Optimal. Leaf size=151 \[ -\frac {1}{2} i b \text {Li}_2\left (1-\frac {2}{i c x^2+1}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )+\frac {1}{2} i b \text {Li}_2\left (\frac {2}{i c x^2+1}-1\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )+\tanh ^{-1}\left (1-\frac {2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2-\frac {1}{4} b^2 \text {Li}_3\left (1-\frac {2}{i c x^2+1}\right )+\frac {1}{4} b^2 \text {Li}_3\left (\frac {2}{i c x^2+1}-1\right ) \]
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Rubi [A] time = 0.32, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5031, 4850, 4988, 4884, 4994, 6610} \[ -\frac {1}{2} i b \text {PolyLog}\left (2,1-\frac {2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )+\frac {1}{2} i b \text {PolyLog}\left (2,-1+\frac {2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )-\frac {1}{4} b^2 \text {PolyLog}\left (3,1-\frac {2}{1+i c x^2}\right )+\frac {1}{4} b^2 \text {PolyLog}\left (3,-1+\frac {2}{1+i c x^2}\right )+\tanh ^{-1}\left (1-\frac {2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \]
Antiderivative was successfully verified.
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Rule 4850
Rule 4884
Rule 4988
Rule 4994
Rule 5031
Rule 6610
Rubi steps
\begin {align*} \int \frac {\left (a+b \tan ^{-1}\left (c x^2\right )\right )^2}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\left (a+b \tan ^{-1}(c x)\right )^2}{x} \, dx,x,x^2\right )\\ &=\left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x^2}\right )-(2 b c) \operatorname {Subst}\left (\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,x,x^2\right )\\ &=\left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x^2}\right )+(b c) \operatorname {Subst}\left (\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,x,x^2\right )-(b c) \operatorname {Subst}\left (\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,x,x^2\right )\\ &=\left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x^2}\right )-\frac {1}{2} i b \left (a+b \tan ^{-1}\left (c x^2\right )\right ) \text {Li}_2\left (1-\frac {2}{1+i c x^2}\right )+\frac {1}{2} i b \left (a+b \tan ^{-1}\left (c x^2\right )\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x^2}\right )+\frac {1}{2} \left (i b^2 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (1-\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,x^2\right )-\frac {1}{2} \left (i b^2 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2\left (-1+\frac {2}{1+i c x}\right )}{1+c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1+i c x^2}\right )-\frac {1}{2} i b \left (a+b \tan ^{-1}\left (c x^2\right )\right ) \text {Li}_2\left (1-\frac {2}{1+i c x^2}\right )+\frac {1}{2} i b \left (a+b \tan ^{-1}\left (c x^2\right )\right ) \text {Li}_2\left (-1+\frac {2}{1+i c x^2}\right )-\frac {1}{4} b^2 \text {Li}_3\left (1-\frac {2}{1+i c x^2}\right )+\frac {1}{4} b^2 \text {Li}_3\left (-1+\frac {2}{1+i c x^2}\right )\\ \end {align*}
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Mathematica [A] time = 0.11, size = 165, normalized size = 1.09 \[ \frac {1}{4} b \left (2 i \text {Li}_2\left (\frac {c x^2+i}{i-c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )-2 i \text {Li}_2\left (\frac {c x^2+i}{c x^2-i}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )+b \left (\text {Li}_3\left (\frac {c x^2+i}{i-c x^2}\right )-\text {Li}_3\left (\frac {c x^2+i}{c x^2-i}\right )\right )\right )+\tanh ^{-1}\left (1+\frac {2 i}{c x^2-i}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \]
Antiderivative was successfully verified.
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fricas [F] time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \arctan \left (c x^{2}\right )^{2} + 2 \, a b \arctan \left (c x^{2}\right ) + a^{2}}{x}, x\right ) \]
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 \[ \int \frac {{\left (b \arctan \left (c x^{2}\right ) + a\right )}^{2}}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.36, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arctan \left (c \,x^{2}\right )\right )^{2}}{x}\, dx \]
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 \[ a^{2} \log \relax (x) + \frac {1}{16} \, \int \frac {12 \, b^{2} \arctan \left (c x^{2}\right )^{2} + b^{2} \log \left (c^{2} x^{4} + 1\right )^{2} + 32 \, a b \arctan \left (c x^{2}\right )}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {atan}\left (c\,x^2\right )\right )}^2}{x} \,d x \]
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 \[ \int \frac {\left (a + b \operatorname {atan}{\left (c x^{2} \right )}\right )^{2}}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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