\(\int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx\) [27]

Optimal result

Integrand size = 36, antiderivative size = 189 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx=\frac {1}{16} a^2 (7 A-2 B) c^4 x+\frac {a^2 (7 A-2 B) c^4 \cos ^5(e+f x)}{30 f}+\frac {a^2 (7 A-2 B) c^4 \cos (e+f x) \sin (e+f x)}{16 f}+\frac {a^2 (7 A-2 B) c^4 \cos ^3(e+f x) \sin (e+f x)}{24 f}-\frac {a^2 B \cos ^5(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{7 f}+\frac {a^2 (7 A-2 B) \cos ^5(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{42 f} \] Output:

1/16*a^2*(7*A-2*B)*c^4*x+1/30*a^2*(7*A-2*B)*c^4*cos(f*x+e)^5/f+1/16*a^2*(7 
*A-2*B)*c^4*cos(f*x+e)*sin(f*x+e)/f+1/24*a^2*(7*A-2*B)*c^4*cos(f*x+e)^3*si 
n(f*x+e)/f-1/7*a^2*B*cos(f*x+e)^5*(c^2-c^2*sin(f*x+e))^2/f+1/42*a^2*(7*A-2 
*B)*cos(f*x+e)^5*(c^4-c^4*sin(f*x+e))/f
 

Mathematica [A] (verified)

Time = 8.25 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.86 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx=\frac {a^2 c^4 (2940 A e-840 B e+2940 A f x-840 B f x+105 (16 A-11 B) \cos (e+f x)+105 (8 A-5 B) \cos (3 (e+f x))+168 A \cos (5 (e+f x))-63 B \cos (5 (e+f x))+15 B \cos (7 (e+f x))+1785 A \sin (2 (e+f x))-210 B \sin (2 (e+f x))+105 A \sin (4 (e+f x))+210 B \sin (4 (e+f x))-35 A \sin (6 (e+f x))+70 B \sin (6 (e+f x)))}{6720 f} \] Input:

Integrate[(a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x]) 
^4,x]
 

Output:

(a^2*c^4*(2940*A*e - 840*B*e + 2940*A*f*x - 840*B*f*x + 105*(16*A - 11*B)* 
Cos[e + f*x] + 105*(8*A - 5*B)*Cos[3*(e + f*x)] + 168*A*Cos[5*(e + f*x)] - 
 63*B*Cos[5*(e + f*x)] + 15*B*Cos[7*(e + f*x)] + 1785*A*Sin[2*(e + f*x)] - 
 210*B*Sin[2*(e + f*x)] + 105*A*Sin[4*(e + f*x)] + 210*B*Sin[4*(e + f*x)] 
- 35*A*Sin[6*(e + f*x)] + 70*B*Sin[6*(e + f*x)]))/(6720*f)
 

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.81, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.361, Rules used = {3042, 3446, 3042, 3339, 3042, 3157, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3339

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3157

\(\displaystyle a^2 c^2 \left (\frac {1}{7} (7 A-2 B) \left (\frac {7}{6} c \int \cos ^4(e+f x) (c-c \sin (e+f x))dx+\frac {\cos ^5(e+f x) \left (c^2-c^2 \sin (e+f x)\right )}{6 f}\right )-\frac {B \cos ^5(e+f x) (c-c \sin (e+f x))^2}{7 f}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \left (\frac {1}{7} (7 A-2 B) \left (\frac {7}{6} c \int \cos (e+f x)^4 (c-c \sin (e+f x))dx+\frac {\cos ^5(e+f x) \left (c^2-c^2 \sin (e+f x)\right )}{6 f}\right )-\frac {B \cos ^5(e+f x) (c-c \sin (e+f x))^2}{7 f}\right )\)

\(\Big \downarrow \) 3148

\(\displaystyle a^2 c^2 \left (\frac {1}{7} (7 A-2 B) \left (\frac {7}{6} c \left (c \int \cos ^4(e+f x)dx+\frac {c \cos ^5(e+f x)}{5 f}\right )+\frac {\cos ^5(e+f x) \left (c^2-c^2 \sin (e+f x)\right )}{6 f}\right )-\frac {B \cos ^5(e+f x) (c-c \sin (e+f x))^2}{7 f}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \left (\frac {1}{7} (7 A-2 B) \left (\frac {7}{6} c \left (c \int \sin \left (e+f x+\frac {\pi }{2}\right )^4dx+\frac {c \cos ^5(e+f x)}{5 f}\right )+\frac {\cos ^5(e+f x) \left (c^2-c^2 \sin (e+f x)\right )}{6 f}\right )-\frac {B \cos ^5(e+f x) (c-c \sin (e+f x))^2}{7 f}\right )\)

\(\Big \downarrow \) 3115

\(\displaystyle a^2 c^2 \left (\frac {1}{7} (7 A-2 B) \left (\frac {7}{6} c \left (c \left (\frac {3}{4} \int \cos ^2(e+f x)dx+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f}\right )+\frac {c \cos ^5(e+f x)}{5 f}\right )+\frac {\cos ^5(e+f x) \left (c^2-c^2 \sin (e+f x)\right )}{6 f}\right )-\frac {B \cos ^5(e+f x) (c-c \sin (e+f x))^2}{7 f}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle a^2 c^2 \left (\frac {1}{7} (7 A-2 B) \left (\frac {7}{6} c \left (c \left (\frac {3}{4} \int \sin \left (e+f x+\frac {\pi }{2}\right )^2dx+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f}\right )+\frac {c \cos ^5(e+f x)}{5 f}\right )+\frac {\cos ^5(e+f x) \left (c^2-c^2 \sin (e+f x)\right )}{6 f}\right )-\frac {B \cos ^5(e+f x) (c-c \sin (e+f x))^2}{7 f}\right )\)

\(\Big \downarrow \) 3115

\(\displaystyle a^2 c^2 \left (\frac {1}{7} (7 A-2 B) \left (\frac {7}{6} c \left (c \left (\frac {3}{4} \left (\frac {\int 1dx}{2}+\frac {\sin (e+f x) \cos (e+f x)}{2 f}\right )+\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f}\right )+\frac {c \cos ^5(e+f x)}{5 f}\right )+\frac {\cos ^5(e+f x) \left (c^2-c^2 \sin (e+f x)\right )}{6 f}\right )-\frac {B \cos ^5(e+f x) (c-c \sin (e+f x))^2}{7 f}\right )\)

\(\Big \downarrow \) 24

\(\displaystyle a^2 c^2 \left (\frac {1}{7} (7 A-2 B) \left (\frac {\cos ^5(e+f x) \left (c^2-c^2 \sin (e+f x)\right )}{6 f}+\frac {7}{6} c \left (\frac {c \cos ^5(e+f x)}{5 f}+c \left (\frac {\sin (e+f x) \cos ^3(e+f x)}{4 f}+\frac {3}{4} \left (\frac {\sin (e+f x) \cos (e+f x)}{2 f}+\frac {x}{2}\right )\right )\right )\right )-\frac {B \cos ^5(e+f x) (c-c \sin (e+f x))^2}{7 f}\right )\)

Input:

Int[(a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^4,x]
 

Output:

a^2*c^2*(-1/7*(B*Cos[e + f*x]^5*(c - c*Sin[e + f*x])^2)/f + ((7*A - 2*B)*( 
(Cos[e + f*x]^5*(c^2 - c^2*Sin[e + f*x]))/(6*f) + (7*c*((c*Cos[e + f*x]^5) 
/(5*f) + c*((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))/7)
 

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 3157

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] && GtQ[m, 0] && NeQ[m + p, 0] && Integers 
Q[2*m, 2*p]
 

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. \(462\) vs. \(2(177)=354\).

Time = 0.11 (sec) , antiderivative size = 463, normalized size of antiderivative = 2.45

\[\frac {a^{2} A \,c^{4} \left (f x +e \right )-\frac {4 a^{2} A \,c^{4} \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}-\frac {a^{2} B \,c^{4} \left (\frac {16}{5}+\sin \left (f x +e \right )^{6}+\frac {6 \sin \left (f x +e \right )^{4}}{5}+\frac {8 \sin \left (f x +e \right )^{2}}{5}\right ) \cos \left (f x +e \right )}{7}-2 a^{2} B \,c^{4} \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^{2} B \,c^{4} \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}+4 a^{2} B \,c^{4} \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 )-a^{2} B \,c^{4} \cos \left (f x +e \right )-a^{2} A \,c^{4} \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )+\frac {a^{2} B \,c^{4} \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}+2 a^{2} A \,c^{4} \cos \left (f x +e \right )-2 a^{2} B \,c^{4} \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^{4} \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 {2 a^{2} A \,c^{4} \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^{2} A \,c^{4} \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 )}{f}\]

Input:

int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^4,x)
 

Output:

1/f*(a^2*A*c^4*(f*x+e)-4/3*a^2*A*c^4*(2+sin(f*x+e)^2)*cos(f*x+e)-1/7*a^2*B 
*c^4*(16/5+sin(f*x+e)^6+6/5*sin(f*x+e)^4+8/5*sin(f*x+e)^2)*cos(f*x+e)-2*a^ 
2*B*c^4*(-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^2*B*c^4*(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+ 
e)+4*a^2*B*c^4*(-1/4*(sin(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8* 
e)-a^2*B*c^4*cos(f*x+e)-a^2*A*c^4*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2* 
e)+1/3*a^2*B*c^4*(2+sin(f*x+e)^2)*cos(f*x+e)+2*a^2*A*c^4*cos(f*x+e)-2*a^2* 
B*c^4*(-1/2*sin(f*x+e)*cos(f*x+e)+1/2*f*x+1/2*e)+a^2*A*c^4*(-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)+2/5*a^2 
*A*c^4*(8/3+sin(f*x+e)^4+4/3*sin(f*x+e)^2)*cos(f*x+e)-a^2*A*c^4*(-1/4*(sin 
(f*x+e)^3+3/2*sin(f*x+e))*cos(f*x+e)+3/8*f*x+3/8*e))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.71 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx=\frac {240 \, B a^{2} c^{4} \cos \left (f x + e\right )^{7} + 672 \, {\left (A - B\right )} a^{2} c^{4} \cos \left (f x + e\right )^{5} + 105 \, {\left (7 \, A - 2 \, B\right )} a^{2} c^{4} f x - 35 \, {\left (8 \, {\left (A - 2 \, B\right )} a^{2} c^{4} \cos \left (f x + e\right )^{5} - 2 \, {\left (7 \, A - 2 \, B\right )} a^{2} c^{4} \cos \left (f x + e\right )^{3} - 3 \, {\left (7 \, A - 2 \, B\right )} a^{2} c^{4} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{1680 \, f} \] Input:

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^4,x, algori 
thm="fricas")
 

Output:

1/1680*(240*B*a^2*c^4*cos(f*x + e)^7 + 672*(A - B)*a^2*c^4*cos(f*x + e)^5 
+ 105*(7*A - 2*B)*a^2*c^4*f*x - 35*(8*(A - 2*B)*a^2*c^4*cos(f*x + e)^5 - 2 
*(7*A - 2*B)*a^2*c^4*cos(f*x + e)^3 - 3*(7*A - 2*B)*a^2*c^4*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. 1210 vs. \(2 (172) = 344\).

Time = 0.66 (sec) , antiderivative size = 1210, normalized size of antiderivative = 6.40 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx=\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**4,x)
 

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 460 vs. \(2 (179) = 358\).

Time = 0.05 (sec) , antiderivative size = 460, normalized size of antiderivative = 2.43 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx=\frac {896 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} A a^{2} c^{4} + 8960 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a^{2} c^{4} + 35 \, {\left (4 \, \sin \left (2 \, f x + 2 \, e\right )^{3} + 60 \, f x + 60 \, e + 9 \, \sin \left (4 \, f x + 4 \, e\right ) - 48 \, \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} c^{4} - 210 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} c^{4} - 1680 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a^{2} c^{4} + 6720 \, {\left (f x + e\right )} A a^{2} c^{4} + 192 \, {\left (5 \, \cos \left (f x + e\right )^{7} - 21 \, \cos \left (f x + e\right )^{5} + 35 \, \cos \left (f x + e\right )^{3} - 35 \, \cos \left (f x + e\right )\right )} B a^{2} c^{4} + 448 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} B a^{2} c^{4} - 2240 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a^{2} c^{4} - 70 \, {\left (4 \, \sin \left (2 \, f x + 2 \, e\right )^{3} + 60 \, f x + 60 \, e + 9 \, \sin \left (4 \, f x + 4 \, e\right ) - 48 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} c^{4} + 840 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} c^{4} - 3360 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a^{2} c^{4} + 13440 \, A a^{2} c^{4} \cos \left (f x + e\right ) - 6720 \, B a^{2} c^{4} \cos \left (f x + e\right )}{6720 \, f} \] Input:

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^4,x, algori 
thm="maxima")
 

Output:

1/6720*(896*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*A*a^2 
*c^4 + 8960*(cos(f*x + e)^3 - 3*cos(f*x + e))*A*a^2*c^4 + 35*(4*sin(2*f*x 
+ 2*e)^3 + 60*f*x + 60*e + 9*sin(4*f*x + 4*e) - 48*sin(2*f*x + 2*e))*A*a^2 
*c^4 - 210*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*A*a^2*c 
^4 - 1680*(2*f*x + 2*e - sin(2*f*x + 2*e))*A*a^2*c^4 + 6720*(f*x + e)*A*a^ 
2*c^4 + 192*(5*cos(f*x + e)^7 - 21*cos(f*x + e)^5 + 35*cos(f*x + e)^3 - 35 
*cos(f*x + e))*B*a^2*c^4 + 448*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15* 
cos(f*x + e))*B*a^2*c^4 - 2240*(cos(f*x + e)^3 - 3*cos(f*x + e))*B*a^2*c^4 
 - 70*(4*sin(2*f*x + 2*e)^3 + 60*f*x + 60*e + 9*sin(4*f*x + 4*e) - 48*sin( 
2*f*x + 2*e))*B*a^2*c^4 + 840*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2* 
f*x + 2*e))*B*a^2*c^4 - 3360*(2*f*x + 2*e - sin(2*f*x + 2*e))*B*a^2*c^4 + 
13440*A*a^2*c^4*cos(f*x + e) - 6720*B*a^2*c^4*cos(f*x + e))/f
 

Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.25 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx=\frac {B a^{2} c^{4} \cos \left (7 \, f x + 7 \, e\right )}{448 \, f} + \frac {1}{16} \, {\left (7 \, A a^{2} c^{4} - 2 \, B a^{2} c^{4}\right )} x + \frac {{\left (8 \, A a^{2} c^{4} - 3 \, B a^{2} c^{4}\right )} \cos \left (5 \, f x + 5 \, e\right )}{320 \, f} + \frac {{\left (8 \, A a^{2} c^{4} - 5 \, B a^{2} c^{4}\right )} \cos \left (3 \, f x + 3 \, e\right )}{64 \, f} + \frac {{\left (16 \, A a^{2} c^{4} - 11 \, B a^{2} c^{4}\right )} \cos \left (f x + e\right )}{64 \, f} - \frac {{\left (A a^{2} c^{4} - 2 \, B a^{2} c^{4}\right )} \sin \left (6 \, f x + 6 \, e\right )}{192 \, f} + \frac {{\left (A a^{2} c^{4} + 2 \, B a^{2} c^{4}\right )} \sin \left (4 \, f x + 4 \, e\right )}{64 \, f} + \frac {{\left (17 \, A a^{2} c^{4} - 2 \, B a^{2} c^{4}\right )} \sin \left (2 \, f x + 2 \, e\right )}{64 \, f} \] Input:

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^4,x, algori 
thm="giac")
 

Output:

1/448*B*a^2*c^4*cos(7*f*x + 7*e)/f + 1/16*(7*A*a^2*c^4 - 2*B*a^2*c^4)*x + 
1/320*(8*A*a^2*c^4 - 3*B*a^2*c^4)*cos(5*f*x + 5*e)/f + 1/64*(8*A*a^2*c^4 - 
 5*B*a^2*c^4)*cos(3*f*x + 3*e)/f + 1/64*(16*A*a^2*c^4 - 11*B*a^2*c^4)*cos( 
f*x + e)/f - 1/192*(A*a^2*c^4 - 2*B*a^2*c^4)*sin(6*f*x + 6*e)/f + 1/64*(A* 
a^2*c^4 + 2*B*a^2*c^4)*sin(4*f*x + 4*e)/f + 1/64*(17*A*a^2*c^4 - 2*B*a^2*c 
^4)*sin(2*f*x + 2*e)/f
 

Mupad [B] (verification not implemented)

Time = 37.93 (sec) , antiderivative size = 553, normalized size of antiderivative = 2.93 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx =\text {Too large to display} \] Input:

int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))^2*(c - c*sin(e + f*x))^4,x)
                                                                                    
                                                                                    
 

Output:

(tan(e/2 + (f*x)/2)^12*(4*A*a^2*c^4 - 2*B*a^2*c^4) + tan(e/2 + (f*x)/2)^8* 
(12*A*a^2*c^4 - 2*B*a^2*c^4) + tan(e/2 + (f*x)/2)^10*(8*A*a^2*c^4 - 8*B*a^ 
2*c^4) + tan(e/2 + (f*x)/2)^2*((8*A*a^2*c^4)/5 - (8*B*a^2*c^4)/5) - tan(e/ 
2 + (f*x)/2)^13*((9*A*a^2*c^4)/8 + (B*a^2*c^4)/4) + tan(e/2 + (f*x)/2)^6*( 
16*A*a^2*c^4 - 16*B*a^2*c^4) + tan(e/2 + (f*x)/2)^3*((29*A*a^2*c^4)/6 - (1 
1*B*a^2*c^4)/3) - tan(e/2 + (f*x)/2)^11*((29*A*a^2*c^4)/6 - (11*B*a^2*c^4) 
/3) + tan(e/2 + (f*x)/2)^4*((44*A*a^2*c^4)/5 - (14*B*a^2*c^4)/5) + tan(e/2 
 + (f*x)/2)^5*((23*A*a^2*c^4)/24 + (31*B*a^2*c^4)/12) - tan(e/2 + (f*x)/2) 
^9*((23*A*a^2*c^4)/24 + (31*B*a^2*c^4)/12) + tan(e/2 + (f*x)/2)*((9*A*a^2* 
c^4)/8 + (B*a^2*c^4)/4) + (4*A*a^2*c^4)/5 - (18*B*a^2*c^4)/35)/(f*(7*tan(e 
/2 + (f*x)/2)^2 + 21*tan(e/2 + (f*x)/2)^4 + 35*tan(e/2 + (f*x)/2)^6 + 35*t 
an(e/2 + (f*x)/2)^8 + 21*tan(e/2 + (f*x)/2)^10 + 7*tan(e/2 + (f*x)/2)^12 + 
 tan(e/2 + (f*x)/2)^14 + 1)) + (a^2*c^4*atan((a^2*c^4*tan(e/2 + (f*x)/2)*( 
7*A - 2*B))/(8*((7*A*a^2*c^4)/8 - (B*a^2*c^4)/4)))*(7*A - 2*B))/(8*f)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.21 \[ \int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx=\frac {a^{2} c^{4} \left (-240 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{6} b -280 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{5} a +560 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{5} b +672 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a +48 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} b +70 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} a -980 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b -1344 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a +624 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b +945 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a +210 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b +672 \cos \left (f x +e \right ) a -432 \cos \left (f x +e \right ) b +735 a f x -672 a -210 b f x +432 b \right )}{1680 f} \] Input:

int((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^4,x)
 

Output:

(a**2*c**4*( - 240*cos(e + f*x)*sin(e + f*x)**6*b - 280*cos(e + f*x)*sin(e 
 + f*x)**5*a + 560*cos(e + f*x)*sin(e + f*x)**5*b + 672*cos(e + f*x)*sin(e 
 + f*x)**4*a + 48*cos(e + f*x)*sin(e + f*x)**4*b + 70*cos(e + f*x)*sin(e + 
 f*x)**3*a - 980*cos(e + f*x)*sin(e + f*x)**3*b - 1344*cos(e + f*x)*sin(e 
+ f*x)**2*a + 624*cos(e + f*x)*sin(e + f*x)**2*b + 945*cos(e + f*x)*sin(e 
+ f*x)*a + 210*cos(e + f*x)*sin(e + f*x)*b + 672*cos(e + f*x)*a - 432*cos( 
e + f*x)*b + 735*a*f*x - 672*a - 210*b*f*x + 432*b))/(1680*f)