Integrand size = 18, antiderivative size = 100 \[ \int F^{c (a+b x)} \sec ^n(d+e x) \, dx=\frac {\left (1+e^{2 i (d+e x)}\right )^n F^{a c+b c x} \operatorname {Hypergeometric2F1}\left (n,\frac {e n-i b c \log (F)}{2 e},\frac {1}{2} \left (2+n-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right ) \sec ^n(d+e x)}{i e n+b c \log (F)} \] Output:
(1+exp(2*I*(e*x+d)))^n*F^(b*c*x+a*c)*hypergeom([n, 1/2*(e*n-I*b*c*ln(F))/e ],[1+1/2*n-1/2*I*b*c*ln(F)/e],-exp(2*I*(e*x+d)))*sec(e*x+d)^n/(I*e*n+b*c*l n(F))
Time = 0.07 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.02 \[ \int F^{c (a+b x)} \sec ^n(d+e x) \, dx=-\frac {i \left (1+e^{2 i (d+e x)}\right )^n F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (n,\frac {e n-i b c \log (F)}{2 e},\frac {1}{2} \left (2+n-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right ) \sec ^n(d+e x)}{e n-i b c \log (F)} \] Input:
Integrate[F^(c*(a + b*x))*Sec[d + e*x]^n,x]
Output:
((-I)*(1 + E^((2*I)*(d + e*x)))^n*F^(c*(a + b*x))*Hypergeometric2F1[n, (e* n - I*b*c*Log[F])/(2*e), (2 + n - (I*b*c*Log[F])/e)/2, -E^((2*I)*(d + e*x) )]*Sec[d + e*x]^n)/(e*n - I*b*c*Log[F])
Time = 0.36 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.26, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4954, 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.
\(\displaystyle \int F^{c (a+b x)} \sec ^n(d+e x) \, dx\) |
\(\Big \downarrow \) 4954 |
\(\displaystyle e^{-i n (d+e x)} \left (1+e^{2 i (d+e x)}\right )^n \sec ^n(d+e x) \int e^{i d n+i e x n} \left (1+e^{2 i (d+e x)}\right )^{-n} F^{a c+b x c}dx\) |
\(\Big \downarrow \) 2689 |
\(\displaystyle \frac {e^{-i n (d+e x)+i d n+i e n x} \left (1+e^{2 i (d+e x)}\right )^n F^{a c+b c x} \sec ^n(d+e x) \operatorname {Hypergeometric2F1}\left (n,\frac {e n-i b c \log (F)}{2 e},\frac {1}{2} \left (n-\frac {i b c \log (F)}{e}+2\right ),-e^{2 i (d+e x)}\right )}{b c \log (F)+i e n}\) |
Input:
Int[F^(c*(a + b*x))*Sec[d + e*x]^n,x]
Output:
(E^(I*d*n + I*e*n*x - I*n*(d + e*x))*(1 + E^((2*I)*(d + e*x)))^n*F^(a*c + b*c*x)*Hypergeometric2F1[n, (e*n - I*b*c*Log[F])/(2*e), (2 + n - (I*b*c*Lo g[F])/e)/2, -E^((2*I)*(d + e*x))]*Sec[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[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_.), x_Symb ol] :> Simp[(1 + E^(2*I*(d + e*x)))^n*(Sec[d + e*x]^n/E^(I*n*(d + e*x))) Int[SimplifyIntegrand[F^(c*(a + b*x))*(E^(I*n*(d + e*x))/(1 + E^(2*I*(d + e *x)))^n), x], x], x] /; FreeQ[{F, a, b, c, d, e}, x] && !IntegerQ[n]
\[\int F^{c \left (b x +a \right )} \sec \left (e x +d \right )^{n}d x\]
Input:
int(F^(c*(b*x+a))*sec(e*x+d)^n,x)
Output:
int(F^(c*(b*x+a))*sec(e*x+d)^n,x)
\[ \int F^{c (a+b x)} \sec ^n(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \sec \left (e x + d\right )^{n} \,d x } \] Input:
integrate(F^(c*(b*x+a))*sec(e*x+d)^n,x, algorithm="fricas")
Output:
integral(F^(b*c*x + a*c)*sec(e*x + d)^n, x)
\[ \int F^{c (a+b x)} \sec ^n(d+e x) \, dx=\int F^{c \left (a + b x\right )} \sec ^{n}{\left (d + e x \right )}\, dx \] Input:
integrate(F**(c*(b*x+a))*sec(e*x+d)**n,x)
Output:
Integral(F**(c*(a + b*x))*sec(d + e*x)**n, x)
\[ \int F^{c (a+b x)} \sec ^n(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \sec \left (e x + d\right )^{n} \,d x } \] Input:
integrate(F^(c*(b*x+a))*sec(e*x+d)^n,x, algorithm="maxima")
Output:
integrate(F^((b*x + a)*c)*sec(e*x + d)^n, x)
\[ \int F^{c (a+b x)} \sec ^n(d+e x) \, dx=\int { F^{{\left (b x + a\right )} c} \sec \left (e x + d\right )^{n} \,d x } \] Input:
integrate(F^(c*(b*x+a))*sec(e*x+d)^n,x, algorithm="giac")
Output:
integrate(F^((b*x + a)*c)*sec(e*x + d)^n, x)
Timed out. \[ \int F^{c (a+b x)} \sec ^n(d+e x) \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\left (\frac {1}{\cos \left (d+e\,x\right )}\right )}^n \,d x \] Input:
int(F^(c*(a + b*x))*(1/cos(d + e*x))^n,x)
Output:
int(F^(c*(a + b*x))*(1/cos(d + e*x))^n, x)
\[ \int F^{c (a+b x)} \sec ^n(d+e x) \, dx=f^{a c} \left (\int f^{b c x} \sec \left (e x +d \right )^{n}d x \right ) \] Input:
int(F^(c*(b*x+a))*sec(e*x+d)^n,x)
Output:
f**(a*c)*int(f**(b*c*x)*sec(d + e*x)**n,x)