Optimal. Leaf size=128 \[ \frac{x \text{PolyLog}\left (3,-(d+1) e^{2 a+2 b x}\right )}{4 b^2}-\frac{\text{PolyLog}\left (4,-(d+1) e^{2 a+2 b x}\right )}{8 b^3}-\frac{x^2 \text{PolyLog}\left (2,-(d+1) e^{2 a+2 b x}\right )}{4 b}-\frac{1}{6} x^3 \log \left ((d+1) e^{2 a+2 b x}+1\right )+\frac{1}{3} x^3 \coth ^{-1}(d \tanh (a+b x)+d+1)+\frac{b x^4}{12} \]
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Rubi [A] time = 0.269003, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {6240, 2184, 2190, 2531, 6609, 2282, 6589} \[ \frac{x \text{PolyLog}\left (3,-(d+1) e^{2 a+2 b x}\right )}{4 b^2}-\frac{\text{PolyLog}\left (4,-(d+1) e^{2 a+2 b x}\right )}{8 b^3}-\frac{x^2 \text{PolyLog}\left (2,-(d+1) e^{2 a+2 b x}\right )}{4 b}-\frac{1}{6} x^3 \log \left ((d+1) e^{2 a+2 b x}+1\right )+\frac{1}{3} x^3 \coth ^{-1}(d \tanh (a+b x)+d+1)+\frac{b x^4}{12} \]
Antiderivative was successfully verified.
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Rule 6240
Rule 2184
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x^2 \coth ^{-1}(1+d+d \tanh (a+b x)) \, dx &=\frac{1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))+\frac{1}{3} b \int \frac{x^3}{1+(1+d) e^{2 a+2 b x}} \, dx\\ &=\frac{b x^4}{12}+\frac{1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac{1}{3} (b (1+d)) \int \frac{e^{2 a+2 b x} x^3}{1+(1+d) e^{2 a+2 b x}} \, dx\\ &=\frac{b x^4}{12}+\frac{1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac{1}{6} x^3 \log \left (1+(1+d) e^{2 a+2 b x}\right )+\frac{1}{2} \int x^2 \log \left (1+(1+d) e^{2 a+2 b x}\right ) \, dx\\ &=\frac{b x^4}{12}+\frac{1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac{1}{6} x^3 \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac{x^2 \text{Li}_2\left (-(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{\int x \text{Li}_2\left (-(1+d) e^{2 a+2 b x}\right ) \, dx}{2 b}\\ &=\frac{b x^4}{12}+\frac{1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac{1}{6} x^3 \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac{x^2 \text{Li}_2\left (-(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{x \text{Li}_3\left (-(1+d) e^{2 a+2 b x}\right )}{4 b^2}-\frac{\int \text{Li}_3\left ((-1-d) e^{2 a+2 b x}\right ) \, dx}{4 b^2}\\ &=\frac{b x^4}{12}+\frac{1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac{1}{6} x^3 \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac{x^2 \text{Li}_2\left (-(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{x \text{Li}_3\left (-(1+d) e^{2 a+2 b x}\right )}{4 b^2}-\frac{\operatorname{Subst}\left (\int \frac{\text{Li}_3((-1-d) x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{8 b^3}\\ &=\frac{b x^4}{12}+\frac{1}{3} x^3 \coth ^{-1}(1+d+d \tanh (a+b x))-\frac{1}{6} x^3 \log \left (1+(1+d) e^{2 a+2 b x}\right )-\frac{x^2 \text{Li}_2\left (-(1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{x \text{Li}_3\left (-(1+d) e^{2 a+2 b x}\right )}{4 b^2}-\frac{\text{Li}_4\left (-(1+d) e^{2 a+2 b x}\right )}{8 b^3}\\ \end{align*}
Mathematica [A] time = 0.113944, size = 118, normalized size = 0.92 \[ \frac{1}{24} \left (\frac{6 x \text{PolyLog}\left (3,-\frac{e^{-2 (a+b x)}}{d+1}\right )}{b^2}+\frac{3 \text{PolyLog}\left (4,-\frac{e^{-2 (a+b x)}}{d+1}\right )}{b^3}+\frac{6 x^2 \text{PolyLog}\left (2,-\frac{e^{-2 (a+b x)}}{d+1}\right )}{b}-4 x^3 \log \left (\frac{e^{-2 (a+b x)}}{d+1}+1\right )+8 x^3 \coth ^{-1}(d \tanh (a+b x)+d+1)\right ) \]
Antiderivative was successfully verified.
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Maple [C] time = 37.152, size = 1667, normalized size = 13. \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 [A] time = 3.46662, size = 169, normalized size = 1.32 \begin{align*} \frac{1}{3} \, x^{3} \operatorname{arcoth}\left (d \tanh \left (b x + a\right ) + d + 1\right ) + \frac{1}{36} \,{\left (\frac{3 \, x^{4}}{d} - \frac{2 \,{\left (4 \, b^{3} x^{3} \log \left ({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2}{\rm Li}_2\left (-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x{\rm Li}_{3}(-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) + 3 \,{\rm Li}_{4}(-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{4} d}\right )} b d \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 1.76501, size = 1160, normalized size = 9.06 \begin{align*} \frac{b^{4} x^{4} + 2 \, b^{3} x^{3} \log \left (\frac{{\left (d + 2\right )} \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) - 6 \, b^{2} x^{2}{\rm Li}_2\left (\frac{1}{2} \, \sqrt{-4 \, d - 4}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 6 \, b^{2} x^{2}{\rm Li}_2\left (-\frac{1}{2} \, \sqrt{-4 \, d - 4}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 2 \, a^{3} \log \left (2 \,{\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \,{\left (d + 1\right )} \sinh \left (b x + a\right ) + \sqrt{-4 \, d - 4}\right ) + 2 \, a^{3} \log \left (2 \,{\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \,{\left (d + 1\right )} \sinh \left (b x + a\right ) - \sqrt{-4 \, d - 4}\right ) + 12 \, b x{\rm polylog}\left (3, \frac{1}{2} \, \sqrt{-4 \, d - 4}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 12 \, b x{\rm polylog}\left (3, -\frac{1}{2} \, \sqrt{-4 \, d - 4}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 2 \,{\left (b^{3} x^{3} + a^{3}\right )} \log \left (\frac{1}{2} \, \sqrt{-4 \, d - 4}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 2 \,{\left (b^{3} x^{3} + a^{3}\right )} \log \left (-\frac{1}{2} \, \sqrt{-4 \, d - 4}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 12 \,{\rm polylog}\left (4, \frac{1}{2} \, \sqrt{-4 \, d - 4}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 12 \,{\rm polylog}\left (4, -\frac{1}{2} \, \sqrt{-4 \, d - 4}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{12 \, b^{3}} \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{arcoth}\left (d \tanh \left (b x + a\right ) + d + 1\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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