Optimal. Leaf size=279 \[ -\frac{b d \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e^2}+\frac{b d \left (a+b \tanh ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^2}-\frac{b^2 d \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 e^2}+\frac{b^2 d \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 e^2}-\frac{b^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{c e}+\frac{d \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{e^2}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^2}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c e}-\frac{2 b \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c e} \]
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Rubi [A] time = 0.257585, antiderivative size = 279, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.368, Rules used = {5940, 5910, 5984, 5918, 2402, 2315, 5922} \[ -\frac{b d \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e^2}+\frac{b d \left (a+b \tanh ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^2}-\frac{b^2 d \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 e^2}+\frac{b^2 d \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 e^2}-\frac{b^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{c e}+\frac{d \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{e^2}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^2}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c e}-\frac{2 b \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c e} \]
Antiderivative was successfully verified.
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Rule 5940
Rule 5910
Rule 5984
Rule 5918
Rule 2402
Rule 2315
Rule 5922
Rubi steps
\begin{align*} \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{d+e x} \, dx &=\int \left (\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{e}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2}{e (d+e x)}\right ) \, dx\\ &=\frac{\int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{e}-\frac{d \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d+e x} \, dx}{e}\\ &=\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}+\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{e^2}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac{b d \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{e^2}+\frac{b d \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac{b^2 d \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 e^2}+\frac{b^2 d \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^2}-\frac{(2 b c) \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{e}\\ &=\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c e}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}+\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{e^2}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac{b d \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{e^2}+\frac{b d \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac{b^2 d \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 e^2}+\frac{b^2 d \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^2}-\frac{(2 b) \int \frac{a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{e}\\ &=\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c e}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c e}+\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{e^2}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac{b d \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{e^2}+\frac{b d \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac{b^2 d \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 e^2}+\frac{b^2 d \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^2}+\frac{\left (2 b^2\right ) \int \frac{\log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{e}\\ &=\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c e}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c e}+\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{e^2}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac{b d \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{e^2}+\frac{b d \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac{b^2 d \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 e^2}+\frac{b^2 d \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^2}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c x}\right )}{c e}\\ &=\frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{c e}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}-\frac{2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1-c x}\right )}{c e}+\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{e^2}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac{b^2 \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{c e}-\frac{b d \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{e^2}+\frac{b d \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac{b^2 d \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 e^2}+\frac{b^2 d \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^2}\\ \end{align*}
Mathematica [C] time = 15.0144, size = 1036, normalized size = 3.71 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 1.167, size = 13923, normalized size = 49.9 \begin{align*} \text{output too large to display} \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{b^{2} x \log \left (-c x + 1\right )^{2}}{4 \, e} + a^{2}{\left (\frac{x}{e} - \frac{d \log \left (e x + d\right )}{e^{2}}\right )} - \int -\frac{{\left (b^{2} c e x^{2} - b^{2} e x\right )} \log \left (c x + 1\right )^{2} + 4 \,{\left (a b c e x^{2} - a b e x\right )} \log \left (c x + 1\right ) - 2 \,{\left ({\left (2 \, a b c e + b^{2} c e\right )} x^{2} +{\left (b^{2} c d - 2 \, a b e\right )} x +{\left (b^{2} c e x^{2} - b^{2} e x\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{4 \,{\left (c e^{2} x^{2} - d e +{\left (c d e - e^{2}\right )} x\right )}}\,{d x} \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{b^{2} x \operatorname{artanh}\left (c x\right )^{2} + 2 \, a b x \operatorname{artanh}\left (c x\right ) + a^{2} x}{e x + d}, 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{x \left (a + b \operatorname{atanh}{\left (c x \right )}\right )^{2}}{d + e 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 (b \operatorname{artanh}\left (c x\right ) + a\right )}^{2} x}{e x + d}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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