Optimal. Leaf size=231 \[ -\frac{\text{PolyLog}\left (3,-\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{8 b^2}+\frac{\text{PolyLog}\left (3,-\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{8 b^2}+\frac{x \text{PolyLog}\left (2,-\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b}-\frac{x \text{PolyLog}\left (2,-\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b}+\frac{1}{4} x^2 \log \left (\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}+1\right )-\frac{1}{4} x^2 \log \left (\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}+1\right )+\frac{1}{2} x^2 \tanh ^{-1}(d \tanh (a+b x)+c) \]
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Rubi [A] time = 0.370669, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {6243, 2190, 2531, 2282, 6589} \[ -\frac{\text{PolyLog}\left (3,-\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{8 b^2}+\frac{\text{PolyLog}\left (3,-\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{8 b^2}+\frac{x \text{PolyLog}\left (2,-\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b}-\frac{x \text{PolyLog}\left (2,-\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b}+\frac{1}{4} x^2 \log \left (\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}+1\right )-\frac{1}{4} x^2 \log \left (\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}+1\right )+\frac{1}{2} x^2 \tanh ^{-1}(d \tanh (a+b x)+c) \]
Antiderivative was successfully verified.
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Rule 6243
Rule 2190
Rule 2531
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x \tanh ^{-1}(c+d \tanh (a+b x)) \, dx &=\frac{1}{2} x^2 \tanh ^{-1}(c+d \tanh (a+b x))+\frac{1}{2} (b (1-c-d)) \int \frac{e^{2 a+2 b x} x^2}{1-c+d+(1-c-d) e^{2 a+2 b x}} \, dx-\frac{1}{2} (b (1+c+d)) \int \frac{e^{2 a+2 b x} x^2}{1+c-d+(1+c+d) e^{2 a+2 b x}} \, dx\\ &=\frac{1}{2} x^2 \tanh ^{-1}(c+d \tanh (a+b x))+\frac{1}{4} x^2 \log \left (1+\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac{1}{4} x^2 \log \left (1+\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )-\frac{1}{2} \int x \log \left (1+\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right ) \, dx+\frac{1}{2} \int x \log \left (1+\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right ) \, dx\\ &=\frac{1}{2} x^2 \tanh ^{-1}(c+d \tanh (a+b x))+\frac{1}{4} x^2 \log \left (1+\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac{1}{4} x^2 \log \left (1+\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac{x \text{Li}_2\left (-\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac{x \text{Li}_2\left (-\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac{\int \text{Li}_2\left (-\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right ) \, dx}{4 b}+\frac{\int \text{Li}_2\left (-\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right ) \, dx}{4 b}\\ &=\frac{1}{2} x^2 \tanh ^{-1}(c+d \tanh (a+b x))+\frac{1}{4} x^2 \log \left (1+\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac{1}{4} x^2 \log \left (1+\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac{x \text{Li}_2\left (-\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac{x \text{Li}_2\left (-\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{(-1+c+d) x}{-1+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^2}+\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-\frac{(1+c+d) x}{1+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^2}\\ &=\frac{1}{2} x^2 \tanh ^{-1}(c+d \tanh (a+b x))+\frac{1}{4} x^2 \log \left (1+\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac{1}{4} x^2 \log \left (1+\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac{x \text{Li}_2\left (-\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac{x \text{Li}_2\left (-\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac{\text{Li}_3\left (-\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{8 b^2}+\frac{\text{Li}_3\left (-\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{8 b^2}\\ \end{align*}
Mathematica [A] time = 9.42936, size = 259, normalized size = 1.12 \[ \frac{-2 b x \text{PolyLog}\left (2,\frac{(c-d-1) (\sinh (2 (a+b x))-\cosh (2 (a+b x)))}{c+d-1}\right )+2 b x \text{PolyLog}\left (2,\frac{(c-d+1) (\sinh (2 (a+b x))-\cosh (2 (a+b x)))}{c+d+1}\right )-\text{PolyLog}\left (3,\frac{(c-d-1) (\sinh (2 (a+b x))-\cosh (2 (a+b x)))}{c+d-1}\right )+\text{PolyLog}\left (3,\frac{(c-d+1) (\sinh (2 (a+b x))-\cosh (2 (a+b x)))}{c+d+1}\right )+2 b^2 x^2 \log \left (\frac{(c-d-1) (\cosh (2 (a+b x))-\sinh (2 (a+b x)))}{c+d-1}+1\right )-2 b^2 x^2 \log \left (\frac{(c-d+1) (\cosh (2 (a+b x))-\sinh (2 (a+b x)))}{c+d+1}+1\right )}{8 b^2}+\frac{1}{2} x^2 \tanh ^{-1}(d \tanh (a+b x)+c) \]
Antiderivative was successfully verified.
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Maple [C] time = 5.313, size = 5062, normalized size = 21.9 \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 [A] time = 2.1098, size = 290, normalized size = 1.26 \begin{align*} -\frac{1}{8} \, b d{\left (\frac{2 \, b^{2} x^{2} \log \left (\frac{{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1} + 1\right ) + 2 \, b x{\rm Li}_2\left (-\frac{{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}\right ) -{\rm Li}_{3}(-\frac{{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1})}{b^{3} d} - \frac{2 \, b^{2} x^{2} \log \left (\frac{{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1} + 1\right ) + 2 \, b x{\rm Li}_2\left (-\frac{{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}\right ) -{\rm Li}_{3}(-\frac{{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1})}{b^{3} d}\right )} + \frac{1}{2} \, x^{2} \operatorname{artanh}\left (d \tanh \left (b x + a\right ) + c\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.2615, size = 2115, normalized size = 9.16 \begin{align*} \text{result too large to display} \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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x \operatorname{artanh}\left (d \tanh \left (b x + a\right ) + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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