\(\int F^{c (a+b x)} \sin ^n(d+e x) \, dx\) [1]

Optimal result

Integrand size = 18, antiderivative size = 107 \[ \int F^{c (a+b x)} \sin ^n(d+e x) \, dx=-\frac {\left (1-e^{2 i (d+e x)}\right )^{-n} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (-n,-\frac {e n+i b c \log (F)}{2 e},\frac {1}{2} \left (2-n-\frac {i b c \log (F)}{e}\right ),e^{2 i (d+e x)}\right ) \sin ^n(d+e x)}{i e n-b c \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.03 \[ \int F^{c (a+b x)} \sin ^n(d+e x) \, dx=\frac {\left (1-e^{2 i (d+e x)}\right )^{-n} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (-n,-\frac {i (-i e n+b c \log (F))}{2 e},1-\frac {i (-i e n+b c \log (F))}{2 e},e^{2 i (d+e x)}\right ) \sin ^n(d+e x)}{-i e n+b c \log (F)} \] Input:

Integrate[F^(c*(a + b*x))*Sin[d + e*x]^n,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {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.

\(\displaystyle \int F^{c (a+b x)} \sin ^n(d+e x) \, dx\)

\(\Big \downarrow \) 4940

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

\(\Big \downarrow \) 2689

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

Input:

Int[F^(c*(a + b*x))*Sin[d + e*x]^n,x]
 

Output:

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

Defintions of rubi rules used

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]
 

Maple [F]

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

Input:

int(F^(c*(b*x+a))*sin(e*x+d)^n,x)
 

Output:

int(F^(c*(b*x+a))*sin(e*x+d)^n,x)
 

Fricas [F]

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

integrate(F^(c*(b*x+a))*sin(e*x+d)^n,x, algorithm="fricas")
 

Output:

integral(F^(b*c*x + a*c)*sin(e*x + d)^n, x)
 

Sympy [F]

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

integrate(F**(c*(b*x+a))*sin(e*x+d)**n,x)
 

Output:

Integral(F**(c*(a + b*x))*sin(d + e*x)**n, x)
 

Maxima [F]

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

integrate(F^(c*(b*x+a))*sin(e*x+d)^n,x, algorithm="maxima")
 

Output:

integrate(F^((b*x + a)*c)*sin(e*x + d)^n, x)
 

Giac [F]

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

integrate(F^(c*(b*x+a))*sin(e*x+d)^n,x, algorithm="giac")
 

Output:

integrate(F^((b*x + a)*c)*sin(e*x + d)^n, x)
 

Mupad [F(-1)]

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

int(F^(c*(a + b*x))*sin(d + e*x)^n,x)
 

Output:

int(F^(c*(a + b*x))*sin(d + e*x)^n, x)
 

Reduce [F]

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

int(F^(c*(b*x+a))*sin(e*x+d)^n,x)
 

Output:

(f**(a*c)*(f**(b*c*x)*sin(d + e*x)**n - int((f**(b*c*x)*sin(d + e*x)**n*co 
s(d + e*x))/sin(d + e*x),x)*e*n))/(log(f)*b*c)