\(\int F^{c (a+b x)} \sec ^2(d+e x) \, dx\) [15]

Optimal result

Integrand size = 18, antiderivative size = 80 \[ \int F^{c (a+b x)} \sec ^2(d+e x) \, dx=\frac {4 e^{2 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{2 e},2-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{2 i e+b c \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} \sec ^2(d+e x) \, dx=\frac {4 e^{2 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{2 e},2-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{2 i e+b c \log (F)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {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.

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

\(\Big \downarrow \) 4951

\(\displaystyle \frac {4 e^{2 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{2 e},2-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{b c \log (F)+2 i e}\)

Input:

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

Output:

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

Defintions of rubi rules used

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]

\[\int F^{c \left (b x +a \right )} \sec \left (e x +d \right )^{2}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(F^(b*c*x + a*c)*sec(e*x + d)^2, x)
 

Sympy [F]

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

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

Output:

Integral(F**(c*(a + b*x))*sec(d + e*x)**2, x)
 

Maxima [F]

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

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

Output:

4*(24*F^(b*c*x)*F^(a*c)*b*c*e^2*log(F) + 2*(F^(a*c)*b^3*c^3*log(F)^3 + 16* 
F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*cos(2*e*x + 2*d)^2 + 2*(F^(a*c)*b^3*c^3* 
log(F)^3 + 16*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(2*e*x + 2*d)^2 + (F^(a 
*c)*b^3*c^3*log(F)^3 + 64*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*cos(2*e*x + 2* 
d) - 2*(5*F^(a*c)*b^2*c^2*e*log(F)^2 - 16*F^(a*c)*e^3)*F^(b*c*x)*sin(2*e*x 
 + 2*d) + (24*F^(b*c*x)*F^(a*c)*b*c*e^2*log(F) + (F^(a*c)*b^3*c^3*log(F)^3 
 + 16*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*cos(2*e*x + 2*d) - 2*(F^(a*c)*b^2* 
c^2*e*log(F)^2 + 16*F^(a*c)*e^3)*F^(b*c*x)*sin(2*e*x + 2*d))*cos(4*e*x + 4 
*d) - 4*(F^(a*c)*b^5*c^5*e*log(F)^5 + 20*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 64 
*F^(a*c)*b*c*e^5*log(F) + (F^(a*c)*b^5*c^5*e*log(F)^5 + 20*F^(a*c)*b^3*c^3 
*e^3*log(F)^3 + 64*F^(a*c)*b*c*e^5*log(F))*cos(4*e*x + 4*d)^2 + 4*(F^(a*c) 
*b^5*c^5*e*log(F)^5 + 20*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 64*F^(a*c)*b*c*e^5 
*log(F))*cos(2*e*x + 2*d)^2 + (F^(a*c)*b^5*c^5*e*log(F)^5 + 20*F^(a*c)*b^3 
*c^3*e^3*log(F)^3 + 64*F^(a*c)*b*c*e^5*log(F))*sin(4*e*x + 4*d)^2 + 4*(F^( 
a*c)*b^5*c^5*e*log(F)^5 + 20*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 64*F^(a*c)*b*c 
*e^5*log(F))*sin(4*e*x + 4*d)*sin(2*e*x + 2*d) + 4*(F^(a*c)*b^5*c^5*e*log( 
F)^5 + 20*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 64*F^(a*c)*b*c*e^5*log(F))*sin(2* 
e*x + 2*d)^2 + 2*(F^(a*c)*b^5*c^5*e*log(F)^5 + 20*F^(a*c)*b^3*c^3*e^3*log( 
F)^3 + 64*F^(a*c)*b*c*e^5*log(F) + 2*(F^(a*c)*b^5*c^5*e*log(F)^5 + 20*F^(a 
*c)*b^3*c^3*e^3*log(F)^3 + 64*F^(a*c)*b*c*e^5*log(F))*cos(2*e*x + 2*d))...
 

Giac [F]

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

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

Output:

integrate(F^((b*x + a)*c)*sec(e*x + d)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \sec ^2(d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\cos \left (d+e\,x\right )}^2} \,d x \] Input:

int(F^(c*(a + b*x))/cos(d + e*x)^2,x)
 

Output:

int(F^(c*(a + b*x))/cos(d + e*x)^2, x)
 

Reduce [F]

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

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

Output:

f**(a*c)*int(f**(b*c*x)*sec(d + e*x)**2,x)