3.925 \(\int e^{2 (a+b x)} \cosh ^2(a+b x) \sinh ^3(a+b x) \, dx\)

Optimal. Leaf size=100 \[ \frac{e^{-3 a-3 b x}}{96 b}-\frac{e^{-a-b x}}{32 b}+\frac{e^{a+b x}}{16 b}-\frac{e^{3 a+3 b x}}{48 b}-\frac{e^{5 a+5 b x}}{160 b}+\frac{e^{7 a+7 b x}}{224 b} \]

[Out]

E^(-3*a - 3*b*x)/(96*b) - E^(-a - b*x)/(32*b) + E^(a + b*x)/(16*b) - E^(3*a + 3*b*x)/(48*b) - E^(5*a + 5*b*x)/
(160*b) + E^(7*a + 7*b*x)/(224*b)

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Rubi [A]  time = 0.0778496, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2282, 12, 448} \[ \frac{e^{-3 a-3 b x}}{96 b}-\frac{e^{-a-b x}}{32 b}+\frac{e^{a+b x}}{16 b}-\frac{e^{3 a+3 b x}}{48 b}-\frac{e^{5 a+5 b x}}{160 b}+\frac{e^{7 a+7 b x}}{224 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-3*a - 3*b*x)/(96*b) - E^(-a - b*x)/(32*b) + E^(a + b*x)/(16*b) - E^(3*a + 3*b*x)/(48*b) - E^(5*a + 5*b*x)/
(160*b) + E^(7*a + 7*b*x)/(224*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 ^2(a+b x) \sinh ^3(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^3 \left (1+x^2\right )^2}{32 x^4} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^3 \left (1+x^2\right )^2}{x^4} \, dx,x,e^{a+b x}\right )}{32 b}\\ &=\frac{\operatorname{Subst}\left (\int \left (2-\frac{1}{x^4}+\frac{1}{x^2}-2 x^2-x^4+x^6\right ) \, dx,x,e^{a+b x}\right )}{32 b}\\ &=\frac{e^{-3 a-3 b x}}{96 b}-\frac{e^{-a-b x}}{32 b}+\frac{e^{a+b x}}{16 b}-\frac{e^{3 a+3 b x}}{48 b}-\frac{e^{5 a+5 b x}}{160 b}+\frac{e^{7 a+7 b x}}{224 b}\\ \end{align*}

Mathematica [A]  time = 0.102981, size = 73, normalized size = 0.73 \[ \frac{e^{-3 (a+b x)} \left (-105 e^{2 (a+b x)}+210 e^{4 (a+b x)}-70 e^{6 (a+b x)}-21 e^{8 (a+b x)}+15 e^{10 (a+b x)}+35\right )}{3360 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35 - 105*E^(2*(a + b*x)) + 210*E^(4*(a + b*x)) - 70*E^(6*(a + b*x)) - 21*E^(8*(a + b*x)) + 15*E^(10*(a + b*x)
))/(3360*b*E^(3*(a + b*x)))

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Maple [A]  time = 0.016, size = 108, normalized size = 1.1 \begin{align*}{\frac{3\,\sinh \left ( bx+a \right ) }{32\,b}}-{\frac{\sinh \left ( 3\,bx+3\,a \right ) }{32\,b}}-{\frac{\sinh \left ( 5\,bx+5\,a \right ) }{160\,b}}+{\frac{\sinh \left ( 7\,bx+7\,a \right ) }{224\,b}}+{\frac{\cosh \left ( bx+a \right ) }{32\,b}}-{\frac{\cosh \left ( 3\,bx+3\,a \right ) }{96\,b}}-{\frac{\cosh \left ( 5\,bx+5\,a \right ) }{160\,b}}+{\frac{\cosh \left ( 7\,bx+7\,a \right ) }{224\,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)^2*sinh(b*x+a)^3,x)

[Out]

3/32*sinh(b*x+a)/b-1/32/b*sinh(3*b*x+3*a)-1/160/b*sinh(5*b*x+5*a)+1/224/b*sinh(7*b*x+7*a)+1/32*cosh(b*x+a)/b-1
/96*cosh(3*b*x+3*a)/b-1/160*cosh(5*b*x+5*a)/b+1/224*cosh(7*b*x+7*a)/b

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Maxima [A]  time = 1.00432, size = 105, normalized size = 1.05 \begin{align*} -\frac{{\left (21 \, e^{\left (-2 \, b x - 2 \, a\right )} + 70 \, e^{\left (-4 \, b x - 4 \, a\right )} - 210 \, e^{\left (-6 \, b x - 6 \, a\right )} - 15\right )} e^{\left (7 \, b x + 7 \, a\right )}}{3360 \, b} - \frac{3 \, e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}}{96 \, 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)^2*sinh(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/3360*(21*e^(-2*b*x - 2*a) + 70*e^(-4*b*x - 4*a) - 210*e^(-6*b*x - 6*a) - 15)*e^(7*b*x + 7*a)/b - 1/96*(3*e^
(-b*x - a) - e^(-3*b*x - 3*a))/b

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Fricas [B]  time = 1.76445, size = 501, normalized size = 5.01 \begin{align*} \frac{25 \, \cosh \left (b x + a\right )^{5} + 125 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{4} - 10 \, \sinh \left (b x + a\right )^{5} - 2 \,{\left (50 \, \cosh \left (b x + a\right )^{2} - 21\right )} \sinh \left (b x + a\right )^{3} - 63 \, \cosh \left (b x + a\right )^{3} +{\left (250 \, \cosh \left (b x + a\right )^{3} - 189 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 2 \,{\left (25 \, \cosh \left (b x + a\right )^{4} - 63 \, \cosh \left (b x + a\right )^{2} + 70\right )} \sinh \left (b x + a\right ) + 70 \, \cosh \left (b x + a\right )}{1680 \,{\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)^2*sinh(b*x+a)^3,x, algorithm="fricas")

[Out]

1/1680*(25*cosh(b*x + a)^5 + 125*cosh(b*x + a)*sinh(b*x + a)^4 - 10*sinh(b*x + a)^5 - 2*(50*cosh(b*x + a)^2 -
21)*sinh(b*x + a)^3 - 63*cosh(b*x + a)^3 + (250*cosh(b*x + a)^3 - 189*cosh(b*x + a))*sinh(b*x + a)^2 - 2*(25*c
osh(b*x + a)^4 - 63*cosh(b*x + a)^2 + 70)*sinh(b*x + a) + 70*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 [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*cosh(b*x+a)**2*sinh(b*x+a)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.21009, size = 108, normalized size = 1.08 \begin{align*} -\frac{35 \,{\left (3 \, e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-3 \, b x - 3 \, a\right )} -{\left (15 \, e^{\left (7 \, b x + 28 \, a\right )} - 21 \, e^{\left (5 \, b x + 26 \, a\right )} - 70 \, e^{\left (3 \, b x + 24 \, a\right )} + 210 \, e^{\left (b x + 22 \, a\right )}\right )} e^{\left (-21 \, a\right )}}{3360 \, 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)^2*sinh(b*x+a)^3,x, algorithm="giac")

[Out]

-1/3360*(35*(3*e^(2*b*x + 2*a) - 1)*e^(-3*b*x - 3*a) - (15*e^(7*b*x + 28*a) - 21*e^(5*b*x + 26*a) - 70*e^(3*b*
x + 24*a) + 210*e^(b*x + 22*a))*e^(-21*a))/b