### 3.920 $$\int e^{2 (a+b x)} \cosh (a+b x) \sinh ^2(a+b x) \, dx$$

Optimal. Leaf size=66 $-\frac{e^{-a-b x}}{8 b}-\frac{e^{a+b x}}{8 b}-\frac{e^{3 a+3 b x}}{24 b}+\frac{e^{5 a+5 b x}}{40 b}$

[Out]

-E^(-a - b*x)/(8*b) - E^(a + b*x)/(8*b) - E^(3*a + 3*b*x)/(24*b) + E^(5*a + 5*b*x)/(40*b)

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Rubi [A]  time = 0.049802, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {2282, 12, 448} $-\frac{e^{-a-b x}}{8 b}-\frac{e^{a+b x}}{8 b}-\frac{e^{3 a+3 b x}}{24 b}+\frac{e^{5 a+5 b x}}{40 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-E^(-a - b*x)/(8*b) - E^(a + b*x)/(8*b) - E^(3*a + 3*b*x)/(24*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 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

\begin{align*} \int e^{2 (a+b x)} \cosh (a+b x) \sinh ^2(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2 \left (1+x^2\right )}{8 x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2 \left (1+x^2\right )}{x^2} \, dx,x,e^{a+b x}\right )}{8 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (-1+\frac{1}{x^2}-x^2+x^4\right ) \, dx,x,e^{a+b x}\right )}{8 b}\\ &=-\frac{e^{-a-b x}}{8 b}-\frac{e^{a+b x}}{8 b}-\frac{e^{3 a+3 b x}}{24 b}+\frac{e^{5 a+5 b x}}{40 b}\\ \end{align*}

Mathematica [A]  time = 0.0802082, size = 51, normalized size = 0.77 $\frac{3 e^{-a-b x} \left (e^{6 (a+b x)}-5\right )-5 e^{a+b x} \left (e^{2 (a+b x)}+3\right )}{120 b}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(-5*E^(a + b*x)*(3 + E^(2*(a + b*x))) + 3*E^(-a - b*x)*(-5 + E^(6*(a + b*x))))/(120*b)

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Maple [A]  time = 0.01, size = 69, normalized size = 1.1 \begin{align*} -{\frac{\sinh \left ( 3\,bx+3\,a \right ) }{24\,b}}+{\frac{\sinh \left ( 5\,bx+5\,a \right ) }{40\,b}}-{\frac{\cosh \left ( bx+a \right ) }{4\,b}}-{\frac{\cosh \left ( 3\,bx+3\,a \right ) }{24\,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)*cosh(b*x+a)*sinh(b*x+a)^2,x)

[Out]

-1/24/b*sinh(3*b*x+3*a)+1/40/b*sinh(5*b*x+5*a)-1/4*cosh(b*x+a)/b-1/24*cosh(3*b*x+3*a)/b+1/40*cosh(5*b*x+5*a)/b

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Maxima [A]  time = 1.02501, 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 )} - 3\right )} e^{\left (5 \, b x + 5 \, a\right )}}{120 \, 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)*cosh(b*x+a)*sinh(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/120*(5*e^(-2*b*x - 2*a) + 15*e^(-4*b*x - 4*a) - 3)*e^(5*b*x + 5*a)/b - 1/8*e^(-b*x - a)/b

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Fricas [A]  time = 1.87815, size = 292, normalized size = 4.42 \begin{align*} -\frac{6 \, \cosh \left (b x + a\right )^{3} + 18 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - 9 \, \sinh \left (b x + a\right )^{3} -{\left (27 \, \cosh \left (b x + a\right )^{2} + 5\right )} \sinh \left (b x + a\right ) + 10 \, \cosh \left (b x + a\right )}{60 \,{\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)*cosh(b*x+a)*sinh(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/60*(6*cosh(b*x + a)^3 + 18*cosh(b*x + a)*sinh(b*x + a)^2 - 9*sinh(b*x + a)^3 - (27*cosh(b*x + a)^2 + 5)*sin
h(b*x + a) + 10*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 = 30.2939, size = 128, normalized size = 1.94 \begin{align*} \begin{cases} \frac{e^{2 a} e^{2 b x} \sinh ^{3}{\left (a + b x \right )}}{15 b} - \frac{2 e^{2 a} e^{2 b x} \sinh ^{2}{\left (a + b x \right )} \cosh{\left (a + b x \right )}}{15 b} + \frac{8 e^{2 a} e^{2 b x} \sinh{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{15 b} - \frac{4 e^{2 a} e^{2 b x} \cosh ^{3}{\left (a + b x \right )}}{15 b} & \text{for}\: b \neq 0 \\x e^{2 a} \sinh ^{2}{\left (a \right )} \cosh{\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)*cosh(b*x+a)*sinh(b*x+a)**2,x)

[Out]

Piecewise((exp(2*a)*exp(2*b*x)*sinh(a + b*x)**3/(15*b) - 2*exp(2*a)*exp(2*b*x)*sinh(a + b*x)**2*cosh(a + b*x)/
(15*b) + 8*exp(2*a)*exp(2*b*x)*sinh(a + b*x)*cosh(a + b*x)**2/(15*b) - 4*exp(2*a)*exp(2*b*x)*cosh(a + b*x)**3/
(15*b), Ne(b, 0)), (x*exp(2*a)*sinh(a)**2*cosh(a), True))

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Giac [A]  time = 1.12355, size = 74, normalized size = 1.12 \begin{align*} \frac{{\left (3 \, 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 )} - 15 \, e^{\left (-b x - a\right )}}{120 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*((3*e^(5*b*x + 10*a) - 5*e^(3*b*x + 8*a) - 15*e^(b*x + 6*a))*e^(-5*a) - 15*e^(-b*x - a))/b