Optimal. Leaf size=148 \[ -\frac{\sinh (a+b x)}{4 b^3}+\frac{\sinh (3 a+3 b x)}{216 b^3}+\frac{\sinh (5 a+5 b x)}{1000 b^3}+\frac{x \cosh (a+b x)}{4 b^2}-\frac{x \cosh (3 a+3 b x)}{72 b^2}-\frac{x \cosh (5 a+5 b x)}{200 b^2}-\frac{x^2 \sinh (a+b x)}{8 b}+\frac{x^2 \sinh (3 a+3 b x)}{48 b}+\frac{x^2 \sinh (5 a+5 b x)}{80 b} \]
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Rubi [A] time = 0.180373, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {5448, 3296, 2637} \[ -\frac{\sinh (a+b x)}{4 b^3}+\frac{\sinh (3 a+3 b x)}{216 b^3}+\frac{\sinh (5 a+5 b x)}{1000 b^3}+\frac{x \cosh (a+b x)}{4 b^2}-\frac{x \cosh (3 a+3 b x)}{72 b^2}-\frac{x \cosh (5 a+5 b x)}{200 b^2}-\frac{x^2 \sinh (a+b x)}{8 b}+\frac{x^2 \sinh (3 a+3 b x)}{48 b}+\frac{x^2 \sinh (5 a+5 b x)}{80 b} \]
Antiderivative was successfully verified.
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Rule 5448
Rule 3296
Rule 2637
Rubi steps
\begin{align*} \int x^2 \cosh ^3(a+b x) \sinh ^2(a+b x) \, dx &=\int \left (-\frac{1}{8} x^2 \cosh (a+b x)+\frac{1}{16} x^2 \cosh (3 a+3 b x)+\frac{1}{16} x^2 \cosh (5 a+5 b x)\right ) \, dx\\ &=\frac{1}{16} \int x^2 \cosh (3 a+3 b x) \, dx+\frac{1}{16} \int x^2 \cosh (5 a+5 b x) \, dx-\frac{1}{8} \int x^2 \cosh (a+b x) \, dx\\ &=-\frac{x^2 \sinh (a+b x)}{8 b}+\frac{x^2 \sinh (3 a+3 b x)}{48 b}+\frac{x^2 \sinh (5 a+5 b x)}{80 b}-\frac{\int x \sinh (5 a+5 b x) \, dx}{40 b}-\frac{\int x \sinh (3 a+3 b x) \, dx}{24 b}+\frac{\int x \sinh (a+b x) \, dx}{4 b}\\ &=\frac{x \cosh (a+b x)}{4 b^2}-\frac{x \cosh (3 a+3 b x)}{72 b^2}-\frac{x \cosh (5 a+5 b x)}{200 b^2}-\frac{x^2 \sinh (a+b x)}{8 b}+\frac{x^2 \sinh (3 a+3 b x)}{48 b}+\frac{x^2 \sinh (5 a+5 b x)}{80 b}+\frac{\int \cosh (5 a+5 b x) \, dx}{200 b^2}+\frac{\int \cosh (3 a+3 b x) \, dx}{72 b^2}-\frac{\int \cosh (a+b x) \, dx}{4 b^2}\\ &=\frac{x \cosh (a+b x)}{4 b^2}-\frac{x \cosh (3 a+3 b x)}{72 b^2}-\frac{x \cosh (5 a+5 b x)}{200 b^2}-\frac{\sinh (a+b x)}{4 b^3}-\frac{x^2 \sinh (a+b x)}{8 b}+\frac{\sinh (3 a+3 b x)}{216 b^3}+\frac{x^2 \sinh (3 a+3 b x)}{48 b}+\frac{\sinh (5 a+5 b x)}{1000 b^3}+\frac{x^2 \sinh (5 a+5 b x)}{80 b}\\ \end{align*}
Mathematica [A] time = 0.308318, size = 105, normalized size = 0.71 \[ \frac{-6750 \left (\left (b^2 x^2+2\right ) \sinh (a+b x)-2 b x \cosh (a+b x)\right )+125 \left (\left (9 b^2 x^2+2\right ) \sinh (3 (a+b x))-6 b x \cosh (3 (a+b x))\right )+27 \left (\left (25 b^2 x^2+2\right ) \sinh (5 (a+b x))-10 b x \cosh (5 (a+b x))\right )}{54000 b^3} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.008, size = 306, normalized size = 2.1 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{ \left ( bx+a \right ) ^{2}\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{4}}{5}}-{\frac{2\, \left ( bx+a \right ) ^{2}\sinh \left ( bx+a \right ) }{15}}-{\frac{ \left ( bx+a \right ) ^{2}\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{15}}-{\frac{ \left ( 2\,bx+2\,a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2} \left ( \cosh \left ( bx+a \right ) \right ) ^{3}}{25}}-{\frac{ \left ( 8\,bx+8\,a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}\cosh \left ( bx+a \right ) }{225}}+{\frac{ \left ( 52\,bx+52\,a \right ) \cosh \left ( bx+a \right ) }{225}}+{\frac{2\, \left ( \cosh \left ( bx+a \right ) \right ) ^{4}\sinh \left ( bx+a \right ) }{125}}-{\frac{856\,\sinh \left ( bx+a \right ) }{3375}}+{\frac{22\,\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{3375}}-2\,a \left ( 1/5\, \left ( bx+a \right ) \sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{4}-2/15\, \left ( bx+a \right ) \sinh \left ( bx+a \right ) -1/15\, \left ( bx+a \right ) \sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}-1/25\, \left ( \cosh \left ( bx+a \right ) \right ) ^{3} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}-{\frac{4\,\cosh \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{225}}+{\frac{26\,\cosh \left ( bx+a \right ) }{225}} \right ) +{a}^{2} \left ({\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{4}\sinh \left ( bx+a \right ) }{5}}-{\frac{\sinh \left ( bx+a \right ) }{5} \left ({\frac{2}{3}}+{\frac{ \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{3}} \right ) } \right ) \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.06848, size = 252, normalized size = 1.7 \begin{align*} \frac{{\left (25 \, b^{2} x^{2} e^{\left (5 \, a\right )} - 10 \, b x e^{\left (5 \, a\right )} + 2 \, e^{\left (5 \, a\right )}\right )} e^{\left (5 \, b x\right )}}{4000 \, b^{3}} + \frac{{\left (9 \, b^{2} x^{2} e^{\left (3 \, a\right )} - 6 \, b x e^{\left (3 \, a\right )} + 2 \, e^{\left (3 \, a\right )}\right )} e^{\left (3 \, b x\right )}}{864 \, b^{3}} - \frac{{\left (b^{2} x^{2} e^{a} - 2 \, b x e^{a} + 2 \, e^{a}\right )} e^{\left (b x\right )}}{16 \, b^{3}} + \frac{{\left (b^{2} x^{2} + 2 \, b x + 2\right )} e^{\left (-b x - a\right )}}{16 \, b^{3}} - \frac{{\left (9 \, b^{2} x^{2} + 6 \, b x + 2\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{864 \, b^{3}} - \frac{{\left (25 \, b^{2} x^{2} + 10 \, b x + 2\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{4000 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.75437, size = 579, normalized size = 3.91 \begin{align*} -\frac{270 \, b x \cosh \left (b x + a\right )^{5} + 1350 \, b x \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} - 27 \,{\left (25 \, b^{2} x^{2} + 2\right )} \sinh \left (b x + a\right )^{5} + 750 \, b x \cosh \left (b x + a\right )^{3} - 5 \,{\left (225 \, b^{2} x^{2} + 54 \,{\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{2} + 50\right )} \sinh \left (b x + a\right )^{3} - 13500 \, b x \cosh \left (b x + a\right ) + 450 \,{\left (6 \, b x \cosh \left (b x + a\right )^{3} + 5 \, b x \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 15 \,{\left (9 \,{\left (25 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{4} - 450 \, b^{2} x^{2} + 25 \,{\left (9 \, b^{2} x^{2} + 2\right )} \cosh \left (b x + a\right )^{2} - 900\right )} \sinh \left (b x + a\right )}{54000 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 7.97986, size = 182, normalized size = 1.23 \begin{align*} \begin{cases} - \frac{2 x^{2} \sinh ^{5}{\left (a + b x \right )}}{15 b} + \frac{x^{2} \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{3 b} + \frac{4 x \sinh ^{4}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{15 b^{2}} - \frac{26 x \sinh ^{2}{\left (a + b x \right )} \cosh ^{3}{\left (a + b x \right )}}{45 b^{2}} + \frac{52 x \cosh ^{5}{\left (a + b x \right )}}{225 b^{2}} - \frac{856 \sinh ^{5}{\left (a + b x \right )}}{3375 b^{3}} + \frac{338 \sinh ^{3}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{675 b^{3}} - \frac{52 \sinh{\left (a + b x \right )} \cosh ^{4}{\left (a + b x \right )}}{225 b^{3}} & \text{for}\: b \neq 0 \\\frac{x^{3} \sinh ^{2}{\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 = 1.18424, size = 221, normalized size = 1.49 \begin{align*} \frac{{\left (25 \, b^{2} x^{2} - 10 \, b x + 2\right )} e^{\left (5 \, b x + 5 \, a\right )}}{4000 \, b^{3}} + \frac{{\left (9 \, b^{2} x^{2} - 6 \, b x + 2\right )} e^{\left (3 \, b x + 3 \, a\right )}}{864 \, b^{3}} - \frac{{\left (b^{2} x^{2} - 2 \, b x + 2\right )} e^{\left (b x + a\right )}}{16 \, b^{3}} + \frac{{\left (b^{2} x^{2} + 2 \, b x + 2\right )} e^{\left (-b x - a\right )}}{16 \, b^{3}} - \frac{{\left (9 \, b^{2} x^{2} + 6 \, b x + 2\right )} e^{\left (-3 \, b x - 3 \, a\right )}}{864 \, b^{3}} - \frac{{\left (25 \, b^{2} x^{2} + 10 \, b x + 2\right )} e^{\left (-5 \, b x - 5 \, a\right )}}{4000 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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