Optimal. Leaf size=303 \[ -\frac{x \text{PolyLog}\left (3,\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b^2}+\frac{x \text{PolyLog}\left (3,\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b^2}+\frac{\text{PolyLog}\left (4,\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{8 b^3}-\frac{\text{PolyLog}\left (4,\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{8 b^3}+\frac{x^2 \text{PolyLog}\left (2,\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b}-\frac{x^2 \text{PolyLog}\left (2,\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b}+\frac{1}{6} x^3 \log \left (1-\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )-\frac{1}{6} x^3 \log \left (1-\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )+\frac{1}{3} x^3 \tanh ^{-1}(d \coth (a+b x)+c) \]
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Rubi [A] time = 0.456686, antiderivative size = 303, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6245, 2190, 2531, 6609, 2282, 6589} \[ -\frac{x \text{PolyLog}\left (3,\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b^2}+\frac{x \text{PolyLog}\left (3,\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b^2}+\frac{\text{PolyLog}\left (4,\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{8 b^3}-\frac{\text{PolyLog}\left (4,\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{8 b^3}+\frac{x^2 \text{PolyLog}\left (2,\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )}{4 b}-\frac{x^2 \text{PolyLog}\left (2,\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )}{4 b}+\frac{1}{6} x^3 \log \left (1-\frac{(-c-d+1) e^{2 a+2 b x}}{-c+d+1}\right )-\frac{1}{6} x^3 \log \left (1-\frac{(c+d+1) e^{2 a+2 b x}}{c-d+1}\right )+\frac{1}{3} x^3 \tanh ^{-1}(d \coth (a+b x)+c) \]
Antiderivative was successfully verified.
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Rule 6245
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x^2 \tanh ^{-1}(c+d \coth (a+b x)) \, dx &=\frac{1}{3} x^3 \tanh ^{-1}(c+d \coth (a+b x))-\frac{1}{3} (b (1-c-d)) \int \frac{e^{2 a+2 b x} x^3}{1-c+d+(-1+c+d) e^{2 a+2 b x}} \, dx+\frac{1}{3} (b (1+c+d)) \int \frac{e^{2 a+2 b x} x^3}{1+c-d+(-1-c-d) e^{2 a+2 b x}} \, dx\\ &=\frac{1}{3} x^3 \tanh ^{-1}(c+d \coth (a+b x))+\frac{1}{6} x^3 \log \left (1-\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac{1}{6} x^3 \log \left (1-\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac{1}{2} \int x^2 \log \left (1+\frac{(-1-c-d) e^{2 a+2 b x}}{1+c-d}\right ) \, dx-\frac{1}{2} \int x^2 \log \left (1+\frac{(-1+c+d) e^{2 a+2 b x}}{1-c+d}\right ) \, dx\\ &=\frac{1}{3} x^3 \tanh ^{-1}(c+d \coth (a+b x))+\frac{1}{6} x^3 \log \left (1-\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac{1}{6} x^3 \log \left (1-\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac{x^2 \text{Li}_2\left (\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac{x^2 \text{Li}_2\left (\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}+\frac{\int x \text{Li}_2\left (-\frac{(-1-c-d) e^{2 a+2 b x}}{1+c-d}\right ) \, dx}{2 b}-\frac{\int x \text{Li}_2\left (-\frac{(-1+c+d) e^{2 a+2 b x}}{1-c+d}\right ) \, dx}{2 b}\\ &=\frac{1}{3} x^3 \tanh ^{-1}(c+d \coth (a+b x))+\frac{1}{6} x^3 \log \left (1-\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac{1}{6} x^3 \log \left (1-\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac{x^2 \text{Li}_2\left (\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac{x^2 \text{Li}_2\left (\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac{x \text{Li}_3\left (\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac{x \text{Li}_3\left (\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}-\frac{\int \text{Li}_3\left (-\frac{(-1-c-d) e^{2 a+2 b x}}{1+c-d}\right ) \, dx}{4 b^2}+\frac{\int \text{Li}_3\left (-\frac{(-1+c+d) e^{2 a+2 b x}}{1-c+d}\right ) \, dx}{4 b^2}\\ &=\frac{1}{3} x^3 \tanh ^{-1}(c+d \coth (a+b x))+\frac{1}{6} x^3 \log \left (1-\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac{1}{6} x^3 \log \left (1-\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac{x^2 \text{Li}_2\left (\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac{x^2 \text{Li}_2\left (\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac{x \text{Li}_3\left (\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac{x \text{Li}_3\left (\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}+\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\frac{(-1+c+d) x}{-1+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3}-\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\frac{(1+c+d) x}{1+c-d}\right )}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3}\\ &=\frac{1}{3} x^3 \tanh ^{-1}(c+d \coth (a+b x))+\frac{1}{6} x^3 \log \left (1-\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )-\frac{1}{6} x^3 \log \left (1-\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )+\frac{x^2 \text{Li}_2\left (\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b}-\frac{x^2 \text{Li}_2\left (\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b}-\frac{x \text{Li}_3\left (\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{4 b^2}+\frac{x \text{Li}_3\left (\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{4 b^2}+\frac{\text{Li}_4\left (\frac{(1-c-d) e^{2 a+2 b x}}{1-c+d}\right )}{8 b^3}-\frac{\text{Li}_4\left (\frac{(1+c+d) e^{2 a+2 b x}}{1+c-d}\right )}{8 b^3}\\ \end{align*}
Mathematica [A] time = 10.0535, size = 353, normalized size = 1.17 \[ \frac{-6 b^2 x^2 \text{PolyLog}\left (2,\frac{(c-d-1) (\cosh (2 (a+b x))-\sinh (2 (a+b x)))}{c+d-1}\right )+6 b^2 x^2 \text{PolyLog}\left (2,\frac{(c-d+1) (\cosh (2 (a+b x))-\sinh (2 (a+b x)))}{c+d+1}\right )-6 b x \text{PolyLog}\left (3,\frac{(c-d-1) (\cosh (2 (a+b x))-\sinh (2 (a+b x)))}{c+d-1}\right )+6 b x \text{PolyLog}\left (3,\frac{(c-d+1) (\cosh (2 (a+b x))-\sinh (2 (a+b x)))}{c+d+1}\right )-3 \text{PolyLog}\left (4,\frac{(c-d-1) (\cosh (2 (a+b x))-\sinh (2 (a+b x)))}{c+d-1}\right )+3 \text{PolyLog}\left (4,\frac{(c-d+1) (\cosh (2 (a+b x))-\sinh (2 (a+b x)))}{c+d+1}\right )+4 b^3 x^3 \log \left (\frac{2 (\cosh (a+b x)-\sinh (a+b x)) ((c-1) \sinh (a+b x)+d \cosh (a+b x))}{c+d-1}\right )-4 b^3 x^3 \log \left (\frac{(c-d+1) (\sinh (2 (a+b x))-\cosh (2 (a+b x)))}{c+d+1}+1\right )}{24 b^3}+\frac{1}{3} x^3 \tanh ^{-1}(d \coth (a+b x)+c) \]
Warning: Unable to verify antiderivative.
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Maple [C] time = 3.53, size = 5294, normalized size = 17.5 \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 = 2.12903, size = 374, normalized size = 1.23 \begin{align*} \frac{1}{3} \, x^{3} \operatorname{artanh}\left (d \coth \left (b x + a\right ) + c\right ) - \frac{1}{18} \, b d{\left (\frac{4 \, b^{3} x^{3} \log \left (-\frac{{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1} + 1\right ) + 6 \, b^{2} x^{2}{\rm Li}_2\left (\frac{{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}\right ) - 6 \, b x{\rm Li}_{3}(\frac{{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1}) + 3 \,{\rm Li}_{4}(\frac{{\left (c + d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d + 1})}{b^{4} d} - \frac{4 \, b^{3} x^{3} \log \left (-\frac{{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1} + 1\right ) + 6 \, b^{2} x^{2}{\rm Li}_2\left (\frac{{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}\right ) - 6 \, b x{\rm Li}_{3}(\frac{{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1}) + 3 \,{\rm Li}_{4}(\frac{{\left (c + d - 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c - d - 1})}{b^{4} d}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.57225, size = 2561, normalized size = 8.45 \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^{2} \operatorname{artanh}\left (d \coth \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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