3.310 \(\int e^x \sinh ^2(2 x) \, dx\)

Optimal. Leaf size=26 \[ -\frac{1}{12} e^{-3 x}-\frac{e^x}{2}+\frac{e^{5 x}}{20} \]

[Out]

-1/(12*E^(3*x)) - E^x/2 + E^(5*x)/20

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Rubi [A]  time = 0.0197871, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {2282, 12, 270} \[ -\frac{1}{12} e^{-3 x}-\frac{e^x}{2}+\frac{e^{5 x}}{20} \]

Antiderivative was successfully verified.

[In]

Int[E^x*Sinh[2*x]^2,x]

[Out]

-1/(12*E^(3*x)) - E^x/2 + E^(5*x)/20

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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int e^x \sinh ^2(2 x) \, dx &=\operatorname{Subst}\left (\int \frac{\left (1-x^4\right )^2}{4 x^4} \, dx,x,e^x\right )\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{\left (1-x^4\right )^2}{x^4} \, dx,x,e^x\right )\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \left (-2+\frac{1}{x^4}+x^4\right ) \, dx,x,e^x\right )\\ &=-\frac{1}{12} e^{-3 x}-\frac{e^x}{2}+\frac{e^{5 x}}{20}\\ \end{align*}

Mathematica [A]  time = 0.0128874, size = 26, normalized size = 1. \[ -\frac{1}{12} e^{-3 x}-\frac{e^x}{2}+\frac{e^{5 x}}{20} \]

Antiderivative was successfully verified.

[In]

Integrate[E^x*Sinh[2*x]^2,x]

[Out]

-1/(12*E^(3*x)) - E^x/2 + E^(5*x)/20

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Maple [A]  time = 0.019, size = 34, normalized size = 1.3 \begin{align*} -{\frac{\sinh \left ( x \right ) }{2}}+{\frac{\sinh \left ( 3\,x \right ) }{12}}+{\frac{\sinh \left ( 5\,x \right ) }{20}}-{\frac{\cosh \left ( x \right ) }{2}}-{\frac{\cosh \left ( 3\,x \right ) }{12}}+{\frac{\cosh \left ( 5\,x \right ) }{20}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*sinh(2*x)^2,x)

[Out]

-1/2*sinh(x)+1/12*sinh(3*x)+1/20*sinh(5*x)-1/2*cosh(x)-1/12*cosh(3*x)+1/20*cosh(5*x)

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Maxima [A]  time = 1.10359, size = 23, normalized size = 0.88 \begin{align*} \frac{1}{20} \, e^{\left (5 \, x\right )} - \frac{1}{12} \, e^{\left (-3 \, x\right )} - \frac{1}{2} \, e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sinh(2*x)^2,x, algorithm="maxima")

[Out]

1/20*e^(5*x) - 1/12*e^(-3*x) - 1/2*e^x

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Fricas [B]  time = 2.01963, size = 170, normalized size = 6.54 \begin{align*} -\frac{\cosh \left (x\right )^{4} - 16 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} - 16 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 15}{30 \,{\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(2*x)^2,x, algorithm="fricas")

[Out]

-1/30*(cosh(x)^4 - 16*cosh(x)^3*sinh(x) + 6*cosh(x)^2*sinh(x)^2 - 16*cosh(x)*sinh(x)^3 + sinh(x)^4 + 15)/(cosh
(x) - sinh(x))

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Sympy [B]  time = 1.0532, size = 42, normalized size = 1.62 \begin{align*} \frac{7 e^{x} \sinh ^{2}{\left (2 x \right )}}{15} + \frac{4 e^{x} \sinh{\left (2 x \right )} \cosh{\left (2 x \right )}}{15} - \frac{8 e^{x} \cosh ^{2}{\left (2 x \right )}}{15} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sinh(2*x)**2,x)

[Out]

7*exp(x)*sinh(2*x)**2/15 + 4*exp(x)*sinh(2*x)*cosh(2*x)/15 - 8*exp(x)*cosh(2*x)**2/15

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Giac [A]  time = 1.13813, size = 23, normalized size = 0.88 \begin{align*} \frac{1}{20} \, e^{\left (5 \, x\right )} - \frac{1}{12} \, e^{\left (-3 \, x\right )} - \frac{1}{2} \, e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*sinh(2*x)^2,x, algorithm="giac")

[Out]

1/20*e^(5*x) - 1/12*e^(-3*x) - 1/2*e^x