Optimal. Leaf size=275 \[ \frac{\text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 d}+\frac{\tanh ^{-1}(c x) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}+\frac{\text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{2 d}-\frac{\text{PolyLog}\left (3,\frac{2}{1-c x}-1\right )}{2 d}-\frac{\text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 d}-\frac{\tanh ^{-1}(c x) \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{d}+\frac{\tanh ^{-1}(c x) \text{PolyLog}\left (2,\frac{2}{1-c x}-1\right )}{d}-\frac{\tanh ^{-1}(c x) \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right )}{d}-\frac{\tanh ^{-1}(c x)^2 \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}+\frac{2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right ) \tanh ^{-1}(c x)^2}{d}+\frac{\log \left (\frac{2}{c x+1}\right ) \tanh ^{-1}(c x)^2}{d} \]
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Rubi [A] time = 0.335681, antiderivative size = 275, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {5940, 5914, 6052, 5948, 6058, 6610, 5922} \[ \frac{\text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 d}+\frac{\tanh ^{-1}(c x) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}+\frac{\text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{2 d}-\frac{\text{PolyLog}\left (3,\frac{2}{1-c x}-1\right )}{2 d}-\frac{\text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 d}-\frac{\tanh ^{-1}(c x) \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right )}{d}+\frac{\tanh ^{-1}(c x) \text{PolyLog}\left (2,\frac{2}{1-c x}-1\right )}{d}-\frac{\tanh ^{-1}(c x) \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right )}{d}-\frac{\tanh ^{-1}(c x)^2 \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d}+\frac{2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right ) \tanh ^{-1}(c x)^2}{d}+\frac{\log \left (\frac{2}{c x+1}\right ) \tanh ^{-1}(c x)^2}{d} \]
Antiderivative was successfully verified.
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Rule 5940
Rule 5914
Rule 6052
Rule 5948
Rule 6058
Rule 6610
Rule 5922
Rubi steps
\begin{align*} \int \frac{\tanh ^{-1}(c x)^2}{x (d+e x)} \, dx &=\int \left (\frac{\tanh ^{-1}(c x)^2}{d x}-\frac{e \tanh ^{-1}(c x)^2}{d (d+e x)}\right ) \, dx\\ &=\frac{\int \frac{\tanh ^{-1}(c x)^2}{x} \, dx}{d}-\frac{e \int \frac{\tanh ^{-1}(c x)^2}{d+e x} \, dx}{d}\\ &=\frac{2 \tanh ^{-1}(c x)^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d}+\frac{\tanh ^{-1}(c x)^2 \log \left (\frac{2}{1+c x}\right )}{d}-\frac{\tanh ^{-1}(c x)^2 \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{\tanh ^{-1}(c x) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d}+\frac{\tanh ^{-1}(c x) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{\text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d}+\frac{\text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d}-\frac{(4 c) \int \frac{\tanh ^{-1}(c x) \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=\frac{2 \tanh ^{-1}(c x)^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d}+\frac{\tanh ^{-1}(c x)^2 \log \left (\frac{2}{1+c x}\right )}{d}-\frac{\tanh ^{-1}(c x)^2 \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{\tanh ^{-1}(c x) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d}+\frac{\tanh ^{-1}(c x) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{\text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d}+\frac{\text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d}+\frac{(2 c) \int \frac{\tanh ^{-1}(c x) \log \left (\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d}-\frac{(2 c) \int \frac{\tanh ^{-1}(c x) \log \left (2-\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=\frac{2 \tanh ^{-1}(c x)^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d}+\frac{\tanh ^{-1}(c x)^2 \log \left (\frac{2}{1+c x}\right )}{d}-\frac{\tanh ^{-1}(c x)^2 \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{\tanh ^{-1}(c x) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{d}+\frac{\tanh ^{-1}(c x) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{d}-\frac{\tanh ^{-1}(c x) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d}+\frac{\tanh ^{-1}(c x) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{\text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d}+\frac{\text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d}+\frac{c \int \frac{\text{Li}_2\left (1-\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d}-\frac{c \int \frac{\text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d}\\ &=\frac{2 \tanh ^{-1}(c x)^2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right )}{d}+\frac{\tanh ^{-1}(c x)^2 \log \left (\frac{2}{1+c x}\right )}{d}-\frac{\tanh ^{-1}(c x)^2 \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}-\frac{\tanh ^{-1}(c x) \text{Li}_2\left (1-\frac{2}{1-c x}\right )}{d}+\frac{\tanh ^{-1}(c x) \text{Li}_2\left (-1+\frac{2}{1-c x}\right )}{d}-\frac{\tanh ^{-1}(c x) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{d}+\frac{\tanh ^{-1}(c x) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d}+\frac{\text{Li}_3\left (1-\frac{2}{1-c x}\right )}{2 d}-\frac{\text{Li}_3\left (-1+\frac{2}{1-c x}\right )}{2 d}-\frac{\text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 d}+\frac{\text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d}\\ \end{align*}
Mathematica [C] time = 9.24865, size = 733, normalized size = 2.67 \[ \frac{-\frac{24 (c d-e) (c d+e) \left (-6 c d \tanh ^{-1}(c x) \text{PolyLog}\left (2,-\frac{(c d+e) e^{2 \tanh ^{-1}(c x)}}{c d-e}\right )+12 c d \tanh ^{-1}(c x) \text{PolyLog}\left (2,-e^{\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)}\right )+12 c d \tanh ^{-1}(c x) \text{PolyLog}\left (2,e^{\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)}\right )+6 c d \tanh ^{-1}(c x) \text{PolyLog}\left (2,e^{2 \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )+3 c d \text{PolyLog}\left (3,-\frac{(c d+e) e^{2 \tanh ^{-1}(c x)}}{c d-e}\right )-12 c d \text{PolyLog}\left (3,-e^{\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)}\right )-12 c d \text{PolyLog}\left (3,e^{\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)}\right )-3 c d \text{PolyLog}\left (3,e^{2 \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-4 e \sqrt{1-\frac{c^2 d^2}{e^2}} \tanh ^{-1}(c x)^3 e^{-\tanh ^{-1}\left (\frac{c d}{e}\right )}-6 c d \tanh ^{-1}(c x)^2 \log \left (\frac{c (d+e x)}{\sqrt{1-c^2 x^2}}\right )-3 i \pi c d \log \left (1-c^2 x^2\right ) \tanh ^{-1}(c x)-6 c d \tanh ^{-1}(c x)^2 \log \left (\frac{(c d+e) e^{2 \tanh ^{-1}(c x)}}{c d-e}+1\right )+6 c d \tanh ^{-1}(c x)^2 \log \left (1-e^{\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)}\right )+6 c d \tanh ^{-1}(c x)^2 \log \left (e^{\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)}+1\right )+6 c d \tanh ^{-1}(c x)^2 \log \left (1-e^{2 \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )+6 c d \tanh ^{-1}(c x)^2 \log \left (\frac{1}{2} e^{-\tanh ^{-1}(c x)} \left (c d \left (e^{2 \tanh ^{-1}(c x)}+1\right )+e \left (e^{2 \tanh ^{-1}(c x)}-1\right )\right )\right )+12 c d \tanh ^{-1}(c x) \tanh ^{-1}\left (\frac{c d}{e}\right ) \log \left (\frac{1}{2} i e^{-\tanh ^{-1}\left (\frac{c d}{e}\right )-\tanh ^{-1}(c x)} \left (e^{2 \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )}-1\right )\right )-12 c d \tanh ^{-1}(c x) \tanh ^{-1}\left (\frac{c d}{e}\right ) \log \left (i \sinh \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )-6 c d \tanh ^{-1}(c x)^3-6 i \pi c d \tanh ^{-1}(c x) \log \left (\frac{1}{2} \left (e^{-\tanh ^{-1}(c x)}+e^{\tanh ^{-1}(c x)}\right )\right )+2 e \tanh ^{-1}(c x)^3\right )}{6 c^2 d^2-6 e^2}+24 c d \tanh ^{-1}(c x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(c x)}\right )-12 c d \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(c x)}\right )-8 c d \tanh ^{-1}(c x)^3+24 c d \tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )+i \pi ^3 c d-8 e \tanh ^{-1}(c x)^3}{24 c d^2} \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 0.247, size = 1507, normalized size = 5.5 \begin{align*} \text{result 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*} \int \frac{\operatorname{artanh}\left (c x\right )^{2}}{{\left (e x + d\right )} x}\,{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{\operatorname{artanh}\left (c x\right )^{2}}{e x^{2} + d 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{\operatorname{atanh}^{2}{\left (c x \right )}}{x \left (d + e x\right )}\, 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{\operatorname{artanh}\left (c x\right )^{2}}{{\left (e x + d\right )} x}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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