\(\int \frac {F^{c (a+b x)}}{f+f \cos (d+e x)} \, dx\) [36]

Optimal result

Integrand size = 22, antiderivative size = 79 \[ \int \frac {F^{c (a+b x)}}{f+f \cos (d+e x)} \, dx=\frac {2 e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{e},2-\frac {i b c \log (F)}{e},-e^{i (d+e x)}\right )}{f (i e+b c \log (F))} \] Output:

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/(I*e+b*c*ln(F))
 

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.01 \[ \int \frac {F^{c (a+b x)}}{f+f \cos (d+e x)} \, dx=-\frac {2 i e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{e},2-\frac {i b c \log (F)}{e},-e^{i (d+e x)}\right )}{f (e-i b c \log (F))} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.26 (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 = {4957, 4951}

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.

\(\displaystyle \int \frac {F^{c (a+b x)}}{f \cos (d+e x)+f} \, dx\)

\(\Big \downarrow \) 4957

\(\displaystyle \frac {\int F^{c (a+b x)} \sec ^2\left (\frac {d}{2}+\frac {e x}{2}\right )dx}{2 f}\)

\(\Big \downarrow \) 4951

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

Input:

Int[F^(c*(a + b*x))/(f + f*Cos[d + e*x]),x]
 

Output:

(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]))
 

Defintions of rubi rules used

rule 4951

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sec[(d_.) + (e_.)*(x_)]^(n_.), x_Symb 
ol] :> Simp[2^n*E^(I*n*(d + e*x))*(F^(c*(a + b*x))/(I*e*n + b*c*Log[F]))*Hy 
pergeometric2F1[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 4957

Int[(Cos[(d_.) + (e_.)*(x_)]*(g_.) + (f_))^(n_.)*(F_)^((c_.)*((a_.) + (b_.) 
*(x_))), x_Symbol] :> Simp[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] && IL 
tQ[n, 0]
 

Maple [F]

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

Input:

int(F^(c*(b*x+a))/(f+f*cos(e*x+d)),x)
 

Output:

int(F^(c*(b*x+a))/(f+f*cos(e*x+d)),x)
 

Fricas [F]

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

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

Output:

integral(F^(b*c*x + a*c)/(f*cos(e*x + d) + f), x)
 

Sympy [F]

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

integrate(F**(c*(b*x+a))/(f+f*cos(e*x+d)),x)
 

Output:

Integral(F**(a*c + b*c*x)/(cos(d + e*x) + 1), x)/f
 

Maxima [F]

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

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

Output:

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*lo 
g(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*sin(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*lo 
g(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 + ...
 

Giac [F]

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

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

Output:

integrate(F^((b*x + a)*c)/(f*cos(e*x + d) + f), x)
 

Mupad [F(-1)]

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

int(F^(c*(a + b*x))/(f + f*cos(d + e*x)),x)
 

Output:

int(F^(c*(a + b*x))/(f + f*cos(d + e*x)), x)
 

Reduce [F]

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

int(F^(c*(b*x+a))/(f+f*cos(e*x+d)),x)
 

Output:

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