Optimal. Leaf size=64 \[ -\frac {i \text {Li}_2\left (\frac {2}{1-i a x}-1\right )}{2 c}-\frac {i \tan ^{-1}(a x)^2}{2 c}+\frac {\log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{c} \]
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Rubi [A] time = 0.10, antiderivative size = 64, 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 = {4924, 4868, 2447} \[ -\frac {i \text {PolyLog}\left (2,-1+\frac {2}{1-i a x}\right )}{2 c}-\frac {i \tan ^{-1}(a x)^2}{2 c}+\frac {\log \left (2-\frac {2}{1-i a x}\right ) \tan ^{-1}(a x)}{c} \]
Antiderivative was successfully verified.
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Rule 2447
Rule 4868
Rule 4924
Rubi steps
\begin {align*} \int \frac {\tan ^{-1}(a x)}{x \left (c+a^2 c x^2\right )} \, dx &=-\frac {i \tan ^{-1}(a x)^2}{2 c}+\frac {i \int \frac {\tan ^{-1}(a x)}{x (i+a x)} \, dx}{c}\\ &=-\frac {i \tan ^{-1}(a x)^2}{2 c}+\frac {\tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{c}-\frac {a \int \frac {\log \left (2-\frac {2}{1-i a x}\right )}{1+a^2 x^2} \, dx}{c}\\ &=-\frac {i \tan ^{-1}(a x)^2}{2 c}+\frac {\tan ^{-1}(a x) \log \left (2-\frac {2}{1-i a x}\right )}{c}-\frac {i \text {Li}_2\left (-1+\frac {2}{1-i a x}\right )}{2 c}\\ \end {align*}
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Mathematica [A] time = 0.03, size = 103, normalized size = 1.61 \[ \frac {i \text {Li}_2(-i a x)}{2 c}-\frac {i \text {Li}_2(i a x)}{2 c}+\frac {i \text {Li}_2\left (-\frac {a x+i}{i-a x}\right )}{2 c}+\frac {i \tan ^{-1}(a x)^2}{2 c}+\frac {\log \left (\frac {2 i}{-a x+i}\right ) \tan ^{-1}(a x)}{c} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arctan \left (a x\right )}{a^{2} c x^{3} + c x}, x\right ) \]
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 = 0.11, size = 251, normalized size = 3.92 \[ \frac {\arctan \left (a x \right ) \ln \left (a x \right )}{c}-\frac {\arctan \left (a x \right ) \ln \left (a^{2} x^{2}+1\right )}{2 c}+\frac {i \ln \left (a x \right ) \ln \left (i a x +1\right )}{2 c}-\frac {i \ln \left (a x \right ) \ln \left (-i a x +1\right )}{2 c}+\frac {i \dilog \left (i a x +1\right )}{2 c}-\frac {i \dilog \left (-i a x +1\right )}{2 c}-\frac {i \ln \left (a x -i\right ) \ln \left (a^{2} x^{2}+1\right )}{4 c}+\frac {i \ln \left (a x -i\right )^{2}}{8 c}+\frac {i \dilog \left (-\frac {i \left (a x +i\right )}{2}\right )}{4 c}+\frac {i \ln \left (a x -i\right ) \ln \left (-\frac {i \left (a x +i\right )}{2}\right )}{4 c}+\frac {i \ln \left (a x +i\right ) \ln \left (a^{2} x^{2}+1\right )}{4 c}-\frac {i \ln \left (a x +i\right )^{2}}{8 c}-\frac {i \dilog \left (\frac {i \left (a x -i\right )}{2}\right )}{4 c}-\frac {i \ln \left (a x +i\right ) \ln \left (\frac {i \left (a x -i\right )}{2}\right )}{4 c} \]
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 \[ \int \frac {\arctan \left (a x\right )}{{\left (a^{2} c x^{2} + c\right )} x}\,{d x} \]
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 \[ \int \frac {\mathrm {atan}\left (a\,x\right )}{x\,\left (c\,a^2\,x^2+c\right )} \,d x \]
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 \[ \frac {\int \frac {\operatorname {atan}{\left (a x \right )}}{a^{2} x^{3} + x}\, dx}{c} \]
Verification of antiderivative is not currently implemented for this CAS.
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