\(\int e^x \sinh ^2(2 x) \, dx\) [310]

Optimal result

Integrand size = 10, antiderivative size = 26 \[ \int e^x \sinh ^2(2 x) \, dx=-\frac {1}{12} e^{-3 x}-\frac {e^x}{2}+\frac {e^{5 x}}{20} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {2320, 12, 276} \[ \int e^x \sinh ^2(2 x) \, dx=-\frac {1}{12} e^{-3 x}-\frac {e^x}{2}+\frac {e^{5 x}}{20} \]

[In]

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

[Out]

-1/12*1/E^(3*x) - E^x/2 + E^(5*x)/20

Rule 12

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

Rule 276

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]

Rule 2320

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]]]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {\left (1-x^4\right )^2}{4 x^4} \, dx,x,e^x\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \frac {\left (1-x^4\right )^2}{x^4} \, dx,x,e^x\right ) \\ & = \frac {1}{4} \text {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] (verified)

Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int e^x \sinh ^2(2 x) \, dx=-\frac {1}{12} e^{-3 x}-\frac {e^x}{2}+\frac {e^{5 x}}{20} \]

[In]

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

[Out]

-1/12*1/E^(3*x) - E^x/2 + E^(5*x)/20

Maple [A] (verified)

Time = 1.59 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65

method	result	size
parallelrisch	\(-\frac {{\mathrm e}^{x} \left (15+\cosh \left (4 x \right )-4 \sinh \left (4 x \right )\right )}{30}\)	\(17\)
risch	\(\frac {{\mathrm e}^{5 x}}{20}-\frac {{\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{-3 x}}{12}\)	\(18\)
default	\(-\frac {\sinh \left (x \right )}{2}+\frac {\sinh \left (3 x \right )}{12}+\frac {\sinh \left (5 x \right )}{20}-\frac {\cosh \left (x \right )}{2}-\frac {\cosh \left (3 x \right )}{12}+\frac {\cosh \left (5 x \right )}{20}\)	\(34\)

[In]

int(exp(x)*sinh(2*x)^2,x,method=_RETURNVERBOSE)

[Out]

-1/30*exp(x)*(15+cosh(4*x)-4*sinh(4*x))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (17) = 34\).

Time = 0.28 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.81 \[ \int e^x \sinh ^2(2 x) \, dx=-\frac {\cosh \left (x\right )^{4} - 16 \, \cosh \left (x\right )^{3} \sinh \left (x\right ) + 6 \, \cosh \left (x\right )^{2} \sinh \left (x\right )^{2} - 16 \, \cosh \left (x\right ) \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} + 15}{30 \, {\left (\cosh \left (x\right ) - \sinh \left (x\right )\right )}} \]

[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))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (19) = 38\).

Time = 0.16 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.62 \[ \int e^x \sinh ^2(2 x) \, dx=\frac {7 e^{x} \sinh ^{2}{\left (2 x \right )}}{15} + \frac {4 e^{x} \sinh {\left (2 x \right )} \cosh {\left (2 x \right )}}{15} - \frac {8 e^{x} \cosh ^{2}{\left (2 x \right )}}{15} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int e^x \sinh ^2(2 x) \, dx=\frac {1}{20} \, e^{\left (5 \, x\right )} - \frac {1}{12} \, e^{\left (-3 \, x\right )} - \frac {1}{2} \, e^{x} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int e^x \sinh ^2(2 x) \, dx=\frac {1}{20} \, e^{\left (5 \, x\right )} - \frac {1}{12} \, e^{\left (-3 \, x\right )} - \frac {1}{2} \, e^{x} \]

[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

Mupad [B] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int e^x \sinh ^2(2 x) \, dx=\frac {{\mathrm {e}}^{5\,x}}{20}-\frac {{\mathrm {e}}^{-3\,x}}{12}-\frac {{\mathrm {e}}^x}{2} \]

[In]

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

[Out]

exp(5*x)/20 - exp(-3*x)/12 - exp(x)/2