Optimal. Leaf size=59 \[ \frac {i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c d}-\frac {b \text {Li}_2\left (1-\frac {2}{i c x+1}\right )}{2 c d} \]
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Rubi [A] time = 0.05, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4854, 2402, 2315} \[ \frac {i \log \left (\frac {2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c d}-\frac {b \text {PolyLog}\left (2,1-\frac {2}{1+i c x}\right )}{2 c d} \]
Antiderivative was successfully verified.
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Rule 2315
Rule 2402
Rule 4854
Rubi steps
\begin {align*} \int \frac {a+b \tan ^{-1}(c x)}{d+i c d x} \, dx &=\frac {i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {(i b) \int \frac {\log \left (\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{d}\\ &=\frac {i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {b \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+i c x}\right )}{c d}\\ &=\frac {i \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac {2}{1+i c x}\right )}{c d}-\frac {b \text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{2 c d}\\ \end {align*}
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Mathematica [A] time = 0.02, size = 60, normalized size = 1.02 \[ \frac {2 i \log \left (\frac {2 d}{d+i c d x}\right ) \left (a+b \tan ^{-1}(c x)\right )-b \text {Li}_2\left (\frac {c x+i}{c x-i}\right )}{2 c d} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \log \left (-\frac {c x + i}{c x - i}\right ) - 2 i \, a}{2 \, c d x - 2 i \, d}, x\right ) \]
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 [B] time = 0.04, size = 142, normalized size = 2.41 \[ -\frac {i a \ln \left (c^{2} x^{2}+1\right )}{2 c d}+\frac {a \arctan \left (c x \right )}{c d}-\frac {i b \ln \left (i c x +1\right ) \arctan \left (c x \right )}{c d}-\frac {b \ln \left (\frac {1}{2}-\frac {i c x}{2}\right ) \ln \left (i c x +1\right )}{2 c d}+\frac {b \ln \left (\frac {1}{2}-\frac {i c x}{2}\right ) \ln \left (\frac {i c x}{2}+\frac {1}{2}\right )}{2 c d}+\frac {b \dilog \left (\frac {i c x}{2}+\frac {1}{2}\right )}{2 c d}+\frac {b \ln \left (i c x +1\right )^{2}}{4 c d} \]
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 \[ -\frac {{\left (8 i \, c^{2} d \mathit {sage}_{0} x - 4 \, \arctan \left (c x\right )^{2}\right )} b}{8 \, c d} - \frac {i \, a \log \left (i \, c d x + d\right )}{c d} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{d+c\,d\,x\,1{}\mathrm {i}} \,d x \]
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 \[ \frac {b \log {\left (- i c x + 1 \right )} \log {\left (i c x + 1 \right )}}{2 c d} - \frac {i \left (\int \frac {i a}{c^{2} x^{2} + 1}\, dx + \int \frac {a c x}{c^{2} x^{2} + 1}\, dx + \int \left (- \frac {i b c x \log {\left (i c x + 1 \right )}}{c^{2} x^{2} + 1}\right )\, dx\right )}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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