\(\int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^2} \, dx\) [26]

Optimal result

Integrand size = 29, antiderivative size = 108 \[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^2} \, dx=-\frac {4 e^{2 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2} \left (2-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (4-\frac {i b c \log (F)}{e}\right ),\frac {e^{2 i (d+e x)} (f+i g)}{f-i g}\right )}{(f-i g)^2 (2 i e+b c \log (F))} \] Output:

-4*exp(2*I*(e*x+d))*F^(c*(b*x+a))*hypergeom([2, 1-1/2*I*b*c*ln(F)/e],[2-1/ 
2*I*b*c*ln(F)/e],exp(2*I*(e*x+d))*(f+I*g)/(f-I*g))/(f-I*g)^2/(2*I*e+b*c*ln 
(F))
 

Mathematica [F]

\[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^2} \, dx=\int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^2} \, dx \] Input:

Integrate[F^(c*(a + b*x))/(g*Cos[d + e*x] + f*Sin[d + e*x])^2,x]
 

Output:

Integrate[F^(c*(a + b*x))/(g*Cos[d + e*x] + f*Sin[d + e*x])^2, x]
 

Rubi [F]

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.

\(\displaystyle \int \frac {F^{c (a+b x)}}{(f \sin (d+e x)+g \cos (d+e x))^2} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {F^{a c+b c x}}{(f \sin (d+e x)+g \cos (d+e x))^2}dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \frac {F^{a c+b c x}}{(f \sin (d+e x)+g \cos (d+e x))^2}dx\)

Input:

Int[F^(c*(a + b*x))/(g*Cos[d + e*x] + f*Sin[d + e*x])^2,x]
 

Output:

$Aborted
 

Defintions of rubi rules used

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7299

Int[u_, x_] :> CannotIntegrate[u, x]
 

Maple [F]

\[\int \frac {F^{c \left (b x +a \right )}}{\left (g \cos \left (e x +d \right )+f \sin \left (e x +d \right )\right )^{2}}d x\]

Input:

int(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^2,x)
 

Output:

int(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^2,x)
 

Fricas [F]

\[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^2} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (g \cos \left (e x + d\right ) + f \sin \left (e x + d\right )\right )}^{2}} \,d x } \] Input:

integrate(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^2,x, algorithm="fricas 
")
 

Output:

integral(F^(b*c*x + a*c)/(2*f*g*cos(e*x + d)*sin(e*x + d) - (f^2 - g^2)*co 
s(e*x + d)^2 + f^2), x)
 

Sympy [F]

\[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^2} \, dx=\int \frac {F^{c \left (a + b x\right )}}{\left (f \sin {\left (d + e x \right )} + g \cos {\left (d + e x \right )}\right )^{2}}\, dx \] Input:

integrate(F**(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))**2,x)
 

Output:

Integral(F**(c*(a + b*x))/(f*sin(d + e*x) + g*cos(d + e*x))**2, x)
 

Maxima [F]

\[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^2} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (g \cos \left (e x + d\right ) + f \sin \left (e x + d\right )\right )}^{2}} \,d x } \] Input:

integrate(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^2,x, algorithm="maxima 
")
 

Output:

4*(2*((F^(a*c)*b^3*c^3*log(F)^3 + 16*F^(a*c)*b*c*e^2*log(F))*f^4 + 2*(F^(a 
*c)*b^3*c^3*log(F)^3 + 16*F^(a*c)*b*c*e^2*log(F))*f^2*g^2 + (F^(a*c)*b^3*c 
^3*log(F)^3 + 16*F^(a*c)*b*c*e^2*log(F))*g^4)*F^(b*c*x)*cos(2*e*x + 2*d)^2 
 + 2*((F^(a*c)*b^3*c^3*log(F)^3 + 16*F^(a*c)*b*c*e^2*log(F))*f^4 + 2*(F^(a 
*c)*b^3*c^3*log(F)^3 + 16*F^(a*c)*b*c*e^2*log(F))*f^2*g^2 + (F^(a*c)*b^3*c 
^3*log(F)^3 + 16*F^(a*c)*b*c*e^2*log(F))*g^4)*F^(b*c*x)*sin(2*e*x + 2*d)^2 
 - ((F^(a*c)*b^3*c^3*log(F)^3 + 64*F^(a*c)*b*c*e^2*log(F))*f^4 - 4*(5*F^(a 
*c)*b^2*c^2*e*log(F)^2 - 16*F^(a*c)*e^3)*f^3*g - 4*(5*F^(a*c)*b^2*c^2*e*lo 
g(F)^2 - 16*F^(a*c)*e^3)*f*g^3 - (F^(a*c)*b^3*c^3*log(F)^3 + 64*F^(a*c)*b* 
c*e^2*log(F))*g^4)*F^(b*c*x)*cos(2*e*x + 2*d) + 2*((5*F^(a*c)*b^2*c^2*e*lo 
g(F)^2 - 16*F^(a*c)*e^3)*f^4 + (F^(a*c)*b^3*c^3*log(F)^3 + 64*F^(a*c)*b*c* 
e^2*log(F))*f^3*g + (F^(a*c)*b^3*c^3*log(F)^3 + 64*F^(a*c)*b*c*e^2*log(F)) 
*f*g^3 - (5*F^(a*c)*b^2*c^2*e*log(F)^2 - 16*F^(a*c)*e^3)*g^4)*F^(b*c*x)*si 
n(2*e*x + 2*d) + 24*(F^(a*c)*b*c*e^2*f^4*log(F) + 2*F^(a*c)*b*c*e^2*f^2*g^ 
2*log(F) + F^(a*c)*b*c*e^2*g^4*log(F))*F^(b*c*x) - (((F^(a*c)*b^3*c^3*log( 
F)^3 + 16*F^(a*c)*b*c*e^2*log(F))*f^4 + 4*(F^(a*c)*b^2*c^2*e*log(F)^2 + 16 
*F^(a*c)*e^3)*f^3*g + 4*(F^(a*c)*b^2*c^2*e*log(F)^2 + 16*F^(a*c)*e^3)*f*g^ 
3 - (F^(a*c)*b^3*c^3*log(F)^3 + 16*F^(a*c)*b*c*e^2*log(F))*g^4)*F^(b*c*x)* 
cos(2*e*x + 2*d) - 2*((F^(a*c)*b^2*c^2*e*log(F)^2 + 16*F^(a*c)*e^3)*f^4 - 
(F^(a*c)*b^3*c^3*log(F)^3 + 16*F^(a*c)*b*c*e^2*log(F))*f^3*g - (F^(a*c)...
 

Giac [F]

\[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^2} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (g \cos \left (e x + d\right ) + f \sin \left (e x + d\right )\right )}^{2}} \,d x } \] Input:

integrate(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^2,x, algorithm="giac")
 

Output:

integrate(F^((b*x + a)*c)/(g*cos(e*x + d) + f*sin(e*x + d))^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^2} \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (g\,\cos \left (d+e\,x\right )+f\,\sin \left (d+e\,x\right )\right )}^2} \,d x \] Input:

int(F^(c*(a + b*x))/(g*cos(d + e*x) + f*sin(d + e*x))^2,x)
 

Output:

int(F^(c*(a + b*x))/(g*cos(d + e*x) + f*sin(d + e*x))^2, x)
 

Reduce [F]

\[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^2} \, dx=f^{a c} \left (\int \frac {f^{b c x}}{\cos \left (e x +d \right )^{2} g^{2}+2 \cos \left (e x +d \right ) \sin \left (e x +d \right ) f g +\sin \left (e x +d \right )^{2} f^{2}}d x \right ) \] Input:

int(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^2,x)
 

Output:

f**(a*c)*int(f**(b*c*x)/(cos(d + e*x)**2*g**2 + 2*cos(d + e*x)*sin(d + e*x 
)*f*g + sin(d + e*x)**2*f**2),x)