\(\int \frac {F^{c (a+b x)}}{f+f \cos (d+e x)} \, dx\) [141]
Optimal result
Integrand size = 22, antiderivative size = 79 \[
\int \frac {F^{c (a+b x)}}{f+f \cos (d+e x)} \, dx=\frac {2 e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{e},2-\frac {i b c \log (F)}{e},-e^{i (d+e x)}\right )}{f (i e+b c \log (F))}
\]
[Out]
2*exp(I*(e*x+d))*F^(c*(b*x+a))*hypergeom([2, 1-I*b*c*ln(F)/e],[2-I*b*c*ln(F)/e],-exp(I*(e*x+d)))/f/(b*c*ln(F)+
I*e)
Rubi [A] (verified)
Time = 0.07 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number
of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4542, 4536}
\[
\int \frac {F^{c (a+b x)}}{f+f \cos (d+e x)} \, dx=\frac {2 e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{e},2-\frac {i b c \log (F)}{e},-e^{i (d+e x)}\right )}{f (b c \log (F)+i e)}
\]
[In]
Int[F^(c*(a + b*x))/(f + f*Cos[d + e*x]),x]
[Out]
(2*E^(I*(d + e*x))*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 - (I*b*c*Log[F])/e, 2 - (I*b*c*Log[F])/e, -E^(I*(d +
e*x))])/(f*(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]
Rule 4542
Int[(Cos[(d_.) + (e_.)*(x_)]*(g_.) + (f_))^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Dist[2^n*f^n,
Int[F^(c*(a + b*x))*Cos[d/2 + e*(x/2)]^(2*n), x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] && EqQ[f - g, 0] &
& ILtQ[n, 0]
Rubi steps \begin{align*}
\text {integral}& = \frac {\int F^{c (a+b x)} \sec ^2\left (\frac {d}{2}+\frac {e x}{2}\right ) \, dx}{2 f} \\ & = \frac {2 e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{e},2-\frac {i b c \log (F)}{e},-e^{i (d+e x)}\right )}{f (i e+b c \log (F))} \\
\end{align*}
Mathematica [A] (verified)
Time = 0.06 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.01
\[
\int \frac {F^{c (a+b x)}}{f+f \cos (d+e x)} \, dx=-\frac {2 i e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{e},2-\frac {i b c \log (F)}{e},-e^{i (d+e x)}\right )}{f (e-i b c \log (F))}
\]
[In]
Integrate[F^(c*(a + b*x))/(f + f*Cos[d + e*x]),x]
[Out]
((-2*I)*E^(I*(d + e*x))*F^(c*(a + b*x))*Hypergeometric2F1[2, 1 - (I*b*c*Log[F])/e, 2 - (I*b*c*Log[F])/e, -E^(I
*(d + e*x))])/(f*(e - I*b*c*Log[F]))
Maple [F]
\[\int \frac {F^{c \left (x b +a \right )}}{f +f \cos \left (e x +d \right )}d x\]
[In]
int(F^(c*(b*x+a))/(f+f*cos(e*x+d)),x)
[Out]
int(F^(c*(b*x+a))/(f+f*cos(e*x+d)),x)
Fricas [F]
\[
\int \frac {F^{c (a+b x)}}{f+f \cos (d+e x)} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{f \cos \left (e x + d\right ) + f} \,d x }
\]
[In]
integrate(F^(c*(b*x+a))/(f+f*cos(e*x+d)),x, algorithm="fricas")
[Out]
integral(F^(b*c*x + a*c)/(f*cos(e*x + d) + f), x)
Sympy [F]
\[
\int \frac {F^{c (a+b x)}}{f+f \cos (d+e x)} \, dx=\frac {\int \frac {F^{a c + b c x}}{\cos {\left (d + e x \right )} + 1}\, dx}{f}
\]
[In]
integrate(F**(c*(b*x+a))/(f+f*cos(e*x+d)),x)
[Out]
Integral(F**(a*c + b*c*x)/(cos(d + e*x) + 1), x)/f
Maxima [F]
\[
\int \frac {F^{c (a+b x)}}{f+f \cos (d+e x)} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{f \cos \left (e x + d\right ) + f} \,d x }
\]
[In]
integrate(F^(c*(b*x+a))/(f+f*cos(e*x+d)),x, algorithm="maxima")
[Out]
2*(6*F^(b*c*x)*F^(a*c)*b*c*e^2*log(F) + 2*(F^(a*c)*b^3*c^3*log(F)^3 + 4*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*cos(
e*x + d)^2 + 2*(F^(a*c)*b^3*c^3*log(F)^3 + 4*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(e*x + d)^2 + (F^(a*c)*b^3*c
^3*log(F)^3 + 16*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*cos(e*x + d) - (5*F^(a*c)*b^2*c^2*e*log(F)^2 - 4*F^(a*c)*e^
3)*F^(b*c*x)*sin(e*x + d) + (6*F^(b*c*x)*F^(a*c)*b*c*e^2*log(F) + (F^(a*c)*b^3*c^3*log(F)^3 + 4*F^(a*c)*b*c*e^
2*log(F))*F^(b*c*x)*cos(e*x + d) - (F^(a*c)*b^2*c^2*e*log(F)^2 + 4*F^(a*c)*e^3)*F^(b*c*x)*sin(e*x + d))*cos(2*
e*x + 2*d) - 2*((F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*b*c*e^5*log(F))*f*cos
(2*e*x + 2*d)^2 + 4*(F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*b*c*e^5*log(F))*f
*cos(e*x + d)^2 + (F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*b*c*e^5*log(F))*f*s
in(2*e*x + 2*d)^2 + 4*(F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*b*c*e^5*log(F))
*f*sin(2*e*x + 2*d)*sin(e*x + d) + 4*(F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*
b*c*e^5*log(F))*f*sin(e*x + d)^2 + 4*(F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*
b*c*e^5*log(F))*f*cos(e*x + d) + (F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*b*c*
e^5*log(F))*f + 2*(2*(F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*b*c*e^5*log(F))*
f*cos(e*x + d) + (F^(a*c)*b^5*c^5*e*log(F)^5 + 5*F^(a*c)*b^3*c^3*e^3*log(F)^3 + 4*F^(a*c)*b*c*e^5*log(F))*f)*c
os(2*e*x + 2*d))*integrate((3*F^(b*c*x)*b*c*e*cos(3*e*x + 3*d)*log(F) + 9*F^(b*c*x)*b*c*e*cos(2*e*x + 2*d)*log
(F) + 9*F^(b*c*x)*b*c*e*cos(e*x + d)*log(F) + 3*F^(b*c*x)*b*c*e*log(F) - (b^2*c^2*log(F)^2 - 2*e^2)*F^(b*c*x)*
sin(3*e*x + 3*d) - 3*(b^2*c^2*log(F)^2 - 2*e^2)*F^(b*c*x)*sin(2*e*x + 2*d) - 3*(b^2*c^2*log(F)^2 - 2*e^2)*F^(b
*c*x)*sin(e*x + d))/((b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(3*e*x + 3*d)^2 + 9*(b^4*c^4*log
(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(2*e*x + 2*d)^2 + 9*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 +
4*e^4)*f*cos(e*x + d)^2 + (b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(3*e*x + 3*d)^2 + 9*(b^4*c^
4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(2*e*x + 2*d)^2 + 18*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F
)^2 + 4*e^4)*f*sin(2*e*x + 2*d)*sin(e*x + d) + 9*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(e*x
+ d)^2 + 6*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(e*x + d) + (b^4*c^4*log(F)^4 + 5*b^2*c^2
*e^2*log(F)^2 + 4*e^4)*f + 2*(3*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(2*e*x + 2*d) + 3*(b^
4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(e*x + d) + (b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 +
4*e^4)*f)*cos(3*e*x + 3*d) + 6*(3*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(e*x + d) + (b^4*c
^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f)*cos(2*e*x + 2*d) + 6*((b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F
)^2 + 4*e^4)*f*sin(2*e*x + 2*d) + (b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(e*x + d))*sin(3*e*
x + 3*d)), x) + ((F^(a*c)*b^2*c^2*e*log(F)^2 + 4*F^(a*c)*e^3)*F^(b*c*x)*cos(e*x + d) + (F^(a*c)*b^3*c^3*log(F)
^3 + 4*F^(a*c)*b*c*e^2*log(F))*F^(b*c*x)*sin(e*x + d) - 2*(F^(a*c)*b^2*c^2*e*log(F)^2 - 2*F^(a*c)*e^3)*F^(b*c*
x))*sin(2*e*x + 2*d))/((b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(2*e*x + 2*d)^2 + 4*(b^4*c^4*l
og(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(e*x + d)^2 + (b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^
4)*f*sin(2*e*x + 2*d)^2 + 4*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(2*e*x + 2*d)*sin(e*x + d
) + 4*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*sin(e*x + d)^2 + 4*(b^4*c^4*log(F)^4 + 5*b^2*c^2*e
^2*log(F)^2 + 4*e^4)*f*cos(e*x + d) + (b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f + 2*(2*(b^4*c^4*lo
g(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*f*cos(e*x + d) + (b^4*c^4*log(F)^4 + 5*b^2*c^2*e^2*log(F)^2 + 4*e^4)*
f)*cos(2*e*x + 2*d))
Giac [F]
\[
\int \frac {F^{c (a+b x)}}{f+f \cos (d+e x)} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{f \cos \left (e x + d\right ) + f} \,d x }
\]
[In]
integrate(F^(c*(b*x+a))/(f+f*cos(e*x+d)),x, algorithm="giac")
[Out]
integrate(F^((b*x + a)*c)/(f*cos(e*x + d) + f), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {F^{c (a+b x)}}{f+f \cos (d+e x)} \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{f+f\,\cos \left (d+e\,x\right )} \,d x
\]
[In]
int(F^(c*(a + b*x))/(f + f*cos(d + e*x)),x)
[Out]
int(F^(c*(a + b*x))/(f + f*cos(d + e*x)), x)