\(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx\) [125]

Optimal result

Integrand size = 38, antiderivative size = 160 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {32 (A-11 B) c \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}+\frac {8 (A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac {(A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c^3 f} \] Output:

-32/15*(A-11*B)*c*sec(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/a^3/f+8/5*(A-11*B)*s 
ec(f*x+e)^3*(c-c*sin(f*x+e))^(5/2)/a^3/f-1/5*(A-11*B)*sec(f*x+e)^3*(c-c*si 
n(f*x+e))^(7/2)/a^3/c/f-1/5*(A-B)*sec(f*x+e)^5*(c-c*sin(f*x+e))^(11/2)/a^3 
/c^3/f
 

Mathematica [A] (verified)

Time = 10.33 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.82 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {c^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c-c \sin (e+f x)} (58 A-488 B-30 (A-8 B) \cos (2 (e+f x))+5 (8 A-133 B) \sin (e+f x)+15 B \sin (3 (e+f x)))}{30 a^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3} \] Input:

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))/(a + a*Sin[e + 
 f*x])^3,x]
 

Output:

-1/30*(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]* 
(58*A - 488*B - 30*(A - 8*B)*Cos[2*(e + f*x)] + 5*(8*A - 133*B)*Sin[e + f* 
x] + 15*B*Sin[3*(e + f*x)]))/(a^3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])* 
(1 + Sin[e + f*x])^3)
 

Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.93, 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.

\(\displaystyle \int \frac {(c-c \sin (e+f x))^{5/2} (A+B \sin (e+f x))}{(a \sin (e+f x)+a)^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

\(\displaystyle \frac {\int \sec ^6(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{11/2}dx}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{11/2}}{\cos (e+f x)^6}dx}{a^3 c^3}\)

\(\Big \downarrow \) 3334

\(\displaystyle \frac {-\frac {1}{10} c (A-11 B) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{9/2}dx-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{10} c (A-11 B) \int \frac {(c-c \sin (e+f x))^{9/2}}{\cos (e+f x)^4}dx-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{10} c (A-11 B) \left (8 c \int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2}dx+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{10} c (A-11 B) \left (8 c \int \frac {(c-c \sin (e+f x))^{7/2}}{\cos (e+f x)^4}dx+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3153

\(\displaystyle \frac {-\frac {1}{10} c (A-11 B) \left (8 c \left (-4 c \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2}dx-\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {1}{10} c (A-11 B) \left (8 c \left (-4 c \int \frac {(c-c \sin (e+f x))^{5/2}}{\cos (e+f x)^4}dx-\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 f}}{a^3 c^3}\)

\(\Big \downarrow \) 3152

\(\displaystyle \frac {-\frac {1}{10} c (A-11 B) \left (8 c \left (\frac {8 c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 f}-\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{f}\right )+\frac {2 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{f}\right )-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 f}}{a^3 c^3}\)

Input:

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))/(a + a*Sin[e + f*x]) 
^3,x]
 

Output:

(-1/5*((A - B)*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(11/2))/f - ((A - 11*B) 
*c*((2*c*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(7/2))/f + 8*c*((8*c^2*Sec[e 
+ f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(3*f) - (2*c*Sec[e + f*x]^3*(c - c*Si 
n[e + f*x])^(5/2))/f)))/10)/(a^3*c^3)
 

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 = 125.75 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.66

method	result	size
default	\(\frac {2 c^{3} \left (\sin \left (f x +e \right )-1\right ) \left (15 B \cos \left (f x +e \right )^{2} \sin \left (f x +e \right )+\left (-15 A +120 B \right ) \cos \left (f x +e \right )^{2}+\left (10 A -170 B \right ) \sin \left (f x +e \right )+22 A -182 B \right )}{15 a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(105\)

Input:

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x,method=_R 
ETURNVERBOSE)
 

Output:

2/15*c^3/a^3*(sin(f*x+e)-1)/(1+sin(f*x+e))^2*(15*B*cos(f*x+e)^2*sin(f*x+e) 
+(-15*A+120*B)*cos(f*x+e)^2+(10*A-170*B)*sin(f*x+e)+22*A-182*B)/cos(f*x+e) 
/(c-c*sin(f*x+e))^(1/2)/f
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.78 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=-\frac {2 \, {\left (15 \, {\left (A - 8 \, B\right )} c^{2} \cos \left (f x + e\right )^{2} - 2 \, {\left (11 \, A - 91 \, B\right )} c^{2} - 5 \, {\left (3 \, B c^{2} \cos \left (f x + e\right )^{2} + 2 \, {\left (A - 17 \, B\right )} c^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\right )}} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x, al 
gorithm="fricas")
 

Output:

-2/15*(15*(A - 8*B)*c^2*cos(f*x + e)^2 - 2*(11*A - 91*B)*c^2 - 5*(3*B*c^2* 
cos(f*x + e)^2 + 2*(A - 17*B)*c^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c) 
/(a^3*f*cos(f*x + e)^3 - 2*a^3*f*cos(f*x + e)*sin(f*x + e) - 2*a^3*f*cos(f 
*x + e))
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Timed out} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(5/2)/(a+a*sin(f*x+e))**3,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 761 vs. \(2 (144) = 288\).

Time = 0.15 (sec) , antiderivative size = 761, normalized size of antiderivative = 4.76 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x, al 
gorithm="maxima")
 

Output:

2/15*((7*c^(5/2) + 20*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 95*c^(5/2) 
*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 80*c^(5/2)*sin(f*x + e)^3/(cos(f*x 
+ e) + 1)^3 + 250*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 120*c^(5/2 
)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 250*c^(5/2)*sin(f*x + e)^6/(cos(f* 
x + e) + 1)^6 + 80*c^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 95*c^(5/2 
)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 20*c^(5/2)*sin(f*x + e)^9/(cos(f*x 
 + e) + 1)^9 + 7*c^(5/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)*A/((a^3 + 
5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e 
) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e) 
^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*(sin(f* 
x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(5/2)) - 2*(31*c^(5/2) + 155*c^(5/2)*si 
n(f*x + e)/(cos(f*x + e) + 1) + 395*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 
 1)^2 + 680*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1030*c^(5/2)*sin 
(f*x + e)^4/(cos(f*x + e) + 1)^4 + 1050*c^(5/2)*sin(f*x + e)^5/(cos(f*x + 
e) + 1)^5 + 1030*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 680*c^(5/2) 
*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 395*c^(5/2)*sin(f*x + e)^8/(cos(f*x 
 + e) + 1)^8 + 155*c^(5/2)*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 31*c^(5/2 
)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)*B/((a^3 + 5*a^3*sin(f*x + e)/(cos 
(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f 
*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 449 vs. \(2 (144) = 288\).

Time = 0.31 (sec) , antiderivative size = 449, normalized size of antiderivative = 2.81 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x, al 
gorithm="giac")
 

Output:

4/15*sqrt(2)*sqrt(c)*(15*B*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(a^3*(( 
cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 
 1)) + (4*A*c^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 29*B*c^2*sgn(sin(-1/ 
4*pi + 1/2*f*x + 1/2*e)) + 20*A*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*s 
gn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 
130*B*c^2*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 
 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 40*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 - 200*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 
 - 90*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 - 15*B*c^2*(cos(-1/4*p 
i + 1/2*f*x + 1/2*e) - 1)^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4* 
pi + 1/2*f*x + 1/2*e) + 1)^4)/(a^3*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/( 
cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 1)^5))/f
 

Mupad [B] (verification not implemented)

Time = 53.52 (sec) , antiderivative size = 904, normalized size of antiderivative = 5.65 \[ \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(5/2))/(a + a*sin(e + f*x)) 
^3,x)
 

Output:

((c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*( 
(2*B*c^2)/(a^3*f) - (B*c^2*exp(e*1i + f*x*1i)*2i)/(a^3*f)))/(exp(e*1i + f* 
x*1i) - 1i) - (exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (e 
xp(e*1i + f*x*1i)*1i)/2))^(1/2)*((c^2*(A*2i - B*7i)*1i)/(3*a^3*f) - (2*c^2 
*(7*A - 12*B))/(3*a^3*f) + (c^2*(A*23i - B*28i)*2i)/(3*a^3*f) - (c^2*(42*A 
 - 67*B))/(15*a^3*f) + (2*B*c^2)/(3*a^3*f)))/((exp(e*1i + f*x*1i) - 1i)*(e 
xp(e*1i + f*x*1i) + 1i)^3) + (exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x 
*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*((c^2*(A*1i - B*4i)*4i)/(a^ 
3*f) + (4*B*c^2)/(a^3*f)))/((exp(e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) 
+ 1i)) - (exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e* 
1i + f*x*1i)*1i)/2))^(1/2)*((8*c^2*(A*1i - B*1i))/(a^3*f) + (c^2*(A*1i - B 
*3i))/(2*a^3*f) + (c^2*(A*11i - B*1i))/(10*a^3*f) + (c^2*(12*A - 17*B)*1i) 
/(4*a^3*f) + (c^2*(52*A - 47*B)*1i)/(4*a^3*f)))/((exp(e*1i + f*x*1i) - 1i) 
*(exp(e*1i + f*x*1i) + 1i)^4) + (exp(e*1i + f*x*1i)*(c - c*((exp(- e*1i - 
f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*((c^2*(A*1i - B*4i))/(a^ 
3*f) + (c^2*(A*5i - B*4i))/(3*a^3*f) + (c^2*(A - 2*B)*8i)/(a^3*f)))/((exp( 
e*1i + f*x*1i) - 1i)*(exp(e*1i + f*x*1i) + 1i)^2) + (exp(e*1i + f*x*1i)*(c 
 - c*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2))^(1/2)*((c^ 
2*(A*2i - B*5i)*1i)/(5*a^3*f) - (c^2*(4*A - 3*B))/(a^3*f) - (c^2*(2*A - 5* 
B))/(5*a^3*f) + (c^2*(A*4i - B*3i)*1i)/(a^3*f) - (c^2*(10*A - 11*B))/(5...
 

Reduce [F]

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

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^3,x)
 

Output:

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