\(\int F^{c (a+b x)} \cosh ^2(d+e x) \sinh ^3(d+e x) \, dx\) [102]

Optimal result

Integrand size = 26, antiderivative size = 201 \[ \int F^{c (a+b x)} \cosh ^2(d+e x) \sinh ^3(d+e x) \, dx=-\frac {e^{-d-e x} F^{c (a+b x)}}{16 (e-b c \log (F))}-\frac {e^{-3 d-3 e x} F^{c (a+b x)}}{32 (3 e-b c \log (F))}+\frac {e^{-5 d-5 e x} F^{c (a+b x)}}{32 (5 e-b c \log (F))}-\frac {e^{d+e x} F^{c (a+b x)}}{16 (e+b c \log (F))}-\frac {e^{3 d+3 e x} F^{c (a+b x)}}{32 (3 e+b c \log (F))}+\frac {e^{5 d+5 e x} F^{c (a+b x)}}{32 (5 e+b c \log (F))} \] Output:

-1/16*exp(-e*x-d)*F^(c*(b*x+a))/(e-b*c*ln(F))-exp(-3*e*x-3*d)*F^(c*(b*x+a) 
)/(96*e-32*b*c*ln(F))+exp(-5*e*x-5*d)*F^(c*(b*x+a))/(160*e-32*b*c*ln(F))-e 
xp(e*x+d)*F^(c*(b*x+a))/(16*e+16*b*c*ln(F))-exp(3*e*x+3*d)*F^(c*(b*x+a))/( 
96*e+32*b*c*ln(F))+exp(5*e*x+5*d)*F^(c*(b*x+a))/(160*e+32*b*c*ln(F))
 

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.74 \[ \int F^{c (a+b x)} \cosh ^2(d+e x) \sinh ^3(d+e x) \, dx=\frac {1}{16} F^{c (a+b x)} \left (\frac {-2 e \cosh (d+e x)+2 b c \log (F) \sinh (d+e x)}{(e-b c \log (F)) (e+b c \log (F))}+\frac {-3 e \cosh (3 (d+e x))+b c \log (F) \sinh (3 (d+e x))}{9 e^2-b^2 c^2 \log ^2(F)}+\frac {5 e \cosh (5 (d+e x))-b c \log (F) \sinh (5 (d+e x))}{25 e^2-b^2 c^2 \log ^2(F)}\right ) \] Input:

Integrate[F^(c*(a + b*x))*Cosh[d + e*x]^2*Sinh[d + e*x]^3,x]
 

Output:

(F^(c*(a + b*x))*((-2*e*Cosh[d + e*x] + 2*b*c*Log[F]*Sinh[d + e*x])/((e - 
b*c*Log[F])*(e + b*c*Log[F])) + (-3*e*Cosh[3*(d + e*x)] + b*c*Log[F]*Sinh[ 
3*(d + e*x)])/(9*e^2 - b^2*c^2*Log[F]^2) + (5*e*Cosh[5*(d + e*x)] - b*c*Lo 
g[F]*Sinh[5*(d + e*x)])/(25*e^2 - b^2*c^2*Log[F]^2)))/16
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.28, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {6035, 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.

\(\displaystyle \int \sinh ^3(d+e x) \cosh ^2(d+e x) F^{c (a+b x)} \, dx\)

\(\Big \downarrow \) 6035

\(\displaystyle \int \left (-\frac {1}{8} \sinh (d+e x) F^{c (a+b x)}-\frac {1}{16} \sinh (3 d+3 e x) F^{c (a+b x)}+\frac {1}{16} \sinh (5 d+5 e x) F^{c (a+b x)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b c \log (F) \sinh (d+e x) F^{c (a+b x)}}{8 \left (e^2-b^2 c^2 \log ^2(F)\right )}+\frac {b c \log (F) \sinh (3 d+3 e x) F^{c (a+b x)}}{16 \left (9 e^2-b^2 c^2 \log ^2(F)\right )}-\frac {b c \log (F) \sinh (5 d+5 e x) F^{c (a+b x)}}{16 \left (25 e^2-b^2 c^2 \log ^2(F)\right )}-\frac {e \cosh (d+e x) F^{c (a+b x)}}{8 \left (e^2-b^2 c^2 \log ^2(F)\right )}-\frac {3 e \cosh (3 d+3 e x) F^{c (a+b x)}}{16 \left (9 e^2-b^2 c^2 \log ^2(F)\right )}+\frac {5 e \cosh (5 d+5 e x) F^{c (a+b x)}}{16 \left (25 e^2-b^2 c^2 \log ^2(F)\right )}\)

Input:

Int[F^(c*(a + b*x))*Cosh[d + e*x]^2*Sinh[d + e*x]^3,x]
 

Output:

-1/8*(e*F^(c*(a + b*x))*Cosh[d + e*x])/(e^2 - b^2*c^2*Log[F]^2) - (3*e*F^( 
c*(a + b*x))*Cosh[3*d + 3*e*x])/(16*(9*e^2 - b^2*c^2*Log[F]^2)) + (5*e*F^( 
c*(a + b*x))*Cosh[5*d + 5*e*x])/(16*(25*e^2 - b^2*c^2*Log[F]^2)) + (b*c*F^ 
(c*(a + b*x))*Log[F]*Sinh[d + e*x])/(8*(e^2 - b^2*c^2*Log[F]^2)) + (b*c*F^ 
(c*(a + b*x))*Log[F]*Sinh[3*d + 3*e*x])/(16*(9*e^2 - b^2*c^2*Log[F]^2)) - 
(b*c*F^(c*(a + b*x))*Log[F]*Sinh[5*d + 5*e*x])/(16*(25*e^2 - b^2*c^2*Log[F 
]^2))
 

Defintions of rubi rules used

rule 2009

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

rule 6035

Int[Cosh[(f_.) + (g_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sinh[( 
d_.) + (e_.)*(x_)]^(m_.), x_Symbol] :> Int[ExpandTrigReduce[F^(c*(a + b*x)) 
, Sinh[d + e*x]^m*Cosh[f + g*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e, f, g} 
, x] && IGtQ[m, 0] && IGtQ[n, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(767\) vs. \(2(186)=372\).

Time = 0.81 (sec) , antiderivative size = 768, normalized size of antiderivative = 3.82

\[\frac {\left (450 e^{5} {\mathrm e}^{4 e x +4 d}-45 e^{5}+26 \ln \left (F \right )^{3} b^{3} c^{3} e^{2} {\mathrm e}^{8 e x +8 d}+50 \ln \left (F \right )^{2} b^{2} c^{2} e^{3} {\mathrm e}^{10 e x +10 d}+2 \ln \left (F \right )^{4} b^{4} c^{4} e \,{\mathrm e}^{4 e x +4 d}+68 \ln \left (F \right )^{3} b^{3} c^{3} e^{2} {\mathrm e}^{6 e x +6 d}-78 \ln \left (F \right )^{2} b^{2} c^{2} e^{3} {\mathrm e}^{8 e x +8 d}+\ln \left (F \right )^{5} b^{5} c^{5} {\mathrm e}^{10 e x +10 d}-\ln \left (F \right )^{5} b^{5} c^{5} {\mathrm e}^{8 e x +8 d}-2 \ln \left (F \right )^{5} b^{5} c^{5} {\mathrm e}^{6 e x +6 d}+2 \ln \left (F \right )^{5} b^{5} c^{5} {\mathrm e}^{4 e x +4 d}+\ln \left (F \right )^{5} b^{5} c^{5} {\mathrm e}^{2 e x +2 d}-c^{5} b^{5} \ln \left (F \right )^{5}+9 \ln \left (F \right ) b c \,e^{4} {\mathrm e}^{10 e x +10 d}+3 \ln \left (F \right )^{4} b^{4} c^{4} e \,{\mathrm e}^{2 e x +2 d}-68 \ln \left (F \right )^{3} b^{3} c^{3} e^{2} {\mathrm e}^{4 e x +4 d}-68 \ln \left (F \right )^{2} b^{2} c^{2} e^{3} {\mathrm e}^{6 e x +6 d}-25 \ln \left (F \right ) b c \,e^{4} {\mathrm e}^{8 e x +8 d}-5 \ln \left (F \right )^{4} b^{4} c^{4} e +50 \ln \left (F \right )^{2} b^{2} c^{2} e^{3}-26 \ln \left (F \right )^{3} b^{3} c^{3} e^{2} {\mathrm e}^{2 e x +2 d}-68 \ln \left (F \right )^{2} b^{2} c^{2} e^{3} {\mathrm e}^{4 e x +4 d}-450 \ln \left (F \right ) b c \,e^{4} {\mathrm e}^{6 e x +6 d}-78 \ln \left (F \right )^{2} b^{2} c^{2} e^{3} {\mathrm e}^{2 e x +2 d}+450 \ln \left (F \right ) b c \,e^{4} {\mathrm e}^{4 e x +4 d}+25 \ln \left (F \right ) b c \,e^{4} {\mathrm e}^{2 e x +2 d}-5 \ln \left (F \right )^{4} b^{4} c^{4} e \,{\mathrm e}^{10 e x +10 d}+3 \ln \left (F \right )^{4} b^{4} c^{4} e \,{\mathrm e}^{8 e x +8 d}-10 \ln \left (F \right )^{3} b^{3} c^{3} e^{2} {\mathrm e}^{10 e x +10 d}+2 \ln \left (F \right )^{4} b^{4} c^{4} e \,{\mathrm e}^{6 e x +6 d}+75 e^{5} {\mathrm e}^{2 e x +2 d}+450 e^{5} {\mathrm e}^{6 e x +6 d}+75 e^{5} {\mathrm e}^{8 e x +8 d}-45 e^{5} {\mathrm e}^{10 e x +10 d}+10 c^{3} b^{3} \ln \left (F \right )^{3} e^{2}-9 c b \ln \left (F \right ) e^{4}\right ) {\mathrm e}^{-5 e x -5 d} F^{c \left (b x +a \right )}}{32 \left (b c \ln \left (F \right )-e \right ) \left (b c \ln \left (F \right )-3 e \right ) \left (b c \ln \left (F \right )-5 e \right ) \left (e +b c \ln \left (F \right )\right ) \left (b c \ln \left (F \right )+3 e \right ) \left (b c \ln \left (F \right )+5 e \right )}\]

Input:

int(F^(c*(b*x+a))*cosh(e*x+d)^2*sinh(e*x+d)^3,x)
 

Output:

1/32*(450*e^5*exp(4*e*x+4*d)-45*e^5+26*ln(F)^3*b^3*c^3*e^2*exp(8*e*x+8*d)+ 
50*ln(F)^2*b^2*c^2*e^3*exp(10*e*x+10*d)+2*ln(F)^4*b^4*c^4*e*exp(4*e*x+4*d) 
+68*ln(F)^3*b^3*c^3*e^2*exp(6*e*x+6*d)-78*ln(F)^2*b^2*c^2*e^3*exp(8*e*x+8* 
d)+ln(F)^5*b^5*c^5*exp(10*e*x+10*d)-ln(F)^5*b^5*c^5*exp(8*e*x+8*d)-2*ln(F) 
^5*b^5*c^5*exp(6*e*x+6*d)+2*ln(F)^5*b^5*c^5*exp(4*e*x+4*d)+ln(F)^5*b^5*c^5 
*exp(2*e*x+2*d)-c^5*b^5*ln(F)^5+9*ln(F)*b*c*e^4*exp(10*e*x+10*d)+3*ln(F)^4 
*b^4*c^4*e*exp(2*e*x+2*d)-68*ln(F)^3*b^3*c^3*e^2*exp(4*e*x+4*d)-68*ln(F)^2 
*b^2*c^2*e^3*exp(6*e*x+6*d)-25*ln(F)*b*c*e^4*exp(8*e*x+8*d)-5*ln(F)^4*b^4* 
c^4*e+50*ln(F)^2*b^2*c^2*e^3-26*ln(F)^3*b^3*c^3*e^2*exp(2*e*x+2*d)-68*ln(F 
)^2*b^2*c^2*e^3*exp(4*e*x+4*d)-450*ln(F)*b*c*e^4*exp(6*e*x+6*d)-78*ln(F)^2 
*b^2*c^2*e^3*exp(2*e*x+2*d)+450*ln(F)*b*c*e^4*exp(4*e*x+4*d)+25*ln(F)*b*c* 
e^4*exp(2*e*x+2*d)-5*ln(F)^4*b^4*c^4*e*exp(10*e*x+10*d)+3*ln(F)^4*b^4*c^4* 
e*exp(8*e*x+8*d)-10*ln(F)^3*b^3*c^3*e^2*exp(10*e*x+10*d)+2*ln(F)^4*b^4*c^4 
*e*exp(6*e*x+6*d)+75*e^5*exp(2*e*x+2*d)+450*e^5*exp(6*e*x+6*d)+75*e^5*exp( 
8*e*x+8*d)-45*e^5*exp(10*e*x+10*d)+10*c^3*b^3*ln(F)^3*e^2-9*c*b*ln(F)*e^4) 
/(b*c*ln(F)-e)*exp(-5*e*x-5*d)/(b*c*ln(F)-3*e)/(b*c*ln(F)-5*e)/(e+b*c*ln(F 
))/(b*c*ln(F)+3*e)/(b*c*ln(F)+5*e)*F^(c*(b*x+a))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 7798 vs. \(2 (182) = 364\).

Time = 0.64 (sec) , antiderivative size = 7798, normalized size of antiderivative = 38.80 \[ \int F^{c (a+b x)} \cosh ^2(d+e x) \sinh ^3(d+e x) \, dx=\text {Too large to display} \] Input:

integrate(F^(c*(b*x+a))*cosh(e*x+d)^2*sinh(e*x+d)^3,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \cosh ^2(d+e x) \sinh ^3(d+e x) \, dx=\text {Timed out} \] Input:

integrate(F**(c*(b*x+a))*cosh(e*x+d)**2*sinh(e*x+d)**3,x)
 

Output:

Timed out
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.02 \[ \int F^{c (a+b x)} \cosh ^2(d+e x) \sinh ^3(d+e x) \, dx=\frac {F^{a c} e^{\left (b c x \log \left (F\right ) + 5 \, e x + 5 \, d\right )}}{32 \, {\left (b c \log \left (F\right ) + 5 \, e\right )}} - \frac {F^{a c} e^{\left (b c x \log \left (F\right ) + 3 \, e x + 3 \, d\right )}}{32 \, {\left (b c \log \left (F\right ) + 3 \, e\right )}} - \frac {F^{a c} e^{\left (b c x \log \left (F\right ) + e x + d\right )}}{16 \, {\left (b c \log \left (F\right ) + e\right )}} + \frac {F^{a c} e^{\left (b c x \log \left (F\right ) - e x\right )}}{16 \, {\left (b c e^{d} \log \left (F\right ) - e e^{d}\right )}} + \frac {F^{a c} e^{\left (b c x \log \left (F\right ) - 3 \, e x\right )}}{32 \, {\left (b c e^{\left (3 \, d\right )} \log \left (F\right ) - 3 \, e e^{\left (3 \, d\right )}\right )}} - \frac {F^{a c} e^{\left (b c x \log \left (F\right ) - 5 \, e x\right )}}{32 \, {\left (b c e^{\left (5 \, d\right )} \log \left (F\right ) - 5 \, e e^{\left (5 \, d\right )}\right )}} \] Input:

integrate(F^(c*(b*x+a))*cosh(e*x+d)^2*sinh(e*x+d)^3,x, algorithm="maxima")
 

Output:

1/32*F^(a*c)*e^(b*c*x*log(F) + 5*e*x + 5*d)/(b*c*log(F) + 5*e) - 1/32*F^(a 
*c)*e^(b*c*x*log(F) + 3*e*x + 3*d)/(b*c*log(F) + 3*e) - 1/16*F^(a*c)*e^(b* 
c*x*log(F) + e*x + d)/(b*c*log(F) + e) + 1/16*F^(a*c)*e^(b*c*x*log(F) - e* 
x)/(b*c*e^d*log(F) - e*e^d) + 1/32*F^(a*c)*e^(b*c*x*log(F) - 3*e*x)/(b*c*e 
^(3*d)*log(F) - 3*e*e^(3*d)) - 1/32*F^(a*c)*e^(b*c*x*log(F) - 5*e*x)/(b*c* 
e^(5*d)*log(F) - 5*e*e^(5*d))
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.20 (sec) , antiderivative size = 1823, normalized size of antiderivative = 9.07 \[ \int F^{c (a+b x)} \cosh ^2(d+e x) \sinh ^3(d+e x) \, dx=\text {Too large to display} \] Input:

integrate(F^(c*(b*x+a))*cosh(e*x+d)^2*sinh(e*x+d)^3,x, algorithm="giac")
 

Output:

1/16*(2*(b*c*log(abs(F)) + 5*e)*cos(-1/2*pi*b*c*x*sgn(F) + 1/2*pi*b*c*x - 
1/2*pi*a*c*sgn(F) + 1/2*pi*a*c)/((pi*b*c*sgn(F) - pi*b*c)^2 + 4*(b*c*log(a 
bs(F)) + 5*e)^2) - (pi*b*c*sgn(F) - pi*b*c)*sin(-1/2*pi*b*c*x*sgn(F) + 1/2 
*pi*b*c*x - 1/2*pi*a*c*sgn(F) + 1/2*pi*a*c)/((pi*b*c*sgn(F) - pi*b*c)^2 + 
4*(b*c*log(abs(F)) + 5*e)^2))*e^(a*c*log(abs(F)) + (b*c*log(abs(F)) + 5*e) 
*x + 5*d) + I*(I*e^(1/2*I*pi*b*c*x*sgn(F) - 1/2*I*pi*b*c*x + 1/2*I*pi*a*c* 
sgn(F) - 1/2*I*pi*a*c)/(32*I*pi*b*c*sgn(F) - 32*I*pi*b*c + 64*b*c*log(abs( 
F)) + 320*e) - I*e^(-1/2*I*pi*b*c*x*sgn(F) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c 
*sgn(F) + 1/2*I*pi*a*c)/(-32*I*pi*b*c*sgn(F) + 32*I*pi*b*c + 64*b*c*log(ab 
s(F)) + 320*e))*e^(a*c*log(abs(F)) + (b*c*log(abs(F)) + 5*e)*x + 5*d) - 1/ 
16*(2*(b*c*log(abs(F)) + 3*e)*cos(-1/2*pi*b*c*x*sgn(F) + 1/2*pi*b*c*x - 1/ 
2*pi*a*c*sgn(F) + 1/2*pi*a*c)/((pi*b*c*sgn(F) - pi*b*c)^2 + 4*(b*c*log(abs 
(F)) + 3*e)^2) - (pi*b*c*sgn(F) - pi*b*c)*sin(-1/2*pi*b*c*x*sgn(F) + 1/2*p 
i*b*c*x - 1/2*pi*a*c*sgn(F) + 1/2*pi*a*c)/((pi*b*c*sgn(F) - pi*b*c)^2 + 4* 
(b*c*log(abs(F)) + 3*e)^2))*e^(a*c*log(abs(F)) + (b*c*log(abs(F)) + 3*e)*x 
 + 3*d) + I*(-I*e^(1/2*I*pi*b*c*x*sgn(F) - 1/2*I*pi*b*c*x + 1/2*I*pi*a*c*s 
gn(F) - 1/2*I*pi*a*c)/(32*I*pi*b*c*sgn(F) - 32*I*pi*b*c + 64*b*c*log(abs(F 
)) + 192*e) + I*e^(-1/2*I*pi*b*c*x*sgn(F) + 1/2*I*pi*b*c*x - 1/2*I*pi*a*c* 
sgn(F) + 1/2*I*pi*a*c)/(-32*I*pi*b*c*sgn(F) + 32*I*pi*b*c + 64*b*c*log(abs 
(F)) + 192*e))*e^(a*c*log(abs(F)) + (b*c*log(abs(F)) + 3*e)*x + 3*d) - ...
 

Mupad [B] (verification not implemented)

Time = 4.72 (sec) , antiderivative size = 411, normalized size of antiderivative = 2.04 \[ \int F^{c (a+b x)} \cosh ^2(d+e x) \sinh ^3(d+e x) \, dx=-\frac {F^{a\,c+b\,c\,x}\,\left (b^5\,c^5\,{\mathrm {cosh}\left (d+e\,x\right )}^2\,{\mathrm {sinh}\left (d+e\,x\right )}^3\,{\ln \left (F\right )}^5-3\,b^4\,c^4\,e\,{\mathrm {cosh}\left (d+e\,x\right )}^3\,{\mathrm {sinh}\left (d+e\,x\right )}^2\,{\ln \left (F\right )}^4-2\,b^4\,c^4\,e\,\mathrm {cosh}\left (d+e\,x\right )\,{\mathrm {sinh}\left (d+e\,x\right )}^4\,{\ln \left (F\right )}^4+6\,b^3\,c^3\,e^2\,{\mathrm {cosh}\left (d+e\,x\right )}^4\,\mathrm {sinh}\left (d+e\,x\right )\,{\ln \left (F\right )}^3-18\,b^3\,c^3\,e^2\,{\mathrm {cosh}\left (d+e\,x\right )}^2\,{\mathrm {sinh}\left (d+e\,x\right )}^3\,{\ln \left (F\right )}^3+2\,b^3\,c^3\,e^2\,{\mathrm {sinh}\left (d+e\,x\right )}^5\,{\ln \left (F\right )}^3-6\,b^2\,c^2\,e^3\,{\mathrm {cosh}\left (d+e\,x\right )}^5\,{\ln \left (F\right )}^2+30\,b^2\,c^2\,e^3\,{\mathrm {cosh}\left (d+e\,x\right )}^3\,{\mathrm {sinh}\left (d+e\,x\right )}^2\,{\ln \left (F\right )}^2+26\,b^2\,c^2\,e^3\,\mathrm {cosh}\left (d+e\,x\right )\,{\mathrm {sinh}\left (d+e\,x\right )}^4\,{\ln \left (F\right )}^2-30\,b\,c\,e^4\,{\mathrm {cosh}\left (d+e\,x\right )}^4\,\mathrm {sinh}\left (d+e\,x\right )\,\ln \left (F\right )+65\,b\,c\,e^4\,{\mathrm {cosh}\left (d+e\,x\right )}^2\,{\mathrm {sinh}\left (d+e\,x\right )}^3\,\ln \left (F\right )-26\,b\,c\,e^4\,{\mathrm {sinh}\left (d+e\,x\right )}^5\,\ln \left (F\right )+30\,e^5\,{\mathrm {cosh}\left (d+e\,x\right )}^5-75\,e^5\,{\mathrm {cosh}\left (d+e\,x\right )}^3\,{\mathrm {sinh}\left (d+e\,x\right )}^2\right )}{-b^6\,c^6\,{\ln \left (F\right )}^6+35\,b^4\,c^4\,e^2\,{\ln \left (F\right )}^4-259\,b^2\,c^2\,e^4\,{\ln \left (F\right )}^2+225\,e^6} \] Input:

int(F^(c*(a + b*x))*cosh(d + e*x)^2*sinh(d + e*x)^3,x)
 

Output:

-(F^(a*c + b*c*x)*(30*e^5*cosh(d + e*x)^5 - 75*e^5*cosh(d + e*x)^3*sinh(d 
+ e*x)^2 - 26*b*c*e^4*sinh(d + e*x)^5*log(F) - 6*b^2*c^2*e^3*cosh(d + e*x) 
^5*log(F)^2 + b^5*c^5*cosh(d + e*x)^2*sinh(d + e*x)^3*log(F)^5 + 2*b^3*c^3 
*e^2*sinh(d + e*x)^5*log(F)^3 + 65*b*c*e^4*cosh(d + e*x)^2*sinh(d + e*x)^3 
*log(F) + 30*b^2*c^2*e^3*cosh(d + e*x)^3*sinh(d + e*x)^2*log(F)^2 - 18*b^3 
*c^3*e^2*cosh(d + e*x)^2*sinh(d + e*x)^3*log(F)^3 - 2*b^4*c^4*e*cosh(d + e 
*x)*sinh(d + e*x)^4*log(F)^4 - 30*b*c*e^4*cosh(d + e*x)^4*sinh(d + e*x)*lo 
g(F) + 26*b^2*c^2*e^3*cosh(d + e*x)*sinh(d + e*x)^4*log(F)^2 + 6*b^3*c^3*e 
^2*cosh(d + e*x)^4*sinh(d + e*x)*log(F)^3 - 3*b^4*c^4*e*cosh(d + e*x)^3*si 
nh(d + e*x)^2*log(F)^4))/(225*e^6 - b^6*c^6*log(F)^6 - 259*b^2*c^2*e^4*log 
(F)^2 + 35*b^4*c^4*e^2*log(F)^4)
 

Reduce [F]

\[ \int F^{c (a+b x)} \cosh ^2(d+e x) \sinh ^3(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \cosh \left (e x +d \right )^{2} \sinh \left (e x +d \right )^{3}d x \right ) \] Input:

int(F^(c*(b*x+a))*cosh(e*x+d)^2*sinh(e*x+d)^3,x)
 

Output:

f**(a*c)*int(f**(b*c*x)*cosh(d + e*x)**2*sinh(d + e*x)**3,x)