Integrand size = 30, antiderivative size = 236 \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\cos (e+f x)}{4 f (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}}-\frac {\cos (e+f x)}{2 a f (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{5/2}}+\frac {3 \cos (e+f x)}{8 a^2 f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{5/2}}+\frac {3 \cos (e+f x)}{8 a^2 c f \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{3/2}}+\frac {3 \text {arctanh}(\sin (e+f x)) \cos (e+f x)}{8 a^2 c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \] Output:
-1/4*cos(f*x+e)/f/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(5/2)-1/2*cos(f* x+e)/a/f/(a+a*sin(f*x+e))^(3/2)/(c-c*sin(f*x+e))^(5/2)+3/8*cos(f*x+e)/a^2/ f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(5/2)+3/8*cos(f*x+e)/a^2/c/f/(a+ a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(3/2)+3/8*arctanh(sin(f*x+e))*cos(f*x +e)/a^2/c^2/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)
Time = 2.80 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.96 \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\frac {\sec ^4(e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (-9 \log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-12 \cos (2 (e+f x)) \left (\log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )-3 \cos (4 (e+f x)) \left (\log \left (1-\tan \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )\right )+9 \log \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )+22 \sin (e+f x)+6 \sin (3 (e+f x))\right )}{64 a^2 c^2 f \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)}} \] Input:
Integrate[1/((a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(5/2)),x]
Output:
(Sec[e + f*x]^4*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-9*Log[1 - Tan[(e + f*x)/2]] - 12*Cos[2*(e + f*x)]*(Log [1 - Tan[(e + f*x)/2]] - Log[1 + Tan[(e + f*x)/2]]) - 3*Cos[4*(e + f*x)]*( Log[1 - Tan[(e + f*x)/2]] - Log[1 + Tan[(e + f*x)/2]]) + 9*Log[1 + Tan[(e + f*x)/2]] + 22*Sin[e + f*x] + 6*Sin[3*(e + f*x)]))/(64*a^2*c^2*f*Sqrt[a*( 1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]])
Time = 1.28 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3222, 3042, 3222, 3042, 3222, 3042, 3222, 3042, 3220, 3042, 4257}
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.
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3222 |
| \(\displaystyle \frac {\int \frac {1}{(\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{5/2}}dx}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {1}{(\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{5/2}}dx}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3222 |
| \(\displaystyle \frac {\frac {3 \int \frac {1}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{5/2}}dx}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \int \frac {1}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{5/2}}dx}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3222 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{3/2}}dx}{2 c}+\frac {\cos (e+f x)}{4 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{3/2}}dx}{2 c}+\frac {\cos (e+f x)}{4 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3222 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{2 c}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}}{2 c}+\frac {\cos (e+f x)}{4 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {\int \frac {1}{\sqrt {\sin (e+f x) a+a} \sqrt {c-c \sin (e+f x)}}dx}{2 c}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}}{2 c}+\frac {\cos (e+f x)}{4 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3220 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {\cos (e+f x) \int \sec (e+f x)dx}{2 c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}}{2 c}+\frac {\cos (e+f x)}{4 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {\cos (e+f x) \int \csc \left (e+f x+\frac {\pi }{2}\right )dx}{2 c \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}}{2 c}+\frac {\cos (e+f x)}{4 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\frac {\cos (e+f x) \text {arctanh}(\sin (e+f x))}{2 c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {\cos (e+f x)}{2 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{3/2}}}{2 c}+\frac {\cos (e+f x)}{4 f \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{5/2}}\right )}{2 a}-\frac {\cos (e+f x)}{2 f (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{5/2}}}{a}-\frac {\cos (e+f x)}{4 f (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{5/2}}\) |
Input:
Int[1/((a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(5/2)),x]
Output:
-1/4*Cos[e + f*x]/(f*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(5/2) ) + (-1/2*Cos[e + f*x]/(f*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^ (5/2)) + (3*(Cos[e + f*x]/(4*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x ])^(5/2)) + (Cos[e + f*x]/(2*f*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x ])^(3/2)) + (ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(2*c*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]))/(2*c)))/(2*a))/a
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_ .) + (f_.)*(x_)]]), x_Symbol] :> Simp[Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x] ]*Sqrt[c + d*Sin[e + f*x]]) Int[1/Cos[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*( (c + d*Sin[e + f*x])^n/(a*f*(2*m + 1))), x] + Simp[(m + n + 1)/(a*(2*m + 1) ) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; Free Q[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + 1], 0] && NeQ[m, -2^(-1)] && (SumSimplerQ[m, 1] || !SumSimplerQ[n, 1])
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.87 (sec) , antiderivative size = 321, normalized size of antiderivative = 1.36
| method | result | size |
| default | \(-\frac {\sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3} \csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3} \left (192 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4} \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4} \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+1\right )+192 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4} \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4} \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )-1\right )-192 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4} \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4} \ln \left (-\cot \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )+\csc \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )\right )+29 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{8}-154 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{6}+173 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}-32 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}-8\right ) \sqrt {4}}{1024 f \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, a^{2} c^{2}}\) | \(321\) |
Input:
int(1/(a+sin(f*x+e)*a)^(5/2)/(c-c*sin(f*x+e))^(5/2),x,method=_RETURNVERBOS E)
Output:
-1/1024/f*sec(1/4*Pi+1/2*f*x+1/2*e)^3*csc(1/4*Pi+1/2*f*x+1/2*e)^3*(192*cos (1/4*Pi+1/2*f*x+1/2*e)^4*sin(1/4*Pi+1/2*f*x+1/2*e)^4*ln(-cot(1/4*Pi+1/2*f* x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e)+1)+192*cos(1/4*Pi+1/2*f*x+1/2*e)^4*sin( 1/4*Pi+1/2*f*x+1/2*e)^4*ln(-cot(1/4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1 /2*e)-1)-192*cos(1/4*Pi+1/2*f*x+1/2*e)^4*sin(1/4*Pi+1/2*f*x+1/2*e)^4*ln(-c ot(1/4*Pi+1/2*f*x+1/2*e)+csc(1/4*Pi+1/2*f*x+1/2*e))+29*cos(1/4*Pi+1/2*f*x+ 1/2*e)^8-154*cos(1/4*Pi+1/2*f*x+1/2*e)^6+173*cos(1/4*Pi+1/2*f*x+1/2*e)^4-3 2*cos(1/4*Pi+1/2*f*x+1/2*e)^2-8)/(a*sin(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/(c* cos(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/a^2/c^2*4^(1/2)
Time = 0.13 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\left [\frac {3 \, \sqrt {a c} \cos \left (f x + e\right )^{5} \log \left (-\frac {a c \cos \left (f x + e\right )^{3} - 2 \, a c \cos \left (f x + e\right ) - 2 \, \sqrt {a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{\cos \left (f x + e\right )^{3}}\right ) + 2 \, {\left (3 \, \cos \left (f x + e\right )^{2} + 2\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{16 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}}, -\frac {3 \, \sqrt {-a c} \arctan \left (\frac {\sqrt {-a c} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{a c \cos \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{5} - {\left (3 \, \cos \left (f x + e\right )^{2} + 2\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c} \sin \left (f x + e\right )}{8 \, a^{3} c^{3} f \cos \left (f x + e\right )^{5}}\right ] \] Input:
integrate(1/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="fr icas")
Output:
[1/16*(3*sqrt(a*c)*cos(f*x + e)^5*log(-(a*c*cos(f*x + e)^3 - 2*a*c*cos(f*x + e) - 2*sqrt(a*c)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)*sin (f*x + e))/cos(f*x + e)^3) + 2*(3*cos(f*x + e)^2 + 2)*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)^5), - 1/8*(3*sqrt(-a*c)*arctan(sqrt(-a*c)*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f *x + e) + c)*sin(f*x + e)/(a*c*cos(f*x + e)))*cos(f*x + e)^5 - (3*cos(f*x + e)^2 + 2)*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)^5)]
Timed out. \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:
integrate(1/(a+a*sin(f*x+e))**(5/2)/(c-c*sin(f*x+e))**(5/2),x)
Output:
Timed out
\[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="ma xima")
Output:
integrate(1/((a*sin(f*x + e) + a)^(5/2)*(-c*sin(f*x + e) + c)^(5/2)), x)
Exception generated. \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(1/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(5/2),x, algorithm="gi ac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {1}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:
int(1/((a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^(5/2)),x)
Output:
int(1/((a + a*sin(e + f*x))^(5/2)*(c - c*sin(e + f*x))^(5/2)), x)
\[ \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {c}\, \sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{6}-3 \sin \left (f x +e \right )^{4}+3 \sin \left (f x +e \right )^{2}-1}d x \right )}{a^{3} c^{3}} \] Input:
int(1/(a+a*sin(f*x+e))^(5/2)/(c-c*sin(f*x+e))^(5/2),x)
Output:
( - sqrt(c)*sqrt(a)*int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)) /(sin(e + f*x)**6 - 3*sin(e + f*x)**4 + 3*sin(e + f*x)**2 - 1),x))/(a**3*c **3)