Optimal. Leaf size=137 \[ \frac{1}{2} i b d^2 \text{PolyLog}(2,-i c x)-\frac{1}{2} i b d^2 \text{PolyLog}(2,i c x)+d e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+a d^2 \log (x)+\frac{b d e \tan ^{-1}(c x)}{c^2}+\frac{b e^2 x}{4 c^3}-\frac{b e^2 \tan ^{-1}(c x)}{4 c^4}-\frac{b d e x}{c}-\frac{b e^2 x^3}{12 c} \]
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Rubi [A] time = 0.179517, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {4980, 4848, 2391, 4852, 321, 203, 302} \[ \frac{1}{2} i b d^2 \text{PolyLog}(2,-i c x)-\frac{1}{2} i b d^2 \text{PolyLog}(2,i c x)+d e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+a d^2 \log (x)+\frac{b d e \tan ^{-1}(c x)}{c^2}+\frac{b e^2 x}{4 c^3}-\frac{b e^2 \tan ^{-1}(c x)}{4 c^4}-\frac{b d e x}{c}-\frac{b e^2 x^3}{12 c} \]
Antiderivative was successfully verified.
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Rule 4980
Rule 4848
Rule 2391
Rule 4852
Rule 321
Rule 203
Rule 302
Rubi steps
\begin{align*} \int \frac{\left (d+e x^2\right )^2 \left (a+b \tan ^{-1}(c x)\right )}{x} \, dx &=\int \left (\frac{d^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+2 d e x \left (a+b \tan ^{-1}(c x)\right )+e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )\right ) \, dx\\ &=d^2 \int \frac{a+b \tan ^{-1}(c x)}{x} \, dx+(2 d e) \int x \left (a+b \tan ^{-1}(c x)\right ) \, dx+e^2 \int x^3 \left (a+b \tan ^{-1}(c x)\right ) \, dx\\ &=d e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+a d^2 \log (x)+\frac{1}{2} \left (i b d^2\right ) \int \frac{\log (1-i c x)}{x} \, dx-\frac{1}{2} \left (i b d^2\right ) \int \frac{\log (1+i c x)}{x} \, dx-(b c d e) \int \frac{x^2}{1+c^2 x^2} \, dx-\frac{1}{4} \left (b c e^2\right ) \int \frac{x^4}{1+c^2 x^2} \, dx\\ &=-\frac{b d e x}{c}+d e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+a d^2 \log (x)+\frac{1}{2} i b d^2 \text{Li}_2(-i c x)-\frac{1}{2} i b d^2 \text{Li}_2(i c x)+\frac{(b d e) \int \frac{1}{1+c^2 x^2} \, dx}{c}-\frac{1}{4} \left (b c e^2\right ) \int \left (-\frac{1}{c^4}+\frac{x^2}{c^2}+\frac{1}{c^4 \left (1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac{b d e x}{c}+\frac{b e^2 x}{4 c^3}-\frac{b e^2 x^3}{12 c}+\frac{b d e \tan ^{-1}(c x)}{c^2}+d e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+a d^2 \log (x)+\frac{1}{2} i b d^2 \text{Li}_2(-i c x)-\frac{1}{2} i b d^2 \text{Li}_2(i c x)-\frac{\left (b e^2\right ) \int \frac{1}{1+c^2 x^2} \, dx}{4 c^3}\\ &=-\frac{b d e x}{c}+\frac{b e^2 x}{4 c^3}-\frac{b e^2 x^3}{12 c}+\frac{b d e \tan ^{-1}(c x)}{c^2}-\frac{b e^2 \tan ^{-1}(c x)}{4 c^4}+d e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+a d^2 \log (x)+\frac{1}{2} i b d^2 \text{Li}_2(-i c x)-\frac{1}{2} i b d^2 \text{Li}_2(i c x)\\ \end{align*}
Mathematica [A] time = 0.101077, size = 123, normalized size = 0.9 \[ \frac{1}{2} i b d^2 \text{PolyLog}(2,-i c x)-\frac{1}{2} i b d^2 \text{PolyLog}(2,i c x)+d e x^2 \left (a+b \tan ^{-1}(c x)\right )+\frac{1}{4} e^2 x^4 \left (a+b \tan ^{-1}(c x)\right )+a d^2 \log (x)-\frac{b d e \left (c x-\tan ^{-1}(c x)\right )}{c^2}-\frac{b e^2 \left (c^3 x^3-3 c x+3 \tan ^{-1}(c x)\right )}{12 c^4} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.051, size = 187, normalized size = 1.4 \begin{align*}{\frac{a{x}^{4}{e}^{2}}{4}}+a{x}^{2}de+a{d}^{2}\ln \left ( cx \right ) +{\frac{b\arctan \left ( cx \right ){x}^{4}{e}^{2}}{4}}+b\arctan \left ( cx \right ){x}^{2}de+b\arctan \left ( cx \right ){d}^{2}\ln \left ( cx \right ) -{\frac{b{e}^{2}{x}^{3}}{12\,c}}-{\frac{bdex}{c}}+{\frac{b{e}^{2}x}{4\,{c}^{3}}}+{\frac{bde\arctan \left ( cx \right ) }{{c}^{2}}}-{\frac{b{e}^{2}\arctan \left ( cx \right ) }{4\,{c}^{4}}}+{\frac{i}{2}}b{d}^{2}\ln \left ( cx \right ) \ln \left ( 1+icx \right ) -{\frac{i}{2}}b{d}^{2}\ln \left ( cx \right ) \ln \left ( 1-icx \right ) +{\frac{i}{2}}b{d}^{2}{\it dilog} \left ( 1+icx \right ) -{\frac{i}{2}}b{d}^{2}{\it dilog} \left ( 1-icx \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 2.14796, size = 251, normalized size = 1.83 \begin{align*} \frac{1}{4} \, a e^{2} x^{4} + a d e x^{2} + a d^{2} \log \left (x\right ) - \frac{b c^{3} e^{2} x^{3} + 3 \, \pi b c^{4} d^{2} \log \left (c^{2} x^{2} + 1\right ) - 12 \, b c^{4} d^{2} \arctan \left (c x\right ) \log \left (x{\left | c \right |}\right ) + 6 i \, b c^{4} d^{2}{\rm Li}_2\left (i \, c x + 1\right ) - 6 i \, b c^{4} d^{2}{\rm Li}_2\left (-i \, c x + 1\right ) + 3 \,{\left (4 \, b c^{3} d e - b c e^{2}\right )} x -{\left (3 \, b c^{4} e^{2} x^{4} + 12 \, b c^{4} d e x^{2} + 12 i \, b c^{4} d^{2} \arctan \left (0, c\right ) + 12 \, b c^{2} d e - 3 \, b e^{2}\right )} \arctan \left (c x\right )}{12 \, c^{4}} \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{a e^{2} x^{4} + 2 \, a d e x^{2} + a d^{2} +{\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \arctan \left (c x\right )}{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 \frac{\left (a + b \operatorname{atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{2}}{x}\, 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 \frac{{\left (e x^{2} + d\right )}^{2}{\left (b \arctan \left (c x\right ) + a\right )}}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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