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

Optimal. Leaf size=49 \[ -\frac{e^{-a-b x}}{4 b}-\frac{e^{a+b x}}{2 b}+\frac{e^{3 a+3 b x}}{12 b} \]

[Out]

-E^(-a - b*x)/(4*b) - E^(a + b*x)/(2*b) + E^(3*a + 3*b*x)/(12*b)

________________________________________________________________________________________

Rubi [A]  time = 0.0312144, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {2282, 12, 270} \[ -\frac{e^{-a-b x}}{4 b}-\frac{e^{a+b x}}{2 b}+\frac{e^{3 a+3 b x}}{12 b} \]

Antiderivative was successfully verified.

[In]

Int[E^(a + b*x)*Sinh[a + b*x]^2,x]

[Out]

-E^(-a - b*x)/(4*b) - E^(a + b*x)/(2*b) + E^(3*a + 3*b*x)/(12*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]]

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

Mathematica [A]  time = 0.0205688, size = 39, normalized size = 0.8 \[ \frac{e^{-a-b x} \left (-6 e^{2 (a+b x)}+e^{4 (a+b x)}-3\right )}{12 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(-a - b*x)*(-3 - 6*E^(2*(a + b*x)) + E^(4*(a + b*x))))/(12*b)

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 49, normalized size = 1. \begin{align*}{\frac{1}{b} \left ( \left ( -{\frac{2}{3}}+{\frac{ \left ( \sinh \left ( bx+a \right ) \right ) ^{2}}{3}} \right ) \cosh \left ( bx+a \right ) +{\frac{\sinh \left ( bx+a \right ) \left ( \cosh \left ( bx+a \right ) \right ) ^{2}}{3}}-{\frac{\sinh \left ( bx+a \right ) }{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.05769, size = 54, normalized size = 1.1 \begin{align*} \frac{e^{\left (3 \, b x + 3 \, a\right )}}{12 \, b} - \frac{e^{\left (b x + a\right )}}{2 \, b} - \frac{e^{\left (-b x - a\right )}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*e^(3*b*x + 3*a)/b - 1/2*e^(b*x + a)/b - 1/4*e^(-b*x - a)/b

________________________________________________________________________________________

Fricas [A]  time = 1.99496, size = 154, normalized size = 3.14 \begin{align*} -\frac{\cosh \left (b x + a\right )^{2} - 4 \, \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(b*x+a)*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/6*(cosh(b*x + a)^2 - 4*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 + 3)/(b*cosh(b*x + a) - b*sinh(b*x + a
))

________________________________________________________________________________________

Sympy [A]  time = 7.02315, size = 78, normalized size = 1.59 \begin{align*} \begin{cases} \frac{e^{a} e^{b x} \sinh ^{2}{\left (a + b x \right )}}{3 b} + \frac{2 e^{a} e^{b x} \sinh{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{3 b} - \frac{2 e^{a} e^{b x} \cosh ^{2}{\left (a + b x \right )}}{3 b} & \text{for}\: b \neq 0 \\x e^{a} \sinh ^{2}{\left (a \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.17573, size = 46, normalized size = 0.94 \begin{align*} \frac{e^{\left (3 \, b x + 3 \, a\right )} - 6 \, e^{\left (b x + a\right )} - 3 \, e^{\left (-b x - a\right )}}{12 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(e^(3*b*x + 3*a) - 6*e^(b*x + a) - 3*e^(-b*x - a))/b