3.315 \(\int e^x \sinh (3 x) \, dx\)

Optimal. Leaf size=19 \[ \frac{e^{-2 x}}{4}+\frac{e^{4 x}}{8} \]

[Out]

1/(4*E^(2*x)) + E^(4*x)/8

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Rubi [A]  time = 0.0119808, 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} \[ \frac{e^{-2 x}}{4}+\frac{e^{4 x}}{8} \]

Antiderivative was successfully verified.

[In]

Int[E^x*Sinh[3*x],x]

[Out]

1/(4*E^(2*x)) + E^(4*x)/8

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

Mathematica [A]  time = 0.0091856, size = 16, normalized size = 0.84 \[ \frac{1}{8} e^{-2 x} \left (e^{6 x}+2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Sinh[3*x],x]

[Out]

(2 + E^(6*x))/(8*E^(2*x))

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Maple [A]  time = 0.016, size = 26, normalized size = 1.4 \begin{align*} -{\frac{\sinh \left ( 2\,x \right ) }{4}}+{\frac{\sinh \left ( 4\,x \right ) }{8}}+{\frac{\cosh \left ( 2\,x \right ) }{4}}+{\frac{\cosh \left ( 4\,x \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*sinh(3*x),x)

[Out]

-1/4*sinh(2*x)+1/8*sinh(4*x)+1/4*cosh(2*x)+1/8*cosh(4*x)

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Maxima [A]  time = 1.10799, size = 18, normalized size = 0.95 \begin{align*} \frac{1}{8} \, e^{\left (4 \, x\right )} + \frac{1}{4} \, e^{\left (-2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sinh(3*x),x, algorithm="maxima")

[Out]

1/8*e^(4*x) + 1/4*e^(-2*x)

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Fricas [B]  time = 1.96551, size = 128, normalized size = 6.74 \begin{align*} \frac{3 \, \cosh \left (x\right )^{3} - 3 \, \cosh \left (x\right )^{2} \sinh \left (x\right ) + 9 \, \cosh \left (x\right ) \sinh \left (x\right )^{2} - \sinh \left (x\right )^{3}}{8 \,{\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(3*x),x, algorithm="fricas")

[Out]

1/8*(3*cosh(x)^3 - 3*cosh(x)^2*sinh(x) + 9*cosh(x)*sinh(x)^2 - sinh(x)^3)/(cosh(x) - sinh(x))

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Sympy [A]  time = 0.329055, size = 20, normalized size = 1.05 \begin{align*} - \frac{e^{x} \sinh{\left (3 x \right )}}{8} + \frac{3 e^{x} \cosh{\left (3 x \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sinh(3*x),x)

[Out]

-exp(x)*sinh(3*x)/8 + 3*exp(x)*cosh(3*x)/8

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Giac [A]  time = 1.11711, size = 18, normalized size = 0.95 \begin{align*} \frac{1}{8} \, e^{\left (4 \, x\right )} + \frac{1}{4} \, e^{\left (-2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sinh(3*x),x, algorithm="giac")

[Out]

1/8*e^(4*x) + 1/4*e^(-2*x)