\(\int F^{c (a+b x)} (g \sec (d+e x))^q (f \sin (d+e x))^p \, dx\) [160]

Optimal result

Integrand size = 30, antiderivative size = 173 \[ \int F^{c (a+b x)} (g \sec (d+e x))^q (f \sin (d+e x))^p \, dx=\frac {i 2^{-1-q} \left (e^{2 i (d+e x)}\right )^{\frac {1}{2} \left (p-q+\frac {i b c \log (F)}{e}\right )} \left (1-e^{2 i (d+e x)}\right ) \left (1+e^{2 i (d+e x)}\right )^q F^{c (a+b x)} \operatorname {AppellF1}\left (1+p,\frac {1}{2} \left (2+p-q+\frac {i b c \log (F)}{e}\right ),q,2+p,1-e^{2 i (d+e x)},\frac {1}{2} \left (1-e^{2 i (d+e x)}\right )\right ) (g \sec (d+e x))^q (f \sin (d+e x))^p}{e (1+p)} \] Output:

I*2^(-1-q)*exp(2*I*(e*x+d))^(1/2*p-1/2*q+1/2*I*b*c*ln(F)/e)*(1-exp(2*I*(e* 
x+d)))*(1+exp(2*I*(e*x+d)))^q*F^(c*(b*x+a))*AppellF1(p+1,1+1/2*p-1/2*q+1/2 
*I*b*c*ln(F)/e,q,2+p,1-exp(2*I*(e*x+d)),1/2-1/2*exp(2*I*(e*x+d)))*(g*sec(e 
*x+d))^q*(f*sin(e*x+d))^p/e/(p+1)
 

Mathematica [F]

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

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

Output:

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

Rubi [F]

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.

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

\(\Big \downarrow \) 7271

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

\(\Big \downarrow \) 7271

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7299

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

Input:

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

Output:

$Aborted
 

Defintions of rubi rules used

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])
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7299

Int[u_, x_] :> CannotIntegrate[u, x]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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