\(\int \frac {F^{c (a+b x)}}{f+f \sin (d+e x)} \, dx\) [30]

Optimal result

Integrand size = 22, antiderivative size = 80 \[ \int \frac {F^{c (a+b x)}}{f+f \sin (d+e x)} \, dx=-\frac {2 e^{i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,1-\frac {i b c \log (F)}{e},2-\frac {i b c \log (F)}{e},i e^{i (d+e x)}\right )}{f (e-i b c \log (F))} \] 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],I*exp(I*(e*x+d)))/f/(e-I*b*c*ln(F))
 

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.60 \[ \int \frac {F^{c (a+b x)}}{f+f \sin (d+e x)} \, dx=\frac {2 F^{c (a+b x)} \left (-i \operatorname {Hypergeometric2F1}\left (1,-\frac {i b c \log (F)}{e},1-\frac {i b c \log (F)}{e},i \cos (d+e x)-\sin (d+e x)\right )-\frac {1}{\cos (d)+i (1+\sin (d))}+\frac {\sin \left (\frac {e x}{2}\right )}{\left (\cos \left (\frac {d}{2}\right )+\sin \left (\frac {d}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (d+e x)\right )+\sin \left (\frac {1}{2} (d+e x)\right )\right )}\right )}{e f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 80, 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 = {4956, 4952}

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.

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

\(\Big \downarrow \) 4956

\(\displaystyle \frac {\int F^{c (a+b x)} \csc ^2\left (\frac {d}{2}+\frac {e x}{2}+\frac {\pi }{4}\right )dx}{2 f}\)

\(\Big \downarrow \) 4952

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

Input:

Int[F^(c*(a + b*x))/(f + f*Sin[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, I*E^(I*(d + e*x))])/(f*(e - I*b*c*Log[F]))
 

Defintions of rubi rules used

rule 4952

Int[Csc[(d_.) + Pi*(k_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_ 
))), x_Symbol] :> Simp[(-2*I)^n*E^(I*k*n*Pi)*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*k*Pi)*E^(2*I*(d + e*x))], x] /; Fre 
eQ[{F, a, b, c, d, e}, x] && IntegerQ[4*k] && IntegerQ[n]
 

rule 4956

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_) + (g_.)*Sin[(d_.) + (e_.)*(x_)] 
)^(n_.), x_Symbol] :> Simp[2^n*f^n   Int[F^(c*(a + b*x))*Cos[d/2 - f*(Pi/(4 
*g)) + e*(x/2)]^(2*n), x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] && EqQ[ 
f^2 - g^2, 0] && ILtQ[n, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(F^(c*(b*x+a))/(f+f*sin(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 + (5*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*l 
og(F)^3 + 16*F^(a*c)*b*c*e^2*log(F))*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^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))*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 + 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)*sin(2*e*x + 2*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*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*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*sin(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*sin(e*x + d) + (F^(a*c)*b^5*c^5*e*log(F)^5 + ...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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