3.319 \(\int e^x \sinh (4 x) \, dx\)

Optimal. Leaf size=19 \[ \frac{e^{-3 x}}{6}+\frac{e^{5 x}}{10} \]

[Out]

1/(6*E^(3*x)) + E^(5*x)/10

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Rubi [A]  time = 0.0124461, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {2282, 12, 14} \[ \frac{e^{-3 x}}{6}+\frac{e^{5 x}}{10} \]

Antiderivative was successfully verified.

[In]

Int[E^x*Sinh[4*x],x]

[Out]

1/(6*E^(3*x)) + E^(5*x)/10

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int e^x \sinh (4 x) \, dx &=\operatorname{Subst}\left (\int \frac{-1+x^8}{2 x^4} \, dx,x,e^x\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{-1+x^8}{x^4} \, dx,x,e^x\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{1}{x^4}+x^4\right ) \, dx,x,e^x\right )\\ &=\frac{e^{-3 x}}{6}+\frac{e^{5 x}}{10}\\ \end{align*}

Mathematica [A]  time = 0.0091856, size = 19, normalized size = 1. \[ \frac{e^{-3 x}}{6}+\frac{e^{5 x}}{10} \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Sinh[4*x],x]

[Out]

1/(6*E^(3*x)) + E^(5*x)/10

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Maple [A]  time = 0.006, size = 26, normalized size = 1.4 \begin{align*} -{\frac{\sinh \left ( 3\,x \right ) }{6}}+{\frac{\sinh \left ( 5\,x \right ) }{10}}+{\frac{\cosh \left ( 3\,x \right ) }{6}}+{\frac{\cosh \left ( 5\,x \right ) }{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*sinh(4*x),x)

[Out]

-1/6*sinh(3*x)+1/10*sinh(5*x)+1/6*cosh(3*x)+1/10*cosh(5*x)

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Maxima [A]  time = 1.09995, size = 18, normalized size = 0.95 \begin{align*} \frac{1}{10} \, e^{\left (5 \, x\right )} + \frac{1}{6} \, e^{\left (-3 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sinh(4*x),x, algorithm="maxima")

[Out]

1/10*e^(5*x) + 1/6*e^(-3*x)

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Fricas [B]  time = 2.03866, size = 154, normalized size = 8.11 \begin{align*} \frac{4 \,{\left (\cosh \left (x\right )^{4} - \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} - \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4}\right )}}{15 \,{\left (\cosh \left (x\right ) - \sinh \left (x\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sinh(4*x),x, algorithm="fricas")

[Out]

4/15*(cosh(x)^4 - cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 - cosh(x)*sinh(x)^3 + sinh(x)^4)/(cosh(x) - sinh(x
))

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Sympy [A]  time = 0.350168, size = 20, normalized size = 1.05 \begin{align*} - \frac{e^{x} \sinh{\left (4 x \right )}}{15} + \frac{4 e^{x} \cosh{\left (4 x \right )}}{15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sinh(4*x),x)

[Out]

-exp(x)*sinh(4*x)/15 + 4*exp(x)*cosh(4*x)/15

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Giac [A]  time = 1.09391, size = 18, normalized size = 0.95 \begin{align*} \frac{1}{10} \, e^{\left (5 \, x\right )} + \frac{1}{6} \, e^{\left (-3 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sinh(4*x),x, algorithm="giac")

[Out]

1/10*e^(5*x) + 1/6*e^(-3*x)