Optimal. Leaf size=101 \[ \frac{i b^2 \text{PolyLog}\left (2,1-\frac{2}{1+i c x^2}\right )}{2 c}+\frac{1}{2} x^2 \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2+\frac{i \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2}{2 c}+\frac{b \log \left (\frac{2}{1+i c x^2}\right ) \left (a+b \tan ^{-1}\left (c x^2\right )\right )}{c} \]
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Rubi [B] time = 0.553597, antiderivative size = 255, normalized size of antiderivative = 2.52, number of steps used = 28, number of rules used = 12, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.857, Rules used = {5035, 2454, 2389, 2296, 2295, 6715, 2430, 43, 2416, 2394, 2393, 2391} \[ -\frac{i b^2 \text{PolyLog}\left (2,\frac{1}{2} \left (1-i c x^2\right )\right )}{4 c}+\frac{i b^2 \text{PolyLog}\left (2,\frac{1}{2} \left (1+i c x^2\right )\right )}{4 c}+\frac{i b \log \left (\frac{1}{2} \left (1+i c x^2\right )\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )}{4 c}-\frac{1}{4} b x^2 \log \left (1+i c x^2\right ) \left (2 i a-b \log \left (1-i c x^2\right )\right )+\frac{i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 c}+\frac{i b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 c}+\frac{i b^2 \log \left (\frac{1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{4 c} \]
Warning: Unable to verify antiderivative.
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Rule 5035
Rule 2454
Rule 2389
Rule 2296
Rule 2295
Rule 6715
Rule 2430
Rule 43
Rule 2416
Rule 2394
Rule 2393
Rule 2391
Rubi steps
\begin{align*} \int x \left (a+b \tan ^{-1}\left (c x^2\right )\right )^2 \, dx &=\int \left (\frac{1}{4} x \left (2 a+i b \log \left (1-i c x^2\right )\right )^2+\frac{1}{2} b x \left (-2 i a+b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )-\frac{1}{4} b^2 x \log ^2\left (1+i c x^2\right )\right ) \, dx\\ &=\frac{1}{4} \int x \left (2 a+i b \log \left (1-i c x^2\right )\right )^2 \, dx+\frac{1}{2} b \int x \left (-2 i a+b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right ) \, dx-\frac{1}{4} b^2 \int x \log ^2\left (1+i c x^2\right ) \, dx\\ &=\frac{1}{8} \operatorname{Subst}\left (\int (2 a+i b \log (1-i c x))^2 \, dx,x,x^2\right )+\frac{1}{4} b \operatorname{Subst}\left (\int (-2 i a+b \log (1-i c x)) \log (1+i c x) \, dx,x,x^2\right )-\frac{1}{8} b^2 \operatorname{Subst}\left (\int \log ^2(1+i c x) \, dx,x,x^2\right )\\ &=-\frac{1}{4} b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac{i \operatorname{Subst}\left (\int (2 a+i b \log (x))^2 \, dx,x,1-i c x^2\right )}{8 c}+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \log ^2(x) \, dx,x,1+i c x^2\right )}{8 c}-\frac{1}{4} (i b c) \operatorname{Subst}\left (\int \frac{x (-2 i a+b \log (1-i c x))}{1+i c x} \, dx,x,x^2\right )+\frac{1}{4} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \frac{x \log (1+i c x)}{1-i c x} \, dx,x,x^2\right )\\ &=\frac{i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 c}-\frac{1}{4} b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac{i b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 c}+\frac{b \operatorname{Subst}\left (\int (2 a+i b \log (x)) \, dx,x,1-i c x^2\right )}{4 c}-\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \log (x) \, dx,x,1+i c x^2\right )}{4 c}-\frac{1}{4} (i b c) \operatorname{Subst}\left (\int \left (-\frac{i (-2 i a+b \log (1-i c x))}{c}+\frac{-2 i a+b \log (1-i c x)}{c (-i+c x)}\right ) \, dx,x,x^2\right )+\frac{1}{4} \left (i b^2 c\right ) \operatorname{Subst}\left (\int \left (\frac{i \log (1+i c x)}{c}+\frac{\log (1+i c x)}{c (i+c x)}\right ) \, dx,x,x^2\right )\\ &=-\frac{1}{2} i a b x^2-\frac{b^2 x^2}{4}+\frac{i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 c}-\frac{i b^2 \left (1+i c x^2\right ) \log \left (1+i c x^2\right )}{4 c}-\frac{1}{4} b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac{i b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 c}-\frac{1}{4} (i b) \operatorname{Subst}\left (\int \frac{-2 i a+b \log (1-i c x)}{-i+c x} \, dx,x,x^2\right )-\frac{1}{4} b \operatorname{Subst}\left (\int (-2 i a+b \log (1-i c x)) \, dx,x,x^2\right )+\frac{1}{4} \left (i b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+i c x)}{i+c x} \, dx,x,x^2\right )-\frac{1}{4} b^2 \operatorname{Subst}\left (\int \log (1+i c x) \, dx,x,x^2\right )+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \log (x) \, dx,x,1-i c x^2\right )}{4 c}\\ &=-\frac{1}{2} b^2 x^2+\frac{i b^2 \left (1-i c x^2\right ) \log \left (1-i c x^2\right )}{4 c}+\frac{i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 c}+\frac{i b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+i c x^2\right )\right )}{4 c}-\frac{i b^2 \left (1+i c x^2\right ) \log \left (1+i c x^2\right )}{4 c}+\frac{i b^2 \log \left (\frac{1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{4 c}-\frac{1}{4} b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac{i b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 c}-\frac{1}{4} b^2 \operatorname{Subst}\left (\int \log (1-i c x) \, dx,x,x^2\right )+\frac{1}{4} b^2 \operatorname{Subst}\left (\int \frac{\log \left (\frac{1}{2} i (-i+c x)\right )}{1-i c x} \, dx,x,x^2\right )+\frac{1}{4} b^2 \operatorname{Subst}\left (\int \frac{\log \left (-\frac{1}{2} i (i+c x)\right )}{1+i c x} \, dx,x,x^2\right )+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \log (x) \, dx,x,1+i c x^2\right )}{4 c}\\ &=-\frac{1}{4} b^2 x^2+\frac{i b^2 \left (1-i c x^2\right ) \log \left (1-i c x^2\right )}{4 c}+\frac{i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 c}+\frac{i b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+i c x^2\right )\right )}{4 c}+\frac{i b^2 \log \left (\frac{1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{4 c}-\frac{1}{4} b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac{i b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 c}+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right )}{x} \, dx,x,1-i c x^2\right )}{4 c}-\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right )}{x} \, dx,x,1+i c x^2\right )}{4 c}-\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \log (x) \, dx,x,1-i c x^2\right )}{4 c}\\ &=\frac{i \left (1-i c x^2\right ) \left (2 a+i b \log \left (1-i c x^2\right )\right )^2}{8 c}+\frac{i b \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+i c x^2\right )\right )}{4 c}+\frac{i b^2 \log \left (\frac{1}{2} \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )}{4 c}-\frac{1}{4} b x^2 \left (2 i a-b \log \left (1-i c x^2\right )\right ) \log \left (1+i c x^2\right )+\frac{i b^2 \left (1+i c x^2\right ) \log ^2\left (1+i c x^2\right )}{8 c}-\frac{i b^2 \text{Li}_2\left (\frac{1}{2} \left (1-i c x^2\right )\right )}{4 c}+\frac{i b^2 \text{Li}_2\left (\frac{1}{2} \left (1+i c x^2\right )\right )}{4 c}\\ \end{align*}
Mathematica [A] time = 0.0869276, size = 107, normalized size = 1.06 \[ \frac{-i b^2 \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}\left (c x^2\right )}\right )+a \left (a c x^2-b \log \left (c^2 x^4+1\right )\right )+2 b \tan ^{-1}\left (c x^2\right ) \left (a c x^2+b \log \left (1+e^{2 i \tan ^{-1}\left (c x^2\right )}\right )\right )+b^2 \left (c x^2-i\right ) \tan ^{-1}\left (c x^2\right )^2}{2 c} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.087, size = 146, normalized size = 1.5 \begin{align*}{\frac{{x}^{2}{b}^{2} \left ( \arctan \left ( c{x}^{2} \right ) \right ) ^{2}}{2}}+{x}^{2}ab\arctan \left ( c{x}^{2} \right ) -{\frac{{\frac{i}{2}} \left ( \arctan \left ( c{x}^{2} \right ) \right ) ^{2}{b}^{2}}{c}}+{\frac{{a}^{2}{x}^{2}}{2}}+{\frac{\arctan \left ( c{x}^{2} \right ){b}^{2}}{c}\ln \left ({\frac{ \left ( 1+ic{x}^{2} \right ) ^{2}}{{c}^{2}{x}^{4}+1}}+1 \right ) }-{\frac{{\frac{i}{2}}{b}^{2}}{c}{\it polylog} \left ( 2,-{\frac{ \left ( 1+ic{x}^{2} \right ) ^{2}}{{c}^{2}{x}^{4}+1}} \right ) }-{\frac{ab\ln \left ({c}^{2}{x}^{4}+1 \right ) }{2\,c}} \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*} \frac{1}{2} \, a^{2} x^{2} + \frac{1}{32} \,{\left (4 \, x^{2} \arctan \left (c x^{2}\right )^{2} - x^{2} \log \left (c^{2} x^{4} + 1\right )^{2} + 384 \, c^{2} \int \frac{x^{5} \arctan \left (c x^{2}\right )^{2}}{16 \,{\left (c^{2} x^{4} + 1\right )}}\,{d x} + 32 \, c^{2} \int \frac{x^{5} \log \left (c^{2} x^{4} + 1\right )^{2}}{16 \,{\left (c^{2} x^{4} + 1\right )}}\,{d x} + 128 \, c^{2} \int \frac{x^{5} \log \left (c^{2} x^{4} + 1\right )}{16 \,{\left (c^{2} x^{4} + 1\right )}}\,{d x} + \frac{4 \, \arctan \left (c x^{2}\right )^{3}}{c} - 256 \, c \int \frac{x^{3} \arctan \left (c x^{2}\right )}{16 \,{\left (c^{2} x^{4} + 1\right )}}\,{d x} + 32 \, \int \frac{x \log \left (c^{2} x^{4} + 1\right )^{2}}{16 \,{\left (c^{2} x^{4} + 1\right )}}\,{d x}\right )} b^{2} + \frac{{\left (2 \, c x^{2} \arctan \left (c x^{2}\right ) - \log \left (c^{2} x^{4} + 1\right )\right )} a b}{2 \, c} \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 (b^{2} x \arctan \left (c x^{2}\right )^{2} + 2 \, a b x \arctan \left (c x^{2}\right ) + a^{2} 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*} \int x \left (a + b \operatorname{atan}{\left (c x^{2} \right )}\right )^{2}\, dx \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{\left (b \arctan \left (c x^{2}\right ) + a\right )}^{2} x\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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