\(\int e^{c (a+b x)} \cosh ^2(a c+b c x)^{5/2} \, dx\) [293]

Optimal result

Integrand size = 25, antiderivative size = 250 \[ \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{5/2} \, dx=-\frac {e^{-4 c (a+b x)} \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)}{128 b c}-\frac {5 e^{-2 c (a+b x)} \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)}{64 b c}+\frac {5 e^{2 c (a+b x)} \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)}{32 b c}+\frac {5 e^{4 c (a+b x)} \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)}{128 b c}+\frac {e^{6 c (a+b x)} \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)}{192 b c}+\frac {5}{16} x \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x) \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.44 \[ \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{5/2} \, dx=\frac {\left (-\frac {1}{128} e^{-4 c (a+b x)}-\frac {5}{64} e^{-2 c (a+b x)}+\frac {5}{32} e^{2 c (a+b x)}+\frac {5}{128} e^{4 c (a+b x)}+\frac {1}{192} e^{6 c (a+b x)}+\frac {5 b c x}{16}\right ) \cosh ^2(c (a+b x))^{5/2} \text {sech}^5(c (a+b x))}{b c} \] Input:

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

Output:

((-1/128*1/E^(4*c*(a + b*x)) - 5/(64*E^(2*c*(a + b*x))) + (5*E^(2*c*(a + b 
*x)))/32 + (5*E^(4*c*(a + b*x)))/128 + E^(6*c*(a + b*x))/192 + (5*b*c*x)/1 
6)*(Cosh[c*(a + b*x)]^2)^(5/2)*Sech[c*(a + b*x)]^5)/(b*c)
 

Rubi [A] (warning: unable to verify)

Time = 0.47 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.41, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {7271, 2720, 27, 243, 49, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 7271

\(\displaystyle \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x) \int e^{c (a+b x)} \cosh ^5(a c+b x c)dx\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x) \int \frac {1}{32} e^{-5 c (a+b x)} \left (1+e^{2 c (a+b x)}\right )^5de^{c (a+b x)}}{b c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x) \int e^{-5 c (a+b x)} \left (1+e^{2 c (a+b x)}\right )^5de^{c (a+b x)}}{32 b c}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x) \int e^{-3 c (a+b x)} \left (1+e^{2 c (a+b x)}\right )^5de^{2 c (a+b x)}}{64 b c}\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {\sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x) \int \left (10+e^{-3 c (a+b x)}+5 e^{-2 c (a+b x)}+10 e^{-c (a+b x)}+6 e^{2 c (a+b x)}\right )de^{2 c (a+b x)}}{64 b c}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (-\frac {1}{2} e^{-2 c (a+b x)}-5 e^{-c (a+b x)}+\frac {25}{2} e^{2 c (a+b x)}+\frac {1}{3} e^{3 c (a+b x)}+10 \log \left (e^{2 c (a+b x)}\right )\right ) \sqrt {\cosh ^2(a c+b c x)} \text {sech}(a c+b c x)}{64 b c}\)

Input:

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

Output:

(Sqrt[Cosh[a*c + b*c*x]^2]*(-1/2*1/E^(2*c*(a + b*x)) - 5/E^(c*(a + b*x)) + 
 (25*E^(2*c*(a + b*x)))/2 + E^(3*c*(a + b*x))/3 + 10*Log[E^(2*c*(a + b*x)) 
])*Sech[a*c + b*c*x])/(64*b*c)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 49

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}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 7271

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[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]) &&  !(Eq 
Q[v, x] && EqQ[m, 1])
 

Maple [A] (verified)

Time = 8.35 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.30

method	result	size
risch	\(\frac {5 x \sqrt {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{c \left (b x +a \right )}}{16 \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )}+\frac {\sqrt {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{7 c \left (b x +a \right )}}{192 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )}+\frac {5 \sqrt {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{5 c \left (b x +a \right )}}{128 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )}+\frac {5 \sqrt {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{3 c \left (b x +a \right )}}{32 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )}-\frac {5 \sqrt {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{-c \left (b x +a \right )}}{64 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )}-\frac {\sqrt {\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2} {\mathrm e}^{-2 c \left (b x +a \right )}}\, {\mathrm e}^{-3 c \left (b x +a \right )}}{128 b c \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )}\)	\(326\)

Input:

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

Output:

5/16*x*((1+exp(2*c*(b*x+a)))^2*exp(-2*c*(b*x+a)))^(1/2)/(1+exp(2*c*(b*x+a) 
))*exp(c*(b*x+a))+1/192/b/c*((1+exp(2*c*(b*x+a)))^2*exp(-2*c*(b*x+a)))^(1/ 
2)/(1+exp(2*c*(b*x+a)))*exp(7*c*(b*x+a))+5/128/b/c*((1+exp(2*c*(b*x+a)))^2 
*exp(-2*c*(b*x+a)))^(1/2)/(1+exp(2*c*(b*x+a)))*exp(5*c*(b*x+a))+5/32/b/c*( 
(1+exp(2*c*(b*x+a)))^2*exp(-2*c*(b*x+a)))^(1/2)/(1+exp(2*c*(b*x+a)))*exp(3 
*c*(b*x+a))-5/64/b/c*((1+exp(2*c*(b*x+a)))^2*exp(-2*c*(b*x+a)))^(1/2)/(1+e 
xp(2*c*(b*x+a)))*exp(-c*(b*x+a))-1/128/b/c*((1+exp(2*c*(b*x+a)))^2*exp(-2* 
c*(b*x+a)))^(1/2)/(1+exp(2*c*(b*x+a)))*exp(-3*c*(b*x+a))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.87 \[ \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{5/2} \, dx=-\frac {\cosh \left (b c x + a c\right )^{5} + 5 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{4} - 5 \, \sinh \left (b c x + a c\right )^{5} - 5 \, {\left (10 \, \cosh \left (b c x + a c\right )^{2} + 9\right )} \sinh \left (b c x + a c\right )^{3} + 15 \, \cosh \left (b c x + a c\right )^{3} + 5 \, {\left (2 \, \cosh \left (b c x + a c\right )^{3} + 9 \, \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{2} - 60 \, {\left (2 \, b c x + 1\right )} \cosh \left (b c x + a c\right ) - 5 \, {\left (5 \, \cosh \left (b c x + a c\right )^{4} - 24 \, b c x + 27 \, \cosh \left (b c x + a c\right )^{2} + 12\right )} \sinh \left (b c x + a c\right )}{384 \, {\left (b c \cosh \left (b c x + a c\right ) - b c \sinh \left (b c x + a c\right )\right )}} \] Input:

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

Output:

-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))
 

Sympy [F(-1)]

Timed out. \[ \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{5/2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.45 \[ \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{5/2} \, dx=\frac {5 \, {\left (b c x + a c\right )}}{16 \, b c} + \frac {e^{\left (6 \, b c x + 6 \, a c\right )}}{192 \, b c} + \frac {5 \, e^{\left (4 \, b c x + 4 \, a c\right )}}{128 \, b c} + \frac {5 \, e^{\left (2 \, b c x + 2 \, a c\right )}}{32 \, b c} - \frac {5 \, e^{\left (-2 \, b c x - 2 \, a c\right )}}{64 \, b c} - \frac {e^{\left (-4 \, b c x - 4 \, a c\right )}}{128 \, b c} \] Input:

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

Output:

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)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.40 \[ \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{5/2} \, dx=\frac {{\left (120 \, b c x e^{\left (4 \, a c\right )} - 3 \, {\left (30 \, e^{\left (4 \, b c x + 4 \, a c\right )} + 10 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )} e^{\left (-4 \, b c x\right )} + 2 \, e^{\left (6 \, b c x + 10 \, a c\right )} + 15 \, e^{\left (4 \, b c x + 8 \, a c\right )} + 60 \, e^{\left (2 \, b c x + 6 \, a c\right )}\right )} e^{\left (-4 \, a c\right )}}{384 \, b c} \] Input:

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

Output:

1/384*(120*b*c*x*e^(4*a*c) - 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) + 2*e^(6*b*c*x + 10*a*c) + 15*e^(4*b*c*x + 8*a*c) 
+ 60*e^(2*b*c*x + 6*a*c))*e^(-4*a*c)/(b*c)
 

Mupad [F(-1)]

Timed out. \[ \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{5/2} \, dx=\int {\mathrm {e}}^{c\,\left (a+b\,x\right )}\,{\left ({\mathrm {cosh}\left (a\,c+b\,c\,x\right )}^2\right )}^{5/2} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.39 \[ \int e^{c (a+b x)} \cosh ^2(a c+b c x)^{5/2} \, dx=\frac {2 e^{10 b c x +10 a c}+15 e^{8 b c x +8 a c}+60 e^{6 b c x +6 a c}+120 e^{4 b c x +4 a c} b c x -30 e^{2 b c x +2 a c}-3}{384 e^{4 b c x +4 a c} b c} \] Input:

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

Output:

(2*e**(10*a*c + 10*b*c*x) + 15*e**(8*a*c + 8*b*c*x) + 60*e**(6*a*c + 6*b*c 
*x) + 120*e**(4*a*c + 4*b*c*x)*b*c*x - 30*e**(2*a*c + 2*b*c*x) - 3)/(384*e 
**(4*a*c + 4*b*c*x)*b*c)