\(\int F^{c (a+b x)} \sec (d+e x) \, dx\) [14]

Optimal result

Integrand size = 16, antiderivative size = 84 \[ \int F^{c (a+b x)} \sec (d+e x) \, dx=\frac {2 e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right )}{i e+b c \log (F)} \]

[Out]

2*exp(I*(e*x+d))*F^(c*(b*x+a))*hypergeom([1, 1/2*(e-I*b*c*ln(F))/e],[3/2-1/2*I*b*c*ln(F)/e],-exp(2*I*(e*x+d)))
/(b*c*ln(F)+I*e)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {4536} \[ \int F^{c (a+b x)} \sec (d+e x) \, dx=\frac {2 e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right )}{b c \log (F)+i e} \]

[In]

Int[F^(c*(a + b*x))*Sec[d + e*x],x]

[Out]

(2*E^(I*(d + e*x))*F^(c*(a + b*x))*Hypergeometric2F1[1, (e - I*b*c*Log[F])/(2*e), (3 - (I*b*c*Log[F])/e)/2, -E
^((2*I)*(d + e*x))])/(I*e + b*c*Log[F])

Rule 4536

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Simp[2^n*E^(I*n*(d + e*x))*(
F^(c*(a + b*x))/(I*e*n + b*c*Log[F]))*Hypergeometric2F1[n, n/2 - I*b*c*(Log[F]/(2*e)), 1 + n/2 - I*b*c*(Log[F]
/(2*e)), -E^(2*I*(d + e*x))], x] /; FreeQ[{F, a, b, c, d, e}, x] && IntegerQ[n]

Rubi steps \begin{align*} \text {integral}& = \frac {2 e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {e-i b c \log (F)}{2 e},\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),-e^{2 i (d+e x)}\right )}{i e+b c \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} \sec (d+e x) \, dx=\frac {2 e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2}-\frac {i b c \log (F)}{2 e},\frac {3}{2}-\frac {i b c \log (F)}{2 e},-e^{2 i (d+e x)}\right )}{i e+b c \log (F)} \]

[In]

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

[Out]

(2*E^(I*(d + e*x))*F^(c*(a + b*x))*Hypergeometric2F1[1, 1/2 - ((I/2)*b*c*Log[F])/e, 3/2 - ((I/2)*b*c*Log[F])/e
, -E^((2*I)*(d + e*x))])/(I*e + b*c*Log[F])

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

2*(F^(b*c*x)*F^(a*c)*b*c*cos(e*x + d)*log(F) - F^(b*c*x)*F^(a*c)*e*sin(e*x + d) + (F^(b*c*x)*F^(a*c)*b*c*cos(e
*x + d)*log(F) - F^(b*c*x)*F^(a*c)*e*sin(e*x + d))*cos(2*e*x + 2*d) + 2*(F^(a*c)*b^2*c^2*e*log(F)^2 + F^(a*c)*
e^3 + (F^(a*c)*b^2*c^2*e*log(F)^2 + F^(a*c)*e^3)*cos(2*e*x + 2*d)^2 + (F^(a*c)*b^2*c^2*e*log(F)^2 + F^(a*c)*e^
3)*sin(2*e*x + 2*d)^2 + 2*(F^(a*c)*b^2*c^2*e*log(F)^2 + F^(a*c)*e^3)*cos(2*e*x + 2*d))*integrate((F^(b*c*x)*b*
c*log(F)*sin(e*x + d) + F^(b*c*x)*e*cos(e*x + d) + (F^(b*c*x)*b*c*log(F)*sin(e*x + d) + F^(b*c*x)*e*cos(e*x +
d))*cos(4*e*x + 4*d) + 2*(F^(b*c*x)*b*c*log(F)*sin(e*x + d) + F^(b*c*x)*e*cos(e*x + d))*cos(2*e*x + 2*d) - (F^
(b*c*x)*b*c*cos(e*x + d)*log(F) - F^(b*c*x)*e*sin(e*x + d))*sin(4*e*x + 4*d) - 2*(F^(b*c*x)*b*c*cos(e*x + d)*l
og(F) - F^(b*c*x)*e*sin(e*x + d))*sin(2*e*x + 2*d))/(b^2*c^2*log(F)^2 + (b^2*c^2*log(F)^2 + e^2)*cos(4*e*x + 4
*d)^2 + 4*(b^2*c^2*log(F)^2 + e^2)*cos(2*e*x + 2*d)^2 + (b^2*c^2*log(F)^2 + e^2)*sin(4*e*x + 4*d)^2 + 4*(b^2*c
^2*log(F)^2 + e^2)*sin(4*e*x + 4*d)*sin(2*e*x + 2*d) + 4*(b^2*c^2*log(F)^2 + e^2)*sin(2*e*x + 2*d)^2 + e^2 + 2
*(b^2*c^2*log(F)^2 + e^2 + 2*(b^2*c^2*log(F)^2 + e^2)*cos(2*e*x + 2*d))*cos(4*e*x + 4*d) + 4*(b^2*c^2*log(F)^2
+ e^2)*cos(2*e*x + 2*d)), x) + (F^(b*c*x)*F^(a*c)*b*c*log(F)*sin(e*x + d) + F^(b*c*x)*F^(a*c)*e*cos(e*x + d))
*sin(2*e*x + 2*d))/(b^2*c^2*log(F)^2 + (b^2*c^2*log(F)^2 + e^2)*cos(2*e*x + 2*d)^2 + (b^2*c^2*log(F)^2 + e^2)*
sin(2*e*x + 2*d)^2 + e^2 + 2*(b^2*c^2*log(F)^2 + e^2)*cos(2*e*x + 2*d))

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \sec (d+e x) \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{\cos \left (d+e\,x\right )} \,d x \]

[In]

int(F^(c*(a + b*x))/cos(d + e*x),x)

[Out]

int(F^(c*(a + b*x))/cos(d + e*x), x)