Integrand size = 42, antiderivative size = 291 \[ \int \frac {(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=-\frac {2 (g \cos (e+f x))^{5/2}}{5 f g (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}-\frac {2 (g \cos (e+f x))^{5/2}}{a f g (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac {6 (g \cos (e+f x))^{5/2}}{5 a^2 f g \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {6 (g \cos (e+f x))^{5/2}}{5 a^2 c f g \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}-\frac {6 g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 a^2 c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \]
-2/5*(g*cos(f*x+e))^(5/2)/f/g/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(5/2 )-2*(g*cos(f*x+e))^(5/2)/a/f/g/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/ 2)+6/5*(g*cos(f*x+e))^(5/2)/a^2/f/g/(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e) )^(1/2)+6/5*(g*cos(f*x+e))^(5/2)/a^2/c/f/g/(c-c*sin(f*x+e))^(3/2)/(a+a*sin (f*x+e))^(1/2)-6/5*g*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*Ellip ticE(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)*(g*cos(f*x+e))^(1/2)/a^2 /c^2/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)
Time = 5.12 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.36 \[ \int \frac {(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\frac {(g \cos (e+f x))^{3/2} \sec ^3(e+f x) \left (-12 \cos ^{\frac {5}{2}}(e+f x) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+7 \sin (e+f x)+3 \sin (3 (e+f x))\right )}{10 a^2 c^2 f \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)}} \]
((g*Cos[e + f*x])^(3/2)*Sec[e + f*x]^3*(-12*Cos[e + f*x]^(5/2)*EllipticE[( e + f*x)/2, 2] + 7*Sin[e + f*x] + 3*Sin[3*(e + f*x)]))/(10*a^2*c^2*f*Sqrt[ a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]])
Time = 2.29 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3331, 3042, 3331, 3042, 3331, 3042, 3331, 3042, 3321, 3042, 3121, 3042, 3119}
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.
| \(\displaystyle \int \frac {(g \cos (e+f x))^{3/2}}{(a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(g \cos (e+f x))^{3/2}}{(a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3331 |
| \(\displaystyle \frac {\int \frac {(g \cos (e+f x))^{3/2}}{(\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{5/2}}dx}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(g \cos (e+f x))^{3/2}}{(\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{5/2}}dx}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3331 |
| \(\displaystyle \frac {\frac {3 \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{5/2}}dx}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{5/2}}dx}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3331 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{3/2}}dx}{5 c}+\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{3/2}}dx}{5 c}+\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3331 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{c}}{5 c}+\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {\int \frac {(g \cos (e+f x))^{3/2}}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{c}}{5 c}+\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3321 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {g \cos (e+f x) \int \sqrt {g \cos (e+f x)}dx}{c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{5 c}+\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {g \cos (e+f x) \int \sqrt {g \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{5 c}+\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3121 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} \int \sqrt {\cos (e+f x)}dx}{c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{5 c}+\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {g \sqrt {\cos (e+f x)} \sqrt {g \cos (e+f x)} \int \sqrt {\sin \left (e+f x+\frac {\pi }{2}\right )}dx}{c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{5 c}+\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {\frac {3 \left (\frac {2 (g \cos (e+f x))^{5/2}}{5 f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}+\frac {\frac {2 (g \cos (e+f x))^{5/2}}{f g \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}-\frac {2 g \sqrt {\cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}}{5 c}\right )}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{f g (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {2 (g \cos (e+f x))^{5/2}}{5 f g (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
(-2*(g*Cos[e + f*x])^(5/2))/(5*f*g*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(5/2)) + ((-2*(g*Cos[e + f*x])^(5/2))/(f*g*(a + a*Sin[e + f*x])^( 3/2)*(c - c*Sin[e + f*x])^(5/2)) + (3*((2*(g*Cos[e + f*x])^(5/2))/(5*f*g*S qrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(5/2)) + ((2*(g*Cos[e + f*x]) ^(5/2))/(f*g*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(3/2)) - (2*g*S qrt[Cos[e + f*x]]*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(c*f*Sqr t[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]))/(5*c)))/a)/a
3.2.49.3.1 Defintions of rubi rules used
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) ^n/Sin[c + d*x]^n Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt Q[-1, n, 1] && IntegerQ[2*n]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_ .)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[g* (Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])) Int[(g *Cos[e + f*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && EqQ [b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b* (g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(a* f*g*(2*m + p + 1))), x] + Simp[(m + n + p + 1)/(a*(2*m + p + 1)) Int[(g*C os[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] && !LtQ[m, n, -1] && Integers Q[2*m, 2*n, 2*p]
Result contains complex when optimal does not.
Time = 1.06 (sec) , antiderivative size = 447, normalized size of antiderivative = 1.54
| method | result | size |
| default | \(-\frac {2 \sqrt {g \cos \left (f x +e \right )}\, g \left (3 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, E\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (\cos ^{2}\left (f x +e \right )\right )-3 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \left (\cos ^{2}\left (f x +e \right )\right )+6 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, E\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \cos \left (f x +e \right )-6 i \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right ) \cos \left (f x +e \right )+3 i \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, E\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )-3 i \sqrt {\frac {\cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sqrt {\frac {1}{1+\cos \left (f x +e \right )}}\, F\left (i \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right ), i\right )-3 \sin \left (f x +e \right )-\tan \left (f x +e \right )-\sec \left (f x +e \right ) \tan \left (f x +e \right )\right )}{5 f \left (1+\cos \left (f x +e \right )\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, a^{2} c^{2}}\) | \(447\) |
int((g*cos(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(5/2),x,m ethod=_RETURNVERBOSE)
-2/5/f*(g*cos(f*x+e))^(1/2)*g/(1+cos(f*x+e))/(a*(1+sin(f*x+e)))^(1/2)/(-c* (sin(f*x+e)-1))^(1/2)/a^2/c^2*(3*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1 +cos(f*x+e)))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*cos(f*x+e)^2-3* I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticF(I*( csc(f*x+e)-cot(f*x+e)),I)*cos(f*x+e)^2+6*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f *x+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)*cos(f*x +e)-6*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*Ellipti cF(I*(csc(f*x+e)-cot(f*x+e)),I)*cos(f*x+e)+3*I*(1/(1+cos(f*x+e)))^(1/2)*(c os(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(csc(f*x+e)-cot(f*x+e)),I)-3*I *(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticF(I*(c sc(f*x+e)-cot(f*x+e)),I)-3*sin(f*x+e)-tan(f*x+e)-sec(f*x+e)*tan(f*x+e))
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.54 \[ \int \frac {(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\frac {3 i \, \sqrt {2} \sqrt {a c g} g \cos \left (f x + e\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) - 3 i \, \sqrt {2} \sqrt {a c g} g \cos \left (f x + e\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) + 2 \, {\left (3 \, g \cos \left (f x + e\right )^{2} + g\right )} \sqrt {g \cos \left (f x + e\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{5 \, a^{3} c^{3} f \cos \left (f x + e\right )^{4}} \]
integrate((g*cos(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(5/ 2),x, algorithm="fricas")
1/5*(3*I*sqrt(2)*sqrt(a*c*g)*g*cos(f*x + e)^4*weierstrassZeta(-4, 0, weier strassPInverse(-4, 0, cos(f*x + e) + I*sin(f*x + e))) - 3*I*sqrt(2)*sqrt(a *c*g)*g*cos(f*x + e)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, c os(f*x + e) - I*sin(f*x + e))) + 2*(3*g*cos(f*x + e)^2 + g)*sqrt(g*cos(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*sin(f*x + e))/(a ^3*c^3*f*cos(f*x + e)^4)
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
integrate((g*cos(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(5/ 2),x, algorithm="maxima")
integrate((g*cos(f*x + e))^(3/2)/((a*sin(f*x + e) + a)^(5/2)*(-c*sin(f*x + e) + c)^(5/2)), x)
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \]
integrate((g*cos(f*x+e))^(3/2)/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(5/ 2),x, algorithm="giac")
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2}}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]