Optimal. Leaf size=181 \[ \frac{i b \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 d}+\frac{i b \text{PolyLog}(2,-i c x)}{2 d}-\frac{i b \text{PolyLog}(2,i c x)}{2 d}-\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{d}+\frac{\log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}+\frac{a \log (x)}{d} \]
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Rubi [A] time = 0.186101, antiderivative size = 181, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {4876, 4848, 2391, 4856, 2402, 2315, 2447} \[ \frac{i b \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 d}+\frac{i b \text{PolyLog}(2,-i c x)}{2 d}-\frac{i b \text{PolyLog}(2,i c x)}{2 d}-\frac{i b \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{2 d}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{d}+\frac{\log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{d}+\frac{a \log (x)}{d} \]
Antiderivative was successfully verified.
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Rule 4876
Rule 4848
Rule 2391
Rule 4856
Rule 2402
Rule 2315
Rule 2447
Rubi steps
\begin{align*} \int \frac{a+b \tan ^{-1}(c x)}{x (d+e x)} \, dx &=\int \left (\frac{a+b \tan ^{-1}(c x)}{d x}-\frac{e \left (a+b \tan ^{-1}(c x)\right )}{d (d+e x)}\right ) \, dx\\ &=\frac{\int \frac{a+b \tan ^{-1}(c x)}{x} \, dx}{d}-\frac{e \int \frac{a+b \tan ^{-1}(c x)}{d+e x} \, dx}{d}\\ &=\frac{a \log (x)}{d}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d}+\frac{(i b) \int \frac{\log (1-i c x)}{x} \, dx}{2 d}-\frac{(i b) \int \frac{\log (1+i c x)}{x} \, dx}{2 d}-\frac{(b c) \int \frac{\log \left (\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{d}+\frac{(b c) \int \frac{\log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{1+c^2 x^2} \, dx}{d}\\ &=\frac{a \log (x)}{d}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d}+\frac{i b \text{Li}_2(-i c x)}{2 d}-\frac{i b \text{Li}_2(i c x)}{2 d}+\frac{i b \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d}-\frac{(i b) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-i c x}\right )}{d}\\ &=\frac{a \log (x)}{d}+\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{d}-\frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{d}+\frac{i b \text{Li}_2(-i c x)}{2 d}-\frac{i b \text{Li}_2(i c x)}{2 d}-\frac{i b \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{2 d}+\frac{i b \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 d}\\ \end{align*}
Mathematica [A] time = 0.0950062, size = 169, normalized size = 0.93 \[ \frac{-i b \text{PolyLog}\left (2,\frac{e (1-i c x)}{e+i c d}\right )+i b \text{PolyLog}\left (2,-\frac{e (c x-i)}{c d+i e}\right )+i b \text{PolyLog}(2,-i c x)-i b \text{PolyLog}(2,i c x)-2 a \log (d+e x)+2 a \log (x)-i b \log (1-i c x) \log \left (\frac{c (d+e x)}{c d-i e}\right )+i b \log (1+i c x) \log \left (\frac{c (d+e x)}{c d+i e}\right )}{2 d} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.051, size = 260, normalized size = 1.4 \begin{align*} -{\frac{a\ln \left ( ecx+dc \right ) }{d}}+{\frac{a\ln \left ( cx \right ) }{d}}-{\frac{b\arctan \left ( cx \right ) \ln \left ( ecx+dc \right ) }{d}}+{\frac{b\arctan \left ( cx \right ) \ln \left ( cx \right ) }{d}}+{\frac{{\frac{i}{2}}b\ln \left ( cx \right ) \ln \left ( 1+icx \right ) }{d}}-{\frac{{\frac{i}{2}}b\ln \left ( cx \right ) \ln \left ( 1-icx \right ) }{d}}+{\frac{{\frac{i}{2}}b{\it dilog} \left ( 1+icx \right ) }{d}}-{\frac{{\frac{i}{2}}b{\it dilog} \left ( 1-icx \right ) }{d}}-{\frac{{\frac{i}{2}}b\ln \left ( ecx+dc \right ) }{d}\ln \left ({\frac{ie-ecx}{dc+ie}} \right ) }+{\frac{{\frac{i}{2}}b\ln \left ( ecx+dc \right ) }{d}\ln \left ({\frac{ie+ecx}{ie-dc}} \right ) }-{\frac{{\frac{i}{2}}b}{d}{\it dilog} \left ({\frac{ie-ecx}{dc+ie}} \right ) }+{\frac{{\frac{i}{2}}b}{d}{\it dilog} \left ({\frac{ie+ecx}{ie-dc}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -a{\left (\frac{\log \left (e x + d\right )}{d} - \frac{\log \left (x\right )}{d}\right )} + 2 \, b \int \frac{\arctan \left (c x\right )}{2 \,{\left (e x^{2} + d x\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \arctan \left (c x\right ) + a}{e x^{2} + d x}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \arctan \left (c x\right ) + a}{{\left (e x + d\right )} x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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