Optimal. Leaf size=92 \[ \frac{1}{2} a e \text{PolyLog}\left (2,\frac{g x^2}{f}+1\right )+b e \text{CannotIntegrate}\left (\frac{\tanh ^{-1}(c x) \log \left (f+g x^2\right )}{x},x\right )-\frac{1}{2} b d \text{PolyLog}(2,-c x)+\frac{1}{2} b d \text{PolyLog}(2,c x)+a d \log (x)+\frac{1}{2} a e \log \left (-\frac{g x^2}{f}\right ) \log \left (f+g x^2\right ) \]
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Rubi [A] time = 0.252387, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x} \, dx \]
Verification is Not applicable to the result.
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Rubi steps
\begin{align*} \int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x} \, dx &=d \int \frac{a+b \tanh ^{-1}(c x)}{x} \, dx+e \int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \log \left (f+g x^2\right )}{x} \, dx\\ &=a d \log (x)-\frac{1}{2} b d \text{Li}_2(-c x)+\frac{1}{2} b d \text{Li}_2(c x)+(a e) \int \frac{\log \left (f+g x^2\right )}{x} \, dx+(b e) \int \frac{\tanh ^{-1}(c x) \log \left (f+g x^2\right )}{x} \, dx\\ &=a d \log (x)-\frac{1}{2} b d \text{Li}_2(-c x)+\frac{1}{2} b d \text{Li}_2(c x)+\frac{1}{2} (a e) \operatorname{Subst}\left (\int \frac{\log (f+g x)}{x} \, dx,x,x^2\right )+(b e) \int \frac{\tanh ^{-1}(c x) \log \left (f+g x^2\right )}{x} \, dx\\ &=a d \log (x)+\frac{1}{2} a e \log \left (-\frac{g x^2}{f}\right ) \log \left (f+g x^2\right )-\frac{1}{2} b d \text{Li}_2(-c x)+\frac{1}{2} b d \text{Li}_2(c x)+(b e) \int \frac{\tanh ^{-1}(c x) \log \left (f+g x^2\right )}{x} \, dx-\frac{1}{2} (a e g) \operatorname{Subst}\left (\int \frac{\log \left (-\frac{g x}{f}\right )}{f+g x} \, dx,x,x^2\right )\\ &=a d \log (x)+\frac{1}{2} a e \log \left (-\frac{g x^2}{f}\right ) \log \left (f+g x^2\right )-\frac{1}{2} b d \text{Li}_2(-c x)+\frac{1}{2} b d \text{Li}_2(c x)+\frac{1}{2} a e \text{Li}_2\left (1+\frac{g x^2}{f}\right )+(b e) \int \frac{\tanh ^{-1}(c x) \log \left (f+g x^2\right )}{x} \, dx\\ \end{align*}
Mathematica [A] time = 0.213927, size = 0, normalized size = 0. \[ \int \frac{\left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (f+g x^2\right )\right )}{x} \, dx \]
Verification is Not applicable to the result.
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Maple [A] time = 0.759, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b{\it Artanh} \left ( cx \right ) \right ) \left ( d+e\ln \left ( g{x}^{2}+f \right ) \right ) }{x}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 0., size = 0, normalized size = 0. \begin{align*} a d \log \left (x\right ) + \int \frac{b e{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )} \log \left (g x^{2} + f\right )}{2 \, x} + \frac{b d{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}}{2 \, x} + \frac{a e \log \left (g x^{2} + f\right )}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b d \operatorname{artanh}\left (c x\right ) + a d +{\left (b e \operatorname{artanh}\left (c x\right ) + a e\right )} \log \left (g x^{2} + f\right )}{x}, 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 [A] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}{\left (e \log \left (g x^{2} + f\right ) + d\right )}}{x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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