Optimal. Leaf size=216 \[ -i b d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+i b d \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+b^2 (-d) \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )-\frac{1}{2} b^2 d \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 d \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )-d \left (a+b \tan ^{-1}(c x)\right )^2+i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 i b d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+2 d \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2 \]
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Rubi [A] time = 0.420807, antiderivative size = 216, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 11, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.478, Rules used = {4876, 4846, 4920, 4854, 2402, 2315, 4850, 4988, 4884, 4994, 6610} \[ -i b d \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+i b d \text{PolyLog}\left (2,-1+\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+b^2 (-d) \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right )-\frac{1}{2} b^2 d \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 d \text{PolyLog}\left (3,-1+\frac{2}{1+i c x}\right )-d \left (a+b \tan ^{-1}(c x)\right )^2+i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 i b d \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )+2 d \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2 \]
Antiderivative was successfully verified.
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Rule 4876
Rule 4846
Rule 4920
Rule 4854
Rule 2402
Rule 2315
Rule 4850
Rule 4988
Rule 4884
Rule 4994
Rule 6610
Rubi steps
\begin{align*} \int \frac{(d+i c d x) \left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx &=\int \left (i c d \left (a+b \tan ^{-1}(c x)\right )^2+\frac{d \left (a+b \tan ^{-1}(c x)\right )^2}{x}\right ) \, dx\\ &=d \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx+(i c d) \int \left (a+b \tan ^{-1}(c x)\right )^2 \, dx\\ &=i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )-(4 b c d) \int \frac{\left (a+b \tan ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (2 i b c^2 d\right ) \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{1+c^2 x^2} \, dx\\ &=-d \left (a+b \tan ^{-1}(c x)\right )^2+i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )+(2 i b c d) \int \frac{a+b \tan ^{-1}(c x)}{i-c x} \, dx+(2 b c d) \int \frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx-(2 b c d) \int \frac{\left (a+b \tan ^{-1}(c x)\right ) \log \left (2-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx\\ &=-d \left (a+b \tan ^{-1}(c x)\right )^2+i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )+2 i b d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )-i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )+i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )+\left (i b^2 c d\right ) \int \frac{\text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (i b^2 c d\right ) \int \frac{\text{Li}_2\left (-1+\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx-\left (2 i b^2 c d\right ) \int \frac{\log \left (\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx\\ &=-d \left (a+b \tan ^{-1}(c x)\right )^2+i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )+2 i b d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )-i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )+i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )-\frac{1}{2} b^2 d \text{Li}_3\left (1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 d \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )-\left (2 b^2 d\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+i c x}\right )\\ &=-d \left (a+b \tan ^{-1}(c x)\right )^2+i c d x \left (a+b \tan ^{-1}(c x)\right )^2+2 d \left (a+b \tan ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+i c x}\right )+2 i b d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1+i c x}\right )-b^2 d \text{Li}_2\left (1-\frac{2}{1+i c x}\right )-i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )+i b d \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (-1+\frac{2}{1+i c x}\right )-\frac{1}{2} b^2 d \text{Li}_3\left (1-\frac{2}{1+i c x}\right )+\frac{1}{2} b^2 d \text{Li}_3\left (-1+\frac{2}{1+i c x}\right )\\ \end{align*}
Mathematica [A] time = 0.453498, size = 272, normalized size = 1.26 \[ d \left (i a b (\text{PolyLog}(2,-i c x)-\text{PolyLog}(2,i c x))+b^2 \left (\text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+\tan ^{-1}(c x) \left ((1+i c x) \tan ^{-1}(c x)+2 i \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )\right )\right )+b^2 \left (i \tan ^{-1}(c x) \text{PolyLog}\left (2,e^{-2 i \tan ^{-1}(c x)}\right )+i \tan ^{-1}(c x) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,e^{-2 i \tan ^{-1}(c x)}\right )-\frac{1}{2} \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(c x)}\right )+\frac{2}{3} i \tan ^{-1}(c x)^3+\tan ^{-1}(c x)^2 \log \left (1-e^{-2 i \tan ^{-1}(c x)}\right )-\tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-\frac{i \pi ^3}{24}\right )+i a^2 c x+a^2 \log (c x)+i a b \left (2 c x \tan ^{-1}(c x)-\log \left (c^2 x^2+1\right )\right )\right ) \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.599, size = 7034, normalized size = 32.6 \begin{align*} \text{output too large to display} \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*} \text{result too large to display} \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{4 i \, a^{2} c d x + 4 \, a^{2} d +{\left (-i \, b^{2} c d x - b^{2} d\right )} \log \left (-\frac{c x + i}{c x - i}\right )^{2} - 4 \,{\left (a b c d x - i \, a b d\right )} \log \left (-\frac{c x + i}{c x - i}\right )}{4 \, x}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} d \left (\int \frac{a^{2}}{x}\, dx + \int i a^{2} c\, dx + \int \frac{b^{2} \operatorname{atan}^{2}{\left (c x \right )}}{x}\, dx + \int i b^{2} c \operatorname{atan}^{2}{\left (c x \right )}\, dx + \int \frac{2 a b \operatorname{atan}{\left (c x \right )}}{x}\, dx + \int 2 i a b c \operatorname{atan}{\left (c x \right )}\, dx\right ) \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{{\left (i \, c d x + d\right )}{\left (b \arctan \left (c x\right ) + a\right )}^{2}}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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