Optimal. Leaf size=152 \[ \frac{3 x^2 \text{PolyLog}\left (3,(d+1) e^{2 a+2 b x}\right )}{8 b^2}-\frac{3 x \text{PolyLog}\left (4,(d+1) e^{2 a+2 b x}\right )}{8 b^3}+\frac{3 \text{PolyLog}\left (5,(d+1) e^{2 a+2 b x}\right )}{16 b^4}-\frac{x^3 \text{PolyLog}\left (2,(d+1) e^{2 a+2 b x}\right )}{4 b}-\frac{1}{8} x^4 \log \left (1-(d+1) e^{2 a+2 b x}\right )+\frac{1}{4} x^4 \tanh ^{-1}(d \coth (a+b x)+d+1)+\frac{b x^5}{20} \]
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Rubi [A] time = 0.303812, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {6241, 2184, 2190, 2531, 6609, 2282, 6589} \[ \frac{3 x^2 \text{PolyLog}\left (3,(d+1) e^{2 a+2 b x}\right )}{8 b^2}-\frac{3 x \text{PolyLog}\left (4,(d+1) e^{2 a+2 b x}\right )}{8 b^3}+\frac{3 \text{PolyLog}\left (5,(d+1) e^{2 a+2 b x}\right )}{16 b^4}-\frac{x^3 \text{PolyLog}\left (2,(d+1) e^{2 a+2 b x}\right )}{4 b}-\frac{1}{8} x^4 \log \left (1-(d+1) e^{2 a+2 b x}\right )+\frac{1}{4} x^4 \tanh ^{-1}(d \coth (a+b x)+d+1)+\frac{b x^5}{20} \]
Antiderivative was successfully verified.
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Rule 6241
Rule 2184
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int x^3 \tanh ^{-1}(1+d+d \coth (a+b x)) \, dx &=\frac{1}{4} x^4 \tanh ^{-1}(1+d+d \coth (a+b x))+\frac{1}{4} b \int \frac{x^4}{1+(-1-d) e^{2 a+2 b x}} \, dx\\ &=\frac{b x^5}{20}+\frac{1}{4} x^4 \tanh ^{-1}(1+d+d \coth (a+b x))+\frac{1}{4} (b (1+d)) \int \frac{e^{2 a+2 b x} x^4}{1+(-1-d) e^{2 a+2 b x}} \, dx\\ &=\frac{b x^5}{20}+\frac{1}{4} x^4 \tanh ^{-1}(1+d+d \coth (a+b x))-\frac{1}{8} x^4 \log \left (1-(1+d) e^{2 a+2 b x}\right )+\frac{1}{2} \int x^3 \log \left (1+(-1-d) e^{2 a+2 b x}\right ) \, dx\\ &=\frac{b x^5}{20}+\frac{1}{4} x^4 \tanh ^{-1}(1+d+d \coth (a+b x))-\frac{1}{8} x^4 \log \left (1-(1+d) e^{2 a+2 b x}\right )-\frac{x^3 \text{Li}_2\left ((1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{3 \int x^2 \text{Li}_2\left (-(-1-d) e^{2 a+2 b x}\right ) \, dx}{4 b}\\ &=\frac{b x^5}{20}+\frac{1}{4} x^4 \tanh ^{-1}(1+d+d \coth (a+b x))-\frac{1}{8} x^4 \log \left (1-(1+d) e^{2 a+2 b x}\right )-\frac{x^3 \text{Li}_2\left ((1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{3 x^2 \text{Li}_3\left ((1+d) e^{2 a+2 b x}\right )}{8 b^2}-\frac{3 \int x \text{Li}_3\left ((1+d) e^{2 a+2 b x}\right ) \, dx}{4 b^2}\\ &=\frac{b x^5}{20}+\frac{1}{4} x^4 \tanh ^{-1}(1+d+d \coth (a+b x))-\frac{1}{8} x^4 \log \left (1-(1+d) e^{2 a+2 b x}\right )-\frac{x^3 \text{Li}_2\left ((1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{3 x^2 \text{Li}_3\left ((1+d) e^{2 a+2 b x}\right )}{8 b^2}-\frac{3 x \text{Li}_4\left ((1+d) e^{2 a+2 b x}\right )}{8 b^3}+\frac{3 \int \text{Li}_4\left ((1+d) e^{2 a+2 b x}\right ) \, dx}{8 b^3}\\ &=\frac{b x^5}{20}+\frac{1}{4} x^4 \tanh ^{-1}(1+d+d \coth (a+b x))-\frac{1}{8} x^4 \log \left (1-(1+d) e^{2 a+2 b x}\right )-\frac{x^3 \text{Li}_2\left ((1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{3 x^2 \text{Li}_3\left ((1+d) e^{2 a+2 b x}\right )}{8 b^2}-\frac{3 x \text{Li}_4\left ((1+d) e^{2 a+2 b x}\right )}{8 b^3}+\frac{3 \operatorname{Subst}\left (\int \frac{\text{Li}_4((1+d) x)}{x} \, dx,x,e^{2 a+2 b x}\right )}{16 b^4}\\ &=\frac{b x^5}{20}+\frac{1}{4} x^4 \tanh ^{-1}(1+d+d \coth (a+b x))-\frac{1}{8} x^4 \log \left (1-(1+d) e^{2 a+2 b x}\right )-\frac{x^3 \text{Li}_2\left ((1+d) e^{2 a+2 b x}\right )}{4 b}+\frac{3 x^2 \text{Li}_3\left ((1+d) e^{2 a+2 b x}\right )}{8 b^2}-\frac{3 x \text{Li}_4\left ((1+d) e^{2 a+2 b x}\right )}{8 b^3}+\frac{3 \text{Li}_5\left ((1+d) e^{2 a+2 b x}\right )}{16 b^4}\\ \end{align*}
Mathematica [A] time = 4.29145, size = 141, normalized size = 0.93 \[ \frac{1}{16} \left (\frac{6 x^2 \text{PolyLog}\left (3,\frac{e^{-2 (a+b x)}}{d+1}\right )}{b^2}+\frac{6 x \text{PolyLog}\left (4,\frac{e^{-2 (a+b x)}}{d+1}\right )}{b^3}+\frac{3 \text{PolyLog}\left (5,\frac{e^{-2 (a+b x)}}{d+1}\right )}{b^4}+\frac{4 x^3 \text{PolyLog}\left (2,\frac{e^{-2 (a+b x)}}{d+1}\right )}{b}-2 x^4 \log \left (1-\frac{e^{-2 (a+b x)}}{d+1}\right )+4 x^4 \tanh ^{-1}(d \coth (a+b x)+d+1)\right ) \]
Antiderivative was successfully verified.
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Maple [C] time = 13.915, size = 1741, normalized size = 11.5 \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 [A] time = 3.2687, size = 197, normalized size = 1.3 \begin{align*} \frac{1}{4} \, x^{4} \operatorname{artanh}\left (d \coth \left (b x + a\right ) + d + 1\right ) + \frac{1}{40} \,{\left (\frac{2 \, x^{5}}{d} - \frac{5 \,{\left (2 \, b^{4} x^{4} \log \left (-{\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 4 \, b^{3} x^{3}{\rm Li}_2\left ({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b^{2} x^{2}{\rm Li}_{3}({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) + 6 \, b x{\rm Li}_{4}({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}) - 3 \,{\rm Li}_{5}({\left (d + 1\right )} e^{\left (2 \, b x + 2 \, a\right )})\right )}}{b^{5} d}\right )} b d \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.09132, size = 1281, normalized size = 8.43 \begin{align*} \frac{2 \, b^{5} x^{5} + 5 \, b^{4} x^{4} \log \left (-\frac{d \cosh \left (b x + a\right ) +{\left (d + 2\right )} \sinh \left (b x + a\right )}{d \cosh \left (b x + a\right ) + d \sinh \left (b x + a\right )}\right ) - 20 \, b^{3} x^{3}{\rm Li}_2\left (\sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 20 \, b^{3} x^{3}{\rm Li}_2\left (-\sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 5 \, a^{4} \log \left (2 \,{\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \,{\left (d + 1\right )} \sinh \left (b x + a\right ) + 2 \, \sqrt{d + 1}\right ) - 5 \, a^{4} \log \left (2 \,{\left (d + 1\right )} \cosh \left (b x + a\right ) + 2 \,{\left (d + 1\right )} \sinh \left (b x + a\right ) - 2 \, \sqrt{d + 1}\right ) + 60 \, b^{2} x^{2}{\rm polylog}\left (3, \sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 60 \, b^{2} x^{2}{\rm polylog}\left (3, -\sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 120 \, b x{\rm polylog}\left (4, \sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 120 \, b x{\rm polylog}\left (4, -\sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) - 5 \,{\left (b^{4} x^{4} - a^{4}\right )} \log \left (\sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) - 5 \,{\left (b^{4} x^{4} - a^{4}\right )} \log \left (-\sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} + 1\right ) + 120 \,{\rm polylog}\left (5, \sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right ) + 120 \,{\rm polylog}\left (5, -\sqrt{d + 1}{\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}\right )}{40 \, b^{4}} \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^{3} \operatorname{artanh}\left (d \coth \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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