Optimal. Leaf size=101 \[ \frac{\cosh ^4(a+b x)}{32 b^3}+\frac{3 \cosh ^2(a+b x)}{32 b^3}-\frac{x \sinh (a+b x) \cosh ^3(a+b x)}{8 b^2}-\frac{3 x \sinh (a+b x) \cosh (a+b x)}{16 b^2}+\frac{x^2 \cosh ^4(a+b x)}{4 b}-\frac{3 x^2}{32 b} \]
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Rubi [A] time = 0.0760245, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {5373, 3310, 30} \[ \frac{\cosh ^4(a+b x)}{32 b^3}+\frac{3 \cosh ^2(a+b x)}{32 b^3}-\frac{x \sinh (a+b x) \cosh ^3(a+b x)}{8 b^2}-\frac{3 x \sinh (a+b x) \cosh (a+b x)}{16 b^2}+\frac{x^2 \cosh ^4(a+b x)}{4 b}-\frac{3 x^2}{32 b} \]
Antiderivative was successfully verified.
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Rule 5373
Rule 3310
Rule 30
Rubi steps
\begin{align*} \int x^2 \cosh ^3(a+b x) \sinh (a+b x) \, dx &=\frac{x^2 \cosh ^4(a+b x)}{4 b}-\frac{\int x \cosh ^4(a+b x) \, dx}{2 b}\\ &=\frac{\cosh ^4(a+b x)}{32 b^3}+\frac{x^2 \cosh ^4(a+b x)}{4 b}-\frac{x \cosh ^3(a+b x) \sinh (a+b x)}{8 b^2}-\frac{3 \int x \cosh ^2(a+b x) \, dx}{8 b}\\ &=\frac{3 \cosh ^2(a+b x)}{32 b^3}+\frac{\cosh ^4(a+b x)}{32 b^3}+\frac{x^2 \cosh ^4(a+b x)}{4 b}-\frac{3 x \cosh (a+b x) \sinh (a+b x)}{16 b^2}-\frac{x \cosh ^3(a+b x) \sinh (a+b x)}{8 b^2}-\frac{3 \int x \, dx}{16 b}\\ &=-\frac{3 x^2}{32 b}+\frac{3 \cosh ^2(a+b x)}{32 b^3}+\frac{\cosh ^4(a+b x)}{32 b^3}+\frac{x^2 \cosh ^4(a+b x)}{4 b}-\frac{3 x \cosh (a+b x) \sinh (a+b x)}{16 b^2}-\frac{x \cosh ^3(a+b x) \sinh (a+b x)}{8 b^2}\\ \end{align*}
Mathematica [A] time = 0.231325, size = 70, normalized size = 0.69 \[ \frac{16 \left (2 b^2 x^2+1\right ) \cosh (2 (a+b x))+\left (8 b^2 x^2+1\right ) \cosh (4 (a+b x))-4 b x (8 \sinh (2 (a+b x))+\sinh (4 (a+b x)))}{256 b^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.006, size = 237, normalized size = 2.4 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{ \left ( bx+a \right ) ^{2} \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{4}}+{\frac{ \left ( bx+a \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{4}}-{\frac{ \left ( bx+a \right ) \sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{8}}-{\frac{ \left ( 3\,bx+3\,a \right ) \cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{16}}-{\frac{3\, \left ( bx+a \right ) ^{2}}{32}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{32}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{8}}-2\,a \left ( 1/4\, \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}+1/4\, \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}-1/16\, \left ( \cosh \left ( bx+a \right ) \right ) ^{3}\sinh \left ( bx+a \right ) -{\frac{3\,\cosh \left ( bx+a \right ) \sinh \left ( bx+a \right ) }{32}}-{\frac{3\,bx}{32}}-{\frac{3\,a}{32}} \right ) +{a}^{2} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{4}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{4}} \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.20604, size = 171, normalized size = 1.69 \begin{align*} \frac{{\left (8 \, b^{2} x^{2} e^{\left (4 \, a\right )} - 4 \, b x e^{\left (4 \, a\right )} + e^{\left (4 \, a\right )}\right )} e^{\left (4 \, b x\right )}}{512 \, b^{3}} + \frac{{\left (2 \, b^{2} x^{2} e^{\left (2 \, a\right )} - 2 \, b x e^{\left (2 \, a\right )} + e^{\left (2 \, a\right )}\right )} e^{\left (2 \, b x\right )}}{32 \, b^{3}} + \frac{{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b^{3}} + \frac{{\left (8 \, b^{2} x^{2} + 4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{512 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.75687, size = 394, normalized size = 3.9 \begin{align*} -\frac{16 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} -{\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{4} -{\left (8 \, b^{2} x^{2} + 1\right )} \sinh \left (b x + a\right )^{4} - 16 \,{\left (2 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{2} - 2 \,{\left (16 \, b^{2} x^{2} + 3 \,{\left (8 \, b^{2} x^{2} + 1\right )} \cosh \left (b x + a\right )^{2} + 8\right )} \sinh \left (b x + a\right )^{2} + 16 \,{\left (b x \cosh \left (b x + a\right )^{3} + 4 \, b x \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )}{256 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 4.30484, size = 150, normalized size = 1.49 \begin{align*} \begin{cases} - \frac{3 x^{2} \sinh ^{4}{\left (a + b x \right )}}{32 b} + \frac{3 x^{2} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{16 b} + \frac{5 x^{2} \cosh ^{4}{\left (a + b x \right )}}{32 b} + \frac{3 x \sinh ^{3}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{16 b^{2}} - \frac{5 x \sinh{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{16 b^{2}} - \frac{3 \sinh ^{4}{\left (a + b x \right )}}{64 b^{3}} + \frac{5 \cosh ^{4}{\left (a + b x \right )}}{64 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \sinh{\left (a \right )} \cosh ^{3}{\left (a \right )}}{3} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.17006, size = 153, normalized size = 1.51 \begin{align*} \frac{{\left (8 \, b^{2} x^{2} - 4 \, b x + 1\right )} e^{\left (4 \, b x + 4 \, a\right )}}{512 \, b^{3}} + \frac{{\left (2 \, b^{2} x^{2} - 2 \, b x + 1\right )} e^{\left (2 \, b x + 2 \, a\right )}}{32 \, b^{3}} + \frac{{\left (2 \, b^{2} x^{2} + 2 \, b x + 1\right )} e^{\left (-2 \, b x - 2 \, a\right )}}{32 \, b^{3}} + \frac{{\left (8 \, b^{2} x^{2} + 4 \, b x + 1\right )} e^{\left (-4 \, b x - 4 \, a\right )}}{512 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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