Optimal. Leaf size=220 \[ -\frac {3 i b^2 d \text {Li}_2\left (1-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}-\frac {3 i b^2 d \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}+\frac {3 b d \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-\frac {3}{2} i b d x \left (a+b \tan ^{-1}(c x)\right )^2-\frac {i d (1+i c x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c}+\frac {3 b d \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c}+\frac {3 b^3 d \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{2 c}+\frac {3 b^3 d \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 c} \]
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Rubi [A] time = 0.34, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4864, 4846, 4920, 4854, 2402, 2315, 1586, 4884, 4992, 6610} \[ -\frac {3 i b^2 d \text {PolyLog}\left (2,1-\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}+\frac {3 b^3 d \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c}+\frac {3 b^3 d \text {PolyLog}\left (3,1-\frac {2}{1-i c x}\right )}{2 c}-\frac {3 i b^2 d \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c}+\frac {3 b d \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-\frac {3}{2} i b d x \left (a+b \tan ^{-1}(c x)\right )^2-\frac {i d (1+i c x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c}+\frac {3 b d \log \left (\frac {2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c} \]
Antiderivative was successfully verified.
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Rule 1586
Rule 2315
Rule 2402
Rule 4846
Rule 4854
Rule 4864
Rule 4884
Rule 4920
Rule 4992
Rule 6610
Rubi steps
\begin {align*} \int (d+i c d x) \left (a+b \tan ^{-1}(c x)\right )^3 \, dx &=-\frac {i d (1+i c x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c}+\frac {(3 i b) \int \left (-d^2 \left (a+b \tan ^{-1}(c x)\right )^2-\frac {2 i \left (i d^2-c d^2 x\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2}\right ) \, dx}{2 d}\\ &=-\frac {i d (1+i c x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c}+\frac {(3 b) \int \frac {\left (i d^2-c d^2 x\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{1+c^2 x^2} \, dx}{d}-\frac {1}{2} (3 i b d) \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=-\frac {3}{2} i b d x \left (a+b \tan ^{-1}(c x)\right )^2-\frac {i d (1+i c x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c}+\frac {(3 b) \int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{-\frac {i}{d^2}-\frac {c x}{d^2}} \, dx}{d}+\left (3 i b^2 c d\right ) \int \frac {x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=\frac {3 b d \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-\frac {3}{2} i b d x \left (a+b \tan ^{-1}(c x)\right )^2-\frac {i d (1+i c x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c}+\frac {3 b d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{c}-\left (3 i b^2 d\right ) \int \frac {a+b \tan ^{-1}(c x)}{i-c x} \, dx-\left (6 b^2 d\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\\ &=\frac {3 b d \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-\frac {3}{2} i b d x \left (a+b \tan ^{-1}(c x)\right )^2-\frac {i d (1+i c x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c}+\frac {3 b d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{c}-\frac {3 i b^2 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c}-\frac {3 i b^2 d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{c}+\left (3 i b^3 d\right ) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx+\left (3 i b^3 d\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{1+c^2 x^2} \, dx\\ &=\frac {3 b d \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-\frac {3}{2} i b d x \left (a+b \tan ^{-1}(c x)\right )^2-\frac {i d (1+i c x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c}+\frac {3 b d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{c}-\frac {3 i b^2 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c}-\frac {3 i b^2 d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{c}+\frac {3 b^3 d \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 c}+\frac {\left (3 b^3 d\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{c}\\ &=\frac {3 b d \left (a+b \tan ^{-1}(c x)\right )^2}{2 c}-\frac {3}{2} i b d x \left (a+b \tan ^{-1}(c x)\right )^2-\frac {i d (1+i c x)^2 \left (a+b \tan ^{-1}(c x)\right )^3}{2 c}+\frac {3 b d \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-i c x}\right )}{c}-\frac {3 i b^2 d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c}-\frac {3 i b^2 d \left (a+b \tan ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-i c x}\right )}{c}+\frac {3 b^3 d \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c}+\frac {3 b^3 d \text {Li}_3\left (1-\frac {2}{1-i c x}\right )}{2 c}\\ \end {align*}
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Mathematica [A] time = 0.49, size = 367, normalized size = 1.67 \[ \frac {i d \left (a^3 c^2 x^2-2 i a^3 c x+3 i a^2 b \log \left (c^2 x^2+1\right )+3 a^2 b c^2 x^2 \tan ^{-1}(c x)-3 a^2 b c x+3 a^2 b \tan ^{-1}(c x)-6 i a^2 b c x \tan ^{-1}(c x)+3 a b^2 \log \left (c^2 x^2+1\right )+3 a b^2 c^2 x^2 \tan ^{-1}(c x)^2-3 b^2 \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right ) \left (2 a+2 b \tan ^{-1}(c x)-i b\right )-3 a b^2 \tan ^{-1}(c x)^2-6 i a b^2 c x \tan ^{-1}(c x)^2-6 a b^2 c x \tan ^{-1}(c x)-12 i a b^2 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+b^3 c^2 x^2 \tan ^{-1}(c x)^3-3 i b^3 \text {Li}_3\left (-e^{2 i \tan ^{-1}(c x)}\right )-b^3 \tan ^{-1}(c x)^3-2 i b^3 c x \tan ^{-1}(c x)^3+3 i b^3 \tan ^{-1}(c x)^2-3 b^3 c x \tan ^{-1}(c x)^2-6 i b^3 \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-6 b^3 \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )}{2 c} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.55, size = 0, normalized size = 0.00 \[ \frac {1}{16} \, {\left (b^{3} c d x^{2} - 2 i \, b^{3} d x\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{3} + {\rm integral}\left (\frac {8 i \, a^{3} c^{3} d x^{3} + 8 \, a^{3} c^{2} d x^{2} + 8 i \, a^{3} c d x + 8 \, a^{3} d + {\left (-6 i \, a b^{2} c^{3} d x^{3} - {\left (6 \, a b^{2} - 3 i \, b^{3}\right )} c^{2} d x^{2} - 6 \, a b^{2} d - 6 \, {\left (i \, a b^{2} - b^{3}\right )} c d x\right )} \log \left (-\frac {c x + i}{c x - i}\right )^{2} - {\left (12 \, a^{2} b c^{3} d x^{3} - 12 i \, a^{2} b c^{2} d x^{2} + 12 \, a^{2} b c d x - 12 i \, a^{2} b d\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{8 \, {\left (c^{2} x^{2} + 1\right )}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 2.54, size = 7451, normalized size = 33.87 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}^3\,\left (d+c\,d\,x\,1{}\mathrm {i}\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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