Integrand size = 29, antiderivative size = 109 \[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^3} \, dx=\frac {8 e^{3 i (d+e x)} F^{c (a+b x)} \operatorname {Hypergeometric2F1}\left (3,\frac {1}{2} \left (3-\frac {i b c \log (F)}{e}\right ),\frac {1}{2} \left (5-\frac {i b c \log (F)}{e}\right ),\frac {e^{2 i (d+e x)} (f+i g)}{f-i g}\right )}{(f-i g)^3 (3 e-i b c \log (F))} \] Output:
8*exp(3*I*(e*x+d))*F^(c*(b*x+a))*hypergeom([3, 3/2-1/2*I*b*c*ln(F)/e],[5/2 -1/2*I*b*c*ln(F)/e],exp(2*I*(e*x+d))*(f+I*g)/(f-I*g))/(f-I*g)^3/(3*e-I*b*c *ln(F))
\[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^3} \, dx=\int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^3} \, dx \] Input:
Integrate[F^(c*(a + b*x))/(g*Cos[d + e*x] + f*Sin[d + e*x])^3,x]
Output:
Integrate[F^(c*(a + b*x))/(g*Cos[d + e*x] + f*Sin[d + e*x])^3, x]
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.
\(\displaystyle \int \frac {F^{c (a+b x)}}{(f \sin (d+e x)+g \cos (d+e x))^3} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {F^{a c+b c x}}{(f \sin (d+e x)+g \cos (d+e x))^3}dx\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \int \frac {F^{a c+b c x}}{(f \sin (d+e x)+g \cos (d+e x))^3}dx\) |
Input:
Int[F^(c*(a + b*x))/(g*Cos[d + e*x] + f*Sin[d + e*x])^3,x]
Output:
$Aborted
\[\int \frac {F^{c \left (b x +a \right )}}{\left (g \cos \left (e x +d \right )+f \sin \left (e x +d \right )\right )^{3}}d x\]
Input:
int(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^3,x)
Output:
int(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^3,x)
\[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^3} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (g \cos \left (e x + d\right ) + f \sin \left (e x + d\right )\right )}^{3}} \,d x } \] Input:
integrate(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^3,x, algorithm="fricas ")
Output:
integral(F^(b*c*x + a*c)/(3*f^2*g*cos(e*x + d) - (3*f^2*g - g^3)*cos(e*x + d)^3 + (f^3 - (f^3 - 3*f*g^2)*cos(e*x + d)^2)*sin(e*x + d)), x)
\[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^3} \, dx=\int \frac {F^{c \left (a + b x\right )}}{\left (f \sin {\left (d + e x \right )} + g \cos {\left (d + e x \right )}\right )^{3}}\, dx \] Input:
integrate(F**(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))**3,x)
Output:
Integral(F**(c*(a + b*x))/(f*sin(d + e*x) + g*cos(d + e*x))**3, x)
Timed out. \[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^3} \, dx=\text {Timed out} \] Input:
integrate(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^3,x, algorithm="maxima ")
Output:
Timed out
\[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^3} \, dx=\int { \frac {F^{{\left (b x + a\right )} c}}{{\left (g \cos \left (e x + d\right ) + f \sin \left (e x + d\right )\right )}^{3}} \,d x } \] Input:
integrate(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^3,x, algorithm="giac")
Output:
integrate(F^((b*x + a)*c)/(g*cos(e*x + d) + f*sin(e*x + d))^3, x)
Timed out. \[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^3} \, dx=\int \frac {F^{c\,\left (a+b\,x\right )}}{{\left (g\,\cos \left (d+e\,x\right )+f\,\sin \left (d+e\,x\right )\right )}^3} \,d x \] Input:
int(F^(c*(a + b*x))/(g*cos(d + e*x) + f*sin(d + e*x))^3,x)
Output:
int(F^(c*(a + b*x))/(g*cos(d + e*x) + f*sin(d + e*x))^3, x)
\[ \int \frac {F^{c (a+b x)}}{(g \cos (d+e x)+f \sin (d+e x))^3} \, dx=f^{a c} \left (\int \frac {f^{b c x}}{\cos \left (e x +d \right )^{3} g^{3}+3 \cos \left (e x +d \right )^{2} \sin \left (e x +d \right ) f \,g^{2}+3 \cos \left (e x +d \right ) \sin \left (e x +d \right )^{2} f^{2} g +\sin \left (e x +d \right )^{3} f^{3}}d x \right ) \] Input:
int(F^(c*(b*x+a))/(g*cos(e*x+d)+f*sin(e*x+d))^3,x)
Output:
f**(a*c)*int(f**(b*c*x)/(cos(d + e*x)**3*g**3 + 3*cos(d + e*x)**2*sin(d + e*x)*f*g**2 + 3*cos(d + e*x)*sin(d + e*x)**2*f**2*g + sin(d + e*x)**3*f**3 ),x)