### 3.872 $$\int e^{2 (a+b x)} \sinh ^3(a+b x) \, dx$$

Optimal. Leaf size=66 $\frac{e^{-a-b x}}{8 b}+\frac{3 e^{a+b x}}{8 b}-\frac{e^{3 a+3 b x}}{8 b}+\frac{e^{5 a+5 b x}}{40 b}$

[Out]

E^(-a - b*x)/(8*b) + (3*E^(a + b*x))/(8*b) - E^(3*a + 3*b*x)/(8*b) + E^(5*a + 5*b*x)/(40*b)

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Rubi [A]  time = 0.0360092, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {2282, 12, 270} $\frac{e^{-a-b x}}{8 b}+\frac{3 e^{a+b x}}{8 b}-\frac{e^{3 a+3 b x}}{8 b}+\frac{e^{5 a+5 b x}}{40 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^(2*(a + b*x))*Sinh[a + b*x]^3,x]

[Out]

E^(-a - b*x)/(8*b) + (3*E^(a + b*x))/(8*b) - E^(3*a + 3*b*x)/(8*b) + E^(5*a + 5*b*x)/(40*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^{2 (a+b x)} \sinh ^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^3}{8 x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^3}{x^2} \, dx,x,e^{a+b x}\right )}{8 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (3-\frac{1}{x^2}-3 x^2+x^4\right ) \, dx,x,e^{a+b x}\right )}{8 b}\\ &=\frac{e^{-a-b x}}{8 b}+\frac{3 e^{a+b x}}{8 b}-\frac{e^{3 a+3 b x}}{8 b}+\frac{e^{5 a+5 b x}}{40 b}\\ \end{align*}

Mathematica [A]  time = 0.0342435, size = 50, normalized size = 0.76 $\frac{e^{-a-b x} \left (15 e^{2 (a+b x)}-5 e^{4 (a+b x)}+e^{6 (a+b x)}+5\right )}{40 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^(2*(a + b*x))*Sinh[a + b*x]^3,x]

[Out]

(E^(-a - b*x)*(5 + 15*E^(2*(a + b*x)) - 5*E^(4*(a + b*x)) + E^(6*(a + b*x))))/(40*b)

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Maple [A]  time = 0.022, size = 80, normalized size = 1.2 \begin{align*}{\frac{\sinh \left ( bx+a \right ) }{4\,b}}-{\frac{\sinh \left ( 3\,bx+3\,a \right ) }{8\,b}}+{\frac{\sinh \left ( 5\,bx+5\,a \right ) }{40\,b}}+{\frac{\cosh \left ( bx+a \right ) }{2\,b}}-{\frac{\cosh \left ( 3\,bx+3\,a \right ) }{8\,b}}+{\frac{\cosh \left ( 5\,bx+5\,a \right ) }{40\,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)^3,x)

[Out]

1/4*sinh(b*x+a)/b-1/8/b*sinh(3*b*x+3*a)+1/40/b*sinh(5*b*x+5*a)+1/2*cosh(b*x+a)/b-1/8*cosh(3*b*x+3*a)/b+1/40*co
sh(5*b*x+5*a)/b

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Maxima [A]  time = 1.23995, size = 72, normalized size = 1.09 \begin{align*} -\frac{{\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} - 15 \, e^{\left (-4 \, b x - 4 \, a\right )} - 1\right )} e^{\left (5 \, b x + 5 \, a\right )}}{40 \, b} + \frac{e^{\left (-b x - a\right )}}{8 \, 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)^3,x, algorithm="maxima")

[Out]

-1/40*(5*e^(-2*b*x - 2*a) - 15*e^(-4*b*x - 4*a) - 1)*e^(5*b*x + 5*a)/b + 1/8*e^(-b*x - a)/b

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Fricas [A]  time = 1.75249, size = 289, normalized size = 4.38 \begin{align*} \frac{3 \, \cosh \left (b x + a\right )^{3} + 9 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - 2 \, \sinh \left (b x + a\right )^{3} - 2 \,{\left (3 \, \cosh \left (b x + a\right )^{2} + 5\right )} \sinh \left (b x + a\right ) + 5 \, \cosh \left (b x + a\right )}{20 \,{\left (b \cosh \left (b x + a\right )^{2} - 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2}\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)^3,x, algorithm="fricas")

[Out]

1/20*(3*cosh(b*x + a)^3 + 9*cosh(b*x + a)*sinh(b*x + a)^2 - 2*sinh(b*x + a)^3 - 2*(3*cosh(b*x + a)^2 + 5)*sinh
(b*x + a) + 5*cosh(b*x + a))/(b*cosh(b*x + a)^2 - 2*b*cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2)

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Sympy [A]  time = 38.2161, size = 124, normalized size = 1.88 \begin{align*} \begin{cases} \frac{2 e^{2 a} e^{2 b x} \sinh ^{3}{\left (a + b x \right )}}{5 b} + \frac{e^{2 a} e^{2 b x} \sinh ^{2}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{5 b} - \frac{4 e^{2 a} e^{2 b x} \sinh{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{5 b} + \frac{2 e^{2 a} e^{2 b x} \cosh ^{3}{\left (a + b x \right )}}{5 b} & \text{for}\: b \neq 0 \\x e^{2 a} \sinh ^{3}{\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)**3,x)

[Out]

Piecewise((2*exp(2*a)*exp(2*b*x)*sinh(a + b*x)**3/(5*b) + exp(2*a)*exp(2*b*x)*sinh(a + b*x)**2*cosh(a + b*x)/(
5*b) - 4*exp(2*a)*exp(2*b*x)*sinh(a + b*x)*cosh(a + b*x)**2/(5*b) + 2*exp(2*a)*exp(2*b*x)*cosh(a + b*x)**3/(5*
b), Ne(b, 0)), (x*exp(2*a)*sinh(a)**3, True))

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Giac [A]  time = 1.13422, size = 72, normalized size = 1.09 \begin{align*} \frac{{\left (e^{\left (5 \, b x + 10 \, a\right )} - 5 \, e^{\left (3 \, b x + 8 \, a\right )} + 15 \, e^{\left (b x + 6 \, a\right )}\right )} e^{\left (-5 \, a\right )} + 5 \, e^{\left (-b x - a\right )}}{40 \, 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)^3,x, algorithm="giac")

[Out]

1/40*((e^(5*b*x + 10*a) - 5*e^(3*b*x + 8*a) + 15*e^(b*x + 6*a))*e^(-5*a) + 5*e^(-b*x - a))/b