\(\int F^{c (a+b x)} (f \cos (d+e x))^p (g \sec (d+e x))^q \, dx\) [154]

Optimal result

Integrand size = 30, antiderivative size = 129 \[ \int F^{c (a+b x)} (f \cos (d+e x))^p (g \sec (d+e x))^q \, dx=-\frac {\left (1+e^{2 i (d+e x)}\right )^{-p+q} F^{c (a+b x)} (f \cos (d+e x))^p \operatorname {Hypergeometric2F1}\left (-p+q,\frac {1}{2} \left (-p+q-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (2-p+q-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right ) (g \sec (d+e x))^q}{i e (p-q)-b c \log (F)} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 13.16 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.63 \[ \int F^{c (a+b x)} (f \cos (d+e x))^p (g \sec (d+e x))^q \, dx=\frac {i 2^{-p+q} \left (\frac {e^{i (d+e x)}}{1+e^{2 i (d+e x)}}\right )^q \left (1+e^{2 i (d+e x)}\right ) \left (e^{-i (d+e x)} \left (1+e^{2 i (d+e x)}\right )\right )^p F^{c (a+b x)} \cos ^{-p}(d+e x) (f \cos (d+e x))^p \operatorname {Hypergeometric2F1}\left (1,\frac {e (2+p-q)-i b c \log (F)}{2 e},\frac {e (2-p+q)-i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right ) \sec ^{-q}(d+e x) (g \sec (d+e x))^q}{e (p-q)+i b c \log (F)} \] Input:

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

Output:

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

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.

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

\(\Big \downarrow \) 7271

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

\(\Big \downarrow \) 7271

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7299

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

Input:

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

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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