\(\int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx\) [147]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [F]
Giac [F(-2)]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 40, antiderivative size = 146 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {(A+B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{8 f (c-c \sin (e+f x))^{9/2}}+\frac {(A-3 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{24 c f (c-c \sin (e+f x))^{7/2}}+\frac {(A-3 B) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{96 c^2 f (c-c \sin (e+f x))^{5/2}} \] Output:

1/8*(A+B)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f/(c-c*sin(f*x+e))^(9/2)+1/24* 
(A-3*B)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/c/f/(c-c*sin(f*x+e))^(7/2)+1/96* 
(A-3*B)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/c^2/f/(c-c*sin(f*x+e))^(5/2)
 

Mathematica [A] (verified)

Time = 12.06 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.84 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (2 A+3 B-3 B \cos (2 (e+f x))+4 A \sin (e+f x))}{12 c^4 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (-1+\sin (e+f x))^4 \sqrt {c-c \sin (e+f x)}} \] Input:

Integrate[((a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + 
 f*x])^(9/2),x]
 

Output:

(a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(2*A + 
 3*B - 3*B*Cos[2*(e + f*x)] + 4*A*Sin[e + f*x]))/(12*c^4*f*(Cos[(e + f*x)/ 
2] + Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^4*Sqrt[c - c*Sin[e + f*x]])
 

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 3451, 3042, 3222, 3042, 3221}

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.

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}}dx\)

\(\Big \downarrow \) 3451

\(\displaystyle \frac {(A-3 B) \int \frac {(\sin (e+f x) a+a)^{3/2}}{(c-c \sin (e+f x))^{7/2}}dx}{4 c}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{8 f (c-c \sin (e+f x))^{9/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A-3 B) \int \frac {(\sin (e+f x) a+a)^{3/2}}{(c-c \sin (e+f x))^{7/2}}dx}{4 c}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{8 f (c-c \sin (e+f x))^{9/2}}\)

\(\Big \downarrow \) 3222

\(\displaystyle \frac {(A-3 B) \left (\frac {\int \frac {(\sin (e+f x) a+a)^{3/2}}{(c-c \sin (e+f x))^{5/2}}dx}{6 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}}\right )}{4 c}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{8 f (c-c \sin (e+f x))^{9/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(A-3 B) \left (\frac {\int \frac {(\sin (e+f x) a+a)^{3/2}}{(c-c \sin (e+f x))^{5/2}}dx}{6 c}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}}\right )}{4 c}+\frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{8 f (c-c \sin (e+f x))^{9/2}}\)

\(\Big \downarrow \) 3221

\(\displaystyle \frac {(A+B) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{8 f (c-c \sin (e+f x))^{9/2}}+\frac {(A-3 B) \left (\frac {\cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{24 c f (c-c \sin (e+f x))^{5/2}}+\frac {\cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{6 f (c-c \sin (e+f x))^{7/2}}\right )}{4 c}\)

Input:

Int[((a + a*Sin[e + f*x])^(3/2)*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x]) 
^(9/2),x]
 

Output:

((A + B)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(8*f*(c - c*Sin[e + f*x] 
)^(9/2)) + ((A - 3*B)*((Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(6*f*(c - 
 c*Sin[e + f*x])^(7/2)) + (Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(24*c* 
f*(c - c*Sin[e + f*x])^(5/2))))/(4*c)
 

Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3221
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] /; FreeQ[{a, b, c, d, e, f, m, 
n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] && Ne 
Q[m, -2^(-1)]
 

rule 3222
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])
 

rule 3451
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]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(306\) vs. \(2(128)=256\).

Time = 7.40 (sec) , antiderivative size = 307, normalized size of antiderivative = 2.10

method result size
parts \(\frac {A \sqrt {4}\, \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+2 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+3\right ) a \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3} \sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}}{192 f \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{4}}+\frac {2 B \sqrt {\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) a}\, \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a}{f \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \left (8 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-8 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-12 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-6 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+12 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \sqrt {-\left (2 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) c}\, c^{4}}\) \(307\)
default \(\frac {\left (\left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {3 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{2}-\frac {3 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )}{4}+\frac {3 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{2}+\frac {1}{8}\right ) \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-\frac {1}{2}\right ) \left (\cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+2 \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}+3\right ) A \tan \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{3} \sec \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{4}+12 B \sin \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) a \sqrt {a \sin \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}}{6 \sqrt {c \cos \left (\frac {\pi }{4}+\frac {f x}{2}+\frac {e}{2}\right )^{2}}\, c^{4} f \left (-1+2 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right ) \left (8 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{5} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-8 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \sin \left (\frac {f x}{2}+\frac {e}{2}\right )-12 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-6 \sin \left (\frac {f x}{2}+\frac {e}{2}\right ) \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+12 \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right )}\) \(349\)

Input:

int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x,metho 
d=_RETURNVERBOSE)
 

Output:

1/192*A/f*4^(1/2)*(a*sin(1/4*Pi+1/2*f*x+1/2*e)^2)^(1/2)*(cos(1/4*Pi+1/2*f* 
x+1/2*e)^4+2*cos(1/4*Pi+1/2*f*x+1/2*e)^2+3)*a/(c*cos(1/4*Pi+1/2*f*x+1/2*e) 
^2)^(1/2)/c^4*tan(1/4*Pi+1/2*f*x+1/2*e)^3*sec(1/4*Pi+1/2*f*x+1/2*e)^4+2*B/ 
f*((2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)+1)*a)^(1/2)*sin(1/2*f*x+1/2*e) 
^2*cos(1/2*f*x+1/2*e)^2*a/(-1+2*cos(1/2*f*x+1/2*e)^2)/(8*cos(1/2*f*x+1/2*e 
)^5*sin(1/2*f*x+1/2*e)-8*cos(1/2*f*x+1/2*e)^3*sin(1/2*f*x+1/2*e)-12*cos(1/ 
2*f*x+1/2*e)^4-6*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)+12*cos(1/2*f*x+1/2* 
e)^2+1)/(-(2*sin(1/2*f*x+1/2*e)*cos(1/2*f*x+1/2*e)-1)*c)^(1/2)/c^4
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.92 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=-\frac {{\left (3 \, B a \cos \left (f x + e\right )^{2} - 2 \, A a \sin \left (f x + e\right ) - {\left (A + 3 \, B\right )} a\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{6 \, {\left (c^{5} f \cos \left (f x + e\right )^{5} - 8 \, c^{5} f \cos \left (f x + e\right )^{3} + 8 \, c^{5} f \cos \left (f x + e\right ) + 4 \, {\left (c^{5} f \cos \left (f x + e\right )^{3} - 2 \, c^{5} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )}} \] Input:

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),x 
, algorithm="fricas")
 

Output:

-1/6*(3*B*a*cos(f*x + e)^2 - 2*A*a*sin(f*x + e) - (A + 3*B)*a)*sqrt(a*sin( 
f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c^5*f*cos(f*x + e)^5 - 8*c^5*f*co 
s(f*x + e)^3 + 8*c^5*f*cos(f*x + e) + 4*(c^5*f*cos(f*x + e)^3 - 2*c^5*f*co 
s(f*x + e))*sin(f*x + e))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(9/2) 
,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\int { \frac {{\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}} \,d x } \] Input:

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),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)^(9/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),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
 

Mupad [B] (verification not implemented)

Time = 45.21 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.68 \[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=\frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {8\,a\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\left (2\,A+3\,B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^5\,f}+\frac {32\,A\,a\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{3\,c^5\,f}-\frac {8\,B\,a\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{c^5\,f}\right )}{84\,\cos \left (e+f\,x\right )\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}-54\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (3\,e+3\,f\,x\right )+2\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\cos \left (5\,e+5\,f\,x\right )-96\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (2\,e+2\,f\,x\right )+16\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,\sin \left (4\,e+4\,f\,x\right )} \] Input:

int(((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^(3/2))/(c - c*sin(e + f*x)) 
^(9/2),x)
 

Output:

((c - c*sin(e + f*x))^(1/2)*((8*a*exp(e*5i + f*x*5i)*(2*A + 3*B)*(a + a*si 
n(e + f*x))^(1/2))/(3*c^5*f) + (32*A*a*exp(e*5i + f*x*5i)*sin(e + f*x)*(a 
+ a*sin(e + f*x))^(1/2))/(3*c^5*f) - (8*B*a*exp(e*5i + f*x*5i)*cos(2*e + 2 
*f*x)*(a + a*sin(e + f*x))^(1/2))/(c^5*f)))/(84*cos(e + f*x)*exp(e*5i + f* 
x*5i) - 54*exp(e*5i + f*x*5i)*cos(3*e + 3*f*x) + 2*exp(e*5i + f*x*5i)*cos( 
5*e + 5*f*x) - 96*exp(e*5i + f*x*5i)*sin(2*e + 2*f*x) + 16*exp(e*5i + f*x* 
5i)*sin(4*e + 4*f*x))
 

Reduce [F]

\[ \int \frac {(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{9/2}} \, dx=-\frac {\sqrt {c}\, \sqrt {a}\, a \left (\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )^{2}}{\sin \left (f x +e \right )^{5}-5 \sin \left (f x +e \right )^{4}+10 \sin \left (f x +e \right )^{3}-10 \sin \left (f x +e \right )^{2}+5 \sin \left (f x +e \right )-1}d x \right ) b +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{5}-5 \sin \left (f x +e \right )^{4}+10 \sin \left (f x +e \right )^{3}-10 \sin \left (f x +e \right )^{2}+5 \sin \left (f x +e \right )-1}d x \right ) a +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right )^{5}-5 \sin \left (f x +e \right )^{4}+10 \sin \left (f x +e \right )^{3}-10 \sin \left (f x +e \right )^{2}+5 \sin \left (f x +e \right )-1}d x \right ) b +\left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right )^{5}-5 \sin \left (f x +e \right )^{4}+10 \sin \left (f x +e \right )^{3}-10 \sin \left (f x +e \right )^{2}+5 \sin \left (f x +e \right )-1}d x \right ) a \right )}{c^{5}} \] Input:

int((a+a*sin(f*x+e))^(3/2)*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(9/2),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)**5 - 5*sin(e + f*x)**4 + 10*sin(e + f*x) 
**3 - 10*sin(e + f*x)**2 + 5*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)**5 - 5*sin(e 
 + f*x)**4 + 10*sin(e + f*x)**3 - 10*sin(e + f*x)**2 + 5*sin(e + f*x) - 1) 
,x)*a + int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*sin(e + f*x) 
)/(sin(e + f*x)**5 - 5*sin(e + f*x)**4 + 10*sin(e + f*x)**3 - 10*sin(e + f 
*x)**2 + 5*sin(e + f*x) - 1),x)*b + int((sqrt(sin(e + f*x) + 1)*sqrt( - si 
n(e + f*x) + 1))/(sin(e + f*x)**5 - 5*sin(e + f*x)**4 + 10*sin(e + f*x)**3 
 - 10*sin(e + f*x)**2 + 5*sin(e + f*x) - 1),x)*a))/c**5