\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx\) [117]

Optimal result

Integrand size = 38, antiderivative size = 154 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {32 (A-3 B) c^2 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{3 a^2 f}-\frac {8 (A-3 B) c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}-\frac {(A-3 B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f}-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 a^2 c^2 f} \] Output:

32/3*(A-3*B)*c^2*sec(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a^2/f-8/3*(A-3*B)*c*sec 
(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a^2/f-1/3*(A-3*B)*sec(f*x+e)*(c-c*sin(f*x+e 
))^(5/2)/a^2/f-1/3*(A-B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(9/2)/a^2/c^2/f
 

Mathematica [A] (verified)

Time = 9.19 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.84 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx=-\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} (-50 A+160 B+6 (A-4 B) \cos (2 (e+f x))+(-72 A+201 B) \sin (e+f x)+B \sin (3 (e+f x)))}{6 a^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2} \] Input:

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))/(a + a*Sin[e + 
 f*x])^2,x]
 

Output:

-1/6*(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*( 
-50*A + 160*B + 6*(A - 4*B)*Cos[2*(e + f*x)] + (-72*A + 201*B)*Sin[e + f*x 
] + B*Sin[3*(e + f*x)]))/(a^2*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + 
 Sin[e + f*x])^2)
 

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {3042, 3446, 3042, 3334, 3042, 3153, 3042, 3153, 3042, 3152}

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.

\(\displaystyle \int \frac {(c-c \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c-c \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^2}dx\)

\(\Big \downarrow \) 3446

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3334

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{7/2}dx-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \int \frac {(c-c \sin (e+f x))^{7/2}}{\cos (e+f x)^2}dx-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \left (\frac {8}{3} c \int \sec ^2(e+f x) (c-c \sin (e+f x))^{5/2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \left (\frac {8}{3} c \int \frac {(c-c \sin (e+f x))^{5/2}}{\cos (e+f x)^2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \left (\frac {8}{3} c \left (4 c \int \sec ^2(e+f x) (c-c \sin (e+f x))^{3/2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \left (\frac {8}{3} c \left (4 c \int \frac {(c-c \sin (e+f x))^{3/2}}{\cos (e+f x)^2}dx+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}}{a^2 c^2}\)

\(\Big \downarrow \) 3152

\(\displaystyle \frac {-\frac {1}{2} c (A-3 B) \left (\frac {8}{3} c \left (\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{f}-\frac {8 c^2 \sec (e+f x) \sqrt {c-c \sin (e+f x)}}{f}\right )+\frac {2 c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 f}\right )-\frac {(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{3 f}}{a^2 c^2}\)

Input:

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))/(a + a*Sin[e + f*x]) 
^2,x]
 

Output:

(-1/3*((A - B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(9/2))/f - ((A - 3*B)*c 
*((2*c*Sec[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(3*f) + (8*c*((-8*c^2*Sec[ 
e + f*x]*Sqrt[c - c*Sin[e + f*x]])/f + (2*c*Sec[e + f*x]*(c - c*Sin[e + f* 
x])^(3/2))/f))/3))/2)/(a^2*c^2)
 

Defintions of rubi rules used

rule 3042

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

rule 3152

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x 
])^(m - 1)/(f*g*(m - 1))), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 
 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
 

rule 3153

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[(-b)*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + 
f*x])^(m - 1)/(f*g*(m + p))), x] + Simp[a*((2*m + p - 1)/(m + p))   Int[(g* 
Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, 
g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p - 1)/2], 0] && 
NeQ[m + p, 0]
 

rule 3334

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(b* 
c + a*d))*(g*Cos[e + f*x])^(p + 1)*((a + b*Sin[e + f*x])^m/(a*f*g*(p + 1))) 
, x] + Simp[b*((a*d*m + b*c*(m + p + 1))/(a*g^2*(p + 1)))   Int[(g*Cos[e + 
f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, 
 f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]
 

rule 3446

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + 
 (f_.)*(x_)])*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si 
mp[a^m*c^m   Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B*Sin 
[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a* 
d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] 
&& GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))
 

Maple [A] (verified)

Time = 20.71 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.68

method	result	size
default	\(-\frac {2 c^{3} \left (\sin \left (f x +e \right )-1\right ) \left (-B \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )+\left (-3 A +12 B \right ) \cos \left (f x +e \right )^{2}+\left (18 A -50 B \right ) \sin \left (f x +e \right )+14 A -46 B \right )}{3 a^{2} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(105\)

Input:

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^2,x,method=_R 
ETURNVERBOSE)
 

Output:

-2/3*c^3/a^2*(sin(f*x+e)-1)/(1+sin(f*x+e))*(-B*cos(f*x+e)^2*sin(f*x+e)+(-3 
*A+12*B)*cos(f*x+e)^2+(18*A-50*B)*sin(f*x+e)+14*A-46*B)/cos(f*x+e)/(c-c*si 
n(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.71 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx=-\frac {2 \, {\left (3 \, {\left (A - 4 \, B\right )} c^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (7 \, A - 23 \, B\right )} c^{2} + {\left (B c^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (9 \, A - 25 \, B\right )} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, {\left (a^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} f \cos \left (f x + e\right )\right )}} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^2,x, al 
gorithm="fricas")
 

Output:

-2/3*(3*(A - 4*B)*c^2*cos(f*x + e)^2 - 2*(7*A - 23*B)*c^2 + (B*c^2*cos(f*x 
 + e)^2 - 2*(9*A - 25*B)*c^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(a^2 
*f*cos(f*x + e)*sin(f*x + e) + a^2*f*cos(f*x + e))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx=\text {Timed out} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(5/2)/(a+a*sin(f*x+e))**2,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 577 vs. \(2 (138) = 276\).

Time = 0.14 (sec) , antiderivative size = 577, normalized size of antiderivative = 3.75 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx =\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^2,x, al 
gorithm="maxima")
 

Output:

-2/3*((11*c^(5/2) + 36*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 56*c^(5/2 
)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 108*c^(5/2)*sin(f*x + e)^3/(cos(f* 
x + e) + 1)^3 + 90*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 108*c^(5/ 
2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 56*c^(5/2)*sin(f*x + e)^6/(cos(f* 
x + e) + 1)^6 + 36*c^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 11*c^(5/2 
)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)*A/((a^2 + 3*a^2*sin(f*x + e)/(cos(f 
*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e 
)^3/(cos(f*x + e) + 1)^3)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(5/2)) 
 - 2*(17*c^(5/2) + 51*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 92*c^(5/2) 
*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 149*c^(5/2)*sin(f*x + e)^3/(cos(f*x 
 + e) + 1)^3 + 150*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 149*c^(5/ 
2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 92*c^(5/2)*sin(f*x + e)^6/(cos(f* 
x + e) + 1)^6 + 51*c^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 17*c^(5/2 
)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)*B/((a^2 + 3*a^2*sin(f*x + e)/(cos(f 
*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e 
)^3/(cos(f*x + e) + 1)^3)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(5/2)) 
)/f
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (138) = 276\).

Time = 0.34 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.90 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx=\frac {32 \, \sqrt {2} {\left (A c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - 3 \, B c^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \frac {3 \, A c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} + \frac {9 \, B c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} + \frac {2 \, A c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} + \frac {2 \, B c^{2} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}}\right )} \sqrt {c}}{3 \, a^{2} f {\left (\frac {{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} - 1\right )}^{3}} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^2,x, al 
gorithm="giac")
 

Output:

32/3*sqrt(2)*(A*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 3*B*c^2*sgn(sin( 
-1/4*pi + 1/2*f*x + 1/2*e)) - 3*A*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1) 
^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1 
)^2 + 9*B*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2 
*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 + 2*A*c^2*(cos(-1/4* 
pi + 1/2*f*x + 1/2*e) - 1)^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4 
*pi + 1/2*f*x + 1/2*e) + 1)^3 + 2*B*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 
1)^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 
 1)^3)*sqrt(c)/(a^2*f*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi 
 + 1/2*f*x + 1/2*e) + 1)^2 - 1)^3)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^2} \, dx=\int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2} \,d x \] Input:

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(5/2))/(a + a*sin(e + f*x)) 
^2,x)
 

Output:

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(5/2))/(a + a*sin(e + f*x)) 
^2, x)
 

Reduce [F]

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

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^2,x)
 

Output:

(sqrt(c)*c**2*(int(sqrt( - sin(e + f*x) + 1)/(sin(e + f*x)**2 + 2*sin(e + 
f*x) + 1),x)*a + int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x)**3)/(sin(e + 
f*x)**2 + 2*sin(e + f*x) + 1),x)*b + int((sqrt( - sin(e + f*x) + 1)*sin(e 
+ f*x)**2)/(sin(e + f*x)**2 + 2*sin(e + f*x) + 1),x)*a - 2*int((sqrt( - si 
n(e + f*x) + 1)*sin(e + f*x)**2)/(sin(e + f*x)**2 + 2*sin(e + f*x) + 1),x) 
*b - 2*int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)**2 + 2*s 
in(e + f*x) + 1),x)*a + int((sqrt( - sin(e + f*x) + 1)*sin(e + f*x))/(sin( 
e + f*x)**2 + 2*sin(e + f*x) + 1),x)*b))/a**2