\(\int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^{3/2}} \, dx\) [31]

Optimal result

Integrand size = 31, antiderivative size = 183 \[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^{3/2}} \, dx=\frac {2 F^{c (a+b x)} \sqrt {1-\frac {e^{2 i (d+e x)} (f+i g)}{f-i g}} \left (f-e^{2 i (d+e x)} (f+i g)-i g\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1}{4} \left (3-\frac {2 i b c \log (F)}{e}\right ),\frac {1}{4} \left (7-\frac {2 i b c \log (F)}{e}\right ),\frac {e^{2 i (d+e x)} (f+i g)}{f-i g}\right )}{(i f+g) (3 e-2 i b c \log (F)) (g \cos (d+e x)+f \sin (d+e x))^{3/2}} \] Output:

2*F^(c*(b*x+a))*(1-exp(2*I*(e*x+d))*(f+I*g)/(f-I*g))^(1/2)*(f-exp(2*I*(e*x 
+d))*(f+I*g)-I*g)*hypergeom([3/2, 3/4-1/2*I*b*c*ln(F)/e],[7/4-1/2*I*b*c*ln 
(F)/e],exp(2*I*(e*x+d))*(f+I*g)/(f-I*g))/(I*f+g)/(3*e-2*I*b*c*ln(F))/(g*co 
s(e*x+d)+f*sin(e*x+d))^(3/2)
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(433\) vs. \(2(183)=366\).

Time = 4.56 (sec) , antiderivative size = 433, normalized size of antiderivative = 2.37 \[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^{3/2}} \, dx=-\frac {\sqrt {2} e^{-\frac {1}{2} i (d+e x)} F^{c \left (a-\frac {b d}{e}\right )} \left (\left (-1+e^{2 i (d+e x)}\right ) f+i \left (1+e^{2 i (d+e x)}\right ) g\right ) (e-2 i b c \log (F)) \left (e^{(d+e x) \left (-\frac {i}{2}+\frac {b c \log (F)}{e}\right )} \operatorname {Hypergeometric2F1}\left (1,\frac {e-2 i b c \log (F)}{4 e},\frac {3}{4}-\frac {i b c \log (F)}{2 e},\frac {e^{2 i (d+e x)} (f+i g)}{f-i g}\right ) (3 e-2 i b c \log (F))+e^{(d+e x) \left (\frac {3 i}{2}+\frac {b c \log (F)}{e}\right )} \operatorname {Hypergeometric2F1}\left (1,\frac {5}{4}-\frac {i b c \log (F)}{2 e},\frac {7}{4}-\frac {i b c \log (F)}{2 e},\frac {e^{2 i (d+e x)} (f+i g)}{f-i g}\right ) (e+2 i b c \log (F))\right )}{e (f-i g) \sqrt {e^{-i (d+e x)} \left (-i \left (-1+e^{2 i (d+e x)}\right ) f+\left (1+e^{2 i (d+e x)}\right ) g\right )} (3 e-2 i b c \log (F)) (e f+2 b c g \log (F))}-\frac {2 F^{c (a+b x)} (e \cos (d+e x)-2 b c \log (F) \sin (d+e x))}{e (e f+2 b c g \log (F)) \sqrt {g \cos (d+e x)+f \sin (d+e x)}} \] Input:

Integrate[F^(c*(a + b*x))/(g*Cos[d + e*x] + f*Sin[d + e*x])^(3/2),x]
 

Output:

-((Sqrt[2]*F^(c*(a - (b*d)/e))*((-1 + E^((2*I)*(d + e*x)))*f + I*(1 + E^(( 
2*I)*(d + e*x)))*g)*(e - (2*I)*b*c*Log[F])*(E^((d + e*x)*(-1/2*I + (b*c*Lo 
g[F])/e))*Hypergeometric2F1[1, (e - (2*I)*b*c*Log[F])/(4*e), 3/4 - ((I/2)* 
b*c*Log[F])/e, (E^((2*I)*(d + e*x))*(f + I*g))/(f - I*g)]*(3*e - (2*I)*b*c 
*Log[F]) + E^((d + e*x)*((3*I)/2 + (b*c*Log[F])/e))*Hypergeometric2F1[1, 5 
/4 - ((I/2)*b*c*Log[F])/e, 7/4 - ((I/2)*b*c*Log[F])/e, (E^((2*I)*(d + e*x) 
)*(f + I*g))/(f - I*g)]*(e + (2*I)*b*c*Log[F])))/(e*E^((I/2)*(d + e*x))*(f 
 - I*g)*Sqrt[((-I)*(-1 + E^((2*I)*(d + e*x)))*f + (1 + E^((2*I)*(d + e*x)) 
)*g)/E^(I*(d + e*x))]*(3*e - (2*I)*b*c*Log[F])*(e*f + 2*b*c*g*Log[F]))) - 
(2*F^(c*(a + b*x))*(e*Cos[d + e*x] - 2*b*c*Log[F]*Sin[d + e*x]))/(e*(e*f + 
 2*b*c*g*Log[F])*Sqrt[g*Cos[d + e*x] + f*Sin[d + e*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.

\(\displaystyle \int \frac {F^{c (a+b x)}}{(f \sin (d+e x)+g \cos (d+e x))^{3/2}} \, dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \int \frac {F^{a c+b c x}}{(f \sin (d+e x)+g \cos (d+e x))^{3/2}}dx\)

\(\Big \downarrow \) 7299

\(\displaystyle \int \frac {F^{a c+b c x}}{(f \sin (d+e x)+g \cos (d+e x))^{3/2}}dx\)

Input:

Int[F^(c*(a + b*x))/(g*Cos[d + e*x] + f*Sin[d + e*x])^(3/2),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]

\[\int \frac {F^{c \left (b x +a \right )}}{\left (g \cos \left (e x +d \right )+f \sin \left (e x +d \right )\right )^{\frac {3}{2}}}d x\]

Input:

int(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^(3/2),x)
 

Output:

int(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^(3/2),x)
 

Fricas [F]

\[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^{3/2}} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (g \cos \left (e x + d\right ) + f \sin \left (e x + d\right )\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^(3/2),x, algorithm="fr 
icas")
 

Output:

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

Sympy [F]

\[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^{3/2}} \, dx=\int \frac {F^{c \left (a + b x\right )}}{\left (f \sin {\left (d + e x \right )} + g \cos {\left (d + e x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(F**(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))**(3/2),x)
 

Output:

Integral(F**(c*(a + b*x))/(f*sin(d + e*x) + g*cos(d + e*x))**(3/2), x)
 

Maxima [F]

\[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^{3/2}} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (g \cos \left (e x + d\right ) + f \sin \left (e x + d\right )\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^(3/2),x, algorithm="ma 
xima")
 

Output:

integrate(F^((b*x + a)*c)/(g*cos(e*x + d) + f*sin(e*x + d))^(3/2), x)
 

Giac [F]

\[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^{3/2}} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (g \cos \left (e x + d\right ) + f \sin \left (e x + d\right )\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^(3/2),x, algorithm="gi 
ac")
 

Output:

integrate(F^((b*x + a)*c)/(g*cos(e*x + d) + f*sin(e*x + d))^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^{3/2}} \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (g\,\cos \left (d+e\,x\right )+f\,\sin \left (d+e\,x\right )\right )}^{3/2}} \,d x \] Input:

int(F^(c*(a + b*x))/(g*cos(d + e*x) + f*sin(d + e*x))^(3/2),x)
 

Output:

int(F^(c*(a + b*x))/(g*cos(d + e*x) + f*sin(d + e*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^{3/2}} \, dx=f^{a c} \left (\int \frac {f^{b c x} \sqrt {\cos \left (e x +d \right ) g +\sin \left (e x +d \right ) f}}{\cos \left (e x +d \right )^{2} g^{2}+2 \cos \left (e x +d \right ) \sin \left (e x +d \right ) f g +\sin \left (e x +d \right )^{2} f^{2}}d x \right ) \] Input:

int(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^(3/2),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)**2*g**2 + 2*cos(d + e*x)*sin(d + e*x)*f*g + sin(d + e*x)**2*f**2),x)