\(\int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx\) [17]

Optimal result

Integrand size = 34, antiderivative size = 182 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx=\frac {7}{16} a (2 A-B) c^4 x+\frac {7 a (2 A-B) c^4 \cos ^3(e+f x)}{24 f}+\frac {7 a (2 A-B) c^4 \cos (e+f x) \sin (e+f x)}{16 f}-\frac {a B c \cos ^3(e+f x) (c-c \sin (e+f x))^3}{6 f}+\frac {a (2 A-B) \cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )^2}{10 f}+\frac {7 a (2 A-B) \cos ^3(e+f x) \left (c^4-c^4 \sin (e+f x)\right )}{40 f} \] Output:

7/16*a*(2*A-B)*c^4*x+7/24*a*(2*A-B)*c^4*cos(f*x+e)^3/f+7/16*a*(2*A-B)*c^4* 
cos(f*x+e)*sin(f*x+e)/f-1/6*a*B*c*cos(f*x+e)^3*(c-c*sin(f*x+e))^3/f+1/10*a 
*(2*A-B)*cos(f*x+e)^3*(c^2-c^2*sin(f*x+e))^2/f+7/40*a*(2*A-B)*cos(f*x+e)^3 
*(c^4-c^4*sin(f*x+e))/f
 

Mathematica [A] (verified)

Time = 2.59 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.77 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx=\frac {a c^4 \cos (e+f x) \left (272 A-176 B-\frac {210 (2 A-B) \arcsin \left (\frac {\sqrt {1-\sin (e+f x)}}{\sqrt {2}}\right )}{\sqrt {\cos ^2(e+f x)}}+15 (2 A+7 B) \sin (e+f x)-32 (7 A-B) \sin ^2(e+f x)+10 (18 A-17 B) \sin ^3(e+f x)-48 (A-3 B) \sin ^4(e+f x)-40 B \sin ^5(e+f x)\right )}{240 f} \] Input:

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

Output:

(a*c^4*Cos[e + f*x]*(272*A - 176*B - (210*(2*A - B)*ArcSin[Sqrt[1 - Sin[e 
+ f*x]]/Sqrt[2]])/Sqrt[Cos[e + f*x]^2] + 15*(2*A + 7*B)*Sin[e + f*x] - 32* 
(7*A - B)*Sin[e + f*x]^2 + 10*(18*A - 17*B)*Sin[e + f*x]^3 - 48*(A - 3*B)* 
Sin[e + f*x]^4 - 40*B*Sin[e + f*x]^5))/(240*f)
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.87, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.382, Rules used = {3042, 3446, 3042, 3339, 3042, 3157, 3042, 3157, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3446

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3339

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3157

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3157

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3148

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 24

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

Input:

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

Output:

a*c*(-1/6*(B*Cos[e + f*x]^3*(c - c*Sin[e + f*x])^3)/f + ((2*A - B)*((c*Cos 
[e + f*x]^3*(c - c*Sin[e + f*x])^2)/(5*f) + (7*c*((Cos[e + f*x]^3*(c^2 - c 
^2*Sin[e + f*x]))/(4*f) + (5*c*((c*Cos[e + f*x]^3)/(3*f) + c*(x/2 + (Cos[e 
 + f*x]*Sin[e + f*x])/(2*f))))/4))/5))/2)
 

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. \(341\) vs. \(2(170)=340\).

Time = 0.10 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.88

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.68 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx=-\frac {48 \, {\left (A - 3 \, B\right )} a c^{4} \cos \left (f x + e\right )^{5} - 320 \, {\left (A - B\right )} a c^{4} \cos \left (f x + e\right )^{3} - 105 \, {\left (2 \, A - B\right )} a c^{4} f x + 5 \, {\left (8 \, B a c^{4} \cos \left (f x + e\right )^{5} + 2 \, {\left (18 \, A - 25 \, B\right )} a c^{4} \cos \left (f x + e\right )^{3} - 21 \, {\left (2 \, A - B\right )} a c^{4} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{240 \, f} \] Input:

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

Output:

-1/240*(48*(A - 3*B)*a*c^4*cos(f*x + e)^5 - 320*(A - B)*a*c^4*cos(f*x + e) 
^3 - 105*(2*A - B)*a*c^4*f*x + 5*(8*B*a*c^4*cos(f*x + e)^5 + 2*(18*A - 25* 
B)*a*c^4*cos(f*x + e)^3 - 21*(2*A - B)*a*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. 853 vs. \(2 (163) = 326\).

Time = 0.50 (sec) , antiderivative size = 853, normalized size of antiderivative = 4.69 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx =\text {Too large to display} \] Input:

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

Output:

Piecewise((-9*A*a*c**4*x*sin(e + f*x)**4/8 - 9*A*a*c**4*x*sin(e + f*x)**2* 
cos(e + f*x)**2/4 + A*a*c**4*x*sin(e + f*x)**2 - 9*A*a*c**4*x*cos(e + f*x) 
**4/8 + A*a*c**4*x*cos(e + f*x)**2 + A*a*c**4*x - A*a*c**4*sin(e + f*x)**4 
*cos(e + f*x)/f + 15*A*a*c**4*sin(e + f*x)**3*cos(e + f*x)/(8*f) - 4*A*a*c 
**4*sin(e + f*x)**2*cos(e + f*x)**3/(3*f) - 2*A*a*c**4*sin(e + f*x)**2*cos 
(e + f*x)/f + 9*A*a*c**4*sin(e + f*x)*cos(e + f*x)**3/(8*f) - A*a*c**4*sin 
(e + f*x)*cos(e + f*x)/f - 8*A*a*c**4*cos(e + f*x)**5/(15*f) - 4*A*a*c**4* 
cos(e + f*x)**3/(3*f) + 3*A*a*c**4*cos(e + f*x)/f + 5*B*a*c**4*x*sin(e + f 
*x)**6/16 + 15*B*a*c**4*x*sin(e + f*x)**4*cos(e + f*x)**2/16 + 3*B*a*c**4* 
x*sin(e + f*x)**4/4 + 15*B*a*c**4*x*sin(e + f*x)**2*cos(e + f*x)**4/16 + 3 
*B*a*c**4*x*sin(e + f*x)**2*cos(e + f*x)**2/2 - 3*B*a*c**4*x*sin(e + f*x)* 
*2/2 + 5*B*a*c**4*x*cos(e + f*x)**6/16 + 3*B*a*c**4*x*cos(e + f*x)**4/4 - 
3*B*a*c**4*x*cos(e + f*x)**2/2 - 11*B*a*c**4*sin(e + f*x)**5*cos(e + f*x)/ 
(16*f) + 3*B*a*c**4*sin(e + f*x)**4*cos(e + f*x)/f - 5*B*a*c**4*sin(e + f* 
x)**3*cos(e + f*x)**3/(6*f) - 5*B*a*c**4*sin(e + f*x)**3*cos(e + f*x)/(4*f 
) + 4*B*a*c**4*sin(e + f*x)**2*cos(e + f*x)**3/f - 2*B*a*c**4*sin(e + f*x) 
**2*cos(e + f*x)/f - 5*B*a*c**4*sin(e + f*x)*cos(e + f*x)**5/(16*f) - 3*B* 
a*c**4*sin(e + f*x)*cos(e + f*x)**3/(4*f) + 3*B*a*c**4*sin(e + f*x)*cos(e 
+ f*x)/(2*f) + 8*B*a*c**4*cos(e + f*x)**5/(5*f) - 4*B*a*c**4*cos(e + f*x)* 
*3/(3*f) - B*a*c**4*cos(e + f*x)/f, Ne(f, 0)), (x*(A + B*sin(e))*(a*sin...
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.85 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx=-\frac {64 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} A a c^{4} - 640 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a c^{4} + 90 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} A a c^{4} - 480 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} A a c^{4} - 960 \, {\left (f x + e\right )} A a c^{4} - 192 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} B a c^{4} - 640 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} B a c^{4} - 5 \, {\left (4 \, \sin \left (2 \, f x + 2 \, e\right )^{3} + 60 \, f x + 60 \, e + 9 \, \sin \left (4 \, f x + 4 \, e\right ) - 48 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a c^{4} - 60 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a c^{4} + 720 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a c^{4} - 2880 \, A a c^{4} \cos \left (f x + e\right ) + 960 \, B a c^{4} \cos \left (f x + e\right )}{960 \, f} \] Input:

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

Output:

-1/960*(64*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*A*a*c^ 
4 - 640*(cos(f*x + e)^3 - 3*cos(f*x + e))*A*a*c^4 + 90*(12*f*x + 12*e + si 
n(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*A*a*c^4 - 480*(2*f*x + 2*e - sin(2*f* 
x + 2*e))*A*a*c^4 - 960*(f*x + e)*A*a*c^4 - 192*(3*cos(f*x + e)^5 - 10*cos 
(f*x + e)^3 + 15*cos(f*x + e))*B*a*c^4 - 640*(cos(f*x + e)^3 - 3*cos(f*x + 
 e))*B*a*c^4 - 5*(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*c^4 - 60*(12*f*x + 12*e + sin(4*f*x + 4*e) - 
8*sin(2*f*x + 2*e))*B*a*c^4 + 720*(2*f*x + 2*e - sin(2*f*x + 2*e))*B*a*c^4 
 - 2880*A*a*c^4*cos(f*x + e) + 960*B*a*c^4*cos(f*x + e))/f
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.98 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx=-\frac {B a c^{4} \sin \left (6 \, f x + 6 \, e\right )}{192 \, f} + \frac {7}{16} \, {\left (2 \, A a c^{4} - B a c^{4}\right )} x - \frac {{\left (A a c^{4} - 3 \, B a c^{4}\right )} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac {{\left (13 \, A a c^{4} - 7 \, B a c^{4}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} + \frac {{\left (7 \, A a c^{4} - 5 \, B a c^{4}\right )} \cos \left (f x + e\right )}{8 \, f} - \frac {{\left (6 \, A a c^{4} - 7 \, B a c^{4}\right )} \sin \left (4 \, f x + 4 \, e\right )}{64 \, f} + \frac {{\left (16 \, A a c^{4} + B a c^{4}\right )} \sin \left (2 \, f x + 2 \, e\right )}{64 \, f} \] Input:

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

Output:

-1/192*B*a*c^4*sin(6*f*x + 6*e)/f + 7/16*(2*A*a*c^4 - B*a*c^4)*x - 1/80*(A 
*a*c^4 - 3*B*a*c^4)*cos(5*f*x + 5*e)/f + 1/48*(13*A*a*c^4 - 7*B*a*c^4)*cos 
(3*f*x + 3*e)/f + 1/8*(7*A*a*c^4 - 5*B*a*c^4)*cos(f*x + e)/f - 1/64*(6*A*a 
*c^4 - 7*B*a*c^4)*sin(4*f*x + 4*e)/f + 1/64*(16*A*a*c^4 + B*a*c^4)*sin(2*f 
*x + 2*e)/f
 

Mupad [B] (verification not implemented)

Time = 37.34 (sec) , antiderivative size = 454, normalized size of antiderivative = 2.49 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx=\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {A\,a\,c^4}{4}+\frac {7\,B\,a\,c^4}{8}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}\,\left (6\,A\,a\,c^4-2\,B\,a\,c^4\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (12\,A\,a\,c^4-4\,B\,a\,c^4\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{11}\,\left (\frac {A\,a\,c^4}{4}+\frac {7\,B\,a\,c^4}{8}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (22\,A\,a\,c^4-18\,B\,a\,c^4\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,\left (\frac {13\,A\,a\,c^4}{2}-\frac {37\,B\,a\,c^4}{4}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (\frac {13\,A\,a\,c^4}{2}-\frac {37\,B\,a\,c^4}{4}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {38\,A\,a\,c^4}{5}-\frac {34\,B\,a\,c^4}{5}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (\frac {68\,A\,a\,c^4}{3}-\frac {44\,B\,a\,c^4}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {27\,A\,a\,c^4}{4}-\frac {73\,B\,a\,c^4}{24}\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9\,\left (\frac {27\,A\,a\,c^4}{4}-\frac {73\,B\,a\,c^4}{24}\right )+\frac {34\,A\,a\,c^4}{15}-\frac {22\,B\,a\,c^4}{15}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}+\frac {7\,a\,c^4\,\mathrm {atan}\left (\frac {7\,a\,c^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,A-B\right )}{8\,\left (\frac {7\,A\,a\,c^4}{4}-\frac {7\,B\,a\,c^4}{8}\right )}\right )\,\left (2\,A-B\right )}{8\,f} \] Input:

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

Output:

(tan(e/2 + (f*x)/2)*((A*a*c^4)/4 + (7*B*a*c^4)/8) + tan(e/2 + (f*x)/2)^10* 
(6*A*a*c^4 - 2*B*a*c^4) + tan(e/2 + (f*x)/2)^4*(12*A*a*c^4 - 4*B*a*c^4) - 
tan(e/2 + (f*x)/2)^11*((A*a*c^4)/4 + (7*B*a*c^4)/8) + tan(e/2 + (f*x)/2)^8 
*(22*A*a*c^4 - 18*B*a*c^4) + tan(e/2 + (f*x)/2)^5*((13*A*a*c^4)/2 - (37*B* 
a*c^4)/4) - tan(e/2 + (f*x)/2)^7*((13*A*a*c^4)/2 - (37*B*a*c^4)/4) + tan(e 
/2 + (f*x)/2)^2*((38*A*a*c^4)/5 - (34*B*a*c^4)/5) + tan(e/2 + (f*x)/2)^6*( 
(68*A*a*c^4)/3 - (44*B*a*c^4)/3) + tan(e/2 + (f*x)/2)^3*((27*A*a*c^4)/4 - 
(73*B*a*c^4)/24) - tan(e/2 + (f*x)/2)^9*((27*A*a*c^4)/4 - (73*B*a*c^4)/24) 
 + (34*A*a*c^4)/15 - (22*B*a*c^4)/15)/(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*t 
an(e/2 + (f*x)/2)^10 + tan(e/2 + (f*x)/2)^12 + 1)) + (7*a*c^4*atan((7*a*c^ 
4*tan(e/2 + (f*x)/2)*(2*A - B))/(8*((7*A*a*c^4)/4 - (7*B*a*c^4)/8)))*(2*A 
- B))/(8*f)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.06 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^4 \, dx=\frac {a \,c^{4} \left (-40 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{5} b -48 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a +144 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} b +180 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} a -170 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b -224 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a +32 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b +30 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a +105 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b +272 \cos \left (f x +e \right ) a -176 \cos \left (f x +e \right ) b +210 a f x -272 a -105 b f x +176 b \right )}{240 f} \] Input:

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

Output:

(a*c**4*( - 40*cos(e + f*x)*sin(e + f*x)**5*b - 48*cos(e + f*x)*sin(e + f* 
x)**4*a + 144*cos(e + f*x)*sin(e + f*x)**4*b + 180*cos(e + f*x)*sin(e + f* 
x)**3*a - 170*cos(e + f*x)*sin(e + f*x)**3*b - 224*cos(e + f*x)*sin(e + f* 
x)**2*a + 32*cos(e + f*x)*sin(e + f*x)**2*b + 30*cos(e + f*x)*sin(e + f*x) 
*a + 105*cos(e + f*x)*sin(e + f*x)*b + 272*cos(e + f*x)*a - 176*cos(e + f* 
x)*b + 210*a*f*x - 272*a - 105*b*f*x + 176*b))/(240*f)