Integrand size = 40, antiderivative size = 149 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {a B \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{c f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 B \cos (e+f x) \log (1-\sin (e+f x))}{c^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}} \] Output:
1/4*(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f/(c-c*sin(f*x+e))^(5/2)-a*B*c os(f*x+e)*(a+a*sin(f*x+e))^(1/2)/c/f/(c-c*sin(f*x+e))^(3/2)-a^2*B*cos(f*x+ e)*ln(1-sin(f*x+e))/c^2/f/(a+a*sin(f*x+e))^(1/2)/(c-c*sin(f*x+e))^(1/2)
Time = 11.68 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.33 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} \left (B \cos (2 (e+f x)) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-B \left (2+3 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+\left (A+3 B+4 B \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )\right ) \sin (e+f x)\right )}{c^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^2 \sqrt {c-c \sin (e+f x)}} \] Input:
Integrate[((a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^(5/2),x]
Output:
(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(B*Cos [2*(e + f*x)]*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - B*(2 + 3*Log[Cos[ (e + f*x)/2] - Sin[(e + f*x)/2]]) + (A + 3*B + 4*B*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]])*Sin[e + f*x]))/(c^2*f*(Cos[(e + f*x)/2] + Sin[(e + f*x) /2])*(-1 + Sin[e + f*x])^2*Sqrt[c - c*Sin[e + f*x]])
Time = 0.91 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.01, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.225, Rules used = {3042, 3451, 3042, 3218, 3042, 3216, 3042, 3146, 16}
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.
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 3451 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {B \int \frac {(\sin (e+f x) a+a)^{3/2}}{(c-c \sin (e+f x))^{3/2}}dx}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {B \int \frac {(\sin (e+f x) a+a)^{3/2}}{(c-c \sin (e+f x))^{3/2}}dx}{c}\) |
| \(\Big \downarrow \) 3218 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {B \left (\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f (c-c \sin (e+f x))^{3/2}}-\frac {a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx}{c}\right )}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {B \left (\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f (c-c \sin (e+f x))^{3/2}}-\frac {a \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {c-c \sin (e+f x)}}dx}{c}\right )}{c}\) |
| \(\Big \downarrow \) 3216 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {B \left (\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 \cos (e+f x) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {B \left (\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f (c-c \sin (e+f x))^{3/2}}-\frac {a^2 \cos (e+f x) \int \frac {\cos (e+f x)}{c-c \sin (e+f x)}dx}{\sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}\right )}{c}\) |
| \(\Big \downarrow \) 3146 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {B \left (\frac {a^2 \cos (e+f x) \int \frac {1}{c-c \sin (e+f x)}d(-c \sin (e+f x))}{c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f (c-c \sin (e+f x))^{3/2}}\right )}{c}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{4 f (c-c \sin (e+f x))^{5/2}}-\frac {B \left (\frac {a^2 \cos (e+f x) \log (c-c \sin (e+f x))}{c f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}+\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{f (c-c \sin (e+f x))^{3/2}}\right )}{c}\) |
Input:
Int[((a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x]) ^(5/2),x]
Output:
((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(4*f*(c - c*Sin[e + f*x] )^(5/2)) - (B*((a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(f*(c - c*Sin[e + f*x])^(3/2)) + (a^2*Cos[e + f*x]*Log[c - c*Sin[e + f*x]])/(c*f*Sqrt[a + a *Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])))/c
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x )^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && I ntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] || !IntegerQ[m + 1/ 2])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[a*c*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x ]]*Sqrt[c + d*Sin[e + f*x]])) Int[Cos[e + f*x]/(c + d*Sin[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[-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^ (m - 1)*((c + d*Sin[e + f*x])^n/(f*(2*n + 1))), x] - Simp[b*((2*m - 1)/(d*( 2*n + 1))) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b ^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] && !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim p[(A*b - a*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[(a*B*(m - n) + A*b*(m + n + 1))/(a*b*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[ {a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0 ] && (LtQ[m, -2^(-1)] || (ILtQ[m + n, 0] && !SumSimplerQ[n, 1])) && NeQ[2* m + 1, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(380\) vs. \(2(135)=270\).
Time = 6.85 (sec) , antiderivative size = 381, normalized size of antiderivative = 2.56
| method | result | size |
| parts | \(\frac {A \sqrt {4}\, a \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3}}{8 f \,c^{2} \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}+\frac {2 B \left (\left (-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-4 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \ln \left (\frac {2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )+\left (4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+4 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right ) \ln \left (\frac {2}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )+\cos \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right ) \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, a}{f \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}\, c^{2} \left (4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}\) | \(381\) |
| default | \(\frac {A \sqrt {4}\, a \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3}}{8 f \,c^{2} \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}-\frac {2 B \left (\left (4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+4 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right ) \ln \left (-\frac {2 \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )-\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )+\left (-4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-4 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \ln \left (\frac {2}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )+\cos \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right ) \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\sin \left (\frac {f x}{2}+\frac {e}{2}\right )\right )\right ) \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, a}{f \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}\, c^{2} \left (4 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}\) | \(383\) |
Input:
int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x,metho d=_RETURNVERBOSE)
Output:
1/8*A/f*4^(1/2)*a*(a*sin(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)/c^2/(c*cos(1/4*Pi+ 1/2*f*x+1/2*e)^2)^(1/2)*tan(1/4*Pi+1/2*f*x+1/2*e)^3+2*B/f*((-4*cos(1/2*f*x +1/2*e)^4+4*cos(1/2*f*x+1/2*e)^2-4*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)+1 )*ln(2*(sin(1/2*f*x+1/2*e)-cos(1/2*f*x+1/2*e))/(cos(1/2*f*x+1/2*e)+1))+(4* cos(1/2*f*x+1/2*e)^4+4*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)-4*cos(1/2*f*x +1/2*e)^2-1)*ln(2/(cos(1/2*f*x+1/2*e)+1))+cos(1/2*f*x+1/2*e)*(-4*cos(1/2*f *x+1/2*e)*sin(1/2*f*x+1/2*e)^2+sin(1/2*f*x+1/2*e)))*((2*sin(1/2*f*x+1/2*e) *cos(1/2*f*x+1/2*e)+1)*a)^(1/2)*a/(-(2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2* e)-1)*c)^(1/2)/c^2/(4*cos(1/2*f*x+1/2*e)^3*sin(1/2*f*x+1/2*e)-2*cos(1/2*f* x+1/2*e)^2-2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)+1)
\[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x , algorithm="fricas")
Output:
integral((B*a*cos(f*x + e)^2 - (A + B)*a*sin(f*x + e) - (A + B)*a)*sqrt(a* sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(3*c^3*cos(f*x + e)^2 - 4*c^3 - (c^3*cos(f*x + e)^2 - 4*c^3)*sin(f*x + e)), x)
\[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \left (A + B \sin {\left (e + f x \right )}\right )}{\left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((a+a*sin(f*x+e))**(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(5/2) ,x)
Output:
Integral((a*(sin(e + f*x) + 1))**(3/2)*(A + B*sin(e + f*x))/(-c*(sin(e + f *x) - 1))**(5/2), x)
\[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x , algorithm="maxima")
Output:
integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^(3/2)/(-c*sin(f*x + e) + c)^(5/2), x)
Exception generated. \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x , algorithm="giac")
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. \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(3/2))/(c - c*sin(e + f*x)) ^(5/2),x)
Output:
int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(3/2))/(c - c*sin(e + f*x)) ^(5/2), x)
\[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx=-\frac {\sqrt {c}\, \sqrt {a}\, a \left (\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right ) b +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right ) a +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right ) b +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{3}-3 \sin \left (f x +e \right )^{2}+3 \sin \left (f x +e \right )-1}d x \right ) a \right )}{c^{3}} \] Input:
int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x)
Output:
( - sqrt(c)*sqrt(a)*a*(int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)**3 - 3*sin(e + f*x)**2 + 3*sin(e + f*x) - 1),x)*b + int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**3 - 3*sin(e + f*x)**2 + 3*sin(e + f*x) - 1),x)*a + in t((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**3 - 3*sin(e + f*x)**2 + 3*sin(e + f*x) - 1),x)*b + int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1))/(sin(e + f*x)**3 - 3*sin(e + f*x)** 2 + 3*sin(e + f*x) - 1),x)*a))/c**3