\(\int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx\) [42]

Optimal result

Integrand size = 36, antiderivative size = 138 \[ \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\frac {1}{16} a^3 (6 A+B) c^2 x-\frac {a^3 (6 A+B) c^2 \cos ^5(e+f x)}{30 f}+\frac {a^3 (6 A+B) c^2 \cos (e+f x) \sin (e+f x)}{16 f}+\frac {a^3 (6 A+B) c^2 \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac {B c^2 \cos ^5(e+f x) \left (a^3+a^3 \sin (e+f x)\right )}{6 f} \] Output:

1/16*a^3*(6*A+B)*c^2*x-1/30*a^3*(6*A+B)*c^2*cos(f*x+e)^5/f+1/16*a^3*(6*A+B 
)*c^2*cos(f*x+e)*sin(f*x+e)/f+1/24*a^3*(6*A+B)*c^2*cos(f*x+e)^3*sin(f*x+e) 
/f-1/6*B*c^2*cos(f*x+e)^5*(a^3+a^3*sin(f*x+e))/f
 

Mathematica [A] (verified)

Time = 7.65 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.96 \[ \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\frac {a^3 c^2 (360 A e+60 B e+360 A f x+60 B f x-120 (A+B) \cos (e+f x)-60 (A+B) \cos (3 (e+f x))-12 A \cos (5 (e+f x))-12 B \cos (5 (e+f x))+240 A \sin (2 (e+f x))+15 B \sin (2 (e+f x))+30 A \sin (4 (e+f x))-15 B \sin (4 (e+f x))-5 B \sin (6 (e+f x)))}{960 f} \] Input:

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

Output:

(a^3*c^2*(360*A*e + 60*B*e + 360*A*f*x + 60*B*f*x - 120*(A + B)*Cos[e + f* 
x] - 60*(A + B)*Cos[3*(e + f*x)] - 12*A*Cos[5*(e + f*x)] - 12*B*Cos[5*(e + 
 f*x)] + 240*A*Sin[2*(e + f*x)] + 15*B*Sin[2*(e + f*x)] + 30*A*Sin[4*(e + 
f*x)] - 15*B*Sin[4*(e + f*x)] - 5*B*Sin[6*(e + f*x)]))/(960*f)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.82, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.306, Rules used = {3042, 3446, 3042, 3339, 3042, 3148, 3042, 3115, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int (a \sin (e+f x)+a)^3 (c-c \sin (e+f x))^2 (A+B \sin (e+f x))dx\)

\(\Big \downarrow \) 3446

\(\displaystyle a^2 c^2 \int \cos ^4(e+f x) (\sin (e+f x) a+a) (A+B \sin (e+f x))dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \int \cos (e+f x)^4 (\sin (e+f x) a+a) (A+B \sin (e+f x))dx\)

\(\Big \downarrow \) 3339

\(\displaystyle a^2 c^2 \left (\frac {1}{6} (6 A+B) \int \cos ^4(e+f x) (\sin (e+f x) a+a)dx-\frac {B \cos ^5(e+f x) (a \sin (e+f x)+a)}{6 f}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \left (\frac {1}{6} (6 A+B) \int \cos (e+f x)^4 (\sin (e+f x) a+a)dx-\frac {B \cos ^5(e+f x) (a \sin (e+f x)+a)}{6 f}\right )\)

\(\Big \downarrow \) 3148

\(\displaystyle a^2 c^2 \left (\frac {1}{6} (6 A+B) \left (a \int \cos ^4(e+f x)dx-\frac {a \cos ^5(e+f x)}{5 f}\right )-\frac {B \cos ^5(e+f x) (a \sin (e+f x)+a)}{6 f}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \left (\frac {1}{6} (6 A+B) \left (a \int \sin \left (e+f x+\frac {\pi }{2}\right )^4dx-\frac {a \cos ^5(e+f x)}{5 f}\right )-\frac {B \cos ^5(e+f x) (a \sin (e+f x)+a)}{6 f}\right )\)

\(\Big \downarrow \) 3115

\(\displaystyle a^2 c^2 \left (\frac {1}{6} (6 A+B) \left (a \left (\frac {3}{4} \int \cos ^2(e+f x)dx+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f}\right )-\frac {a \cos ^5(e+f x)}{5 f}\right )-\frac {B \cos ^5(e+f x) (a \sin (e+f x)+a)}{6 f}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \left (\frac {1}{6} (6 A+B) \left (a \left (\frac {3}{4} \int \sin \left (e+f x+\frac {\pi }{2}\right )^2dx+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f}\right )-\frac {a \cos ^5(e+f x)}{5 f}\right )-\frac {B \cos ^5(e+f x) (a \sin (e+f x)+a)}{6 f}\right )\)

\(\Big \downarrow \) 3115

\(\displaystyle a^2 c^2 \left (\frac {1}{6} (6 A+B) \left (a \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (e+f x) \cos (e+f x)}{2 f}\right )+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f}\right )-\frac {a \cos ^5(e+f x)}{5 f}\right )-\frac {B \cos ^5(e+f x) (a \sin (e+f x)+a)}{6 f}\right )\)

\(\Big \downarrow \) 24

\(\displaystyle a^2 c^2 \left (\frac {1}{6} (6 A+B) \left (a \left (\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f}+\frac {3}{4} \left (\frac {\sin (e+f x) \cos (e+f x)}{2 f}+\frac {x}{2}\right )\right )-\frac {a \cos ^5(e+f x)}{5 f}\right )-\frac {B \cos ^5(e+f x) (a \sin (e+f x)+a)}{6 f}\right )\)

Input:

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

Output:

a^2*c^2*(-1/6*(B*Cos[e + f*x]^5*(a + a*Sin[e + f*x]))/f + ((6*A + B)*(-1/5 
*(a*Cos[e + f*x]^5)/f + a*((Cos[e + f*x]^3*Sin[e + f*x])/(4*f) + (3*(x/2 + 
 (Cos[e + f*x]*Sin[e + f*x])/(2*f)))/4)))/6)
 

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. \(363\) vs. \(2(128)=256\).

Time = 0.15 (sec) , antiderivative size = 364, normalized size of antiderivative = 2.64

\[\frac {-\frac {a^{3} A \,c^{2} \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5}+a^{3} A \,c^{2} \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+\frac {2 a^{3} A \,c^{2} \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}+a^{3} B \,c^{2} \left (-\frac {\left (\sin \left (f x +e \right )^{5}+\frac {5 \sin \left (f x +e \right )^{3}}{4}+\frac {15 \sin \left (f x +e \right )}{8}\right ) \cos \left (f x +e \right )}{6}+\frac {5 f x}{16}+\frac {5 e}{16}\right )-\frac {a^{3} B \,c^{2} \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5}-2 a^{3} B \,c^{2} \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-2 a^{3} A \,c^{2} \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+\frac {2 a^{3} B \,c^{2} \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}-a^{3} A \,c^{2} \cos \left (f x +e \right )+a^{3} B \,c^{2} \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+a^{3} A \,c^{2} \left (f x +e \right )-a^{3} B \,c^{2} \cos \left (f x +e \right )}{f}\]

Input:

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

Output:

1/f*(-1/5*a^3*A*c^2*(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e)+a^3*A*c 
^2*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)+2/3*a^3*A 
*c^2*(2+sin(f*x+e)^2)*cos(f*x+e)+a^3*B*c^2*(-1/6*(sin(f*x+e)^5+5/4*sin(f*x 
+e)^3+15/8*sin(f*x+e))*cos(f*x+e)+5/16*f*x+5/16*e)-1/5*a^3*B*c^2*(8/3+sin( 
f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e)-2*a^3*B*c^2*(-1/4*(sin(f*x+e)^3+3/2* 
sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e)-2*a^3*A*c^2*(-1/2*sin(f*x+e)*cos(f*x 
+e)+1/2*f*x+1/2*e)+2/3*a^3*B*c^2*(2+sin(f*x+e)^2)*cos(f*x+e)-a^3*A*c^2*cos 
(f*x+e)+a^3*B*c^2*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)+a^3*A*c^2*(f* 
x+e)-a^3*B*c^2*cos(f*x+e))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.77 \[ \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=-\frac {48 \, {\left (A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{5} - 15 \, {\left (6 \, A + B\right )} a^{3} c^{2} f x + 5 \, {\left (8 \, B a^{3} c^{2} \cos \left (f x + e\right )^{5} - 2 \, {\left (6 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )^{3} - 3 \, {\left (6 \, A + B\right )} a^{3} c^{2} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{240 \, f} \] Input:

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

Output:

-1/240*(48*(A + B)*a^3*c^2*cos(f*x + e)^5 - 15*(6*A + B)*a^3*c^2*f*x + 5*( 
8*B*a^3*c^2*cos(f*x + e)^5 - 2*(6*A + B)*a^3*c^2*cos(f*x + e)^3 - 3*(6*A + 
 B)*a^3*c^2*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. 910 vs. \(2 (128) = 256\).

Time = 0.48 (sec) , antiderivative size = 910, normalized size of antiderivative = 6.59 \[ \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx =\text {Too large to display} \] Input:

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

Output:

Piecewise((3*A*a**3*c**2*x*sin(e + f*x)**4/8 + 3*A*a**3*c**2*x*sin(e + f*x 
)**2*cos(e + f*x)**2/4 - A*a**3*c**2*x*sin(e + f*x)**2 + 3*A*a**3*c**2*x*c 
os(e + f*x)**4/8 - A*a**3*c**2*x*cos(e + f*x)**2 + A*a**3*c**2*x - A*a**3* 
c**2*sin(e + f*x)**4*cos(e + f*x)/f - 5*A*a**3*c**2*sin(e + f*x)**3*cos(e 
+ f*x)/(8*f) - 4*A*a**3*c**2*sin(e + f*x)**2*cos(e + f*x)**3/(3*f) + 2*A*a 
**3*c**2*sin(e + f*x)**2*cos(e + f*x)/f - 3*A*a**3*c**2*sin(e + f*x)*cos(e 
 + f*x)**3/(8*f) + A*a**3*c**2*sin(e + f*x)*cos(e + f*x)/f - 8*A*a**3*c**2 
*cos(e + f*x)**5/(15*f) + 4*A*a**3*c**2*cos(e + f*x)**3/(3*f) - A*a**3*c** 
2*cos(e + f*x)/f + 5*B*a**3*c**2*x*sin(e + f*x)**6/16 + 15*B*a**3*c**2*x*s 
in(e + f*x)**4*cos(e + f*x)**2/16 - 3*B*a**3*c**2*x*sin(e + f*x)**4/4 + 15 
*B*a**3*c**2*x*sin(e + f*x)**2*cos(e + f*x)**4/16 - 3*B*a**3*c**2*x*sin(e 
+ f*x)**2*cos(e + f*x)**2/2 + B*a**3*c**2*x*sin(e + f*x)**2/2 + 5*B*a**3*c 
**2*x*cos(e + f*x)**6/16 - 3*B*a**3*c**2*x*cos(e + f*x)**4/4 + B*a**3*c**2 
*x*cos(e + f*x)**2/2 - 11*B*a**3*c**2*sin(e + f*x)**5*cos(e + f*x)/(16*f) 
- B*a**3*c**2*sin(e + f*x)**4*cos(e + f*x)/f - 5*B*a**3*c**2*sin(e + f*x)* 
*3*cos(e + f*x)**3/(6*f) + 5*B*a**3*c**2*sin(e + f*x)**3*cos(e + f*x)/(4*f 
) - 4*B*a**3*c**2*sin(e + f*x)**2*cos(e + f*x)**3/(3*f) + 2*B*a**3*c**2*si 
n(e + f*x)**2*cos(e + f*x)/f - 5*B*a**3*c**2*sin(e + f*x)*cos(e + f*x)**5/ 
(16*f) + 3*B*a**3*c**2*sin(e + f*x)*cos(e + f*x)**3/(4*f) - B*a**3*c**2*si 
n(e + f*x)*cos(e + f*x)/(2*f) - 8*B*a**3*c**2*cos(e + f*x)**5/(15*f) + ...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (128) = 256\).

Time = 0.04 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.61 \[ \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=-\frac {64 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} A a^{3} c^{2} + 640 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a^{3} c^{2} - 30 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{3} c^{2} + 480 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{3} c^{2} - 960 \, {\left (f x + e\right )} A a^{3} c^{2} + 64 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} B a^{3} c^{2} + 640 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{3} c^{2} - 5 \, {\left (4 \, \sin \left (2 \, f x + 2 \, e\right )^{3} + 60 \, f x + 60 \, e + 9 \, \sin \left (4 \, f x + 4 \, e\right ) - 48 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{3} c^{2} + 60 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{3} c^{2} - 240 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{3} c^{2} + 960 \, A a^{3} c^{2} \cos \left (f x + e\right ) + 960 \, B a^{3} c^{2} \cos \left (f x + e\right )}{960 \, f} \] Input:

integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2,x, algori 
thm="maxima")
 

Output:

-1/960*(64*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*A*a^3* 
c^2 + 640*(cos(f*x + e)^3 - 3*cos(f*x + e))*A*a^3*c^2 - 30*(12*f*x + 12*e 
+ sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*A*a^3*c^2 + 480*(2*f*x + 2*e - si 
n(2*f*x + 2*e))*A*a^3*c^2 - 960*(f*x + e)*A*a^3*c^2 + 64*(3*cos(f*x + e)^5 
 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*B*a^3*c^2 + 640*(cos(f*x + e)^3 - 
3*cos(f*x + e))*B*a^3*c^2 - 5*(4*sin(2*f*x + 2*e)^3 + 60*f*x + 60*e + 9*si 
n(4*f*x + 4*e) - 48*sin(2*f*x + 2*e))*B*a^3*c^2 + 60*(12*f*x + 12*e + sin( 
4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*B*a^3*c^2 - 240*(2*f*x + 2*e - sin(2*f* 
x + 2*e))*B*a^3*c^2 + 960*A*a^3*c^2*cos(f*x + e) + 960*B*a^3*c^2*cos(f*x + 
 e))/f
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.43 \[ \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=-\frac {B a^{3} c^{2} \sin \left (6 \, f x + 6 \, e\right )}{192 \, f} + \frac {1}{16} \, {\left (6 \, A a^{3} c^{2} + B a^{3} c^{2}\right )} x - \frac {{\left (A a^{3} c^{2} + B a^{3} c^{2}\right )} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} - \frac {{\left (A a^{3} c^{2} + B a^{3} c^{2}\right )} \cos \left (3 \, f x + 3 \, e\right )}{16 \, f} - \frac {{\left (A a^{3} c^{2} + B a^{3} c^{2}\right )} \cos \left (f x + e\right )}{8 \, f} + \frac {{\left (2 \, A a^{3} c^{2} - B a^{3} c^{2}\right )} \sin \left (4 \, f x + 4 \, e\right )}{64 \, f} + \frac {{\left (16 \, A a^{3} c^{2} + B a^{3} c^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{64 \, f} \] Input:

integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^2,x, algori 
thm="giac")
 

Output:

-1/192*B*a^3*c^2*sin(6*f*x + 6*e)/f + 1/16*(6*A*a^3*c^2 + B*a^3*c^2)*x - 1 
/80*(A*a^3*c^2 + B*a^3*c^2)*cos(5*f*x + 5*e)/f - 1/16*(A*a^3*c^2 + B*a^3*c 
^2)*cos(3*f*x + 3*e)/f - 1/8*(A*a^3*c^2 + B*a^3*c^2)*cos(f*x + e)/f + 1/64 
*(2*A*a^3*c^2 - B*a^3*c^2)*sin(4*f*x + 4*e)/f + 1/64*(16*A*a^3*c^2 + B*a^3 
*c^2)*sin(2*f*x + 2*e)/f
 

Mupad [B] (verification not implemented)

Time = 36.92 (sec) , antiderivative size = 536, normalized size of antiderivative = 3.88 \[ \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx =\text {Too large to display} \] Input:

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

Output:

(a^3*c^2*atan((a^3*c^2*tan(e/2 + (f*x)/2)*(6*A + B))/(8*((3*A*a^3*c^2)/4 + 
 (B*a^3*c^2)/8)))*(6*A + B))/(8*f) - (tan(e/2 + (f*x)/2)^4*(4*A*a^3*c^2 + 
4*B*a^3*c^2) + tan(e/2 + (f*x)/2)^8*(2*A*a^3*c^2 + 2*B*a^3*c^2) + tan(e/2 
+ (f*x)/2)^6*(4*A*a^3*c^2 + 4*B*a^3*c^2) + tan(e/2 + (f*x)/2)^10*(2*A*a^3* 
c^2 + 2*B*a^3*c^2) + tan(e/2 + (f*x)/2)^2*((2*A*a^3*c^2)/5 + (2*B*a^3*c^2) 
/5) - tan(e/2 + (f*x)/2)^5*((A*a^3*c^2)/2 - (13*B*a^3*c^2)/4) + tan(e/2 + 
(f*x)/2)^7*((A*a^3*c^2)/2 - (13*B*a^3*c^2)/4) + tan(e/2 + (f*x)/2)^11*((5* 
A*a^3*c^2)/4 - (B*a^3*c^2)/8) - tan(e/2 + (f*x)/2)^3*((7*A*a^3*c^2)/4 + (4 
7*B*a^3*c^2)/24) + tan(e/2 + (f*x)/2)^9*((7*A*a^3*c^2)/4 + (47*B*a^3*c^2)/ 
24) - tan(e/2 + (f*x)/2)*((5*A*a^3*c^2)/4 - (B*a^3*c^2)/8) + (2*A*a^3*c^2) 
/5 + (2*B*a^3*c^2)/5)/(f*(6*tan(e/2 + (f*x)/2)^2 + 15*tan(e/2 + (f*x)/2)^4 
 + 20*tan(e/2 + (f*x)/2)^6 + 15*tan(e/2 + (f*x)/2)^8 + 6*tan(e/2 + (f*x)/2 
)^10 + tan(e/2 + (f*x)/2)^12 + 1)) - (a^3*c^2*(6*A + B)*(atan(tan(e/2 + (f 
*x)/2)) - (f*x)/2))/(8*f)
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.41 \[ \int (a+a \sin (e+f x))^3 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx=\frac {a^{3} c^{2} \left (-40 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{5} b -48 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a -48 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} b -60 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} a +70 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b +96 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a +96 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b +150 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a -15 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b -48 \cos \left (f x +e \right ) a -48 \cos \left (f x +e \right ) b +90 a f x +48 a +15 b f x +48 b \right )}{240 f} \] Input:

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

Output:

(a**3*c**2*( - 40*cos(e + f*x)*sin(e + f*x)**5*b - 48*cos(e + f*x)*sin(e + 
 f*x)**4*a - 48*cos(e + f*x)*sin(e + f*x)**4*b - 60*cos(e + f*x)*sin(e + f 
*x)**3*a + 70*cos(e + f*x)*sin(e + f*x)**3*b + 96*cos(e + f*x)*sin(e + f*x 
)**2*a + 96*cos(e + f*x)*sin(e + f*x)**2*b + 150*cos(e + f*x)*sin(e + f*x) 
*a - 15*cos(e + f*x)*sin(e + f*x)*b - 48*cos(e + f*x)*a - 48*cos(e + f*x)* 
b + 90*a*f*x + 48*a + 15*b*f*x + 48*b))/(240*f)