\(\int F^{c (a+b x)} (f+f \cosh (d+e x))^2 \, dx\) [7]

Optimal result

Integrand size = 22, antiderivative size = 189 \[ \int F^{c (a+b x)} (f+f \cosh (d+e x))^2 \, dx=\frac {f^2 F^{a c+b c x}}{b c \log (F)}+\frac {2 e^2 f^2 F^{a c+b c x}}{b c \log (F) \left (4 e^2-b^2 c^2 \log ^2(F)\right )}-\frac {f^2 F^{a c+b c x} \cosh (d+e x) (b c \cosh (d+e x) \log (F)-2 e \sinh (d+e x))}{4 e^2-b^2 c^2 \log ^2(F)}-\frac {2 f^2 F^{a c+b c x} (b c \cosh (d+e x) \log (F)-e \sinh (d+e x))}{e^2-b^2 c^2 \log ^2(F)} \] Output:

f^2*F^(b*c*x+a*c)/b/c/ln(F)+2*e^2*f^2*F^(b*c*x+a*c)/b/c/ln(F)/(4*e^2-b^2*c 
^2*ln(F)^2)-f^2*F^(b*c*x+a*c)*cosh(e*x+d)*(b*c*cosh(e*x+d)*ln(F)-2*e*sinh( 
e*x+d))/(4*e^2-b^2*c^2*ln(F)^2)-2*f^2*F^(b*c*x+a*c)*(b*c*cosh(e*x+d)*ln(F) 
-e*sinh(e*x+d))/(e^2-b^2*c^2*ln(F)^2)
 

Mathematica [A] (verified)

Time = 0.69 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.22 \[ \int F^{c (a+b x)} (f+f \cosh (d+e x))^2 \, dx=\frac {f^2 F^{c (a+b x)} \left (12 e^4-15 b^2 c^2 e^2 \log ^2(F)+3 b^4 c^4 \log ^4(F)+4 \cosh (d+e x) \left (-4 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)\right )+\cosh (2 (d+e x)) \left (-b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)\right )+16 b c e^3 \log (F) \sinh (d+e x)-4 b^3 c^3 e \log ^3(F) \sinh (d+e x)+2 b c e^3 \log (F) \sinh (2 (d+e x))-2 b^3 c^3 e \log ^3(F) \sinh (2 (d+e x))\right )}{2 \left (4 b c e^4 \log (F)-5 b^3 c^3 e^2 \log ^3(F)+b^5 c^5 \log ^5(F)\right )} \] Input:

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

Output:

(f^2*F^(c*(a + b*x))*(12*e^4 - 15*b^2*c^2*e^2*Log[F]^2 + 3*b^4*c^4*Log[F]^ 
4 + 4*Cosh[d + e*x]*(-4*b^2*c^2*e^2*Log[F]^2 + b^4*c^4*Log[F]^4) + Cosh[2* 
(d + e*x)]*(-(b^2*c^2*e^2*Log[F]^2) + b^4*c^4*Log[F]^4) + 16*b*c*e^3*Log[F 
]*Sinh[d + e*x] - 4*b^3*c^3*e*Log[F]^3*Sinh[d + e*x] + 2*b*c*e^3*Log[F]*Si 
nh[2*(d + e*x)] - 2*b^3*c^3*e*Log[F]^3*Sinh[2*(d + e*x)]))/(2*(4*b*c*e^4*L 
og[F] - 5*b^3*c^3*e^2*Log[F]^3 + b^5*c^5*Log[F]^5))
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.25, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {7292, 27, 7293, 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.

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

\(\Big \downarrow \) 7292

\(\displaystyle \int f^2 (\cosh (d+e x)+1)^2 F^{a c+b c x}dx\)

\(\Big \downarrow \) 27

\(\displaystyle f^2 \int F^{a c+b x c} (\cosh (d+e x)+1)^2dx\)

\(\Big \downarrow \) 7293

\(\displaystyle f^2 \int \left (\cosh ^2(d+e x) F^{a c+b x c}+2 \cosh (d+e x) F^{a c+b x c}+F^{a c+b x c}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle f^2 \left (\frac {2 e \sinh (d+e x) F^{a c+b c x}}{e^2-b^2 c^2 \log ^2(F)}-\frac {b c \log (F) \cosh ^2(d+e x) F^{a c+b c x}}{4 e^2-b^2 c^2 \log ^2(F)}-\frac {2 b c \log (F) \cosh (d+e x) F^{a c+b c x}}{e^2-b^2 c^2 \log ^2(F)}+\frac {2 e \sinh (d+e x) \cosh (d+e x) F^{a c+b c x}}{4 e^2-b^2 c^2 \log ^2(F)}+\frac {2 e^2 F^{a c+b c x}}{b c \log (F) \left (4 e^2-b^2 c^2 \log ^2(F)\right )}+\frac {F^{a c+b c x}}{b c \log (F)}\right )\)

Input:

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

Output:

f^2*(F^(a*c + b*c*x)/(b*c*Log[F]) - (2*b*c*F^(a*c + b*c*x)*Cosh[d + e*x]*L 
og[F])/(e^2 - b^2*c^2*Log[F]^2) + (2*e^2*F^(a*c + b*c*x))/(b*c*Log[F]*(4*e 
^2 - b^2*c^2*Log[F]^2)) - (b*c*F^(a*c + b*c*x)*Cosh[d + e*x]^2*Log[F])/(4* 
e^2 - b^2*c^2*Log[F]^2) + (2*e*F^(a*c + b*c*x)*Sinh[d + e*x])/(e^2 - b^2*c 
^2*Log[F]^2) + (2*e*F^(a*c + b*c*x)*Cosh[d + e*x]*Sinh[d + e*x])/(4*e^2 - 
b^2*c^2*Log[F]^2))
 

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 2009

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

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [A] (verified)

Time = 1.31 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.02

method	result	size
parallelrisch	\(\frac {2 f^{2} \left (\frac {\left (\ln \left (F \right )^{4} b^{4} c^{4}-\ln \left (F \right )^{2} b^{2} c^{2} e^{2}\right ) \cosh \left (2 e x +2 d \right )}{4}+\frac {\left (-\ln \left (F \right )^{3} b^{3} c^{3} e +\ln \left (F \right ) b c \,e^{3}\right ) \sinh \left (2 e x +2 d \right )}{2}+\left (2 e +b c \ln \left (F \right )\right ) \left (b c \ln \left (F \right )-2 e \right ) \left (\cosh \left (e x +d \right ) b^{2} c^{2} \ln \left (F \right )^{2}+\frac {3 b^{2} c^{2} \ln \left (F \right )^{2}}{4}-\sinh \left (e x +d \right ) b c e \ln \left (F \right )-\frac {3 e^{2}}{4}\right )\right ) F^{c \left (b x +a \right )}}{c^{5} b^{5} \ln \left (F \right )^{5}-5 c^{3} b^{3} \ln \left (F \right )^{3} e^{2}+4 c b \ln \left (F \right ) e^{4}}\)	\(192\)
risch	\(\frac {f^{2} \left (\ln \left (F \right )^{4} b^{4} c^{4} {\mathrm e}^{4 e x +4 d}+4 \ln \left (F \right )^{4} b^{4} c^{4} {\mathrm e}^{3 e x +3 d}+6 \ln \left (F \right )^{4} b^{4} c^{4} {\mathrm e}^{2 e x +2 d}-2 \ln \left (F \right )^{3} b^{3} c^{3} e \,{\mathrm e}^{4 e x +4 d}+4 \ln \left (F \right )^{4} b^{4} c^{4} {\mathrm e}^{e x +d}-4 \ln \left (F \right )^{3} b^{3} c^{3} e \,{\mathrm e}^{3 e x +3 d}+\ln \left (F \right )^{4} b^{4} c^{4}-\ln \left (F \right )^{2} b^{2} c^{2} e^{2} {\mathrm e}^{4 e x +4 d}+4 \ln \left (F \right )^{3} b^{3} c^{3} e \,{\mathrm e}^{e x +d}-16 \ln \left (F \right )^{2} b^{2} c^{2} e^{2} {\mathrm e}^{3 e x +3 d}+2 \ln \left (F \right )^{3} b^{3} c^{3} e -30 \ln \left (F \right )^{2} b^{2} c^{2} e^{2} {\mathrm e}^{2 e x +2 d}+2 \ln \left (F \right ) b c \,e^{3} {\mathrm e}^{4 e x +4 d}-16 \ln \left (F \right )^{2} b^{2} c^{2} e^{2} {\mathrm e}^{e x +d}+16 \ln \left (F \right ) b c \,e^{3} {\mathrm e}^{3 e x +3 d}-\ln \left (F \right )^{2} b^{2} c^{2} e^{2}-16 \ln \left (F \right ) b c \,e^{3} {\mathrm e}^{e x +d}-2 \ln \left (F \right ) b c \,e^{3}+24 e^{4} {\mathrm e}^{2 e x +2 d}\right ) {\mathrm e}^{-2 e x -2 d} F^{c \left (b x +a \right )}}{4 b c \ln \left (F \right ) \left (b c \ln \left (F \right )-e \right ) \left (b c \ln \left (F \right )-2 e \right ) \left (e +b c \ln \left (F \right )\right ) \left (2 e +b c \ln \left (F \right )\right )}\)	\(426\)
orering	\(\text {Expression too large to display}\)	\(971\)

Input:

int(F^(c*(b*x+a))*(f+f*cosh(e*x+d))^2,x,method=_RETURNVERBOSE)
 

Output:

2*f^2*(1/4*(ln(F)^4*b^4*c^4-ln(F)^2*b^2*c^2*e^2)*cosh(2*e*x+2*d)+1/2*(-ln( 
F)^3*b^3*c^3*e+ln(F)*b*c*e^3)*sinh(2*e*x+2*d)+(2*e+b*c*ln(F))*(b*c*ln(F)-2 
*e)*(cosh(e*x+d)*b^2*c^2*ln(F)^2+3/4*b^2*c^2*ln(F)^2-sinh(e*x+d)*b*c*e*ln( 
F)-3/4*e^2))*F^(c*(b*x+a))/(c^5*b^5*ln(F)^5-5*c^3*b^3*ln(F)^3*e^2+4*c*b*ln 
(F)*e^4)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2340 vs. \(2 (187) = 374\).

Time = 0.22 (sec) , antiderivative size = 2340, normalized size of antiderivative = 12.38 \[ \int F^{c (a+b x)} (f+f \cosh (d+e x))^2 \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*((24*e^4*f^2*cosh(e*x + d)^2 + (b^4*c^4*f^2*cosh(e*x + d)^4 + 4*b^4*c^ 
4*f^2*cosh(e*x + d)^3 + 6*b^4*c^4*f^2*cosh(e*x + d)^2 + 4*b^4*c^4*f^2*cosh 
(e*x + d) + b^4*c^4*f^2)*log(F)^4 + (b^4*c^4*f^2*log(F)^4 - 2*b^3*c^3*e*f^ 
2*log(F)^3 - b^2*c^2*e^2*f^2*log(F)^2 + 2*b*c*e^3*f^2*log(F))*sinh(e*x + d 
)^4 - 2*(b^3*c^3*e*f^2*cosh(e*x + d)^4 + 2*b^3*c^3*e*f^2*cosh(e*x + d)^3 - 
 2*b^3*c^3*e*f^2*cosh(e*x + d) - b^3*c^3*e*f^2)*log(F)^3 + 4*((b^4*c^4*f^2 
*cosh(e*x + d) + b^4*c^4*f^2)*log(F)^4 - (2*b^3*c^3*e*f^2*cosh(e*x + d) + 
b^3*c^3*e*f^2)*log(F)^3 - (b^2*c^2*e^2*f^2*cosh(e*x + d) + 4*b^2*c^2*e^2*f 
^2)*log(F)^2 + 2*(b*c*e^3*f^2*cosh(e*x + d) + 2*b*c*e^3*f^2)*log(F))*sinh( 
e*x + d)^3 - (b^2*c^2*e^2*f^2*cosh(e*x + d)^4 + 16*b^2*c^2*e^2*f^2*cosh(e* 
x + d)^3 + 30*b^2*c^2*e^2*f^2*cosh(e*x + d)^2 + 16*b^2*c^2*e^2*f^2*cosh(e* 
x + d) + b^2*c^2*e^2*f^2)*log(F)^2 + 6*(4*e^4*f^2 + (b^4*c^4*f^2*cosh(e*x 
+ d)^2 + 2*b^4*c^4*f^2*cosh(e*x + d) + b^4*c^4*f^2)*log(F)^4 - 2*(b^3*c^3* 
e*f^2*cosh(e*x + d)^2 + b^3*c^3*e*f^2*cosh(e*x + d))*log(F)^3 - (b^2*c^2*e 
^2*f^2*cosh(e*x + d)^2 + 8*b^2*c^2*e^2*f^2*cosh(e*x + d) + 5*b^2*c^2*e^2*f 
^2)*log(F)^2 + 2*(b*c*e^3*f^2*cosh(e*x + d)^2 + 4*b*c*e^3*f^2*cosh(e*x + d 
))*log(F))*sinh(e*x + d)^2 + 2*(b*c*e^3*f^2*cosh(e*x + d)^4 + 8*b*c*e^3*f^ 
2*cosh(e*x + d)^3 - 8*b*c*e^3*f^2*cosh(e*x + d) - b*c*e^3*f^2)*log(F) + 4* 
(12*e^4*f^2*cosh(e*x + d) + (b^4*c^4*f^2*cosh(e*x + d)^3 + 3*b^4*c^4*f^2*c 
osh(e*x + d)^2 + 3*b^4*c^4*f^2*cosh(e*x + d) + b^4*c^4*f^2)*log(F)^4 - ...
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2312 vs. \(2 (175) = 350\).

Time = 1.99 (sec) , antiderivative size = 2312, normalized size of antiderivative = 12.23 \[ \int F^{c (a+b x)} (f+f \cosh (d+e x))^2 \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((x*(f*cosh(d) + f)**2, Eq(F, 1) & Eq(b, 0) & Eq(c, 0) & Eq(e, 0) 
), (-f**2*x*sinh(d + e*x)**2/2 + f**2*x*cosh(d + e*x)**2/2 + f**2*x + f**2 
*sinh(d + e*x)*cosh(d + e*x)/(2*e) + 2*f**2*sinh(d + e*x)/e, Eq(F, 1)), (F 
**(a*c)*(-f**2*x*sinh(d + e*x)**2/2 + f**2*x*cosh(d + e*x)**2/2 + f**2*x + 
 f**2*sinh(d + e*x)*cosh(d + e*x)/(2*e) + 2*f**2*sinh(d + e*x)/e), Eq(b, 0 
)), (-f**2*x*sinh(d + e*x)**2/2 + f**2*x*cosh(d + e*x)**2/2 + f**2*x + f** 
2*sinh(d + e*x)*cosh(d + e*x)/(2*e) + 2*f**2*sinh(d + e*x)/e, Eq(c, 0)), ( 
-F**(a*c + b*c*x)*f**2*x*sinh(b*c*x*log(F) - d) + F**(a*c + b*c*x)*f**2*x* 
cosh(b*c*x*log(F) - d) - 2*F**(a*c + b*c*x)*f**2*sinh(b*c*x*log(F) - d)**2 
/(3*b*c*log(F)) + 2*F**(a*c + b*c*x)*f**2*sinh(b*c*x*log(F) - d)*cosh(b*c* 
x*log(F) - d)/(3*b*c*log(F)) + F**(a*c + b*c*x)*f**2*sinh(b*c*x*log(F) - d 
)/(b*c*log(F)) + F**(a*c + b*c*x)*f**2*cosh(b*c*x*log(F) - d)**2/(3*b*c*lo 
g(F)) + F**(a*c + b*c*x)*f**2/(b*c*log(F)), Eq(e, -b*c*log(F))), (F**(a*c 
+ b*c*x)*f**2*x*sinh(b*c*x*log(F)/2 - d)**2/4 - F**(a*c + b*c*x)*f**2*x*si 
nh(b*c*x*log(F)/2 - d)*cosh(b*c*x*log(F)/2 - d)/2 + F**(a*c + b*c*x)*f**2* 
x*cosh(b*c*x*log(F)/2 - d)**2/4 - F**(a*c + b*c*x)*f**2*sinh(b*c*x*log(F)/ 
2 - d)**2/(b*c*log(F)) + 3*F**(a*c + b*c*x)*f**2*sinh(b*c*x*log(F)/2 - d)* 
cosh(b*c*x*log(F)/2 - d)/(2*b*c*log(F)) - 4*F**(a*c + b*c*x)*f**2*sinh(b*c 
*x*log(F)/2 - d)/(3*b*c*log(F)) + 8*F**(a*c + b*c*x)*f**2*cosh(b*c*x*log(F 
)/2 - d)/(3*b*c*log(F)) + F**(a*c + b*c*x)*f**2/(b*c*log(F)), Eq(e, -b*...
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.99 \[ \int F^{c (a+b x)} (f+f \cosh (d+e x))^2 \, dx=\frac {1}{4} \, f^{2} {\left (\frac {F^{a c} e^{\left (b c x \log \left (F\right ) + 2 \, e x + 2 \, d\right )}}{b c \log \left (F\right ) + 2 \, e} + \frac {F^{a c} e^{\left (b c x \log \left (F\right ) - 2 \, e x\right )}}{b c e^{\left (2 \, d\right )} \log \left (F\right ) - 2 \, e e^{\left (2 \, d\right )}} + \frac {2 \, F^{b c x + a c}}{b c \log \left (F\right )}\right )} + f^{2} {\left (\frac {F^{a c} e^{\left (b c x \log \left (F\right ) + e x + d\right )}}{b c \log \left (F\right ) + e} + \frac {F^{a c} e^{\left (b c x \log \left (F\right ) - e x\right )}}{b c e^{d} \log \left (F\right ) - e e^{d}}\right )} + \frac {F^{b c x + a c} f^{2}}{b c \log \left (F\right )} \] Input:

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

Output:

1/4*f^2*(F^(a*c)*e^(b*c*x*log(F) + 2*e*x + 2*d)/(b*c*log(F) + 2*e) + F^(a* 
c)*e^(b*c*x*log(F) - 2*e*x)/(b*c*e^(2*d)*log(F) - 2*e*e^(2*d)) + 2*F^(b*c* 
x + a*c)/(b*c*log(F))) + f^2*(F^(a*c)*e^(b*c*x*log(F) + e*x + d)/(b*c*log( 
F) + e) + F^(a*c)*e^(b*c*x*log(F) - e*x)/(b*c*e^d*log(F) - e*e^d)) + F^(b* 
c*x + a*c)*f^2/(b*c*log(F))
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.18 (sec) , antiderivative size = 1548, normalized size of antiderivative = 8.19 \[ \int F^{c (a+b x)} (f+f \cosh (d+e x))^2 \, dx=\text {Too large to display} \] Input:

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

Output:

3*(2*b*c*f^2*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)*log(abs(F))/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sgn(F) - pi*b* 
c)^2) - (pi*b*c*sgn(F) - pi*b*c)*f^2*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)/(4*b^2*c^2*log(abs(F))^2 + (pi*b*c*sg 
n(F) - pi*b*c)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + 3*I*(I*f^2*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)/(2*I*pi*b*c*sgn(F) - 2*I*pi*b*c + 4*b*c*log(abs(F))) - I*f^2*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)/(- 
2*I*pi*b*c*sgn(F) + 2*I*pi*b*c + 4*b*c*log(abs(F))))*e^(b*c*x*log(abs(F)) 
+ a*c*log(abs(F))) + 1/2*(2*(b*c*log(abs(F)) + 2*e)*f^2*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)) + 2*e)^2) - (pi*b*c*sgn(F) - pi*b*c)*f^2*si 
n(-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)) + 2*e)^2))*e^(a*c*log(abs(F 
)) + (b*c*log(abs(F)) + 2*e)*x + 2*d) + I*(I*f^2*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)/(4*I*pi*b*c*sgn(F) 
- 4*I*pi*b*c + 8*b*c*log(abs(F)) + 16*e) - I*f^2*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)/(-4*I*pi*b*c*sgn(F 
) + 4*I*pi*b*c + 8*b*c*log(abs(F)) + 16*e))*e^(a*c*log(abs(F)) + (b*c*log( 
abs(F)) + 2*e)*x + 2*d) + 2*(2*(b*c*log(abs(F)) + e)*f^2*cos(-1/2*pi*b*...
 

Mupad [B] (verification not implemented)

Time = 2.90 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.52 \[ \int F^{c (a+b x)} (f+f \cosh (d+e x))^2 \, dx=\frac {2\,F^{b\,c\,x}\,F^{a\,c}\,e\,f^2\,\mathrm {sinh}\left (d+e\,x\right )}{e^2-b^2\,c^2\,{\ln \left (F\right )}^2}+\frac {F^{b\,c\,x}\,F^{a\,c}\,f^2}{b\,c\,\ln \left (F\right )}+\frac {2\,F^{b\,c\,x}\,F^{a\,c}\,e\,f^2\,\mathrm {cosh}\left (d+e\,x\right )\,\mathrm {sinh}\left (d+e\,x\right )}{4\,e^2-b^2\,c^2\,{\ln \left (F\right )}^2}-\frac {2\,F^{b\,c\,x}\,F^{a\,c}\,b\,c\,f^2\,\mathrm {cosh}\left (d+e\,x\right )\,\ln \left (F\right )}{e^2-b^2\,c^2\,{\ln \left (F\right )}^2}-\frac {2\,F^{b\,c\,x}\,F^{a\,c}\,e^2\,f^2\,{\mathrm {sinh}\left (d+e\,x\right )}^2}{b\,c\,\ln \left (F\right )\,\left (4\,e^2-b^2\,c^2\,{\ln \left (F\right )}^2\right )}+\frac {F^{b\,c\,x}\,F^{a\,c}\,f^2\,{\mathrm {cosh}\left (d+e\,x\right )}^2\,\left (2\,e^2-b^2\,c^2\,{\ln \left (F\right )}^2\right )}{b\,c\,\ln \left (F\right )\,\left (4\,e^2-b^2\,c^2\,{\ln \left (F\right )}^2\right )} \] Input:

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

Output:

(2*F^(b*c*x)*F^(a*c)*e*f^2*sinh(d + e*x))/(e^2 - b^2*c^2*log(F)^2) + (F^(b 
*c*x)*F^(a*c)*f^2)/(b*c*log(F)) + (2*F^(b*c*x)*F^(a*c)*e*f^2*cosh(d + e*x) 
*sinh(d + e*x))/(4*e^2 - b^2*c^2*log(F)^2) - (2*F^(b*c*x)*F^(a*c)*b*c*f^2* 
cosh(d + e*x)*log(F))/(e^2 - b^2*c^2*log(F)^2) - (2*F^(b*c*x)*F^(a*c)*e^2* 
f^2*sinh(d + e*x)^2)/(b*c*log(F)*(4*e^2 - b^2*c^2*log(F)^2)) + (F^(b*c*x)* 
F^(a*c)*f^2*cosh(d + e*x)^2*(2*e^2 - b^2*c^2*log(F)^2))/(b*c*log(F)*(4*e^2 
 - b^2*c^2*log(F)^2))
 

Reduce [F]

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

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

Output:

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