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

Optimal result

Integrand size = 31, antiderivative size = 147 \[ \int \frac {F^{c (a+b x)}}{\sqrt {g \cos (d+e x)+f \sin (d+e x)}} \, dx=\frac {2 F^{c (a+b x)} \sqrt {1-\frac {e^{2 i (d+e x)} (f+i g)}{f-i g}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {e-2 i b c \log (F)}{4 e},\frac {1}{4} \left (5-\frac {2 i b c \log (F)}{e}\right ),\frac {e^{2 i (d+e x)} (f+i g)}{f-i g}\right )}{(i e+2 b c \log (F)) \sqrt {g \cos (d+e x)+f \sin (d+e x)}} \] Output:

2*F^(c*(b*x+a))*(1-exp(2*I*(e*x+d))*(f+I*g)/(f-I*g))^(1/2)*hypergeom([1/2, 
 1/4*(e-2*I*b*c*ln(F))/e],[5/4-1/2*I*b*c*ln(F)/e],exp(2*I*(e*x+d))*(f+I*g) 
/(f-I*g))/(I*e+2*b*c*ln(F))/(g*cos(e*x+d)+f*sin(e*x+d))^(1/2)
 

Mathematica [A] (verified)

Time = 2.52 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.98 \[ \int \frac {F^{c (a+b x)}}{\sqrt {g \cos (d+e x)+f \sin (d+e x)}} \, dx=-\frac {4 F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (1,\frac {3}{4}-\frac {i b c \log (F)}{2 e},\frac {5}{4}-\frac {i b c \log (F)}{2 e},\frac {(f+i g) (\cos (2 (d+e x))+i \sin (2 (d+e x)))}{f-i g}\right ) (\cos (d+e x)+i \sin (d+e x)) \sqrt {g \cos (d+e x)+f \sin (d+e x)}}{(f-i g) (e-2 i b c \log (F))} \] Input:

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

Output:

(-4*F^(c*(a + b*x))*Hypergeometric2F1[1, 3/4 - ((I/2)*b*c*Log[F])/e, 5/4 - 
 ((I/2)*b*c*Log[F])/e, ((f + I*g)*(Cos[2*(d + e*x)] + I*Sin[2*(d + e*x)])) 
/(f - I*g)]*(Cos[d + e*x] + I*Sin[d + e*x])*Sqrt[g*Cos[d + e*x] + f*Sin[d 
+ e*x]])/((f - I*g)*(e - (2*I)*b*c*Log[F]))
 

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.

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

\(\Big \downarrow \) 7292

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

\(\Big \downarrow \) 7299

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

Input:

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

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

Input:

int(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^(1/2),x)
 

Output:

int(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^(1/2),x)
 

Fricas [F]

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

integrate(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^(1/2),x, algorithm="fr 
icas")
 

Output:

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

Sympy [F]

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

integrate(F**(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))**(1/2),x)
 

Output:

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

Maxima [F]

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

integrate(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^(1/2),x, algorithm="ma 
xima")
 

Output:

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

Giac [F]

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

integrate(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^(1/2),x, algorithm="gi 
ac")
 

Output:

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

Mupad [F(-1)]

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

int(F^(c*(a + b*x))/(g*cos(d + e*x) + f*sin(d + e*x))^(1/2),x)
 

Output:

int(F^(c*(a + b*x))/(g*cos(d + e*x) + f*sin(d + e*x))^(1/2), x)
 

Reduce [F]

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

int(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^(1/2),x)
 

Output:

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