Integrand size = 31, antiderivative size = 183 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x))^{3/2} \, dx=\frac {2 F^{c (a+b x)} (i f+g) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4} \left (-3-\frac {2 i b c \log (F)}{e}\right ),\frac {e-2 i b c \log (F)}{4 e},\frac {e^{2 i (d+e x)} (f+i g)}{f-i g}\right ) (g \cos (d+e x)+f \sin (d+e x))^{3/2}}{\sqrt {1-\frac {e^{2 i (d+e x)} (f+i g)}{f-i g}} \left (f-e^{2 i (d+e x)} (f+i g)-i g\right ) (3 e+2 i b c \log (F))} \] Output:
2*F^(c*(b*x+a))*(I*f+g)*hypergeom([-3/2, -3/4-1/2*I*b*c*ln(F)/e],[1/4*(e-2 *I*b*c*ln(F))/e],exp(2*I*(e*x+d))*(f+I*g)/(f-I*g))*(g*cos(e*x+d)+f*sin(e*x +d))^(3/2)/(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)/(2*I*b*c*ln(F)+3*e)
Time = 3.99 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.13 \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x))^{3/2} \, dx=\frac {2 F^{c (a+b x)} \sqrt {g \cos (d+e x)+f \sin (d+e x)} \left (-\cos (d+e x) (3 e f-2 b c g \log (F))-\frac {6 e^2 (f+i g) \operatorname {Hypergeometric2F1}\left (1,\frac {3}{4}-\frac {i b c \log (F)}{2 e},\frac {5}{4}-\frac {i b c \log (F)}{2 e},\frac {(f+i g) (\cos (2 (d+e x))+i \sin (2 (d+e x)))}{f-i g}\right ) (\cos (d+e x)+i \sin (d+e x))}{e-2 i b c \log (F)}+(3 e g+2 b c f \log (F)) \sin (d+e x)\right )}{9 e^2+4 b^2 c^2 \log ^2(F)} \] Input:
Integrate[F^(c*(a + b*x))*(g*Cos[d + e*x] + f*Sin[d + e*x])^(3/2),x]
Output:
(2*F^(c*(a + b*x))*Sqrt[g*Cos[d + e*x] + f*Sin[d + e*x]]*(-(Cos[d + e*x]*( 3*e*f - 2*b*c*g*Log[F])) - (6*e^2*(f + I*g)*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]))/(e - (2*I)*b*c*Log[F]) + (3*e*g + 2*b*c*f*Log[F])*Sin[d + e*x]))/(9*e^2 + 4*b^2 *c^2*Log[F]^2)
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.
| \(\displaystyle \int F^{c (a+b x)} (f \sin (d+e x)+g \cos (d+e x))^{3/2} \, dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \int F^{a c+b c x} (f \sin (d+e x)+g \cos (d+e x))^{3/2}dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \int F^{a c+b c x} (f \sin (d+e x)+g \cos (d+e x))^{3/2}dx\) |
Input:
Int[F^(c*(a + b*x))*(g*Cos[d + e*x] + f*Sin[d + e*x])^(3/2),x]
Output:
$Aborted
\[\int F^{c \left (b x +a \right )} \left (g \cos \left (e x +d \right )+f \sin \left (e x +d \right )\right )^{\frac {3}{2}}d x\]
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)
\[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x))^{3/2} \, dx=\int { {\left (g \cos \left (e x + d\right ) + f \sin \left (e x + d\right )\right )}^{\frac {3}{2}} F^{{\left (b x + a\right )} c} \,d x } \] 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((g*cos(e*x + d) + f*sin(e*x + d))^(3/2)*F^(b*c*x + a*c), x)
Timed out. \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate(F**(c*(b*x+a))*(g*cos(e*x+d)+f*sin(e*x+d))**(3/2),x)
Output:
Timed out
\[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x))^{3/2} \, dx=\int { {\left (g \cos \left (e x + d\right ) + f \sin \left (e x + d\right )\right )}^{\frac {3}{2}} F^{{\left (b x + a\right )} c} \,d x } \] 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((g*cos(e*x + d) + f*sin(e*x + d))^(3/2)*F^((b*x + a)*c), x)
\[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x))^{3/2} \, dx=\int { {\left (g \cos \left (e x + d\right ) + f \sin \left (e x + d\right )\right )}^{\frac {3}{2}} F^{{\left (b x + a\right )} c} \,d x } \] 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((g*cos(e*x + d) + f*sin(e*x + d))^(3/2)*F^((b*x + a)*c), x)
Timed out. \[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x))^{3/2} \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\left (g\,\cos \left (d+e\,x\right )+f\,\sin \left (d+e\,x\right )\right )}^{3/2} \,d x \] 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)
\[ \int F^{c (a+b x)} (g \cos (d+e x)+f \sin (d+e x))^{3/2} \, dx=\text {too large to display} \] Input:
int(F^(c*(b*x+a))*(g*cos(e*x+d)+f*sin(e*x+d))^(3/2),x)
Output:
(f**(a*c)*(4*f**(b*c*x)*sqrt(cos(d + e*x)*g + sin(d + e*x)*f)*cos(d + e*x) *log(f)*b*c*g**3 - 2*f**(b*c*x)*sqrt(cos(d + e*x)*g + sin(d + e*x)*f)*cos( d + e*x)*e*f*g**2 + 4*f**(b*c*x)*sqrt(cos(d + e*x)*g + sin(d + e*x)*f)*log (f)*sin(d + e*x)*b*c*f*g**2 + 4*f**(b*c*x)*sqrt(cos(d + e*x)*g + sin(d + e *x)*f)*sin(d + e*x)*e*f**2*g + 6*f**(b*c*x)*sqrt(cos(d + e*x)*g + sin(d + e*x)*f)*sin(d + e*x)*e*g**3 + 12*int((f**(b*c*x)*sqrt(cos(d + e*x)*g + sin (d + e*x)*f)*sin(d + e*x)**2)/(4*cos(d + e*x)*log(f)**2*b**2*c**2*g**3 + 4 *cos(d + e*x)*log(f)*b*c*e*f*g**2 + 3*cos(d + e*x)*e**2*f**2*g + 6*cos(d + e*x)*e**2*g**3 + 4*log(f)**2*sin(d + e*x)*b**2*c**2*f*g**2 + 4*log(f)*sin (d + e*x)*b*c*e*f**2*g + 3*sin(d + e*x)*e**2*f**3 + 6*sin(d + e*x)*e**2*f* g**2),x)*log(f)**2*b**2*c**2*e**2*f**4*g**2 + 24*int((f**(b*c*x)*sqrt(cos( d + e*x)*g + sin(d + e*x)*f)*sin(d + e*x)**2)/(4*cos(d + e*x)*log(f)**2*b* *2*c**2*g**3 + 4*cos(d + e*x)*log(f)*b*c*e*f*g**2 + 3*cos(d + e*x)*e**2*f* *2*g + 6*cos(d + e*x)*e**2*g**3 + 4*log(f)**2*sin(d + e*x)*b**2*c**2*f*g** 2 + 4*log(f)*sin(d + e*x)*b*c*e*f**2*g + 3*sin(d + e*x)*e**2*f**3 + 6*sin( d + e*x)*e**2*f*g**2),x)*log(f)**2*b**2*c**2*e**2*f**2*g**4 + 12*int((f**( b*c*x)*sqrt(cos(d + e*x)*g + sin(d + e*x)*f)*sin(d + e*x)**2)/(4*cos(d + e *x)*log(f)**2*b**2*c**2*g**3 + 4*cos(d + e*x)*log(f)*b*c*e*f*g**2 + 3*cos( d + e*x)*e**2*f**2*g + 6*cos(d + e*x)*e**2*g**3 + 4*log(f)**2*sin(d + e*x) *b**2*c**2*f*g**2 + 4*log(f)*sin(d + e*x)*b*c*e*f**2*g + 3*sin(d + e*x)...