Optimal. Leaf size=170 \[ -\frac {1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )+3 i a c d^3 x+a d^3 \log (x)+\frac {1}{6} i b c^2 d^3 x^2-\frac {5}{3} i b d^3 \log \left (c^2 x^2+1\right )+\frac {1}{2} i b d^3 \text {Li}_2(-i c x)-\frac {1}{2} i b d^3 \text {Li}_2(i c x)+\frac {3}{2} b c d^3 x-\frac {3}{2} b d^3 \tan ^{-1}(c x)+3 i b c d^3 x \tan ^{-1}(c x) \]
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Rubi [A] time = 0.17, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4876, 4846, 260, 4848, 2391, 4852, 321, 203, 266, 43} \[ \frac {1}{2} i b d^3 \text {PolyLog}(2,-i c x)-\frac {1}{2} i b d^3 \text {PolyLog}(2,i c x)-\frac {1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )+3 i a c d^3 x+a d^3 \log (x)+\frac {1}{6} i b c^2 d^3 x^2-\frac {5}{3} i b d^3 \log \left (c^2 x^2+1\right )+\frac {3}{2} b c d^3 x-\frac {3}{2} b d^3 \tan ^{-1}(c x)+3 i b c d^3 x \tan ^{-1}(c x) \]
Antiderivative was successfully verified.
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Rule 43
Rule 203
Rule 260
Rule 266
Rule 321
Rule 2391
Rule 4846
Rule 4848
Rule 4852
Rule 4876
Rubi steps
\begin {align*} \int \frac {(d+i c d x)^3 \left (a+b \tan ^{-1}(c x)\right )}{x} \, dx &=\int \left (3 i c d^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}-3 c^2 d^3 x \left (a+b \tan ^{-1}(c x)\right )-i c^3 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )\right ) \, dx\\ &=d^3 \int \frac {a+b \tan ^{-1}(c x)}{x} \, dx+\left (3 i c d^3\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx-\left (3 c^2 d^3\right ) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx-\left (i c^3 d^3\right ) \int x^2 \left (a+b \tan ^{-1}(c x)\right ) \, dx\\ &=3 i a c d^3 x-\frac {3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+a d^3 \log (x)+\frac {1}{2} \left (i b d^3\right ) \int \frac {\log (1-i c x)}{x} \, dx-\frac {1}{2} \left (i b d^3\right ) \int \frac {\log (1+i c x)}{x} \, dx+\left (3 i b c d^3\right ) \int \tan ^{-1}(c x) \, dx+\frac {1}{2} \left (3 b c^3 d^3\right ) \int \frac {x^2}{1+c^2 x^2} \, dx+\frac {1}{3} \left (i b c^4 d^3\right ) \int \frac {x^3}{1+c^2 x^2} \, dx\\ &=3 i a c d^3 x+\frac {3}{2} b c d^3 x+3 i b c d^3 x \tan ^{-1}(c x)-\frac {3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+a d^3 \log (x)+\frac {1}{2} i b d^3 \text {Li}_2(-i c x)-\frac {1}{2} i b d^3 \text {Li}_2(i c x)-\frac {1}{2} \left (3 b c d^3\right ) \int \frac {1}{1+c^2 x^2} \, dx-\left (3 i b c^2 d^3\right ) \int \frac {x}{1+c^2 x^2} \, dx+\frac {1}{6} \left (i b c^4 d^3\right ) \operatorname {Subst}\left (\int \frac {x}{1+c^2 x} \, dx,x,x^2\right )\\ &=3 i a c d^3 x+\frac {3}{2} b c d^3 x-\frac {3}{2} b d^3 \tan ^{-1}(c x)+3 i b c d^3 x \tan ^{-1}(c x)-\frac {3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+a d^3 \log (x)-\frac {3}{2} i b d^3 \log \left (1+c^2 x^2\right )+\frac {1}{2} i b d^3 \text {Li}_2(-i c x)-\frac {1}{2} i b d^3 \text {Li}_2(i c x)+\frac {1}{6} \left (i b c^4 d^3\right ) \operatorname {Subst}\left (\int \left (\frac {1}{c^2}-\frac {1}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=3 i a c d^3 x+\frac {3}{2} b c d^3 x+\frac {1}{6} i b c^2 d^3 x^2-\frac {3}{2} b d^3 \tan ^{-1}(c x)+3 i b c d^3 x \tan ^{-1}(c x)-\frac {3}{2} c^2 d^3 x^2 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{3} i c^3 d^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+a d^3 \log (x)-\frac {5}{3} i b d^3 \log \left (1+c^2 x^2\right )+\frac {1}{2} i b d^3 \text {Li}_2(-i c x)-\frac {1}{2} i b d^3 \text {Li}_2(i c x)\\ \end {align*}
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Mathematica [A] time = 0.14, size = 139, normalized size = 0.82 \[ -\frac {1}{6} i d^3 \left (2 a c^3 x^3-9 i a c^2 x^2-18 a c x+6 i a \log (x)+2 b c^3 x^3 \tan ^{-1}(c x)-b c^2 x^2+10 b \log \left (c^2 x^2+1\right )-9 i b c^2 x^2 \tan ^{-1}(c x)-3 b \text {Li}_2(-i c x)+3 b \text {Li}_2(i c x)+9 i b c x-18 b c x \tan ^{-1}(c x)-9 i b \tan ^{-1}(c x)\right ) \]
Antiderivative was successfully verified.
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fricas [F] time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {-2 i \, a c^{3} d^{3} x^{3} - 6 \, a c^{2} d^{3} x^{2} + 6 i \, a c d^{3} x + 2 \, a d^{3} + {\left (b c^{3} d^{3} x^{3} - 3 i \, b c^{2} d^{3} x^{2} - 3 \, b c d^{3} x + i \, b d^{3}\right )} \log \left (-\frac {c x + i}{c x - i}\right )}{2 \, x}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [A] time = 0.06, size = 220, normalized size = 1.29 \[ \frac {i d^{3} b \ln \left (c x \right ) \ln \left (i c x +1\right )}{2}+\frac {i d^{3} b \dilog \left (i c x +1\right )}{2}-\frac {3 d^{3} a \,c^{2} x^{2}}{2}+d^{3} a \ln \left (c x \right )-\frac {i d^{3} b \arctan \left (c x \right ) c^{3} x^{3}}{3}-\frac {5 i b \,d^{3} \ln \left (c^{2} x^{2}+1\right )}{3}-\frac {3 d^{3} b \arctan \left (c x \right ) c^{2} x^{2}}{2}+d^{3} b \ln \left (c x \right ) \arctan \left (c x \right )+3 i a c \,d^{3} x +3 i b c \,d^{3} x \arctan \left (c x \right )+\frac {i b \,c^{2} d^{3} x^{2}}{6}-\frac {i d^{3} b \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2}+\frac {3 b c \,d^{3} x}{2}-\frac {i d^{3} a \,c^{3} x^{3}}{3}-\frac {i d^{3} b \dilog \left (-i c x +1\right )}{2}-\frac {3 b \,d^{3} \arctan \left (c x \right )}{2} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.62, size = 184, normalized size = 1.08 \[ -\frac {1}{3} i \, a c^{3} d^{3} x^{3} - \frac {3}{2} \, a c^{2} d^{3} x^{2} + \frac {1}{6} i \, b c^{2} d^{3} x^{2} + 3 i \, a c d^{3} x + \frac {3}{2} \, b c d^{3} x - \frac {1}{12} \, {\left (3 \, \pi + 2 i\right )} b d^{3} \log \left (c^{2} x^{2} + 1\right ) + b d^{3} \arctan \left (c x\right ) \log \left (c x\right ) + \frac {3}{2} i \, {\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{3} - \frac {1}{2} i \, b d^{3} {\rm Li}_2\left (i \, c x + 1\right ) + \frac {1}{2} i \, b d^{3} {\rm Li}_2\left (-i \, c x + 1\right ) + a d^{3} \log \relax (x) - \frac {1}{12} \, {\left (4 i \, b c^{3} d^{3} x^{3} + 18 \, b c^{2} d^{3} x^{2} + 18 \, b d^{3}\right )} \arctan \left (c x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.83, size = 196, normalized size = 1.15 \[ \left \{\begin {array}{cl} a\,d^3\,\ln \relax (x) & \text {\ if\ \ }c=0\\ a\,d^3\,\ln \relax (x)-\frac {b\,d^3\,\ln \left (c^2\,x^2+1\right )\,3{}\mathrm {i}}{2}-\frac {b\,d^3\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {b\,d^3\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}-\frac {3\,a\,c^2\,d^3\,x^2}{2}-\frac {a\,c^3\,d^3\,x^3\,1{}\mathrm {i}}{3}+a\,c\,d^3\,x\,3{}\mathrm {i}+\frac {3\,b\,c\,d^3\,x}{2}+\frac {b\,c^2\,d^3\,\left (\frac {x^2}{2}-\frac {\ln \left (c^2\,x^2+1\right )}{2\,c^2}\right )\,1{}\mathrm {i}}{3}-3\,b\,c^2\,d^3\,\mathrm {atan}\left (c\,x\right )\,\left (\frac {1}{2\,c^2}+\frac {x^2}{2}\right )-\frac {b\,c^3\,d^3\,x^3\,\mathrm {atan}\left (c\,x\right )\,1{}\mathrm {i}}{3}+b\,c\,d^3\,x\,\mathrm {atan}\left (c\,x\right )\,3{}\mathrm {i} & \text {\ if\ \ }c\neq 0 \end {array}\right . \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ - i d^{3} \left (\int \left (- 3 a c\right )\, dx + \int \frac {i a}{x}\, dx + \int a c^{3} x^{2}\, dx + \int \left (- 3 b c \operatorname {atan}{\left (c x \right )}\right )\, dx + \int \left (- 3 i a c^{2} x\right )\, dx + \int \frac {i b \operatorname {atan}{\left (c x \right )}}{x}\, dx + \int b c^{3} x^{2} \operatorname {atan}{\left (c x \right )}\, dx + \int \left (- 3 i b c^{2} x \operatorname {atan}{\left (c x \right )}\right )\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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