {4870, 4852, 4924, 4868, 4884, 4992, 6610, 4994, 4998} \[ -\frac{3 i b^2 c \text{PolyLog}\left (2,-1+\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}-\frac{3 i b^2 c \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{2 d}+\frac{3 b c \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{2 d}+\frac{3 b^3 c \text{PolyLog}\left (3,-1+\frac{2}{1-i c x}\right )}{2 d}-\frac{3 b^3 c \text{PolyLog}\left (4,-1+\frac{2}{1+i c x}\right )}{4 d}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{d x}+\frac{3 b c \log \left (2-\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{d}-\frac{i c \log \left (2-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^3}{d} \]
Antiderivative was successfully verified.
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Rule 4870
Rule 4852
Rule 4924
Rule 4868
Rule 4884
Rule 4992
Rule 6610
Rule 4994
Rule 4998
Rubi steps
\begin{align*} \int \frac{\left (a+b \tan ^{-1}(c x)\right )^3}{x^2 (d+i c d x)} \, dx &=-\left ((i c) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^3}{x (d+i c d x)} \, dx\right )+\frac{\int \frac{\left (a+b \tan ^{-1}(c x)\right )^3}{x^2} \, dx}{d}\\ &=-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{d x}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac{2}{1+i c x}\right )}{d}+\frac{(3 b c) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x \left (1+c^2 x^2\right )} \, dx}{d}+\frac{\left (3 i b c^2\right ) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{d x}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac{2}{1+i c x}\right )}{d}+\frac{3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{2 d}+\frac{(3 i b c) \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x (i+c x)} \, dx}{d}-\frac{\left (3 b^2 c^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}{d}\\ &=-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{d x}+\frac{3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1-i c x}\right )}{d}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac{2}{1+i c x}\right )}{d}+\frac{3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{2 d}-\frac{3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )}{2 d}-\frac{\left (6 b^2 c^2\right ) \int \frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{d}+\frac{\left (3 i b^3 c^2\right ) \int \frac{\text{Li}_3\left (-1+\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{2 d}\\ &=-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{d x}+\frac{3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1-i c x}\right )}{d}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac{2}{1+i c x}\right )}{d}-\frac{3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-i c x}\right )}{d}+\frac{3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{2 d}-\frac{3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )}{2 d}-\frac{3 b^3 c \text{Li}_4\left (-1+\frac{2}{1+i c x}\right )}{4 d}+\frac{\left (3 i b^3 c^2\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^3}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right )^3}{d x}+\frac{3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (2-\frac{2}{1-i c x}\right )}{d}-\frac{i c \left (a+b \tan ^{-1}(c x)\right )^3 \log \left (2-\frac{2}{1+i c x}\right )}{d}-\frac{3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1-i c x}\right )}{d}+\frac{3 b c \left (a+b \tan ^{-1}(c x)\right )^2 \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{2 d}+\frac{3 b^3 c \text{Li}_3\left (-1+\frac{2}{1-i c x}\right )}{2 d}-\frac{3 i b^2 c \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )}{2 d}-\frac{3 b^3 c \text{Li}_4\left (-1+\frac{2}{1+i c x}\right )}{4 d}\\ \end{align*}
Mathematica [A] time = 1.38152, size = 436, normalized size = 1.66 \[ -\frac{3 a^2 b c \left (\text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )+2 \left (-\log \left (\frac{c x}{\sqrt{c^2 x^2+1}}\right )+\tan ^{-1}(c x)^2+\tan ^{-1}(c x) \left (\frac{1}{c x}+i \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )\right )\right )\right )+6 i a b^2 c \left (i \tan ^{-1}(c x) \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(c x)}\right )+\text{PolyLog}\left (2,e^{2 i \tan ^{-1}(c x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(c x)}\right )-\frac{i \tan ^{-1}(c x)^2}{c x}+\tan ^{-1}(c x)^2+\tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )+2 i \tan ^{-1}(c x) \log \left (1-e^{2 i \tan ^{-1}(c x)}\right )-\frac{i \pi ^3}{24}\right )+2 i b^3 c \left (\frac{3}{2} i \left (\tan ^{-1}(c x)+2 i\right ) \tan ^{-1}(c x) \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(c x)}\right )+\frac{3}{2} \left (\tan ^{-1}(c x)+i\right ) \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(c x)}\right )-\frac{3}{4} i \text{PolyLog}\left (4,e^{-2 i \tan ^{-1}(c x)}\right )-\frac{i \tan ^{-1}(c x)^3}{c x}-\tan ^{-1}(c x)^3+\tan ^{-1}(c x)^3 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )+3 i \tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-\frac{i \pi ^4}{64}+\frac{\pi ^3}{8}\right )-i a^3 c \log \left (c^2 x^2+1\right )+2 i a^3 c \log (x)+2 a^3 c \tan ^{-1}(c x)+\frac{2 a^3}{x}}{2 d} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.925, size = 11233, normalized size = 42.7 \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} \log \left (-\frac{c x + i}{c x - i}\right )^{3} - 6 i \, a b^{2} \log \left (-\frac{c x + i}{c x - i}\right )^{2} - 12 \, a^{2} b \log \left (-\frac{c x + i}{c x - i}\right ) + 8 i \, a^{3}}{8 \,{\left (c d x^{3} - i \, d x^{2}\right )}}, 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}}{{\left (i \, c d x + d\right )} x^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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