Optimal. Leaf size=327 \[ \frac{3 x^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d^2}+\frac{3 x^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2-b^2}+a}\right )}{b d^2}-\frac{6 x \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d^3}-\frac{6 x \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2-b^2}+a}\right )}{b d^3}+\frac{6 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d^4}+\frac{6 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{\sqrt{a^2-b^2}+a}\right )}{b d^4}+\frac{x^3 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}+1\right )}{b d}+\frac{x^3 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2-b^2}+a}+1\right )}{b d}-\frac{x^4}{4 b} \]
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Rubi [A] time = 0.479493, antiderivative size = 327, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {5562, 2190, 2531, 6609, 2282, 6589} \[ \frac{3 x^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d^2}+\frac{3 x^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2-b^2}+a}\right )}{b d^2}-\frac{6 x \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d^3}-\frac{6 x \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2-b^2}+a}\right )}{b d^3}+\frac{6 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d^4}+\frac{6 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{\sqrt{a^2-b^2}+a}\right )}{b d^4}+\frac{x^3 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}+1\right )}{b d}+\frac{x^3 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2-b^2}+a}+1\right )}{b d}-\frac{x^4}{4 b} \]
Antiderivative was successfully verified.
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Rule 5562
Rule 2190
Rule 2531
Rule 6609
Rule 2282
Rule 6589
Rubi steps
\begin{align*} \int \frac{x^3 \sinh (c+d x)}{a+b \cosh (c+d x)} \, dx &=-\frac{x^4}{4 b}+\int \frac{e^{c+d x} x^3}{a-\sqrt{a^2-b^2}+b e^{c+d x}} \, dx+\int \frac{e^{c+d x} x^3}{a+\sqrt{a^2-b^2}+b e^{c+d x}} \, dx\\ &=-\frac{x^4}{4 b}+\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d}+\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2-b^2}}\right )}{b d}-\frac{3 \int x^2 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right ) \, dx}{b d}-\frac{3 \int x^2 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2-b^2}}\right ) \, dx}{b d}\\ &=-\frac{x^4}{4 b}+\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d}+\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2-b^2}}\right )}{b d}+\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d^2}+\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2-b^2}}\right )}{b d^2}-\frac{6 \int x \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right ) \, dx}{b d^2}-\frac{6 \int x \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2-b^2}}\right ) \, dx}{b d^2}\\ &=-\frac{x^4}{4 b}+\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d}+\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2-b^2}}\right )}{b d}+\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d^2}+\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2-b^2}}\right )}{b d^2}-\frac{6 x \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d^3}-\frac{6 x \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2-b^2}}\right )}{b d^3}+\frac{6 \int \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right ) \, dx}{b d^3}+\frac{6 \int \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2-b^2}}\right ) \, dx}{b d^3}\\ &=-\frac{x^4}{4 b}+\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d}+\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2-b^2}}\right )}{b d}+\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d^2}+\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2-b^2}}\right )}{b d^2}-\frac{6 x \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d^3}-\frac{6 x \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2-b^2}}\right )}{b d^3}+\frac{6 \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\frac{b x}{-a+\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b d^4}+\frac{6 \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (-\frac{b x}{a+\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{c+d x}\right )}{b d^4}\\ &=-\frac{x^4}{4 b}+\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d}+\frac{x^3 \log \left (1+\frac{b e^{c+d x}}{a+\sqrt{a^2-b^2}}\right )}{b d}+\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d^2}+\frac{3 x^2 \text{Li}_2\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2-b^2}}\right )}{b d^2}-\frac{6 x \text{Li}_3\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d^3}-\frac{6 x \text{Li}_3\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2-b^2}}\right )}{b d^3}+\frac{6 \text{Li}_4\left (-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d^4}+\frac{6 \text{Li}_4\left (-\frac{b e^{c+d x}}{a+\sqrt{a^2-b^2}}\right )}{b d^4}\\ \end{align*}
Mathematica [A] time = 0.0326848, size = 326, normalized size = 1. \[ \frac{3 x^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d^2}+\frac{3 x^2 \text{PolyLog}\left (2,-\frac{b e^{c+d x}}{\sqrt{a^2-b^2}+a}\right )}{b d^2}-\frac{6 x \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}\right )}{b d^3}-\frac{6 x \text{PolyLog}\left (3,-\frac{b e^{c+d x}}{\sqrt{a^2-b^2}+a}\right )}{b d^3}+\frac{6 \text{PolyLog}\left (4,\frac{b e^{c+d x}}{\sqrt{a^2-b^2}-a}\right )}{b d^4}+\frac{6 \text{PolyLog}\left (4,-\frac{b e^{c+d x}}{\sqrt{a^2-b^2}+a}\right )}{b d^4}+\frac{x^3 \log \left (\frac{b e^{c+d x}}{a-\sqrt{a^2-b^2}}+1\right )}{b d}+\frac{x^3 \log \left (\frac{b e^{c+d x}}{\sqrt{a^2-b^2}+a}+1\right )}{b d}-\frac{x^4}{4 b} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.227, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{3}\sinh \left ( dx+c \right ) }{a+b\cosh \left ( dx+c \right ) }}\, dx \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*} \frac{x^{4}}{4 \, b} - \frac{1}{2} \, \int \frac{4 \,{\left (a x^{3} e^{\left (d x + c\right )} + b x^{3}\right )}}{b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b e^{\left (d x + c\right )} + b^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [C] time = 2.09096, size = 1585, normalized size = 4.85 \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 \frac{x^{3} \sinh \left (d x + c\right )}{b \cosh \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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