Optimal. Leaf size=225 \[ -\frac {3 i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3 (-c x+i)}-\frac {a+b \tan ^{-1}(c x)}{2 c^4 d^3 (-c x+i)^2}+\frac {3 \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3}+\frac {i a x}{c^3 d^3}+\frac {3 i b \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{2 c^4 d^3}-\frac {11 b}{8 c^4 d^3 (-c x+i)}+\frac {i b}{8 c^4 d^3 (-c x+i)^2}+\frac {11 b \tan ^{-1}(c x)}{8 c^4 d^3}+\frac {i b x \tan ^{-1}(c x)}{c^3 d^3}-\frac {i b \log \left (c^2 x^2+1\right )}{2 c^4 d^3} \]
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Rubi [A] time = 0.24, antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4876, 4846, 260, 4862, 627, 44, 203, 4854, 2402, 2315} \[ \frac {3 i b \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c^4 d^3}-\frac {3 i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3 (-c x+i)}-\frac {a+b \tan ^{-1}(c x)}{2 c^4 d^3 (-c x+i)^2}+\frac {3 \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3}+\frac {i a x}{c^3 d^3}-\frac {i b \log \left (c^2 x^2+1\right )}{2 c^4 d^3}-\frac {11 b}{8 c^4 d^3 (-c x+i)}+\frac {i b}{8 c^4 d^3 (-c x+i)^2}+\frac {i b x \tan ^{-1}(c x)}{c^3 d^3}+\frac {11 b \tan ^{-1}(c x)}{8 c^4 d^3} \]
Antiderivative was successfully verified.
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Rule 44
Rule 203
Rule 260
Rule 627
Rule 2315
Rule 2402
Rule 4846
Rule 4854
Rule 4862
Rule 4876
Rubi steps
\begin {align*} \int \frac {x^3 \left (a+b \tan ^{-1}(c x)\right )}{(d+i c d x)^3} \, dx &=\int \left (\frac {i \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3}+\frac {a+b \tan ^{-1}(c x)}{c^3 d^3 (-i+c x)^3}-\frac {3 i \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3 (-i+c x)^2}-\frac {3 \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3 (-i+c x)}\right ) \, dx\\ &=\frac {i \int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{c^3 d^3}-\frac {(3 i) \int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{c^3 d^3}+\frac {\int \frac {a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx}{c^3 d^3}-\frac {3 \int \frac {a+b \tan ^{-1}(c x)}{-i+c x} \, dx}{c^3 d^3}\\ &=\frac {i a x}{c^3 d^3}-\frac {a+b \tan ^{-1}(c x)}{2 c^4 d^3 (i-c x)^2}-\frac {3 i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3 (i-c x)}+\frac {3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^4 d^3}+\frac {(i b) \int \tan ^{-1}(c x) \, dx}{c^3 d^3}-\frac {(3 i b) \int \frac {1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{c^3 d^3}+\frac {b \int \frac {1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{2 c^3 d^3}-\frac {(3 b) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^3 d^3}\\ &=\frac {i a x}{c^3 d^3}+\frac {i b x \tan ^{-1}(c x)}{c^3 d^3}-\frac {a+b \tan ^{-1}(c x)}{2 c^4 d^3 (i-c x)^2}-\frac {3 i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3 (i-c x)}+\frac {3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^4 d^3}+\frac {(3 i b) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{c^4 d^3}-\frac {(3 i b) \int \frac {1}{(-i+c x)^2 (i+c x)} \, dx}{c^3 d^3}+\frac {b \int \frac {1}{(-i+c x)^3 (i+c x)} \, dx}{2 c^3 d^3}-\frac {(i b) \int \frac {x}{1+c^2 x^2} \, dx}{c^2 d^3}\\ &=\frac {i a x}{c^3 d^3}+\frac {i b x \tan ^{-1}(c x)}{c^3 d^3}-\frac {a+b \tan ^{-1}(c x)}{2 c^4 d^3 (i-c x)^2}-\frac {3 i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3 (i-c x)}+\frac {3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^4 d^3}-\frac {i b \log \left (1+c^2 x^2\right )}{2 c^4 d^3}+\frac {3 i b \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c^4 d^3}-\frac {(3 i b) \int \left (-\frac {i}{2 (-i+c x)^2}+\frac {i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{c^3 d^3}+\frac {b \int \left (-\frac {i}{2 (-i+c x)^3}+\frac {1}{4 (-i+c x)^2}-\frac {1}{4 \left (1+c^2 x^2\right )}\right ) \, dx}{2 c^3 d^3}\\ &=\frac {i a x}{c^3 d^3}+\frac {i b}{8 c^4 d^3 (i-c x)^2}-\frac {11 b}{8 c^4 d^3 (i-c x)}+\frac {i b x \tan ^{-1}(c x)}{c^3 d^3}-\frac {a+b \tan ^{-1}(c x)}{2 c^4 d^3 (i-c x)^2}-\frac {3 i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3 (i-c x)}+\frac {3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^4 d^3}-\frac {i b \log \left (1+c^2 x^2\right )}{2 c^4 d^3}+\frac {3 i b \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c^4 d^3}-\frac {b \int \frac {1}{1+c^2 x^2} \, dx}{8 c^3 d^3}+\frac {(3 b) \int \frac {1}{1+c^2 x^2} \, dx}{2 c^3 d^3}\\ &=\frac {i a x}{c^3 d^3}+\frac {i b}{8 c^4 d^3 (i-c x)^2}-\frac {11 b}{8 c^4 d^3 (i-c x)}+\frac {11 b \tan ^{-1}(c x)}{8 c^4 d^3}+\frac {i b x \tan ^{-1}(c x)}{c^3 d^3}-\frac {a+b \tan ^{-1}(c x)}{2 c^4 d^3 (i-c x)^2}-\frac {3 i \left (a+b \tan ^{-1}(c x)\right )}{c^4 d^3 (i-c x)}+\frac {3 \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c^4 d^3}-\frac {i b \log \left (1+c^2 x^2\right )}{2 c^4 d^3}+\frac {3 i b \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c^4 d^3}\\ \end {align*}
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Mathematica [A] time = 0.86, size = 216, normalized size = 0.96 \[ \frac {-48 a \log \left (c^2 x^2+1\right )+32 i a c x+\frac {96 i a}{c x-i}-\frac {16 a}{(c x-i)^2}-96 i a \tan ^{-1}(c x)+i b \left (-16 \log \left (c^2 x^2+1\right )-48 \text {Li}_2\left (-e^{2 i \tan ^{-1}(c x)}\right )-96 \tan ^{-1}(c x)^2-20 i \sin \left (2 \tan ^{-1}(c x)\right )+i \sin \left (4 \tan ^{-1}(c x)\right )+20 \cos \left (2 \tan ^{-1}(c x)\right )-\cos \left (4 \tan ^{-1}(c x)\right )+4 \tan ^{-1}(c x) \left (8 c x-24 i \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )+10 \sin \left (2 \tan ^{-1}(c x)\right )-\sin \left (4 \tan ^{-1}(c x)\right )+10 i \cos \left (2 \tan ^{-1}(c x)\right )-i \cos \left (4 \tan ^{-1}(c x)\right )\right )\right )}{32 c^4 d^3} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b x^{3} \log \left (-\frac {c x + i}{c x - i}\right ) - 2 i \, a x^{3}}{2 \, c^{3} d^{3} x^{3} - 6 i \, c^{2} d^{3} x^{2} - 6 \, c d^{3} x + 2 i \, d^{3}}, x\right ) \]
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 [A] time = 0.07, size = 375, normalized size = 1.67 \[ -\frac {3 i a \arctan \left (c x \right )}{c^{4} d^{3}}+\frac {3 i b \arctan \left (c x \right )}{c^{4} d^{3} \left (c x -i\right )}-\frac {3 a \ln \left (c^{2} x^{2}+1\right )}{2 c^{4} d^{3}}+\frac {i b x \arctan \left (c x \right )}{c^{3} d^{3}}-\frac {a}{2 c^{4} d^{3} \left (c x -i\right )^{2}}+\frac {3 i b \ln \left (c x -i\right ) \ln \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 c^{4} d^{3}}+\frac {i a x}{c^{3} d^{3}}-\frac {3 b \arctan \left (c x \right ) \ln \left (c x -i\right )}{c^{4} d^{3}}-\frac {b \arctan \left (c x \right )}{2 c^{4} d^{3} \left (c x -i\right )^{2}}+\frac {3 i b \ln \left (c^{4} x^{4}+10 c^{2} x^{2}+9\right )}{64 c^{4} d^{3}}+\frac {3 b \arctan \left (\frac {1}{6} c^{3} x^{3}+\frac {7}{6} c x \right )}{32 c^{4} d^{3}}-\frac {3 b \arctan \left (\frac {c x}{2}\right )}{32 c^{4} d^{3}}+\frac {3 b \arctan \left (\frac {c x}{2}-\frac {i}{2}\right )}{16 c^{4} d^{3}}-\frac {19 i b \ln \left (c^{2} x^{2}+1\right )}{32 c^{4} d^{3}}-\frac {3 i b \ln \left (c x -i\right )^{2}}{4 c^{4} d^{3}}+\frac {19 b \arctan \left (c x \right )}{16 c^{4} d^{3}}+\frac {11 b}{8 c^{4} d^{3} \left (c x -i\right )}+\frac {3 i a}{c^{4} d^{3} \left (c x -i\right )}+\frac {3 i b \dilog \left (-\frac {i \left (c x +i\right )}{2}\right )}{2 c^{4} d^{3}}+\frac {i b}{8 c^{4} d^{3} \left (c x -i\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 327, normalized size = 1.45 \[ -\frac {-16 i \, a c^{3} x^{3} - 32 \, a c^{2} x^{2} + {\left (-32 i \, a - 22 \, b\right )} c x + {\left (12 i \, b c^{2} x^{2} + 24 \, b c x - 12 i \, b\right )} \arctan \left (c x\right )^{2} + {\left (3 i \, b c^{2} x^{2} + 6 \, b c x - 3 i \, b\right )} \log \left (c^{2} x^{2} + 1\right )^{2} + {\left (12 \, b c^{2} x^{2} - 24 i \, b c x - 12 \, b\right )} \arctan \left (c x\right ) \log \left (\frac {1}{4} \, c^{2} x^{2} + \frac {1}{4}\right ) + {\left (-16 i \, b c^{3} x^{3} + {\left (48 i \, a - 51 \, b\right )} c^{2} x^{2} + 6 \, {\left (16 \, a + i \, b\right )} c x - 48 i \, a - 21 \, b\right )} \arctan \left (c x\right ) + {\left (3 \, b c^{2} x^{2} - 6 i \, b c x - 3 \, b\right )} \arctan \left (c x, -1\right ) + {\left (-24 i \, b c^{2} x^{2} - 48 \, b c x + 24 i \, b\right )} {\rm Li}_2\left (\frac {1}{2} i \, c x + \frac {1}{2}\right ) + {\left (8 \, {\left (3 \, a + i \, b\right )} c^{2} x^{2} + {\left (-48 i \, a + 16 \, b\right )} c x + {\left (-6 i \, b c^{2} x^{2} - 12 \, b c x + 6 i \, b\right )} \log \left (\frac {1}{4} \, c^{2} x^{2} + \frac {1}{4}\right ) - 24 \, a - 8 i \, b\right )} \log \left (c^{2} x^{2} + 1\right ) - 40 \, a + 20 i \, b}{16 \, c^{6} d^{3} x^{2} - 32 i \, c^{5} d^{3} x - 16 \, c^{4} d^{3}} \]
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 \[ \int \frac {x^3\,\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^3} \,d x \]
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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