\(\int F^{c (a+b x)} (f \sin (d+e x))^p (g \sin (d+e x))^q \, dx\) [150]

Optimal result

Integrand size = 30, antiderivative size = 138 \[ \int F^{c (a+b x)} (f \sin (d+e x))^p (g \sin (d+e x))^q \, dx=-\frac {\left (1-e^{2 i (d+e x)}\right )^{-p-q} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (-p-q,\frac {-e (p+q)-i b c \log (F)}{2 e},\frac {e (2-p-q)-i b c \log (F)}{2 e},e^{2 i (d+e x)}\right ) (f \sin (d+e x))^p (g \sin (d+e x))^q}{i e (p+q)-b c \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.99 \[ \int F^{c (a+b x)} (f \sin (d+e x))^p (g \sin (d+e x))^q \, dx=\frac {\left (1-e^{2 i (d+e x)}\right )^{-p-q} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (-p-q,-\frac {i (-i e (p+q)+b c \log (F))}{2 e},1-\frac {i (-i e (p+q)+b c \log (F))}{2 e},e^{2 i (d+e x)}\right ) (f \sin (d+e x))^p (g \sin (d+e x))^q}{-i e (p+q)+b c \log (F)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2034, 7271, 4940, 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.

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

\(\Big \downarrow \) 2034

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

\(\Big \downarrow \) 7271

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

\(\Big \downarrow \) 4940

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

\(\Big \downarrow \) 2689

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

Input:

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

Output:

-(((1 - E^((2*I)*(d + e*x)))^(-p - q)*F^(c*(a + b*x))*Hypergeometric2F1[-p 
 - q, -1/2*(e*(p + q) + I*b*c*Log[F])/e, (e*(2 - p - q) - I*b*c*Log[F])/(2 
*e), E^((2*I)*(d + e*x))]*(f*Sin[d + e*x])^p*(g*Sin[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 4940

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(n_), x_Symbo 
l] :> Simp[E^(I*n*(d + e*x))*(Sin[d + e*x]^n/(-1 + E^(2*I*(d + e*x)))^n) 
Int[F^(c*(a + b*x))*((-1 + E^(2*I*(d + e*x)))^n/E^(I*n*(d + e*x))), x], x] 
/; FreeQ[{F, a, b, c, d, e, n}, 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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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