\(\int F^{c (a+b x)} (f-f \sin (d+e x))^n \, dx\) [33]

Optimal result

Integrand size = 23, antiderivative size = 112 \[ \int F^{c (a+b x)} (f-f \sin (d+e x))^n \, dx=-\frac {\left (1+e^{\frac {1}{2} i (2 d+\pi +2 e x)}\right )^{-2 n} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (-2 n,-n-\frac {i b c \log (F)}{e},1-n-\frac {i b c \log (F)}{e},-i e^{i (d+e x)}\right ) (f-f \sin (d+e x))^n}{i e n-b c \log (F)} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 2.32 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.98 \[ \int F^{c (a+b x)} (f-f \sin (d+e x))^n \, dx=\frac {F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (-2 n,-n-\frac {i b c \log (F)}{e},1-n-\frac {i b c \log (F)}{e},i e^{-i (d+e x)}\right ) (f-f \sin (d+e x))^n}{i \left (1-i e^{-i (d+e x)}\right )^{2 n} (e n+i b c \log (F))+\frac {\csc ^2\left (\frac {1}{2} (d+e x)\right ) \operatorname {Hypergeometric2F1}\left (-2 n,-n-\frac {i b c \log (F)}{e},1-n-\frac {i b c \log (F)}{e},i e^{-i (d+e x)}\right ) (e n-b c \log (F)+b c \cos (d+e x) \log (F)+b c \log (F) \sin (d+e x))}{-1+\cot \left (\frac {1}{2} (d+e x)\right )}} \] Input:

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

Output:

(F^(c*(a + b*x))*Hypergeometric2F1[-2*n, -n - (I*b*c*Log[F])/e, 1 - n - (I 
*b*c*Log[F])/e, I/E^(I*(d + e*x))]*(f - f*Sin[d + e*x])^n)/(I*(1 - I/E^(I* 
(d + e*x)))^(2*n)*(e*n + I*b*c*Log[F]) + (Csc[(d + e*x)/2]^2*Hypergeometri 
c2F1[-2*n, -n - (I*b*c*Log[F])/e, 1 - n - (I*b*c*Log[F])/e, I/E^(I*(d + e* 
x))]*(e*n - b*c*Log[F] + b*c*Cos[d + e*x]*Log[F] + b*c*Log[F]*Sin[d + e*x] 
))/(-1 + Cot[(d + e*x)/2]))
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4959, 4941, 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.

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

\(\Big \downarrow \) 4959

\(\displaystyle \cos ^{-2 n}\left (\frac {d}{2}+\frac {e x}{2}+\frac {\pi }{4}\right ) (f-f \sin (d+e x))^n \int F^{c (a+b x)} \cos ^{2 n}\left (\frac {d}{2}+\frac {e x}{2}+\frac {\pi }{4}\right )dx\)

\(\Big \downarrow \) 4941

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

\(\Big \downarrow \) 2689

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

Input:

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

Output:

-((F^(c*(a + b*x))*Hypergeometric2F1[-2*n, -n - (I*b*c*Log[F])/e, 1 - n - 
(I*b*c*Log[F])/e, (-I)*E^(I*(d + e*x))]*(f - f*Sin[d + e*x])^n)/((1 + E^(( 
I/2)*(2*d + Pi + 2*e*x)))^(2*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 4941

Int[Cos[(d_.) + (e_.)*(x_)]^(n_)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbo 
l] :> Simp[E^(I*n*(d + e*x))*(Cos[d + e*x]^n/(1 + E^(2*I*(d + e*x)))^n)   I 
nt[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 4959

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_) + (g_.)*Sin[(d_.) + (e_.)*(x_)] 
)^(n_.), x_Symbol] :> Simp[(f + g*Sin[d + e*x])^n/Cos[d/2 - f*(Pi/(4*g)) + 
e*(x/2)]^(2*n)   Int[F^(c*(a + b*x))*Cos[d/2 - f*(Pi/(4*g)) + e*(x/2)]^(2*n 
), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && EqQ[f^2 - g^2, 0] && 
 !IntegerQ[n]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

\[ \int F^{c (a+b x)} (f-f \sin (d+e x))^n \, dx=\int F^{c \left (a + b x\right )} \left (- f \left (\sin {\left (d + e x \right )} - 1\right )\right )^{n}\, dx \] Input:

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

Output:

Integral(F**(c*(a + b*x))*(-f*(sin(d + e*x) - 1))**n, x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int F^{c (a+b x)} (f-f \sin (d+e x))^n \, dx=\frac {f^{a c} \left (f^{b c x} \left (-\sin \left (e x +d \right ) f +f \right )^{n}-\left (\int \frac {f^{b c x} \left (-\sin \left (e x +d \right ) f +f \right )^{n} \cos \left (e x +d \right )}{\sin \left (e x +d \right )-1}d x \right ) e n \right )}{\mathrm {log}\left (f \right ) b c} \] Input:

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

Output:

(f**(a*c)*(f**(b*c*x)*( - sin(d + e*x)*f + f)**n - int((f**(b*c*x)*( - sin 
(d + e*x)*f + f)**n*cos(d + e*x))/(sin(d + e*x) - 1),x)*e*n))/(log(f)*b*c)