Optimal. Leaf size=304 \[ \frac{b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3}-\frac{i b^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 c^3 d^3}+\frac{7 i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (-c x+i)}+\frac{b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (-c x+i)^2}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (-c x+i)}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (-c x+i)^2}-\frac{7 i \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac{i \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3}+\frac{13 b^2}{16 c^3 d^3 (-c x+i)}-\frac{i b^2}{16 c^3 d^3 (-c x+i)^2}-\frac{13 b^2 \tan ^{-1}(c x)}{16 c^3 d^3} \]
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Rubi [A] time = 0.560428, antiderivative size = 304, normalized size of antiderivative = 1., number of steps used = 26, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4876, 4864, 4862, 627, 44, 203, 4884, 4854, 4994, 6610} \[ \frac{b \text{PolyLog}\left (2,1-\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{c^3 d^3}-\frac{i b^2 \text{PolyLog}\left (3,1-\frac{2}{1+i c x}\right )}{2 c^3 d^3}+\frac{7 i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (-c x+i)}+\frac{b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (-c x+i)^2}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (-c x+i)}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (-c x+i)^2}-\frac{7 i \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^3 d^3}-\frac{i \log \left (\frac{2}{1+i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3}+\frac{13 b^2}{16 c^3 d^3 (-c x+i)}-\frac{i b^2}{16 c^3 d^3 (-c x+i)^2}-\frac{13 b^2 \tan ^{-1}(c x)}{16 c^3 d^3} \]
Antiderivative was successfully verified.
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Rule 4876
Rule 4864
Rule 4862
Rule 627
Rule 44
Rule 203
Rule 4884
Rule 4854
Rule 4994
Rule 6610
Rubi steps
\begin{align*} \int \frac{x^2 \left (a+b \tan ^{-1}(c x)\right )^2}{(d+i c d x)^3} \, dx &=\int \left (-\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d^3 (-i+c x)^3}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d^3 (-i+c x)^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{c^2 d^3 (-i+c x)}\right ) \, dx\\ &=-\frac{i \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^3} \, dx}{c^2 d^3}+\frac{i \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{-i+c x} \, dx}{c^2 d^3}-\frac{2 \int \frac{\left (a+b \tan ^{-1}(c x)\right )^2}{(-i+c x)^2} \, dx}{c^2 d^3}\\ &=\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (i-c x)^2}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^3 d^3}-\frac{(i b) \int \left (-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 (-i+c x)^3}+\frac{a+b \tan ^{-1}(c x)}{4 (-i+c x)^2}-\frac{a+b \tan ^{-1}(c x)}{4 \left (1+c^2 x^2\right )}\right ) \, dx}{c^2 d^3}+\frac{(2 i b) \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}{c^2 d^3}-\frac{(4 b) \int \left (-\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 (-i+c x)^2}+\frac{i \left (a+b \tan ^{-1}(c x)\right )}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{c^2 d^3}\\ &=\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (i-c x)^2}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^3 d^3}+\frac{b \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^3 d^3}-\frac{(i b) \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{4 c^2 d^3}+\frac{(i b) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{4 c^2 d^3}+\frac{(2 i b) \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^2} \, dx}{c^2 d^3}-\frac{(2 i b) \int \frac{a+b \tan ^{-1}(c x)}{1+c^2 x^2} \, dx}{c^2 d^3}-\frac{b \int \frac{a+b \tan ^{-1}(c x)}{(-i+c x)^3} \, dx}{2 c^2 d^3}-\frac{b^2 \int \frac{\text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{1+c^2 x^2} \, dx}{c^2 d^3}\\ &=\frac{b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)^2}+\frac{7 i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)}-\frac{7 i \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^3 d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (i-c x)^2}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^3 d^3}+\frac{b \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^3 d^3}-\frac{i b^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^3 d^3}-\frac{\left (i b^2\right ) \int \frac{1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{4 c^2 d^3}+\frac{\left (2 i b^2\right ) \int \frac{1}{(-i+c x) \left (1+c^2 x^2\right )} \, dx}{c^2 d^3}-\frac{b^2 \int \frac{1}{(-i+c x)^2 \left (1+c^2 x^2\right )} \, dx}{4 c^2 d^3}\\ &=\frac{b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)^2}+\frac{7 i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)}-\frac{7 i \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^3 d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (i-c x)^2}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^3 d^3}+\frac{b \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^3 d^3}-\frac{i b^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^3 d^3}-\frac{\left (i b^2\right ) \int \frac{1}{(-i+c x)^2 (i+c x)} \, dx}{4 c^2 d^3}+\frac{\left (2 i b^2\right ) \int \frac{1}{(-i+c x)^2 (i+c x)} \, dx}{c^2 d^3}-\frac{b^2 \int \frac{1}{(-i+c x)^3 (i+c x)} \, dx}{4 c^2 d^3}\\ &=\frac{b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)^2}+\frac{7 i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)}-\frac{7 i \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^3 d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (i-c x)^2}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^3 d^3}+\frac{b \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^3 d^3}-\frac{i b^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^3 d^3}-\frac{\left (i b^2\right ) \int \left (-\frac{i}{2 (-i+c x)^2}+\frac{i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{4 c^2 d^3}+\frac{\left (2 i b^2\right ) \int \left (-\frac{i}{2 (-i+c x)^2}+\frac{i}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{c^2 d^3}-\frac{b^2 \int \left (-\frac{i}{2 (-i+c x)^3}+\frac{1}{4 (-i+c x)^2}-\frac{1}{4 \left (1+c^2 x^2\right )}\right ) \, dx}{4 c^2 d^3}\\ &=-\frac{i b^2}{16 c^3 d^3 (i-c x)^2}+\frac{13 b^2}{16 c^3 d^3 (i-c x)}+\frac{b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)^2}+\frac{7 i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)}-\frac{7 i \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^3 d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (i-c x)^2}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^3 d^3}+\frac{b \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^3 d^3}-\frac{i b^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^3 d^3}+\frac{b^2 \int \frac{1}{1+c^2 x^2} \, dx}{16 c^2 d^3}+\frac{b^2 \int \frac{1}{1+c^2 x^2} \, dx}{8 c^2 d^3}-\frac{b^2 \int \frac{1}{1+c^2 x^2} \, dx}{c^2 d^3}\\ &=-\frac{i b^2}{16 c^3 d^3 (i-c x)^2}+\frac{13 b^2}{16 c^3 d^3 (i-c x)}-\frac{13 b^2 \tan ^{-1}(c x)}{16 c^3 d^3}+\frac{b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)^2}+\frac{7 i b \left (a+b \tan ^{-1}(c x)\right )}{4 c^3 d^3 (i-c x)}-\frac{7 i \left (a+b \tan ^{-1}(c x)\right )^2}{8 c^3 d^3}+\frac{i \left (a+b \tan ^{-1}(c x)\right )^2}{2 c^3 d^3 (i-c x)^2}-\frac{2 \left (a+b \tan ^{-1}(c x)\right )^2}{c^3 d^3 (i-c x)}-\frac{i \left (a+b \tan ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+i c x}\right )}{c^3 d^3}+\frac{b \left (a+b \tan ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+i c x}\right )}{c^3 d^3}-\frac{i b^2 \text{Li}_3\left (1-\frac{2}{1+i c x}\right )}{2 c^3 d^3}\\ \end{align*}
Mathematica [A] time = 1.22569, size = 431, normalized size = 1.42 \[ \frac{-12 a b \left (16 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+32 \tan ^{-1}(c x)^2+12 i \sin \left (2 \tan ^{-1}(c x)\right )-i \sin \left (4 \tan ^{-1}(c x)\right )-12 \cos \left (2 \tan ^{-1}(c x)\right )+\cos \left (4 \tan ^{-1}(c x)\right )+4 \tan ^{-1}(c x) \left (8 i \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-6 \sin \left (2 \tan ^{-1}(c x)\right )+\sin \left (4 \tan ^{-1}(c x)\right )-6 i \cos \left (2 \tan ^{-1}(c x)\right )+i \cos \left (4 \tan ^{-1}(c x)\right )\right )\right )-b^2 \left (192 \tan ^{-1}(c x) \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+96 i \text{PolyLog}\left (3,-e^{2 i \tan ^{-1}(c x)}\right )+128 \tan ^{-1}(c x)^3+192 i \tan ^{-1}(c x)^2 \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-144 \tan ^{-1}(c x)^2 \sin \left (2 \tan ^{-1}(c x)\right )+24 \tan ^{-1}(c x)^2 \sin \left (4 \tan ^{-1}(c x)\right )+144 i \tan ^{-1}(c x) \sin \left (2 \tan ^{-1}(c x)\right )-12 i \tan ^{-1}(c x) \sin \left (4 \tan ^{-1}(c x)\right )+72 \sin \left (2 \tan ^{-1}(c x)\right )-3 \sin \left (4 \tan ^{-1}(c x)\right )-144 i \tan ^{-1}(c x)^2 \cos \left (2 \tan ^{-1}(c x)\right )+24 i \tan ^{-1}(c x)^2 \cos \left (4 \tan ^{-1}(c x)\right )-144 \tan ^{-1}(c x) \cos \left (2 \tan ^{-1}(c x)\right )+12 \tan ^{-1}(c x) \cos \left (4 \tan ^{-1}(c x)\right )+72 i \cos \left (2 \tan ^{-1}(c x)\right )-3 i \cos \left (4 \tan ^{-1}(c x)\right )\right )+96 i a^2 \log \left (c^2 x^2+1\right )+\frac{384 a^2}{c x-i}+\frac{96 i a^2}{(c x-i)^2}-192 a^2 \tan ^{-1}(c x)}{192 c^3 d^3} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.347, size = 1276, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [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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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{-i \, b^{2} x^{2} \log \left (-\frac{c x + i}{c x - i}\right )^{2} - 4 \, a b x^{2} \log \left (-\frac{c x + i}{c x - i}\right ) + 4 i \, a^{2} x^{2}}{4 \, c^{3} d^{3} x^{3} - 12 i \, c^{2} d^{3} x^{2} - 12 \, c d^{3} x + 4 i \, d^{3}}, 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{{\left (b \arctan \left (c x\right ) + a\right )}^{2} x^{2}}{{\left (i \, c d x + d\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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