\(\int F^{c (a+b x)} \sinh ^4(d+e x) \, dx\) [16]

Optimal result

Integrand size = 18, antiderivative size = 162 \[ \int F^{c (a+b x)} \sinh ^4(d+e x) \, dx=\frac {3 F^{c (a+b x)}}{8 b c \log (F)}+\frac {e^{-2 d-2 e x} F^{c (a+b x)}}{4 (2 e-b c \log (F))}-\frac {e^{-4 d-4 e x} F^{c (a+b x)}}{16 (4 e-b c \log (F))}-\frac {e^{2 d+2 e x} F^{c (a+b x)}}{4 (2 e+b c \log (F))}+\frac {e^{4 d+4 e x} F^{c (a+b x)}}{16 (4 e+b c \log (F))} \] Output:

3/8*F^(c*(b*x+a))/b/c/ln(F)+exp(-2*e*x-2*d)*F^(c*(b*x+a))/(8*e-4*b*c*ln(F) 
)-exp(-4*e*x-4*d)*F^(c*(b*x+a))/(64*e-16*b*c*ln(F))-exp(2*e*x+2*d)*F^(c*(b 
*x+a))/(8*e+4*b*c*ln(F))+exp(4*e*x+4*d)*F^(c*(b*x+a))/(64*e+16*b*c*ln(F))
 

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.25 \[ \int F^{c (a+b x)} \sinh ^4(d+e x) \, dx=\frac {1}{8} F^{c (a+b x)} \left (\frac {3}{b c \log (F)}+\frac {4 \cosh (2 e x) (b c \cosh (2 d) \log (F)-2 e \sinh (2 d))}{4 e^2-b^2 c^2 \log ^2(F)}+\frac {\cosh (4 e x) (-b c \cosh (4 d) \log (F)+4 e \sinh (4 d))}{16 e^2-b^2 c^2 \log ^2(F)}-\frac {4 (2 e \cosh (2 d)-b c \log (F) \sinh (2 d)) \sinh (2 e x)}{4 e^2-b^2 c^2 \log ^2(F)}+\frac {(4 e \cosh (4 d)-b c \log (F) \sinh (4 d)) \sinh (4 e x)}{16 e^2-b^2 c^2 \log ^2(F)}\right ) \] Input:

Integrate[F^(c*(a + b*x))*Sinh[d + e*x]^4,x]
 

Output:

(F^(c*(a + b*x))*(3/(b*c*Log[F]) + (4*Cosh[2*e*x]*(b*c*Cosh[2*d]*Log[F] - 
2*e*Sinh[2*d]))/(4*e^2 - b^2*c^2*Log[F]^2) + (Cosh[4*e*x]*(-(b*c*Cosh[4*d] 
*Log[F]) + 4*e*Sinh[4*d]))/(16*e^2 - b^2*c^2*Log[F]^2) - (4*(2*e*Cosh[2*d] 
 - b*c*Log[F]*Sinh[2*d])*Sinh[2*e*x])/(4*e^2 - b^2*c^2*Log[F]^2) + ((4*e*C 
osh[4*d] - b*c*Log[F]*Sinh[4*d])*Sinh[4*e*x])/(16*e^2 - b^2*c^2*Log[F]^2)) 
)/8
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.52, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {5999, 5999, 2624}

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.

\(\displaystyle \int \sinh ^4(d+e x) F^{c (a+b x)} \, dx\)

\(\Big \downarrow \) 5999

\(\displaystyle -\frac {12 e^2 \int F^{c (a+b x)} \sinh ^2(d+e x)dx}{16 e^2-b^2 c^2 \log ^2(F)}-\frac {b c \log (F) \sinh ^4(d+e x) F^{c (a+b x)}}{16 e^2-b^2 c^2 \log ^2(F)}+\frac {4 e \sinh ^3(d+e x) \cosh (d+e x) F^{c (a+b x)}}{16 e^2-b^2 c^2 \log ^2(F)}\)

\(\Big \downarrow \) 5999

\(\displaystyle -\frac {12 e^2 \left (-\frac {2 e^2 \int F^{c (a+b x)}dx}{4 e^2-b^2 c^2 \log ^2(F)}-\frac {b c \log (F) \sinh ^2(d+e x) F^{c (a+b x)}}{4 e^2-b^2 c^2 \log ^2(F)}+\frac {2 e \sinh (d+e x) \cosh (d+e x) F^{c (a+b x)}}{4 e^2-b^2 c^2 \log ^2(F)}\right )}{16 e^2-b^2 c^2 \log ^2(F)}-\frac {b c \log (F) \sinh ^4(d+e x) F^{c (a+b x)}}{16 e^2-b^2 c^2 \log ^2(F)}+\frac {4 e \sinh ^3(d+e x) \cosh (d+e x) F^{c (a+b x)}}{16 e^2-b^2 c^2 \log ^2(F)}\)

\(\Big \downarrow \) 2624

\(\displaystyle -\frac {b c \log (F) \sinh ^4(d+e x) F^{c (a+b x)}}{16 e^2-b^2 c^2 \log ^2(F)}+\frac {4 e \sinh ^3(d+e x) \cosh (d+e x) F^{c (a+b x)}}{16 e^2-b^2 c^2 \log ^2(F)}-\frac {12 e^2 \left (-\frac {b c \log (F) \sinh ^2(d+e x) F^{c (a+b x)}}{4 e^2-b^2 c^2 \log ^2(F)}+\frac {2 e \sinh (d+e x) \cosh (d+e x) F^{c (a+b x)}}{4 e^2-b^2 c^2 \log ^2(F)}-\frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (4 e^2-b^2 c^2 \log ^2(F)\right )}\right )}{16 e^2-b^2 c^2 \log ^2(F)}\)

Input:

Int[F^(c*(a + b*x))*Sinh[d + e*x]^4,x]
 

Output:

(4*e*F^(c*(a + b*x))*Cosh[d + e*x]*Sinh[d + e*x]^3)/(16*e^2 - b^2*c^2*Log[ 
F]^2) - (b*c*F^(c*(a + b*x))*Log[F]*Sinh[d + e*x]^4)/(16*e^2 - b^2*c^2*Log 
[F]^2) - (12*e^2*((-2*e^2*F^(c*(a + b*x)))/(b*c*Log[F]*(4*e^2 - b^2*c^2*Lo 
g[F]^2)) + (2*e*F^(c*(a + b*x))*Cosh[d + e*x]*Sinh[d + e*x])/(4*e^2 - b^2* 
c^2*Log[F]^2) - (b*c*F^(c*(a + b*x))*Log[F]*Sinh[d + e*x]^2)/(4*e^2 - b^2* 
c^2*Log[F]^2)))/(16*e^2 - b^2*c^2*Log[F]^2)
 

Defintions of rubi rules used

rule 2624

Int[((F_)^(v_))^(n_.), x_Symbol] :> Simp[(F^v)^n/(n*Log[F]*D[v, x]), x] /; 
FreeQ[{F, n}, x] && LinearQ[v, x]
 

rule 5999

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sinh[(d_.) + (e_.)*(x_)]^(n_), x_Symb 
ol] :> Simp[(-b)*c*Log[F]*F^(c*(a + b*x))*(Sinh[d + e*x]^n/(e^2*n^2 - b^2*c 
^2*Log[F]^2)), x] + (Simp[e*n*F^(c*(a + b*x))*Cosh[d + e*x]*(Sinh[d + e*x]^ 
(n - 1)/(e^2*n^2 - b^2*c^2*Log[F]^2)), x] - Simp[n*(n - 1)*(e^2/(e^2*n^2 - 
b^2*c^2*Log[F]^2))   Int[F^(c*(a + b*x))*Sinh[d + e*x]^(n - 2), x], x]) /; 
FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*n^2 - b^2*c^2*Log[F]^2, 0] && GtQ[n 
, 1]
 

Maple [A] (verified)

Time = 2.16 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.20

method	result	size
parallelrisch	\(\frac {F^{c \left (b x +a \right )} \left (\left (-4 \ln \left (F \right )^{3} b^{3} c^{3} e +16 \ln \left (F \right ) b c \,e^{3}\right ) \sinh \left (4 e x +4 d \right )+\left (\ln \left (F \right )^{4} b^{4} c^{4}-4 \ln \left (F \right )^{2} b^{2} c^{2} e^{2}\right ) \cosh \left (4 e x +4 d \right )+8 \left (b c \ln \left (F \right )+4 e \right ) \left (-\frac {c^{2} b^{2} \ln \left (F \right )^{2} \cosh \left (2 e x +2 d \right )}{2}+\frac {3 b^{2} c^{2} \ln \left (F \right )^{2}}{8}+\ln \left (F \right ) b c e \sinh \left (2 e x +2 d \right )-\frac {3 e^{2}}{2}\right ) \left (b c \ln \left (F \right )-4 e \right )\right )}{8 c^{5} b^{5} \ln \left (F \right )^{5}-160 c^{3} b^{3} \ln \left (F \right )^{3} e^{2}+512 c b \ln \left (F \right ) e^{4}}\)	\(195\)
risch	\(\frac {\left (\ln \left (F \right )^{4} b^{4} c^{4} {\mathrm e}^{8 e x +8 d}-4 \ln \left (F \right )^{4} b^{4} c^{4} {\mathrm e}^{6 e x +6 d}-4 \ln \left (F \right )^{3} b^{3} c^{3} e \,{\mathrm e}^{8 e x +8 d}+6 \ln \left (F \right )^{4} b^{4} c^{4} {\mathrm e}^{4 e x +4 d}+8 \ln \left (F \right )^{3} b^{3} c^{3} e \,{\mathrm e}^{6 e x +6 d}-4 \ln \left (F \right )^{2} b^{2} c^{2} e^{2} {\mathrm e}^{8 e x +8 d}-4 \ln \left (F \right )^{4} b^{4} c^{4} {\mathrm e}^{2 e x +2 d}+64 \ln \left (F \right )^{2} b^{2} c^{2} e^{2} {\mathrm e}^{6 e x +6 d}+16 \ln \left (F \right ) b c \,e^{3} {\mathrm e}^{8 e x +8 d}+\ln \left (F \right )^{4} b^{4} c^{4}-8 \ln \left (F \right )^{3} b^{3} c^{3} e \,{\mathrm e}^{2 e x +2 d}-120 \ln \left (F \right )^{2} b^{2} c^{2} e^{2} {\mathrm e}^{4 e x +4 d}-128 \ln \left (F \right ) b c \,e^{3} {\mathrm e}^{6 e x +6 d}+4 \ln \left (F \right )^{3} b^{3} c^{3} e +64 \ln \left (F \right )^{2} b^{2} c^{2} e^{2} {\mathrm e}^{2 e x +2 d}-4 \ln \left (F \right )^{2} b^{2} c^{2} e^{2}+128 \ln \left (F \right ) b c \,e^{3} {\mathrm e}^{2 e x +2 d}+384 e^{4} {\mathrm e}^{4 e x +4 d}-16 \ln \left (F \right ) b c \,e^{3}\right ) {\mathrm e}^{-4 e x -4 d} F^{c \left (b x +a \right )}}{16 \ln \left (F \right ) b c \left (b c \ln \left (F \right )-2 e \right ) \left (b c \ln \left (F \right )-4 e \right ) \left (2 e +b c \ln \left (F \right )\right ) \left (b c \ln \left (F \right )+4 e \right )}\)	\(437\)
orering	\(\text {Expression too large to display}\)	\(910\)

Input:

int(F^(c*(b*x+a))*sinh(e*x+d)^4,x,method=_RETURNVERBOSE)
 

Output:

F^(c*(b*x+a))*((-4*ln(F)^3*b^3*c^3*e+16*ln(F)*b*c*e^3)*sinh(4*e*x+4*d)+(ln 
(F)^4*b^4*c^4-4*ln(F)^2*b^2*c^2*e^2)*cosh(4*e*x+4*d)+8*(b*c*ln(F)+4*e)*(-1 
/2*c^2*b^2*ln(F)^2*cosh(2*e*x+2*d)+3/8*b^2*c^2*ln(F)^2+ln(F)*b*c*e*sinh(2* 
e*x+2*d)-3/2*e^2)*(b*c*ln(F)-4*e))/(8*c^5*b^5*ln(F)^5-160*c^3*b^3*ln(F)^3* 
e^2+512*c*b*ln(F)*e^4)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3686 vs. \(2 (146) = 292\).

Time = 0.22 (sec) , antiderivative size = 3686, normalized size of antiderivative = 22.75 \[ \int F^{c (a+b x)} \sinh ^4(d+e x) \, dx=\text {Too large to display} \] Input:

integrate(F^(c*(b*x+a))*sinh(e*x+d)^4,x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2409 vs. \(2 (143) = 286\).

Time = 25.03 (sec) , antiderivative size = 2409, normalized size of antiderivative = 14.87 \[ \int F^{c (a+b x)} \sinh ^4(d+e x) \, dx=\text {Too large to display} \] Input:

integrate(F**(c*(b*x+a))*sinh(e*x+d)**4,x)
 

Output:

Piecewise((x*sinh(d)**4, Eq(F, 1) & Eq(b, 0) & Eq(c, 0) & Eq(e, 0)), (3*x* 
sinh(d + e*x)**4/8 - 3*x*sinh(d + e*x)**2*cosh(d + e*x)**2/4 + 3*x*cosh(d 
+ e*x)**4/8 + 5*sinh(d + e*x)**3*cosh(d + e*x)/(8*e) - 3*sinh(d + e*x)*cos 
h(d + e*x)**3/(8*e), Eq(F, 1)), (F**(a*c)*(3*x*sinh(d + e*x)**4/8 - 3*x*si 
nh(d + e*x)**2*cosh(d + e*x)**2/4 + 3*x*cosh(d + e*x)**4/8 + 5*sinh(d + e* 
x)**3*cosh(d + e*x)/(8*e) - 3*sinh(d + e*x)*cosh(d + e*x)**3/(8*e)), Eq(b, 
 0)), (3*x*sinh(d + e*x)**4/8 - 3*x*sinh(d + e*x)**2*cosh(d + e*x)**2/4 + 
3*x*cosh(d + e*x)**4/8 + 5*sinh(d + e*x)**3*cosh(d + e*x)/(8*e) - 3*sinh(d 
 + e*x)*cosh(d + e*x)**3/(8*e), Eq(c, 0)), (F**(a*c + b*c*x)*x*sinh(b*c*x* 
log(F)/2 - d)**4/4 - F**(a*c + b*c*x)*x*sinh(b*c*x*log(F)/2 - d)**3*cosh(b 
*c*x*log(F)/2 - d)/2 + F**(a*c + b*c*x)*x*sinh(b*c*x*log(F)/2 - d)*cosh(b* 
c*x*log(F)/2 - d)**3/2 - F**(a*c + b*c*x)*x*cosh(b*c*x*log(F)/2 - d)**4/4 
+ F**(a*c + b*c*x)*sinh(b*c*x*log(F)/2 - d)**4/(24*b*c*log(F)) + 17*F**(a* 
c + b*c*x)*sinh(b*c*x*log(F)/2 - d)**3*cosh(b*c*x*log(F)/2 - d)/(12*b*c*lo 
g(F)) - F**(a*c + b*c*x)*sinh(b*c*x*log(F)/2 - d)**2*cosh(b*c*x*log(F)/2 - 
 d)**2/(b*c*log(F)) - 3*F**(a*c + b*c*x)*sinh(b*c*x*log(F)/2 - d)*cosh(b*c 
*x*log(F)/2 - d)**3/(4*b*c*log(F)) + 5*F**(a*c + b*c*x)*cosh(b*c*x*log(F)/ 
2 - d)**4/(8*b*c*log(F)), Eq(e, -b*c*log(F)/2)), (F**(a*c + b*c*x)*x*sinh( 
b*c*x*log(F)/4 - d)**4/16 - F**(a*c + b*c*x)*x*sinh(b*c*x*log(F)/4 - d)**3 
*cosh(b*c*x*log(F)/4 - d)/4 + 3*F**(a*c + b*c*x)*x*sinh(b*c*x*log(F)/4 ...
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.02 \[ \int F^{c (a+b x)} \sinh ^4(d+e x) \, dx=\frac {F^{a c} e^{\left (b c x \log \left (F\right ) + 4 \, e x + 4 \, d\right )}}{16 \, {\left (b c \log \left (F\right ) + 4 \, e\right )}} - \frac {F^{a c} e^{\left (b c x \log \left (F\right ) + 2 \, e x + 2 \, d\right )}}{4 \, {\left (b c \log \left (F\right ) + 2 \, e\right )}} - \frac {F^{a c} e^{\left (b c x \log \left (F\right ) - 2 \, e x\right )}}{4 \, {\left (b c e^{\left (2 \, d\right )} \log \left (F\right ) - 2 \, e e^{\left (2 \, d\right )}\right )}} + \frac {F^{a c} e^{\left (b c x \log \left (F\right ) - 4 \, e x\right )}}{16 \, {\left (b c e^{\left (4 \, d\right )} \log \left (F\right ) - 4 \, e e^{\left (4 \, d\right )}\right )}} + \frac {3 \, F^{b c x + a c}}{8 \, b c \log \left (F\right )} \] Input:

integrate(F^(c*(b*x+a))*sinh(e*x+d)^4,x, algorithm="maxima")
 

Output:

1/16*F^(a*c)*e^(b*c*x*log(F) + 4*e*x + 4*d)/(b*c*log(F) + 4*e) - 1/4*F^(a* 
c)*e^(b*c*x*log(F) + 2*e*x + 2*d)/(b*c*log(F) + 2*e) - 1/4*F^(a*c)*e^(b*c* 
x*log(F) - 2*e*x)/(b*c*e^(2*d)*log(F) - 2*e*e^(2*d)) + 1/16*F^(a*c)*e^(b*c 
*x*log(F) - 4*e*x)/(b*c*e^(4*d)*log(F) - 4*e*e^(4*d)) + 3/8*F^(b*c*x + a*c 
)/(b*c*log(F))
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.19 (sec) , antiderivative size = 1502, normalized size of antiderivative = 9.27 \[ \int F^{c (a+b x)} \sinh ^4(d+e x) \, dx=\text {Too large to display} \] Input:

integrate(F^(c*(b*x+a))*sinh(e*x+d)^4,x, algorithm="giac")
 

Output:

3/4*(2*b*c*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)*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*sgn(F) - 
 pi*b*c)^2))*e^(b*c*x*log(abs(F)) + a*c*log(abs(F))) + 3*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)/(8*I*p 
i*b*c*sgn(F) - 8*I*pi*b*c + 16*b*c*log(abs(F))) - 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)/(-8*I*pi*b*c*sg 
n(F) + 8*I*pi*b*c + 16*b*c*log(abs(F))))*e^(b*c*x*log(abs(F)) + a*c*log(ab 
s(F))) + 1/8*(2*(b*c*log(abs(F)) + 4*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)) + 4*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)) + 4*e)^2))*e^(a*c*log(abs(F)) + (b*c*log(abs(F) 
) + 4*e)*x + 4*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)/(16*I*pi*b*c*sgn(F) - 16*I*pi*b*c + 32*b*c* 
log(abs(F)) + 128*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)/(-16*I*pi*b*c*sgn(F) + 16*I*pi*b*c + 32*b* 
c*log(abs(F)) + 128*e))*e^(a*c*log(abs(F)) + (b*c*log(abs(F)) + 4*e)*x + 4 
*d) - 1/2*(2*(b*c*log(abs(F)) + 2*e)*cos(-1/2*pi*b*c*x*sgn(F) + 1/2*pi*...
 

Mupad [B] (verification not implemented)

Time = 5.08 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.59 \[ \int F^{c (a+b x)} \sinh ^4(d+e x) \, dx=\frac {F^{a\,c+b\,c\,x}\,\left (24\,e^4+\frac {3\,b^4\,c^4\,{\ln \left (F\right )}^4}{8}-\frac {b^4\,c^4\,{\ln \left (F\right )}^4\,\mathrm {cosh}\left (2\,d+2\,e\,x\right )}{2}+\frac {b^4\,c^4\,{\ln \left (F\right )}^4\,\mathrm {cosh}\left (4\,d+4\,e\,x\right )}{8}-\frac {15\,b^2\,c^2\,e^2\,{\ln \left (F\right )}^2}{2}+b^3\,c^3\,e\,{\ln \left (F\right )}^3\,\mathrm {sinh}\left (2\,d+2\,e\,x\right )-\frac {b^3\,c^3\,e\,{\ln \left (F\right )}^3\,\mathrm {sinh}\left (4\,d+4\,e\,x\right )}{2}-16\,b\,c\,e^3\,\ln \left (F\right )\,\mathrm {sinh}\left (2\,d+2\,e\,x\right )+2\,b\,c\,e^3\,\ln \left (F\right )\,\mathrm {sinh}\left (4\,d+4\,e\,x\right )+8\,b^2\,c^2\,e^2\,{\ln \left (F\right )}^2\,\mathrm {cosh}\left (2\,d+2\,e\,x\right )-\frac {b^2\,c^2\,e^2\,{\ln \left (F\right )}^2\,\mathrm {cosh}\left (4\,d+4\,e\,x\right )}{2}\right )}{b\,c\,\ln \left (F\right )\,\left (b^4\,c^4\,{\ln \left (F\right )}^4-20\,b^2\,c^2\,e^2\,{\ln \left (F\right )}^2+64\,e^4\right )} \] Input:

int(F^(c*(a + b*x))*sinh(d + e*x)^4,x)
 

Output:

(F^(a*c + b*c*x)*(24*e^4 + (3*b^4*c^4*log(F)^4)/8 - (b^4*c^4*log(F)^4*cosh 
(2*d + 2*e*x))/2 + (b^4*c^4*log(F)^4*cosh(4*d + 4*e*x))/8 - (15*b^2*c^2*e^ 
2*log(F)^2)/2 + b^3*c^3*e*log(F)^3*sinh(2*d + 2*e*x) - (b^3*c^3*e*log(F)^3 
*sinh(4*d + 4*e*x))/2 - 16*b*c*e^3*log(F)*sinh(2*d + 2*e*x) + 2*b*c*e^3*lo 
g(F)*sinh(4*d + 4*e*x) + 8*b^2*c^2*e^2*log(F)^2*cosh(2*d + 2*e*x) - (b^2*c 
^2*e^2*log(F)^2*cosh(4*d + 4*e*x))/2))/(b*c*log(F)*(64*e^4 + b^4*c^4*log(F 
)^4 - 20*b^2*c^2*e^2*log(F)^2))
 

Reduce [F]

\[ \int F^{c (a+b x)} \sinh ^4(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \sinh \left (e x +d \right )^{4}d x \right ) \] Input:

int(F^(c*(b*x+a))*sinh(e*x+d)^4,x)
 

Output:

f**(a*c)*int(f**(b*c*x)*sinh(d + e*x)**4,x)