Integrand size = 18, antiderivative size = 107 \[ \int F^{c (a+b x)} \cos ^n(d+e x) \, dx=-\frac {\left (1+e^{2 i (d+e x)}\right )^{-n} F^{c (a+b x)} \cos ^n(d+e x) \operatorname {Hypergeometric2F1}\left (-n,-\frac {e n+i b c \log (F)}{2 e},\frac {1}{2} \left (2-n-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right )}{i e n-b c \log (F)} \] Output:
-F^(c*(b*x+a))*cos(e*x+d)^n*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)))/((1+exp(2*I*(e*x+d)))^n)/(I*e*n- b*c*ln(F))
Time = 0.04 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.03 \[ \int F^{c (a+b x)} \cos ^n(d+e x) \, dx=\frac {\left (1+e^{2 i (d+e x)}\right )^{-n} F^{c (a+b x)} \cos ^n(d+e x) \operatorname {Hypergeometric2F1}\left (-n,-\frac {i (-i e n+b c \log (F))}{2 e},1-\frac {i (-i e n+b c \log (F))}{2 e},-e^{2 i (d+e x)}\right )}{-i e n+b c \log (F)} \] Input:
Integrate[F^(c*(a + b*x))*Cos[d + e*x]^n,x]
Output:
(F^(c*(a + b*x))*Cos[d + e*x]^n*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))])/((1 + E^((2*I)*(d + e*x)))^n*((-I)*e*n + b*c*Log[F]))
Time = 0.32 (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 = {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.
\(\displaystyle \int F^{c (a+b x)} \cos ^n(d+e x) \, dx\) |
\(\Big \downarrow \) 4941 |
\(\displaystyle e^{i n (d+e x)} \left (1+e^{2 i (d+e x)}\right )^{-n} \cos ^n(d+e x) \int e^{-i n (d+e x)} \left (1+e^{2 i (d+e x)}\right )^n F^{c (a+b x)}dx\) |
\(\Big \downarrow \) 2689 |
\(\displaystyle -\frac {\left (1+e^{2 i (d+e x)}\right )^{-n} F^{c (a+b x)} \cos ^n(d+e x) \operatorname {Hypergeometric2F1}\left (-n,-\frac {e n+i b c \log (F)}{2 e},\frac {1}{2} \left (-n-\frac {i b c \log (F)}{e}+2\right ),-e^{2 i (d+e x)}\right )}{-b c \log (F)+i e n}\) |
Input:
Int[F^(c*(a + b*x))*Cos[d + e*x]^n,x]
Output:
-((F^(c*(a + b*x))*Cos[d + e*x]^n*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))])/((1 + E^(( 2*I)*(d + e*x)))^n*(I*e*n - b*c*Log[F])))
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]
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]
\[\int F^{c \left (b x +a \right )} \cos \left (e x +d \right )^{n}d x\]
Input:
int(F^(c*(b*x+a))*cos(e*x+d)^n,x)
Output:
int(F^(c*(b*x+a))*cos(e*x+d)^n,x)
\[ \int F^{c (a+b x)} \cos ^n(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \cos \left (e x + d\right )^{n} \,d x } \] Input:
integrate(F^(c*(b*x+a))*cos(e*x+d)^n,x, algorithm="fricas")
Output:
integral(F^(b*c*x + a*c)*cos(e*x + d)^n, x)
\[ \int F^{c (a+b x)} \cos ^n(d+e x) \, dx=\int F^{c \left (a + b x\right )} \cos ^{n}{\left (d + e x \right )}\, dx \] Input:
integrate(F**(c*(b*x+a))*cos(e*x+d)**n,x)
Output:
Integral(F**(c*(a + b*x))*cos(d + e*x)**n, x)
\[ \int F^{c (a+b x)} \cos ^n(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \cos \left (e x + d\right )^{n} \,d x } \] Input:
integrate(F^(c*(b*x+a))*cos(e*x+d)^n,x, algorithm="maxima")
Output:
integrate(F^((b*x + a)*c)*cos(e*x + d)^n, x)
\[ \int F^{c (a+b x)} \cos ^n(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \cos \left (e x + d\right )^{n} \,d x } \] Input:
integrate(F^(c*(b*x+a))*cos(e*x+d)^n,x, algorithm="giac")
Output:
integrate(F^((b*x + a)*c)*cos(e*x + d)^n, x)
Timed out. \[ \int F^{c (a+b x)} \cos ^n(d+e x) \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\cos \left (d+e\,x\right )}^n \,d x \] Input:
int(F^(c*(a + b*x))*cos(d + e*x)^n,x)
Output:
int(F^(c*(a + b*x))*cos(d + e*x)^n, x)
\[ \int F^{c (a+b x)} \cos ^n(d+e x) \, dx=\frac {f^{a c} \left (f^{b c x} \cos \left (e x +d \right )^{n}+\left (\int \frac {f^{b c x} \cos \left (e x +d \right )^{n} \sin \left (e x +d \right )}{\cos \left (e x +d \right )}d x \right ) e n \right )}{\mathrm {log}\left (f \right ) b c} \] Input:
int(F^(c*(b*x+a))*cos(e*x+d)^n,x)
Output:
(f**(a*c)*(f**(b*c*x)*cos(d + e*x)**n + int((f**(b*c*x)*cos(d + e*x)**n*si n(d + e*x))/cos(d + e*x),x)*e*n))/(log(f)*b*c)