3.874 \(\int e^{2 (a+b x)} \sinh (a+b x) \, dx\)

Optimal. Leaf size=32 \[ \frac{e^{3 a+3 b x}}{6 b}-\frac{e^{a+b x}}{2 b} \]

[Out]

-E^(a + b*x)/(2*b) + E^(3*a + 3*b*x)/(6*b)

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Rubi [A]  time = 0.0173545, antiderivative size = 32, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {2282, 12} \[ \frac{e^{3 a+3 b x}}{6 b}-\frac{e^{a+b x}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[E^(2*(a + b*x))*Sinh[a + b*x],x]

[Out]

-E^(a + b*x)/(2*b) + E^(3*a + 3*b*x)/(6*b)

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]]

Rubi steps

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

Mathematica [A]  time = 0.0127617, size = 25, normalized size = 0.78 \[ \frac{e^{a+b x} \left (e^{2 (a+b x)}-3\right )}{6 b} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(2*(a + b*x))*Sinh[a + b*x],x]

[Out]

(E^(a + b*x)*(-3 + E^(2*(a + b*x))))/(6*b)

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Maple [A]  time = 0.005, size = 52, normalized size = 1.6 \begin{align*} -{\frac{\sinh \left ( bx+a \right ) }{2\,b}}+{\frac{\sinh \left ( 3\,bx+3\,a \right ) }{6\,b}}-{\frac{\cosh \left ( bx+a \right ) }{2\,b}}+{\frac{\cosh \left ( 3\,bx+3\,a \right ) }{6\,b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*b*x+2*a)*sinh(b*x+a),x)

[Out]

-1/2*sinh(b*x+a)/b+1/6/b*sinh(3*b*x+3*a)-1/2*cosh(b*x+a)/b+1/6*cosh(3*b*x+3*a)/b

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Maxima [A]  time = 1.28741, size = 35, normalized size = 1.09 \begin{align*} \frac{e^{\left (3 \, b x + 3 \, a\right )}}{6 \, b} - \frac{e^{\left (b x + a\right )}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*sinh(b*x+a),x, algorithm="maxima")

[Out]

1/6*e^(3*b*x + 3*a)/b - 1/2*e^(b*x + a)/b

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Fricas [B]  time = 1.69578, size = 153, normalized size = 4.78 \begin{align*} \frac{\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 3}{6 \,{\left (b \cosh \left (b x + a\right ) - b \sinh \left (b x + a\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*sinh(b*x+a),x, algorithm="fricas")

[Out]

1/6*(cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 3)/(b*cosh(b*x + a) - b*sinh(b*x + a)
)

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Sympy [A]  time = 2.51426, size = 54, normalized size = 1.69 \begin{align*} \begin{cases} \frac{2 e^{2 a} e^{2 b x} \sinh{\left (a + b x \right )}}{3 b} - \frac{e^{2 a} e^{2 b x} \cosh{\left (a + b x \right )}}{3 b} & \text{for}\: b \neq 0 \\x e^{2 a} \sinh{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*sinh(b*x+a),x)

[Out]

Piecewise((2*exp(2*a)*exp(2*b*x)*sinh(a + b*x)/(3*b) - exp(2*a)*exp(2*b*x)*cosh(a + b*x)/(3*b), Ne(b, 0)), (x*
exp(2*a)*sinh(a), True))

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Giac [A]  time = 1.13691, size = 31, normalized size = 0.97 \begin{align*} \frac{e^{\left (3 \, b x + 3 \, a\right )} - 3 \, e^{\left (b x + a\right )}}{6 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*sinh(b*x+a),x, algorithm="giac")

[Out]

1/6*(e^(3*b*x + 3*a) - 3*e^(b*x + a))/b