Optimal. Leaf size=66 \[ -\frac{e^{-a-b x}}{8 b}-\frac{e^{a+b x}}{8 b}-\frac{e^{3 a+3 b x}}{24 b}+\frac{e^{5 a+5 b x}}{40 b} \]
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Rubi [A] time = 0.049802, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2282, 12, 448} \[ -\frac{e^{-a-b x}}{8 b}-\frac{e^{a+b x}}{8 b}-\frac{e^{3 a+3 b x}}{24 b}+\frac{e^{5 a+5 b x}}{40 b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 12
Rule 448
Rubi steps
\begin{align*} \int e^{2 (a+b x)} \cosh (a+b x) \sinh ^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2 \left (1+x^2\right )}{8 x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2 \left (1+x^2\right )}{x^2} \, dx,x,e^{a+b x}\right )}{8 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-1+\frac{1}{x^2}-x^2+x^4\right ) \, dx,x,e^{a+b x}\right )}{8 b}\\ &=-\frac{e^{-a-b x}}{8 b}-\frac{e^{a+b x}}{8 b}-\frac{e^{3 a+3 b x}}{24 b}+\frac{e^{5 a+5 b x}}{40 b}\\ \end{align*}
Mathematica [A] time = 0.0802082, size = 51, normalized size = 0.77 \[ \frac{3 e^{-a-b x} \left (e^{6 (a+b x)}-5\right )-5 e^{a+b x} \left (e^{2 (a+b x)}+3\right )}{120 b} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.01, size = 69, normalized size = 1.1 \begin{align*} -{\frac{\sinh \left ( 3\,bx+3\,a \right ) }{24\,b}}+{\frac{\sinh \left ( 5\,bx+5\,a \right ) }{40\,b}}-{\frac{\cosh \left ( bx+a \right ) }{4\,b}}-{\frac{\cosh \left ( 3\,bx+3\,a \right ) }{24\,b}}+{\frac{\cosh \left ( 5\,bx+5\,a \right ) }{40\,b}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.02501, size = 72, normalized size = 1.09 \begin{align*} -\frac{{\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} + 15 \, e^{\left (-4 \, b x - 4 \, a\right )} - 3\right )} e^{\left (5 \, b x + 5 \, a\right )}}{120 \, b} - \frac{e^{\left (-b x - a\right )}}{8 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.87815, size = 292, normalized size = 4.42 \begin{align*} -\frac{6 \, \cosh \left (b x + a\right )^{3} + 18 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - 9 \, \sinh \left (b x + a\right )^{3} -{\left (27 \, \cosh \left (b x + a\right )^{2} + 5\right )} \sinh \left (b x + a\right ) + 10 \, \cosh \left (b x + a\right )}{60 \,{\left (b \cosh \left (b x + a\right )^{2} - 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 30.2939, size = 128, normalized size = 1.94 \begin{align*} \begin{cases} \frac{e^{2 a} e^{2 b x} \sinh ^{3}{\left (a + b x \right )}}{15 b} - \frac{2 e^{2 a} e^{2 b x} \sinh ^{2}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{15 b} + \frac{8 e^{2 a} e^{2 b x} \sinh{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{15 b} - \frac{4 e^{2 a} e^{2 b x} \cosh ^{3}{\left (a + b x \right )}}{15 b} & \text{for}\: b \neq 0 \\x e^{2 a} \sinh ^{2}{\left (a \right )} \cosh{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.12355, size = 74, normalized size = 1.12 \begin{align*} \frac{{\left (3 \, e^{\left (5 \, b x + 10 \, a\right )} - 5 \, e^{\left (3 \, b x + 8 \, a\right )} - 15 \, e^{\left (b x + 6 \, a\right )}\right )} e^{\left (-5 \, a\right )} - 15 \, e^{\left (-b x - a\right )}}{120 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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