\(\int F^{c (a+b x)} \sqrt {f \sec (d+e x)} \sqrt {g \sec (d+e x)} \, dx\) [134]

Optimal result

Integrand size = 34, antiderivative size = 109 \[ \int F^{c (a+b x)} \sqrt {f \sec (d+e x)} \sqrt {g \sec (d+e x)} \, dx=\frac {2 e^{i (d+e x)} f F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right ) \sqrt {g \sec (d+e x)}}{(i e+b c \log (F)) \sqrt {f \sec (d+e x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.02 \[ \int F^{c (a+b x)} \sqrt {f \sec (d+e x)} \sqrt {g \sec (d+e x)} \, dx=-\frac {i \left (1+e^{2 i (d+e x)}\right ) F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {3}{2}-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) \sqrt {f \sec (d+e x)} \sqrt {g \sec (d+e x)}}{e-i b c \log (F)} \] Input:

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

Output:

((-I)*(1 + E^((2*I)*(d + e*x)))*F^(c*(a + b*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))]*Sqr 
t[f*Sec[d + e*x]]*Sqrt[g*Sec[d + e*x]])/(e - I*b*c*Log[F])
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {2031, 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.

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

\(\Big \downarrow \) 2031

\(\displaystyle \frac {f \sqrt {g \sec (d+e x)} \int F^{c (a+b x)} \sec (d+e x)dx}{\sqrt {f \sec (d+e x)}}\)

\(\Big \downarrow \) 4951

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

Input:

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

Output:

(2*E^(I*(d + e*x))*f*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))]*Sqrt[g*Sec[d + e 
*x]])/((I*e + b*c*Log[F])*Sqrt[f*Sec[d + e*x]])
 

Defintions of rubi rules used

rule 2031

Int[(Fx_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[a^(m + 1/ 
2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v])   Int[v^(m + n)*Fx, x], x] /; FreeQ[{a 
, b, m}, x] &&  !IntegerQ[m] && IGtQ[n + 1/2, 0] && IntegerQ[m + n]
 

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]

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

Input:

int(F^(c*(b*x+a))*(f*sec(e*x+d))^(1/2)*(g*sec(e*x+d))^(1/2),x)
 

Output:

int(F^(c*(b*x+a))*(f*sec(e*x+d))^(1/2)*(g*sec(e*x+d))^(1/2),x)
 

Fricas [F]

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

integrate(F^(c*(b*x+a))*(f*sec(e*x+d))^(1/2)*(g*sec(e*x+d))^(1/2),x, algor 
ithm="fricas")
 

Output:

integral(sqrt(f*sec(e*x + d))*sqrt(g*sec(e*x + d))*F^(b*c*x + a*c), x)
 

Sympy [F]

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

integrate(F**(c*(b*x+a))*(f*sec(e*x+d))**(1/2)*(g*sec(e*x+d))**(1/2),x)
 

Output:

Integral(F**(c*(a + b*x))*sqrt(f*sec(d + e*x))*sqrt(g*sec(d + e*x)), x)
 

Maxima [F]

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

integrate(F^(c*(b*x+a))*(f*sec(e*x+d))^(1/2)*(g*sec(e*x+d))^(1/2),x, algor 
ithm="maxima")
 

Output:

integrate(sqrt(f*sec(e*x + d))*sqrt(g*sec(e*x + d))*F^((b*x + a)*c), x)
 

Giac [F]

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

integrate(F^(c*(b*x+a))*(f*sec(e*x+d))^(1/2)*(g*sec(e*x+d))^(1/2),x, algor 
ithm="giac")
 

Output:

integrate(sqrt(f*sec(e*x + d))*sqrt(g*sec(e*x + d))*F^((b*x + a)*c), x)
 

Mupad [F(-1)]

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

int(F^(c*(a + b*x))*(f/cos(d + e*x))^(1/2)*(g/cos(d + e*x))^(1/2),x)
 

Output:

int(F^(c*(a + b*x))*(f/cos(d + e*x))^(1/2)*(g/cos(d + e*x))^(1/2), x)
 

Reduce [F]

\[ \int F^{c (a+b x)} \sqrt {f \sec (d+e x)} \sqrt {g \sec (d+e x)} \, dx=\sqrt {g}\, f^{a c +\frac {1}{2}} \left (\int f^{b c x} \sec \left (e x +d \right )d x \right ) \] Input:

int(F^(c*(b*x+a))*(f*sec(e*x+d))^(1/2)*(g*sec(e*x+d))^(1/2),x)
 

Output:

sqrt(g)*f**((2*a*c + 1)/2)*int(f**(b*c*x)*sec(d + e*x),x)