3.292 \(\int e^{c (a+b x)} \cosh ^2(a c+b c x)^{5/2} \, dx\)

Optimal. Leaf size=250 \[ -\frac{e^{-4 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{128 b c}-\frac{5 e^{-2 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{64 b c}+\frac{5 e^{2 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{32 b c}+\frac{5 e^{4 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{128 b c}+\frac{e^{6 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{192 b c}+\frac{5}{16} x \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x) \]

[Out]

-(Sqrt[Cosh[a*c + b*c*x]^2]*Sech[a*c + b*c*x])/(128*b*c*E^(4*c*(a + b*x))) - (5*Sqrt[Cosh[a*c + b*c*x]^2]*Sech
[a*c + b*c*x])/(64*b*c*E^(2*c*(a + b*x))) + (5*E^(2*c*(a + b*x))*Sqrt[Cosh[a*c + b*c*x]^2]*Sech[a*c + b*c*x])/
(32*b*c) + (5*E^(4*c*(a + b*x))*Sqrt[Cosh[a*c + b*c*x]^2]*Sech[a*c + b*c*x])/(128*b*c) + (E^(6*c*(a + b*x))*Sq
rt[Cosh[a*c + b*c*x]^2]*Sech[a*c + b*c*x])/(192*b*c) + (5*x*Sqrt[Cosh[a*c + b*c*x]^2]*Sech[a*c + b*c*x])/16

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Rubi [A]  time = 0.228918, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6720, 2282, 12, 266, 43} \[ -\frac{e^{-4 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{128 b c}-\frac{5 e^{-2 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{64 b c}+\frac{5 e^{2 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{32 b c}+\frac{5 e^{4 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{128 b c}+\frac{e^{6 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{192 b c}+\frac{5}{16} x \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x) \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))*(Cosh[a*c + b*c*x]^2)^(5/2),x]

[Out]

-(Sqrt[Cosh[a*c + b*c*x]^2]*Sech[a*c + b*c*x])/(128*b*c*E^(4*c*(a + b*x))) - (5*Sqrt[Cosh[a*c + b*c*x]^2]*Sech
[a*c + b*c*x])/(64*b*c*E^(2*c*(a + b*x))) + (5*E^(2*c*(a + b*x))*Sqrt[Cosh[a*c + b*c*x]^2]*Sech[a*c + b*c*x])/
(32*b*c) + (5*E^(4*c*(a + b*x))*Sqrt[Cosh[a*c + b*c*x]^2]*Sech[a*c + b*c*x])/(128*b*c) + (E^(6*c*(a + b*x))*Sq
rt[Cosh[a*c + b*c*x]^2]*Sech[a*c + b*c*x])/(192*b*c) + (5*x*Sqrt[Cosh[a*c + b*c*x]^2]*Sech[a*c + b*c*x])/16

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{5/2} \, dx &=\left (\sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)\right ) \int e^{c (a+b x)} \cosh ^5(a c+b c x) \, dx\\ &=\frac{\left (\sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)\right ) \operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^5}{32 x^5} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac{\left (\sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)\right ) \operatorname{Subst}\left (\int \frac{\left (1+x^2\right )^5}{x^5} \, dx,x,e^{c (a+b x)}\right )}{32 b c}\\ &=\frac{\left (\sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)\right ) \operatorname{Subst}\left (\int \frac{(1+x)^5}{x^3} \, dx,x,e^{2 c (a+b x)}\right )}{64 b c}\\ &=\frac{\left (\sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)\right ) \operatorname{Subst}\left (\int \left (10+\frac{1}{x^3}+\frac{5}{x^2}+\frac{10}{x}+5 x+x^2\right ) \, dx,x,e^{2 c (a+b x)}\right )}{64 b c}\\ &=-\frac{e^{-4 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{128 b c}-\frac{5 e^{-2 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{64 b c}+\frac{5 e^{2 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{32 b c}+\frac{5 e^{4 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{128 b c}+\frac{e^{6 c (a+b x)} \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)}{192 b c}+\frac{5}{16} x \sqrt{\cosh ^2(a c+b c x)} \text{sech}(a c+b c x)\\ \end{align*}

Mathematica [A]  time = 0.0973858, size = 106, normalized size = 0.42 \[ \frac{\left (-\frac{1}{2} e^{-4 c (a+b x)}-5 e^{-2 c (a+b x)}+10 e^{2 c (a+b x)}+\frac{5}{2} e^{4 c (a+b x)}+\frac{1}{3} e^{6 c (a+b x)}+20 b c x\right ) \cosh ^2(c (a+b x))^{5/2} \text{sech}^5(c (a+b x))}{64 b c} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(c*(a + b*x))*(Cosh[a*c + b*c*x]^2)^(5/2),x]

[Out]

((-1/(2*E^(4*c*(a + b*x))) - 5/E^(2*c*(a + b*x)) + 10*E^(2*c*(a + b*x)) + (5*E^(4*c*(a + b*x)))/2 + E^(6*c*(a
+ b*x))/3 + 20*b*c*x)*(Cosh[c*(a + b*x)]^2)^(5/2)*Sech[c*(a + b*x)]^5)/(64*b*c)

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{c \left ( bx+a \right ) }} \left ( \left ( \cosh \left ( bcx+ac \right ) \right ) ^{2} \right ) ^{{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(b*x+a))*(cosh(b*c*x+a*c)^2)^(5/2),x)

[Out]

int(exp(c*(b*x+a))*(cosh(b*c*x+a*c)^2)^(5/2),x)

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Maxima [A]  time = 1.09616, size = 151, normalized size = 0.6 \begin{align*} \frac{5 \,{\left (b c x + a c\right )}}{16 \, b c} + \frac{e^{\left (6 \, b c x + 6 \, a c\right )}}{192 \, b c} + \frac{5 \, e^{\left (4 \, b c x + 4 \, a c\right )}}{128 \, b c} + \frac{5 \, e^{\left (2 \, b c x + 2 \, a c\right )}}{32 \, b c} - \frac{5 \, e^{\left (-2 \, b c x - 2 \, a c\right )}}{64 \, b c} - \frac{e^{\left (-4 \, b c x - 4 \, a c\right )}}{128 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*(cosh(b*c*x+a*c)^2)^(5/2),x, algorithm="maxima")

[Out]

5/16*(b*c*x + a*c)/(b*c) + 1/192*e^(6*b*c*x + 6*a*c)/(b*c) + 5/128*e^(4*b*c*x + 4*a*c)/(b*c) + 5/32*e^(2*b*c*x
+ 2*a*c)/(b*c) - 5/64*e^(-2*b*c*x - 2*a*c)/(b*c) - 1/128*e^(-4*b*c*x - 4*a*c)/(b*c)

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Fricas [A]  time = 2.19512, size = 562, normalized size = 2.25 \begin{align*} -\frac{\cosh \left (b c x + a c\right )^{5} + 5 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{4} - 5 \, \sinh \left (b c x + a c\right )^{5} - 5 \,{\left (10 \, \cosh \left (b c x + a c\right )^{2} + 9\right )} \sinh \left (b c x + a c\right )^{3} + 15 \, \cosh \left (b c x + a c\right )^{3} + 5 \,{\left (2 \, \cosh \left (b c x + a c\right )^{3} + 9 \, \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{2} - 60 \,{\left (2 \, b c x + 1\right )} \cosh \left (b c x + a c\right ) - 5 \,{\left (5 \, \cosh \left (b c x + a c\right )^{4} - 24 \, b c x + 27 \, \cosh \left (b c x + a c\right )^{2} + 12\right )} \sinh \left (b c x + a c\right )}{384 \,{\left (b c \cosh \left (b c x + a c\right ) - b c \sinh \left (b c x + a c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*(cosh(b*c*x+a*c)^2)^(5/2),x, algorithm="fricas")

[Out]

-1/384*(cosh(b*c*x + a*c)^5 + 5*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^4 - 5*sinh(b*c*x + a*c)^5 - 5*(10*cosh(b*c
*x + a*c)^2 + 9)*sinh(b*c*x + a*c)^3 + 15*cosh(b*c*x + a*c)^3 + 5*(2*cosh(b*c*x + a*c)^3 + 9*cosh(b*c*x + a*c)
)*sinh(b*c*x + a*c)^2 - 60*(2*b*c*x + 1)*cosh(b*c*x + a*c) - 5*(5*cosh(b*c*x + a*c)^4 - 24*b*c*x + 27*cosh(b*c
*x + a*c)^2 + 12)*sinh(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c) - b*c*sinh(b*c*x + a*c))

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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(c*(b*x+a))*(cosh(b*c*x+a*c)**2)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.21151, size = 136, normalized size = 0.54 \begin{align*} \frac{120 \, b c x - 3 \,{\left (30 \, e^{\left (4 \, b c x + 4 \, a c\right )} + 10 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )} e^{\left (-4 \, b c x - 4 \, a c\right )} +{\left (2 \, e^{\left (6 \, b c x + 18 \, a c\right )} + 15 \, e^{\left (4 \, b c x + 16 \, a c\right )} + 60 \, e^{\left (2 \, b c x + 14 \, a c\right )}\right )} e^{\left (-12 \, a c\right )}}{384 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*(cosh(b*c*x+a*c)^2)^(5/2),x, algorithm="giac")

[Out]

1/384*(120*b*c*x - 3*(30*e^(4*b*c*x + 4*a*c) + 10*e^(2*b*c*x + 2*a*c) + 1)*e^(-4*b*c*x - 4*a*c) + (2*e^(6*b*c*
x + 18*a*c) + 15*e^(4*b*c*x + 16*a*c) + 60*e^(2*b*c*x + 14*a*c))*e^(-12*a*c))/(b*c)