\(\int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx\) [30]

Optimal result

Integrand size = 34, antiderivative size = 98 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=\frac {1}{8} a^2 (4 A+B) c x-\frac {a^2 (4 A+B) c \cos ^3(e+f x)}{12 f}+\frac {a^2 (4 A+B) c \cos (e+f x) \sin (e+f x)}{8 f}-\frac {B c \cos ^3(e+f x) \left (a^2+a^2 \sin (e+f x)\right )}{4 f} \] Output:

1/8*a^2*(4*A+B)*c*x-1/12*a^2*(4*A+B)*c*cos(f*x+e)^3/f+1/8*a^2*(4*A+B)*c*co 
s(f*x+e)*sin(f*x+e)/f-1/4*B*c*cos(f*x+e)^3*(a^2+a^2*sin(f*x+e))/f
 

Mathematica [A] (verified)

Time = 1.32 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.17 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=-\frac {a^2 c \cos (e+f x) \left (12 (4 A+B) \arcsin \left (\frac {\sqrt {1-\sin (e+f x)}}{\sqrt {2}}\right )+\sqrt {\cos ^2(e+f x)} (8 A+8 B+8 (A+B) \cos (2 (e+f x))-3 (8 A+B) \sin (e+f x)+3 B \sin (3 (e+f x)))\right )}{48 f \sqrt {\cos ^2(e+f x)}} \] Input:

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

Output:

-1/48*(a^2*c*Cos[e + f*x]*(12*(4*A + B)*ArcSin[Sqrt[1 - Sin[e + f*x]]/Sqrt 
[2]] + Sqrt[Cos[e + f*x]^2]*(8*A + 8*B + 8*(A + B)*Cos[2*(e + f*x)] - 3*(8 
*A + B)*Sin[e + f*x] + 3*B*Sin[3*(e + f*x)])))/(f*Sqrt[Cos[e + f*x]^2])
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.85, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.265, Rules used = {3042, 3446, 3042, 3339, 3042, 3148, 3042, 3115, 24}

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.

\(\displaystyle \int (a \sin (e+f x)+a)^2 (c-c \sin (e+f x)) (A+B \sin (e+f x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a)^2 (c-c \sin (e+f x)) (A+B \sin (e+f x))dx\)

\(\Big \downarrow \) 3446

\(\displaystyle a c \int \cos ^2(e+f x) (\sin (e+f x) a+a) (A+B \sin (e+f x))dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a c \int \cos (e+f x)^2 (\sin (e+f x) a+a) (A+B \sin (e+f x))dx\)

\(\Big \downarrow \) 3339

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3148

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 24

\(\displaystyle a c \left (\frac {1}{4} (4 A+B) \left (a \left (\frac {\sin (e+f x) \cos (e+f x)}{2 f}+\frac {x}{2}\right )-\frac {a \cos ^3(e+f x)}{3 f}\right )-\frac {B \cos ^3(e+f x) (a \sin (e+f x)+a)}{4 f}\right )\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

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

rule 3115

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3148

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

rule 3339

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

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(185\) vs. \(2(90)=180\).

Time = 0.23 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.90

\[\frac {\frac {a^{2} A c \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}-a^{2} A c \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-a^{2} B c \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+\frac {a^{2} B c \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}-A \cos \left (f x +e \right ) a^{2} c +a^{2} B c \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+a^{2} A c \left (f x +e \right )-B \cos \left (f x +e \right ) a^{2} c}{f}\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.79 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=-\frac {8 \, {\left (A + B\right )} a^{2} c \cos \left (f x + e\right )^{3} - 3 \, {\left (4 \, A + B\right )} a^{2} c f x + 3 \, {\left (2 \, B a^{2} c \cos \left (f x + e\right )^{3} - {\left (4 \, A + B\right )} a^{2} c \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{24 \, f} \] Input:

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e)),x, algorith 
m="fricas")
 

Output:

-1/24*(8*(A + B)*a^2*c*cos(f*x + e)^3 - 3*(4*A + B)*a^2*c*f*x + 3*(2*B*a^2 
*c*cos(f*x + e)^3 - (4*A + B)*a^2*c*cos(f*x + e))*sin(f*x + e))/f
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 396 vs. \(2 (90) = 180\).

Time = 0.21 (sec) , antiderivative size = 396, normalized size of antiderivative = 4.04 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=\begin {cases} - \frac {A a^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} - \frac {A a^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} + A a^{2} c x + \frac {A a^{2} c \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {A a^{2} c \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} + \frac {2 A a^{2} c \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {A a^{2} c \cos {\left (e + f x \right )}}{f} - \frac {3 B a^{2} c x \sin ^{4}{\left (e + f x \right )}}{8} - \frac {3 B a^{2} c x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} + \frac {B a^{2} c x \sin ^{2}{\left (e + f x \right )}}{2} - \frac {3 B a^{2} c x \cos ^{4}{\left (e + f x \right )}}{8} + \frac {B a^{2} c x \cos ^{2}{\left (e + f x \right )}}{2} + \frac {5 B a^{2} c \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} + \frac {B a^{2} c \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {3 B a^{2} c \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac {B a^{2} c \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{2 f} + \frac {2 B a^{2} c \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {B a^{2} c \cos {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (A + B \sin {\left (e \right )}\right ) \left (a \sin {\left (e \right )} + a\right )^{2} \left (- c \sin {\left (e \right )} + c\right ) & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((-A*a**2*c*x*sin(e + f*x)**2/2 - A*a**2*c*x*cos(e + f*x)**2/2 + 
A*a**2*c*x + A*a**2*c*sin(e + f*x)**2*cos(e + f*x)/f + A*a**2*c*sin(e + f* 
x)*cos(e + f*x)/(2*f) + 2*A*a**2*c*cos(e + f*x)**3/(3*f) - A*a**2*c*cos(e 
+ f*x)/f - 3*B*a**2*c*x*sin(e + f*x)**4/8 - 3*B*a**2*c*x*sin(e + f*x)**2*c 
os(e + f*x)**2/4 + B*a**2*c*x*sin(e + f*x)**2/2 - 3*B*a**2*c*x*cos(e + f*x 
)**4/8 + B*a**2*c*x*cos(e + f*x)**2/2 + 5*B*a**2*c*sin(e + f*x)**3*cos(e + 
 f*x)/(8*f) + B*a**2*c*sin(e + f*x)**2*cos(e + f*x)/f + 3*B*a**2*c*sin(e + 
 f*x)*cos(e + f*x)**3/(8*f) - B*a**2*c*sin(e + f*x)*cos(e + f*x)/(2*f) + 2 
*B*a**2*c*cos(e + f*x)**3/(3*f) - B*a**2*c*cos(e + f*x)/f, Ne(f, 0)), (x*( 
A + B*sin(e))*(a*sin(e) + a)**2*(-c*sin(e) + c), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.83 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=-\frac {32 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a^{2} c + 24 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} c - 96 \, {\left (f x + e\right )} A a^{2} c + 32 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{2} c + 3 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} c - 24 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} c + 96 \, A a^{2} c \cos \left (f x + e\right ) + 96 \, B a^{2} c \cos \left (f x + e\right )}{96 \, f} \] Input:

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e)),x, algorith 
m="maxima")
 

Output:

-1/96*(32*(cos(f*x + e)^3 - 3*cos(f*x + e))*A*a^2*c + 24*(2*f*x + 2*e - si 
n(2*f*x + 2*e))*A*a^2*c - 96*(f*x + e)*A*a^2*c + 32*(cos(f*x + e)^3 - 3*co 
s(f*x + e))*B*a^2*c + 3*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 
2*e))*B*a^2*c - 24*(2*f*x + 2*e - sin(2*f*x + 2*e))*B*a^2*c + 96*A*a^2*c*c 
os(f*x + e) + 96*B*a^2*c*cos(f*x + e))/f
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.09 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=-\frac {B a^{2} c \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} + \frac {A a^{2} c \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} + \frac {1}{8} \, {\left (4 \, A a^{2} c + B a^{2} c\right )} x - \frac {{\left (A a^{2} c + B a^{2} c\right )} \cos \left (3 \, f x + 3 \, e\right )}{12 \, f} - \frac {{\left (A a^{2} c + B a^{2} c\right )} \cos \left (f x + e\right )}{4 \, f} \] Input:

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e)),x, algorith 
m="giac")
 

Output:

-1/32*B*a^2*c*sin(4*f*x + 4*e)/f + 1/4*A*a^2*c*sin(2*f*x + 2*e)/f + 1/8*(4 
*A*a^2*c + B*a^2*c)*x - 1/12*(A*a^2*c + B*a^2*c)*cos(3*f*x + 3*e)/f - 1/4* 
(A*a^2*c + B*a^2*c)*cos(f*x + e)/f
 

Mupad [B] (verification not implemented)

Time = 35.70 (sec) , antiderivative size = 339, normalized size of antiderivative = 3.46 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=\frac {a^2\,c\,\mathrm {atan}\left (\frac {a^2\,c\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (4\,A+B\right )}{4\,\left (A\,a^2\,c+\frac {B\,a^2\,c}{4}\right )}\right )\,\left (4\,A+B\right )}{4\,f}-\frac {a^2\,c\,\left (4\,A+B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {f\,x}{2}\right )}{4\,f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (2\,A\,a^2\,c+2\,B\,a^2\,c\right )-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (A\,a^2\,c-\frac {B\,a^2\,c}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (2\,A\,a^2\,c+2\,B\,a^2\,c\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {2\,A\,a^2\,c}{3}+\frac {2\,B\,a^2\,c}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (A\,a^2\,c-\frac {B\,a^2\,c}{4}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (A\,a^2\,c+\frac {7\,B\,a^2\,c}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (A\,a^2\,c+\frac {7\,B\,a^2\,c}{4}\right )+\frac {2\,A\,a^2\,c}{3}+\frac {2\,B\,a^2\,c}{3}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )} \] Input:

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

Output:

(a^2*c*atan((a^2*c*tan(e/2 + (f*x)/2)*(4*A + B))/(4*(A*a^2*c + (B*a^2*c)/4 
)))*(4*A + B))/(4*f) - (a^2*c*(4*A + B)*(atan(tan(e/2 + (f*x)/2)) - (f*x)/ 
2))/(4*f) - (tan(e/2 + (f*x)/2)^4*(2*A*a^2*c + 2*B*a^2*c) - tan(e/2 + (f*x 
)/2)*(A*a^2*c - (B*a^2*c)/4) + tan(e/2 + (f*x)/2)^6*(2*A*a^2*c + 2*B*a^2*c 
) + tan(e/2 + (f*x)/2)^2*((2*A*a^2*c)/3 + (2*B*a^2*c)/3) + tan(e/2 + (f*x) 
/2)^7*(A*a^2*c - (B*a^2*c)/4) - tan(e/2 + (f*x)/2)^3*(A*a^2*c + (7*B*a^2*c 
)/4) + tan(e/2 + (f*x)/2)^5*(A*a^2*c + (7*B*a^2*c)/4) + (2*A*a^2*c)/3 + (2 
*B*a^2*c)/3)/(f*(4*tan(e/2 + (f*x)/2)^2 + 6*tan(e/2 + (f*x)/2)^4 + 4*tan(e 
/2 + (f*x)/2)^6 + tan(e/2 + (f*x)/2)^8 + 1))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.28 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x)) \, dx=\frac {a^{2} c \left (6 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b +8 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a +8 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b +12 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a -3 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b -8 \cos \left (f x +e \right ) a -8 \cos \left (f x +e \right ) b +12 a f x +8 a +3 b f x +8 b \right )}{24 f} \] Input:

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

Output:

(a**2*c*(6*cos(e + f*x)*sin(e + f*x)**3*b + 8*cos(e + f*x)*sin(e + f*x)**2 
*a + 8*cos(e + f*x)*sin(e + f*x)**2*b + 12*cos(e + f*x)*sin(e + f*x)*a - 3 
*cos(e + f*x)*sin(e + f*x)*b - 8*cos(e + f*x)*a - 8*cos(e + f*x)*b + 12*a* 
f*x + 8*a + 3*b*f*x + 8*b))/(24*f)