\(\int F^{c (a+b x)} \sec ^3(d+e x) \, dx\) [125]

Optimal result

Integrand size = 18, antiderivative size = 84 \[ \int F^{c (a+b x)} \sec ^3(d+e x) \, dx=\frac {8 e^{3 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (5-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right )}{3 i e+b c \log (F)} \] Output:

8*exp(3*I*(e*x+d))*F^(c*(b*x+a))*hypergeom([3, 3/2-1/2*I*b*c*ln(F)/e],[5/2 
-1/2*I*b*c*ln(F)/e],-exp(2*I*(e*x+d)))/(3*I*e+b*c*ln(F))
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.33 \[ \int F^{c (a+b x)} \sec ^3(d+e x) \, dx=\frac {F^{c (a+b x)} \left (2 e^{i (d+e x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {3}{2}-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) (-i e+b c \log (F))+\sec (d+e x) (-b c \log (F)+e \tan (d+e x))\right )}{2 e^2} \] Input:

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

Output:

(F^(c*(a + b*x))*(2*E^(I*(d + e*x))*Hypergeometric2F1[1, (e - I*b*c*Log[F] 
)/(2*e), 3/2 - ((I/2)*b*c*Log[F])/e, -E^((2*I)*(d + e*x))]*((-I)*e + b*c*L 
og[F]) + Sec[d + e*x]*(-(b*c*Log[F]) + e*Tan[d + e*x])))/(2*e^2)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.83, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4948, 4951}

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.

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

\(\Big \downarrow \) 4948

\(\displaystyle \frac {1}{2} \left (\frac {b^2 c^2 \log ^2(F)}{e^2}+1\right ) \int F^{c (a+b x)} \sec (d+e x)dx-\frac {b c \log (F) \sec (d+e x) F^{c (a+b x)}}{2 e^2}+\frac {\tan (d+e x) \sec (d+e x) F^{c (a+b x)}}{2 e}\)

\(\Big \downarrow \) 4951

\(\displaystyle \frac {e^{i (d+e x)} F^{c (a+b x)} \left (\frac {b^2 c^2 \log ^2(F)}{e^2}+1\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right )}{b c \log (F)+i e}-\frac {b c \log (F) \sec (d+e x) F^{c (a+b x)}}{2 e^2}+\frac {\tan (d+e x) \sec (d+e x) F^{c (a+b x)}}{2 e}\)

Input:

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

Output:

(E^(I*(d + e*x))*F^(c*(a + b*x))*Hypergeometric2F1[1, (e - I*b*c*Log[F])/( 
2*e), (3 - (I*b*c*Log[F])/e)/2, -E^((2*I)*(d + e*x))]*(1 + (b^2*c^2*Log[F] 
^2)/e^2))/(I*e + b*c*Log[F]) - (b*c*F^(c*(a + b*x))*Log[F]*Sec[d + e*x])/( 
2*e^2) + (F^(c*(a + b*x))*Sec[d + e*x]*Tan[d + e*x])/(2*e)
 

Defintions of rubi rules used

rule 4948

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

rule 4951

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_.), x_Symb 
ol] :> Simp[2^n*E^(I*n*(d + e*x))*(F^(c*(a + b*x))/(I*e*n + b*c*Log[F]))*Hy 
pergeometric2F1[n, n/2 - I*b*c*(Log[F]/(2*e)), 1 + n/2 - I*b*c*(Log[F]/(2*e 
)), -E^(2*I*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]
 

Maple [F]

\[\int F^{c \left (b x +a \right )} \sec \left (e x +d \right )^{3}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(F^(b*c*x + a*c)*sec(e*x + d)^3, x)
 

Sympy [F]

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

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

Output:

Integral(F**(c*(a + b*x))*sec(d + e*x)**3, x)
 

Maxima [F]

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

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

Output:

8*(48*F^(b*c*x)*F^(a*c)*b*c*e^2*cos(e*x + d)*log(F) + 6*(F^(a*c)*b^2*c^2*e 
*log(F)^2 - 15*F^(a*c)*e^3)*F^(b*c*x)*sin(e*x + d) + (48*F^(b*c*x)*F^(a*c) 
*b*c*e^2*cos(e*x + d)*log(F) + (F^(a*c)*b^3*c^3*log(F)^3 + 25*F^(a*c)*b*c* 
e^2*log(F))*F^(b*c*x)*cos(3*e*x + 3*d) - 3*(F^(a*c)*b^2*c^2*e*log(F)^2 + 2 
5*F^(a*c)*e^3)*F^(b*c*x)*sin(3*e*x + 3*d) + 6*(F^(a*c)*b^2*c^2*e*log(F)^2 
- 15*F^(a*c)*e^3)*F^(b*c*x)*sin(e*x + d))*cos(6*e*x + 6*d) + 3*(48*F^(b*c* 
x)*F^(a*c)*b*c*e^2*cos(e*x + d)*log(F) + (F^(a*c)*b^3*c^3*log(F)^3 + 25*F^ 
(a*c)*b*c*e^2*log(F))*F^(b*c*x)*cos(3*e*x + 3*d) - 3*(F^(a*c)*b^2*c^2*e*lo 
g(F)^2 + 25*F^(a*c)*e^3)*F^(b*c*x)*sin(3*e*x + 3*d) + 6*(F^(a*c)*b^2*c^2*e 
*log(F)^2 - 15*F^(a*c)*e^3)*F^(b*c*x)*sin(e*x + d))*cos(4*e*x + 4*d) + (3* 
(F^(a*c)*b^3*c^3*log(F)^3 + 25*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*cos(2*e*x 
 + 2*d) + 9*(F^(a*c)*b^2*c^2*e*log(F)^2 + 25*F^(a*c)*e^3)*F^(b*c*x)*sin(2* 
e*x + 2*d) + (F^(a*c)*b^3*c^3*log(F)^3 + 25*F^(a*c)*b*c*e^2*log(F))*F^(b*c 
*x))*cos(3*e*x + 3*d) + 18*(8*F^(b*c*x)*F^(a*c)*b*c*e^2*cos(e*x + d)*log(F 
) + (F^(a*c)*b^2*c^2*e*log(F)^2 - 15*F^(a*c)*e^3)*F^(b*c*x)*sin(e*x + d))* 
cos(2*e*x + 2*d) - 6*(F^(a*c)*b^5*c^5*e*cos(d)*log(F)^5 - F^(a*c)*b^4*c^4* 
e^2*log(F)^4*sin(d) + 34*F^(a*c)*b^3*c^3*e^3*cos(d)*log(F)^3 - 34*F^(a*c)* 
b^2*c^2*e^4*log(F)^2*sin(d) + 225*F^(a*c)*b*c*e^5*cos(d)*log(F) - 225*F^(a 
*c)*e^6*sin(d) + (F^(a*c)*b^5*c^5*e*cos(d)*log(F)^5 - F^(a*c)*b^4*c^4*e^2* 
log(F)^4*sin(d) + 34*F^(a*c)*b^3*c^3*e^3*cos(d)*log(F)^3 - 34*F^(a*c)*b...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \sec ^3(d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\cos \left (d+e\,x\right )}^3} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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