\(\int F^{c (a+b x)} (f \sec (d+e x))^p (g \sec (d+e x))^q \, dx\) [155]

Optimal result

Integrand size = 30, antiderivative size = 117 \[ \int F^{c (a+b x)} (f \sec (d+e x))^p (g \sec (d+e x))^q \, dx=\frac {\left (1+e^{2 i (d+e x)}\right )^{p+q} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (p+q,\frac {1}{2} \left (p+q-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (2+p+q-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right ) (f \sec (d+e x))^p (g \sec (d+e x))^q}{i e (p+q)+b c \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.08 \[ \int F^{c (a+b x)} (f \sec (d+e x))^p (g \sec (d+e x))^q \, dx=-\frac {i \left (1+e^{2 i (d+e x)}\right )^{p+q} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (p+q,\frac {e (p+q)-i b c \log (F)}{2 e},\frac {e (2+p+q)-i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) (f \sec (d+e x))^p (g \sec (d+e x))^q}{e (p+q)-i b c \log (F)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.33, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2034, 7271, 4954, 2689}

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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 7271

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

\(\Big \downarrow \) 4954

\(\displaystyle e^{-i (p+q) (d+e x)} \left (1+e^{2 i (d+e x)}\right )^{p+q} (f \sec (d+e x))^p (g \sec (d+e x))^q \int e^{i d (p+q)+i e x (p+q)} \left (1+e^{2 i (d+e x)}\right )^{-p-q} F^{a c+b x c}dx\)

\(\Big \downarrow \) 2689

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2034

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

rule 2689

Int[((a_) + (b_.)*(F_)^((e_.)*((c_.) + (d_.)*(x_))))^(p_)*(G_)^((h_.)*((f_. 
) + (g_.)*(x_)))*(H_)^((t_.)*((r_.) + (s_.)*(x_))), x_Symbol] :> Simp[G^(h* 
(f + g*x))*H^(t*(r + s*x))*((a + b*F^(e*(c + d*x)))^p/((g*h*Log[G] + s*t*Lo 
g[H])*((a + b*F^(e*(c + d*x)))/a)^p))*Hypergeometric2F1[-p, (g*h*Log[G] + s 
*t*Log[H])/(d*e*Log[F]), (g*h*Log[G] + s*t*Log[H])/(d*e*Log[F]) + 1, Simpli 
fy[(-b/a)*F^(e*(c + d*x))]], x] /; FreeQ[{F, G, H, a, b, c, d, e, f, g, h, 
r, s, t, p}, x] &&  !IntegerQ[p]
 

rule 4954

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

rule 7271

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a*v^m)^ 
FracPart[p]/v^(m*FracPart[p]))   Int[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, 
x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&  !(Eq 
Q[v, x] && EqQ[m, 1])
 

Maple [F]

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

Input:

int(F^(c*(b*x+a))*(f*sec(e*x+d))^p*(g*sec(e*x+d))^q,x)
 

Output:

int(F^(c*(b*x+a))*(f*sec(e*x+d))^p*(g*sec(e*x+d))^q,x)
 

Fricas [F]

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

integrate(F^(c*(b*x+a))*(f*sec(e*x+d))^p*(g*sec(e*x+d))^q,x, algorithm="fr 
icas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(F^(c*(b*x+a))*(f*sec(e*x+d))^p*(g*sec(e*x+d))^q,x, algorithm="ma 
xima")
 

Output:

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

Giac [F]

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

integrate(F^(c*(b*x+a))*(f*sec(e*x+d))^p*(g*sec(e*x+d))^q,x, algorithm="gi 
ac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(F^(c*(b*x+a))*(f*sec(e*x+d))^p*(g*sec(e*x+d))^q,x)
 

Output:

g**q*f**(a*c + p)*int(f**(b*c*x)*sec(d + e*x)**(p + q),x)