Optimal. Leaf size=66 \[ \frac{e^{-a-b x}}{8 b}+\frac{3 e^{a+b x}}{8 b}-\frac{e^{3 a+3 b x}}{8 b}+\frac{e^{5 a+5 b x}}{40 b} \]
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Rubi [A] time = 0.0360092, antiderivative size = 66, 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 = {2282, 12, 270} \[ \frac{e^{-a-b x}}{8 b}+\frac{3 e^{a+b x}}{8 b}-\frac{e^{3 a+3 b x}}{8 b}+\frac{e^{5 a+5 b x}}{40 b} \]
Antiderivative was successfully verified.
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Rule 2282
Rule 12
Rule 270
Rubi steps
\begin{align*} \int e^{2 (a+b x)} \sinh ^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^3}{8 x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^3}{x^2} \, dx,x,e^{a+b x}\right )}{8 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (3-\frac{1}{x^2}-3 x^2+x^4\right ) \, dx,x,e^{a+b x}\right )}{8 b}\\ &=\frac{e^{-a-b x}}{8 b}+\frac{3 e^{a+b x}}{8 b}-\frac{e^{3 a+3 b x}}{8 b}+\frac{e^{5 a+5 b x}}{40 b}\\ \end{align*}
Mathematica [A] time = 0.0342435, size = 50, normalized size = 0.76 \[ \frac{e^{-a-b x} \left (15 e^{2 (a+b x)}-5 e^{4 (a+b x)}+e^{6 (a+b x)}+5\right )}{40 b} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.022, size = 80, normalized size = 1.2 \begin{align*}{\frac{\sinh \left ( bx+a \right ) }{4\,b}}-{\frac{\sinh \left ( 3\,bx+3\,a \right ) }{8\,b}}+{\frac{\sinh \left ( 5\,bx+5\,a \right ) }{40\,b}}+{\frac{\cosh \left ( bx+a \right ) }{2\,b}}-{\frac{\cosh \left ( 3\,bx+3\,a \right ) }{8\,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 = 1.23995, 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 )} - 1\right )} e^{\left (5 \, b x + 5 \, a\right )}}{40 \, 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 = 1.75249, size = 289, normalized size = 4.38 \begin{align*} \frac{3 \, \cosh \left (b x + a\right )^{3} + 9 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - 2 \, \sinh \left (b x + a\right )^{3} - 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} + 5\right )} \sinh \left (b x + a\right ) + 5 \, \cosh \left (b x + a\right )}{20 \,{\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 = 38.2161, size = 124, normalized size = 1.88 \begin{align*} \begin{cases} \frac{2 e^{2 a} e^{2 b x} \sinh ^{3}{\left (a + b x \right )}}{5 b} + \frac{e^{2 a} e^{2 b x} \sinh ^{2}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{5 b} - \frac{4 e^{2 a} e^{2 b x} \sinh{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{5 b} + \frac{2 e^{2 a} e^{2 b x} \cosh ^{3}{\left (a + b x \right )}}{5 b} & \text{for}\: b \neq 0 \\x e^{2 a} \sinh ^{3}{\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 = 1.13422, size = 72, normalized size = 1.09 \begin{align*} \frac{{\left (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 )} + 5 \, e^{\left (-b x - a\right )}}{40 \, b} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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