Optimal. Leaf size=89 \[ -a c^2 d^2 x-b c^2 d^2 x \text {ArcTan}(c x)-\frac {d^2 (a+b \text {ArcTan}(c x))}{x}+2 i a c d^2 \log (x)+b c d^2 \log (x)-b c d^2 \text {PolyLog}(2,-i c x)+b c d^2 \text {PolyLog}(2,i c x) \]
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Rubi [A]
time = 0.10, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps
used = 13, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {4996, 4930,
266, 4946, 272, 36, 29, 31, 4940, 2438} \begin {gather*} -\frac {d^2 (a+b \text {ArcTan}(c x))}{x}-a c^2 d^2 x+2 i a c d^2 \log (x)-b c^2 d^2 x \text {ArcTan}(c x)-b c d^2 \text {Li}_2(-i c x)+b c d^2 \text {Li}_2(i c x)+b c d^2 \log (x) \end {gather*}
Antiderivative was successfully verified.
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Rule 29
Rule 31
Rule 36
Rule 266
Rule 272
Rule 2438
Rule 4930
Rule 4940
Rule 4946
Rule 4996
Rubi steps
\begin {align*} \int \frac {(d+i c d x)^2 \left (a+b \tan ^{-1}(c x)\right )}{x^2} \, dx &=\int \left (-c^2 d^2 \left (a+b \tan ^{-1}(c x)\right )+\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{x^2}+\frac {2 i c d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d^2 \int \frac {a+b \tan ^{-1}(c x)}{x^2} \, dx+\left (2 i c d^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{x} \, dx-\left (c^2 d^2\right ) \int \left (a+b \tan ^{-1}(c x)\right ) \, dx\\ &=-a c^2 d^2 x-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+2 i a c d^2 \log (x)+\left (b c d^2\right ) \int \frac {1}{x \left (1+c^2 x^2\right )} \, dx-\left (b c d^2\right ) \int \frac {\log (1-i c x)}{x} \, dx+\left (b c d^2\right ) \int \frac {\log (1+i c x)}{x} \, dx-\left (b c^2 d^2\right ) \int \tan ^{-1}(c x) \, dx\\ &=-a c^2 d^2 x-b c^2 d^2 x \tan ^{-1}(c x)-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+2 i a c d^2 \log (x)-b c d^2 \text {Li}_2(-i c x)+b c d^2 \text {Li}_2(i c x)+\frac {1}{2} \left (b c d^2\right ) \text {Subst}\left (\int \frac {1}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )+\left (b c^3 d^2\right ) \int \frac {x}{1+c^2 x^2} \, dx\\ &=-a c^2 d^2 x-b c^2 d^2 x \tan ^{-1}(c x)-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+2 i a c d^2 \log (x)+\frac {1}{2} b c d^2 \log \left (1+c^2 x^2\right )-b c d^2 \text {Li}_2(-i c x)+b c d^2 \text {Li}_2(i c x)+\frac {1}{2} \left (b c d^2\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )-\frac {1}{2} \left (b c^3 d^2\right ) \text {Subst}\left (\int \frac {1}{1+c^2 x} \, dx,x,x^2\right )\\ &=-a c^2 d^2 x-b c^2 d^2 x \tan ^{-1}(c x)-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+2 i a c d^2 \log (x)+b c d^2 \log (x)-b c d^2 \text {Li}_2(-i c x)+b c d^2 \text {Li}_2(i c x)\\ \end {align*}
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Mathematica [A]
time = 0.05, size = 79, normalized size = 0.89 \begin {gather*} -\frac {d^2 \left (a+a c^2 x^2+b \text {ArcTan}(c x)+b c^2 x^2 \text {ArcTan}(c x)-2 i a c x \log (x)-b c x \log (c x)+b c x \text {PolyLog}(2,-i c x)-b c x \text {PolyLog}(2,i c x)\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.07, size = 149, normalized size = 1.67
method | result | size |
derivativedivides | \(c \left (-a c \,d^{2} x -\frac {d^{2} a}{c x}+2 i d^{2} a \ln \left (c x \right )-b \arctan \left (c x \right ) d^{2} c x -\frac {d^{2} b \arctan \left (c x \right )}{c x}+2 i d^{2} b \arctan \left (c x \right ) \ln \left (c x \right )-d^{2} b \ln \left (c x \right ) \ln \left (i c x +1\right )+d^{2} b \ln \left (c x \right ) \ln \left (-i c x +1\right )+d^{2} b \dilog \left (-i c x +1\right )-d^{2} b \dilog \left (i c x +1\right )+d^{2} b \ln \left (c x \right )\right )\) | \(149\) |
default | \(c \left (-a c \,d^{2} x -\frac {d^{2} a}{c x}+2 i d^{2} a \ln \left (c x \right )-b \arctan \left (c x \right ) d^{2} c x -\frac {d^{2} b \arctan \left (c x \right )}{c x}+2 i d^{2} b \arctan \left (c x \right ) \ln \left (c x \right )-d^{2} b \ln \left (c x \right ) \ln \left (i c x +1\right )+d^{2} b \ln \left (c x \right ) \ln \left (-i c x +1\right )+d^{2} b \dilog \left (-i c x +1\right )-d^{2} b \dilog \left (i c x +1\right )+d^{2} b \ln \left (c x \right )\right )\) | \(149\) |
risch | \(-\frac {i c^{2} d^{2} b \ln \left (-i c x +1\right ) x}{2}+\frac {i b \,d^{2} \ln \left (i c x +1\right )}{2 x}-b c \,d^{2}+c \,d^{2} b \dilog \left (-i c x +1\right )+\frac {c \,d^{2} b \ln \left (-i c x \right )}{2}+2 i c \,d^{2} a \ln \left (-i c x \right )-a \,c^{2} d^{2} x -i c \,d^{2} a +\frac {i b \,d^{2} c^{2} \ln \left (i c x +1\right ) x}{2}-\frac {d^{2} a}{x}-\frac {i d^{2} b \ln \left (-i c x +1\right )}{2 x}-b \,d^{2} c \dilog \left (i c x +1\right )+\frac {b \,d^{2} c \ln \left (i c x \right )}{2}\) | \(179\) |
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} - d^{2} \left (\int a c^{2}\, dx + \int \left (- \frac {a}{x^{2}}\right )\, dx + \int b c^{2} \operatorname {atan}{\left (c x \right )}\, dx + \int \left (- \frac {b \operatorname {atan}{\left (c x \right )}}{x^{2}}\right )\, dx + \int \left (- \frac {2 i a c}{x}\right )\, dx + \int \left (- \frac {2 i b c \operatorname {atan}{\left (c x \right )}}{x}\right )\, dx\right ) \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 0.60, size = 141, normalized size = 1.58 \begin {gather*} \left \{\begin {array}{cl} -\frac {a\,d^2}{x} & \text {\ if\ \ }c=0\\ \frac {b\,d^2\,\left (c^2\,\ln \left (x\right )-\frac {c^2\,\ln \left (c^2\,x^2+1\right )}{2}\right )}{c}+b\,c\,d^2\,\left ({\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )-{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\right )+\frac {b\,c\,d^2\,\ln \left (c^2\,x^2+1\right )}{2}-\frac {a\,d^2\,\left (c^2\,x^2+1-c\,x\,\ln \left (x\right )\,2{}\mathrm {i}\right )}{x}-\frac {b\,d^2\,\mathrm {atan}\left (c\,x\right )}{x}-b\,c^2\,d^2\,x\,\mathrm {atan}\left (c\,x\right ) & \text {\ if\ \ }c\neq 0 \end {array}\right . \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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