\(\int F^{c (a+b x)} \cosh ^3(d+e x) \, dx\) [286]

Optimal result

Integrand size = 18, antiderivative size = 202 \[ \int F^{c (a+b x)} \cosh ^3(d+e x) \, dx=-\frac {b c F^{c (a+b x)} \cosh ^3(d+e x) \log (F)}{9 e^2-b^2 c^2 \log ^2(F)}-\frac {6 b c e^2 F^{c (a+b x)} \cosh (d+e x) \log (F)}{9 e^4-10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)}+\frac {3 e F^{c (a+b x)} \cosh ^2(d+e x) \sinh (d+e x)}{9 e^2-b^2 c^2 \log ^2(F)}+\frac {6 e^3 F^{c (a+b x)} \sinh (d+e x)}{9 e^4-10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)} \] Output:

-b*c*F^(c*(b*x+a))*cosh(e*x+d)^3*ln(F)/(9*e^2-b^2*c^2*ln(F)^2)-6*b*c*e^2*F 
^(c*(b*x+a))*cosh(e*x+d)*ln(F)/(9*e^4-10*b^2*c^2*e^2*ln(F)^2+b^4*c^4*ln(F) 
^4)+3*e*F^(c*(b*x+a))*cosh(e*x+d)^2*sinh(e*x+d)/(9*e^2-b^2*c^2*ln(F)^2)+6* 
e^3*F^(c*(b*x+a))*sinh(e*x+d)/(9*e^4-10*b^2*c^2*e^2*ln(F)^2+b^4*c^4*ln(F)^ 
4)
 

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.79 \[ \int F^{c (a+b x)} \cosh ^3(d+e x) \, dx=\frac {F^{c (a+b x)} \left (3 \cosh (d+e x) \left (-9 b c e^2 \log (F)+b^3 c^3 \log ^3(F)\right )+\cosh (3 (d+e x)) \left (-b c e^2 \log (F)+b^3 c^3 \log ^3(F)\right )+6 e \left (5 e^2-b^2 c^2 \log ^2(F)+\cosh (2 (d+e x)) \left (e^2-b^2 c^2 \log ^2(F)\right )\right ) \sinh (d+e x)\right )}{4 \left (9 e^4-10 b^2 c^2 e^2 \log ^2(F)+b^4 c^4 \log ^4(F)\right )} \] Input:

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

Output:

(F^(c*(a + b*x))*(3*Cosh[d + e*x]*(-9*b*c*e^2*Log[F] + b^3*c^3*Log[F]^3) + 
 Cosh[3*(d + e*x)]*(-(b*c*e^2*Log[F]) + b^3*c^3*Log[F]^3) + 6*e*(5*e^2 - b 
^2*c^2*Log[F]^2 + Cosh[2*(d + e*x)]*(e^2 - b^2*c^2*Log[F]^2))*Sinh[d + e*x 
]))/(4*(9*e^4 - 10*b^2*c^2*e^2*Log[F]^2 + b^4*c^4*Log[F]^4))
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.94, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {6000, 5998}

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.

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

\(\Big \downarrow \) 6000

\(\displaystyle \frac {6 e^2 \int F^{c (a+b x)} \cosh (d+e x)dx}{9 e^2-b^2 c^2 \log ^2(F)}-\frac {b c \log (F) \cosh ^3(d+e x) F^{c (a+b x)}}{9 e^2-b^2 c^2 \log ^2(F)}+\frac {3 e \sinh (d+e x) \cosh ^2(d+e x) F^{c (a+b x)}}{9 e^2-b^2 c^2 \log ^2(F)}\)

\(\Big \downarrow \) 5998

\(\displaystyle -\frac {b c \log (F) \cosh ^3(d+e x) F^{c (a+b x)}}{9 e^2-b^2 c^2 \log ^2(F)}+\frac {3 e \sinh (d+e x) \cosh ^2(d+e x) F^{c (a+b x)}}{9 e^2-b^2 c^2 \log ^2(F)}+\frac {6 e^2 \left (\frac {e \sinh (d+e x) F^{c (a+b x)}}{e^2-b^2 c^2 \log ^2(F)}-\frac {b c \log (F) \cosh (d+e x) F^{c (a+b x)}}{e^2-b^2 c^2 \log ^2(F)}\right )}{9 e^2-b^2 c^2 \log ^2(F)}\)

Input:

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

Output:

-((b*c*F^(c*(a + b*x))*Cosh[d + e*x]^3*Log[F])/(9*e^2 - b^2*c^2*Log[F]^2)) 
 + (3*e*F^(c*(a + b*x))*Cosh[d + e*x]^2*Sinh[d + e*x])/(9*e^2 - b^2*c^2*Lo 
g[F]^2) + (6*e^2*(-((b*c*F^(c*(a + b*x))*Cosh[d + e*x]*Log[F])/(e^2 - b^2* 
c^2*Log[F]^2)) + (e*F^(c*(a + b*x))*Sinh[d + e*x])/(e^2 - b^2*c^2*Log[F]^2 
)))/(9*e^2 - b^2*c^2*Log[F]^2)
 

Defintions of rubi rules used

rule 5998

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

rule 6000

Int[Cosh[(d_.) + (e_.)*(x_)]^(n_)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symb 
ol] :> Simp[(-b)*c*Log[F]*F^(c*(a + b*x))*(Cosh[d + e*x]^n/(e^2*n^2 - b^2*c 
^2*Log[F]^2)), x] + (Simp[e*n*F^(c*(a + b*x))*Sinh[d + e*x]*(Cosh[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))*Cosh[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 = 2.32 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.73

method	result	size
parallelrisch	\(\frac {F^{c \left (b x +a \right )} \left (\left (\ln \left (F \right )^{3} b^{3} c^{3}-\ln \left (F \right ) b c \,e^{2}\right ) \cosh \left (3 e x +3 d \right )+\left (-3 \ln \left (F \right )^{2} b^{2} c^{2} e +3 e^{3}\right ) \sinh \left (3 e x +3 d \right )-3 \left (b c \ln \left (F \right )-3 e \right ) \left (b c \ln \left (F \right )+3 e \right ) \left (-\ln \left (F \right ) \cosh \left (e x +d \right ) b c +e \sinh \left (e x +d \right )\right )\right )}{36 e^{4}-40 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+4 b^{4} c^{4} \ln \left (F \right )^{4}}\)	\(148\)
risch	\(\frac {\left (\ln \left (F \right )^{3} b^{3} c^{3} {\mathrm e}^{6 e x +6 d}+3 \ln \left (F \right )^{3} b^{3} c^{3} {\mathrm e}^{4 e x +4 d}-3 \ln \left (F \right )^{2} b^{2} c^{2} e \,{\mathrm e}^{6 e x +6 d}+3 \ln \left (F \right )^{3} b^{3} c^{3} {\mathrm e}^{2 e x +2 d}-3 \ln \left (F \right )^{2} b^{2} c^{2} e \,{\mathrm e}^{4 e x +4 d}-\ln \left (F \right ) b c \,e^{2} {\mathrm e}^{6 e x +6 d}+\ln \left (F \right )^{3} b^{3} c^{3}+3 \ln \left (F \right )^{2} b^{2} c^{2} e \,{\mathrm e}^{2 e x +2 d}-27 \ln \left (F \right ) b c \,e^{2} {\mathrm e}^{4 e x +4 d}+3 e^{3} {\mathrm e}^{6 e x +6 d}+3 \ln \left (F \right )^{2} b^{2} c^{2} e -27 \ln \left (F \right ) b c \,e^{2} {\mathrm e}^{2 e x +2 d}+27 e^{3} {\mathrm e}^{4 e x +4 d}-\ln \left (F \right ) b c \,e^{2}-27 e^{3} {\mathrm e}^{2 e x +2 d}-3 e^{3}\right ) {\mathrm e}^{-3 e x -3 d} F^{c \left (b x +a \right )}}{8 \left (b c \ln \left (F \right )-e \right ) \left (b c \ln \left (F \right )-3 e \right ) \left (e +b c \ln \left (F \right )\right ) \left (b c \ln \left (F \right )+3 e \right )}\)	\(326\)
orering	\(\frac {4 \ln \left (F \right ) b c \left (b^{2} c^{2} \ln \left (F \right )^{2}-5 e^{2}\right ) F^{c \left (b x +a \right )} \cosh \left (e x +d \right )^{3}}{9 e^{4}-10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}-\frac {2 \left (3 b^{2} c^{2} \ln \left (F \right )^{2}-5 e^{2}\right ) \left (F^{c \left (b x +a \right )} b c \ln \left (F \right ) \cosh \left (e x +d \right )^{3}+3 F^{c \left (b x +a \right )} \cosh \left (e x +d \right )^{2} e \sinh \left (e x +d \right )\right )}{9 e^{4}-10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}+\frac {4 b c \ln \left (F \right ) \left (F^{c \left (b x +a \right )} b^{2} c^{2} \ln \left (F \right )^{2} \cosh \left (e x +d \right )^{3}+6 F^{c \left (b x +a \right )} b c \ln \left (F \right ) \cosh \left (e x +d \right )^{2} e \sinh \left (e x +d \right )+6 F^{c \left (b x +a \right )} \cosh \left (e x +d \right ) e^{2} \sinh \left (e x +d \right )^{2}+3 F^{c \left (b x +a \right )} \cosh \left (e x +d \right )^{3} e^{2}\right )}{9 e^{4}-10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}-\frac {F^{c \left (b x +a \right )} b^{3} c^{3} \ln \left (F \right )^{3} \cosh \left (e x +d \right )^{3}+9 F^{c \left (b x +a \right )} b^{2} c^{2} \ln \left (F \right )^{2} \cosh \left (e x +d \right )^{2} e \sinh \left (e x +d \right )+18 F^{c \left (b x +a \right )} b c \ln \left (F \right ) \cosh \left (e x +d \right ) e^{2} \sinh \left (e x +d \right )^{2}+9 F^{c \left (b x +a \right )} b c \ln \left (F \right ) \cosh \left (e x +d \right )^{3} e^{2}+6 F^{c \left (b x +a \right )} e^{3} \sinh \left (e x +d \right )^{3}+21 F^{c \left (b x +a \right )} \cosh \left (e x +d \right )^{2} e^{3} \sinh \left (e x +d \right )}{9 e^{4}-10 b^{2} c^{2} e^{2} \ln \left (F \right )^{2}+b^{4} c^{4} \ln \left (F \right )^{4}}\)	\(537\)

Input:

int(F^(c*(b*x+a))*cosh(e*x+d)^3,x,method=_RETURNVERBOSE)
 

Output:

F^(c*(b*x+a))*((ln(F)^3*b^3*c^3-ln(F)*b*c*e^2)*cosh(3*e*x+3*d)+(-3*ln(F)^2 
*b^2*c^2*e+3*e^3)*sinh(3*e*x+3*d)-3*(b*c*ln(F)-3*e)*(b*c*ln(F)+3*e)*(-ln(F 
)*cosh(e*x+d)*b*c+e*sinh(e*x+d)))/(36*e^4-40*b^2*c^2*e^2*ln(F)^2+4*b^4*c^4 
*ln(F)^4)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2218 vs. \(2 (199) = 398\).

Time = 0.17 (sec) , antiderivative size = 2218, normalized size of antiderivative = 10.98 \[ \int F^{c (a+b x)} \cosh ^3(d+e x) \, dx=\text {Too large to display} \] Input:

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

Output:

1/8*((3*e^3*cosh(e*x + d)^6 + 27*e^3*cosh(e*x + d)^4 + (b^3*c^3*log(F)^3 - 
 3*b^2*c^2*e*log(F)^2 - b*c*e^2*log(F) + 3*e^3)*sinh(e*x + d)^6 + 6*(b^3*c 
^3*cosh(e*x + d)*log(F)^3 - 3*b^2*c^2*e*cosh(e*x + d)*log(F)^2 - b*c*e^2*c 
osh(e*x + d)*log(F) + 3*e^3*cosh(e*x + d))*sinh(e*x + d)^5 - 27*e^3*cosh(e 
*x + d)^2 + 3*(15*e^3*cosh(e*x + d)^2 + (5*b^3*c^3*cosh(e*x + d)^2 + b^3*c 
^3)*log(F)^3 + 9*e^3 - (15*b^2*c^2*e*cosh(e*x + d)^2 + b^2*c^2*e)*log(F)^2 
 - (5*b*c*e^2*cosh(e*x + d)^2 + 9*b*c*e^2)*log(F))*sinh(e*x + d)^4 + (b^3* 
c^3*cosh(e*x + d)^6 + 3*b^3*c^3*cosh(e*x + d)^4 + 3*b^3*c^3*cosh(e*x + d)^ 
2 + b^3*c^3)*log(F)^3 + 4*(15*e^3*cosh(e*x + d)^3 + 27*e^3*cosh(e*x + d) + 
 (5*b^3*c^3*cosh(e*x + d)^3 + 3*b^3*c^3*cosh(e*x + d))*log(F)^3 - 3*(5*b^2 
*c^2*e*cosh(e*x + d)^3 + b^2*c^2*e*cosh(e*x + d))*log(F)^2 - (5*b*c*e^2*co 
sh(e*x + d)^3 + 27*b*c*e^2*cosh(e*x + d))*log(F))*sinh(e*x + d)^3 - 3*e^3 
- 3*(b^2*c^2*e*cosh(e*x + d)^6 + b^2*c^2*e*cosh(e*x + d)^4 - b^2*c^2*e*cos 
h(e*x + d)^2 - b^2*c^2*e)*log(F)^2 + 3*(15*e^3*cosh(e*x + d)^4 + 54*e^3*co 
sh(e*x + d)^2 + (5*b^3*c^3*cosh(e*x + d)^4 + 6*b^3*c^3*cosh(e*x + d)^2 + b 
^3*c^3)*log(F)^3 - 9*e^3 - (15*b^2*c^2*e*cosh(e*x + d)^4 + 6*b^2*c^2*e*cos 
h(e*x + d)^2 - b^2*c^2*e)*log(F)^2 - (5*b*c*e^2*cosh(e*x + d)^4 + 54*b*c*e 
^2*cosh(e*x + d)^2 + 9*b*c*e^2)*log(F))*sinh(e*x + d)^2 - (b*c*e^2*cosh(e* 
x + d)^6 + 27*b*c*e^2*cosh(e*x + d)^4 + 27*b*c*e^2*cosh(e*x + d)^2 + b*c*e 
^2)*log(F) + 6*(3*e^3*cosh(e*x + d)^5 + 18*e^3*cosh(e*x + d)^3 - 9*e^3*...
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1610 vs. \(2 (199) = 398\).

Time = 3.31 (sec) , antiderivative size = 1610, normalized size of antiderivative = 7.97 \[ \int F^{c (a+b x)} \cosh ^3(d+e x) \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((x*cosh(d)**3, Eq(F, 1) & Eq(e, 0)), (F**(a*c)*x*cosh(d)**3, Eq( 
b, 0) & Eq(e, 0)), (x*cosh(d)**3, Eq(c, 0) & Eq(e, 0)), (3*F**(a*c + b*c*x 
)*x*sinh(b*c*x*log(F) - d)**3/8 - 3*F**(a*c + b*c*x)*x*sinh(b*c*x*log(F) - 
 d)**2*cosh(b*c*x*log(F) - d)/8 - 3*F**(a*c + b*c*x)*x*sinh(b*c*x*log(F) - 
 d)*cosh(b*c*x*log(F) - d)**2/8 + 3*F**(a*c + b*c*x)*x*cosh(b*c*x*log(F) - 
 d)**3/8 - 5*F**(a*c + b*c*x)*sinh(b*c*x*log(F) - d)**3/(8*b*c*log(F)) + F 
**(a*c + b*c*x)*sinh(b*c*x*log(F) - d)**2*cosh(b*c*x*log(F) - d)/(4*b*c*lo 
g(F)) + F**(a*c + b*c*x)*sinh(b*c*x*log(F) - d)*cosh(b*c*x*log(F) - d)**2/ 
(b*c*log(F)) - 3*F**(a*c + b*c*x)*cosh(b*c*x*log(F) - d)**3/(8*b*c*log(F)) 
, Eq(e, -b*c*log(F))), (-F**(a*c + b*c*x)*x*sinh(b*c*x*log(F)/3 - d)**3/8 
+ 3*F**(a*c + b*c*x)*x*sinh(b*c*x*log(F)/3 - d)**2*cosh(b*c*x*log(F)/3 - d 
)/8 - 3*F**(a*c + b*c*x)*x*sinh(b*c*x*log(F)/3 - d)*cosh(b*c*x*log(F)/3 - 
d)**2/8 + F**(a*c + b*c*x)*x*cosh(b*c*x*log(F)/3 - d)**3/8 + 11*F**(a*c + 
b*c*x)*sinh(b*c*x*log(F)/3 - d)**3/(8*b*c*log(F)) - 15*F**(a*c + b*c*x)*si 
nh(b*c*x*log(F)/3 - d)**2*cosh(b*c*x*log(F)/3 - d)/(4*b*c*log(F)) + 3*F**( 
a*c + b*c*x)*sinh(b*c*x*log(F)/3 - d)*cosh(b*c*x*log(F)/3 - d)**2/(b*c*log 
(F)) - F**(a*c + b*c*x)*cosh(b*c*x*log(F)/3 - d)**3/(8*b*c*log(F)), Eq(e, 
-b*c*log(F)/3)), (-F**(a*c + b*c*x)*x*sinh(b*c*x*log(F)/3 + d)**3/8 + 3*F* 
*(a*c + b*c*x)*x*sinh(b*c*x*log(F)/3 + d)**2*cosh(b*c*x*log(F)/3 + d)/8 - 
3*F**(a*c + b*c*x)*x*sinh(b*c*x*log(F)/3 + d)*cosh(b*c*x*log(F)/3 + d)*...
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.66 \[ \int F^{c (a+b x)} \cosh ^3(d+e x) \, dx=\frac {F^{a c} e^{\left (b c x \log \left (F\right ) + 3 \, e x + 3 \, d\right )}}{8 \, {\left (b c \log \left (F\right ) + 3 \, e\right )}} + \frac {3 \, F^{a c} e^{\left (b c x \log \left (F\right ) + e x + d\right )}}{8 \, {\left (b c \log \left (F\right ) + e\right )}} + \frac {3 \, F^{a c} e^{\left (b c x \log \left (F\right ) - e x\right )}}{8 \, {\left (b c e^{d} \log \left (F\right ) - e e^{d}\right )}} + \frac {F^{a c} e^{\left (b c x \log \left (F\right ) - 3 \, e x\right )}}{8 \, {\left (b c e^{\left (3 \, d\right )} \log \left (F\right ) - 3 \, e e^{\left (3 \, d\right )}\right )}} \] Input:

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

Output:

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

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 1211, normalized size of antiderivative = 6.00 \[ \int F^{c (a+b x)} \cosh ^3(d+e x) \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(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(ab 
s(F)) + 3*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)) + 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)/(8*I*pi*b*c*sgn(F) - 8*I*pi*b*c + 16*b*c*log(abs(F)) 
 + 48*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)/(-8*I*pi*b*c*sgn(F) + 8*I*pi*b*c + 16*b*c*log(abs(F)) 
+ 48*e))*e^(a*c*log(abs(F)) + (b*c*log(abs(F)) + 3*e)*x + 3*d) + 3/4*(2*(b 
*c*log(abs(F)) + e)*cos(-1/2*pi*b*c*x*sgn(F) + 1/2*pi*b*c*x - 1/2*pi*a*c*s 
gn(F) + 1/2*pi*a*c)/((pi*b*c*sgn(F) - pi*b*c)^2 + 4*(b*c*log(abs(F)) + 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)) + e)^2))*e^(a*c*log(abs(F)) + (b*c*log(abs(F)) + e)*x + d) + 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*pi*b*c*sgn(F) - 8*I*pi*b*c + 16*b*c*log(abs(F)) + 16*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)/(-8*I*pi*b*c*sgn(F) + 8*I*pi*b*c + 16*b*c*log(abs(F)) + 16*e))*e^(a*c*l 
og(abs(F)) + (b*c*log(abs(F)) + e)*x + d) + 3/4*(2*(b*c*log(abs(F)) - e...
 

Mupad [B] (verification not implemented)

Time = 3.00 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.76 \[ \int F^{c (a+b x)} \cosh ^3(d+e x) \, dx=\frac {F^{a\,c+b\,c\,x}\,\left (6\,e^3\,\mathrm {sinh}\left (d+e\,x\right )+3\,e^3\,{\mathrm {cosh}\left (d+e\,x\right )}^2\,\mathrm {sinh}\left (d+e\,x\right )+b^3\,c^3\,{\mathrm {cosh}\left (d+e\,x\right )}^3\,{\ln \left (F\right )}^3-b\,c\,e^2\,{\mathrm {cosh}\left (d+e\,x\right )}^3\,\ln \left (F\right )-6\,b\,c\,e^2\,\mathrm {cosh}\left (d+e\,x\right )\,\ln \left (F\right )-3\,b^2\,c^2\,e\,{\mathrm {cosh}\left (d+e\,x\right )}^2\,\mathrm {sinh}\left (d+e\,x\right )\,{\ln \left (F\right )}^2\right )}{b^4\,c^4\,{\ln \left (F\right )}^4-10\,b^2\,c^2\,e^2\,{\ln \left (F\right )}^2+9\,e^4} \] Input:

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

Output:

(F^(a*c + b*c*x)*(6*e^3*sinh(d + e*x) + 3*e^3*cosh(d + e*x)^2*sinh(d + e*x 
) + b^3*c^3*cosh(d + e*x)^3*log(F)^3 - b*c*e^2*cosh(d + e*x)^3*log(F) - 6* 
b*c*e^2*cosh(d + e*x)*log(F) - 3*b^2*c^2*e*cosh(d + e*x)^2*sinh(d + e*x)*l 
og(F)^2))/(9*e^4 + b^4*c^4*log(F)^4 - 10*b^2*c^2*e^2*log(F)^2)
 

Reduce [F]

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

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

Output:

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