Integrand size = 34, antiderivative size = 142 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^3 \, dx=\frac {1}{8} a (5 A-2 B) c^3 x+\frac {a (5 A-2 B) c^3 \cos ^3(e+f x)}{12 f}+\frac {a (5 A-2 B) c^3 \cos (e+f x) \sin (e+f x)}{8 f}-\frac {a B c \cos ^3(e+f x) (c-c \sin (e+f x))^2}{5 f}+\frac {a (5 A-2 B) \cos ^3(e+f x) \left (c^3-c^3 \sin (e+f x)\right )}{20 f} \] Output:
1/8*a*(5*A-2*B)*c^3*x+1/12*a*(5*A-2*B)*c^3*cos(f*x+e)^3/f+1/8*a*(5*A-2*B)* c^3*cos(f*x+e)*sin(f*x+e)/f-1/5*a*B*c*cos(f*x+e)^3*(c-c*sin(f*x+e))^2/f+1/ 20*a*(5*A-2*B)*cos(f*x+e)^3*(c^3-c^3*sin(f*x+e))/f
Time = 2.47 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.87 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^3 \, dx=\frac {a c^3 \cos (e+f x) \left (80 A-56 B-\frac {30 (5 A-2 B) \arcsin \left (\frac {\sqrt {1-\sin (e+f x)}}{\sqrt {2}}\right )}{\sqrt {\cos ^2(e+f x)}}+15 (3 A+2 B) \sin (e+f x)+(-80 A+32 B) \sin ^2(e+f x)+30 (A-2 B) \sin ^3(e+f x)+24 B \sin ^4(e+f x)\right )}{120 f} \] Input:
Integrate[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^3 ,x]
Output:
(a*c^3*Cos[e + f*x]*(80*A - 56*B - (30*(5*A - 2*B)*ArcSin[Sqrt[1 - Sin[e + f*x]]/Sqrt[2]])/Sqrt[Cos[e + f*x]^2] + 15*(3*A + 2*B)*Sin[e + f*x] + (-80 *A + 32*B)*Sin[e + f*x]^2 + 30*(A - 2*B)*Sin[e + f*x]^3 + 24*B*Sin[e + f*x ]^4))/(120*f)
Time = 0.67 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.87, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.324, Rules used = {3042, 3446, 3042, 3339, 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.
| \(\displaystyle \int (a \sin (e+f x)+a) (c-c \sin (e+f x))^3 (A+B \sin (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a) (c-c \sin (e+f x))^3 (A+B \sin (e+f x))dx\) |
| \(\Big \downarrow \) 3446 |
| \(\displaystyle a c \int \cos ^2(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^2dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \int \cos (e+f x)^2 (A+B \sin (e+f x)) (c-c \sin (e+f x))^2dx\) |
| \(\Big \downarrow \) 3339 |
| \(\displaystyle a c \left (\frac {1}{5} (5 A-2 B) \int \cos ^2(e+f x) (c-c \sin (e+f x))^2dx-\frac {B \cos ^3(e+f x) (c-c \sin (e+f x))^2}{5 f}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {1}{5} (5 A-2 B) \int \cos (e+f x)^2 (c-c \sin (e+f x))^2dx-\frac {B \cos ^3(e+f x) (c-c \sin (e+f x))^2}{5 f}\right )\) |
| \(\Big \downarrow \) 3157 |
| \(\displaystyle a c \left (\frac {1}{5} (5 A-2 B) \left (\frac {5}{4} c \int \cos ^2(e+f x) (c-c \sin (e+f x))dx+\frac {\cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )}{4 f}\right )-\frac {B \cos ^3(e+f x) (c-c \sin (e+f x))^2}{5 f}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {1}{5} (5 A-2 B) \left (\frac {5}{4} c \int \cos (e+f x)^2 (c-c \sin (e+f x))dx+\frac {\cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )}{4 f}\right )-\frac {B \cos ^3(e+f x) (c-c \sin (e+f x))^2}{5 f}\right )\) |
| \(\Big \downarrow \) 3148 |
| \(\displaystyle a c \left (\frac {1}{5} (5 A-2 B) \left (\frac {5}{4} c \left (c \int \cos ^2(e+f x)dx+\frac {c \cos ^3(e+f x)}{3 f}\right )+\frac {\cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )}{4 f}\right )-\frac {B \cos ^3(e+f x) (c-c \sin (e+f x))^2}{5 f}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a c \left (\frac {1}{5} (5 A-2 B) \left (\frac {5}{4} c \left (c \int \sin \left (e+f x+\frac {\pi }{2}\right )^2dx+\frac {c \cos ^3(e+f x)}{3 f}\right )+\frac {\cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )}{4 f}\right )-\frac {B \cos ^3(e+f x) (c-c \sin (e+f x))^2}{5 f}\right )\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle a c \left (\frac {1}{5} (5 A-2 B) \left (\frac {5}{4} c \left (c \left (\frac {\int 1dx}{2}+\frac {\sin (e+f x) \cos (e+f x)}{2 f}\right )+\frac {c \cos ^3(e+f x)}{3 f}\right )+\frac {\cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )}{4 f}\right )-\frac {B \cos ^3(e+f x) (c-c \sin (e+f x))^2}{5 f}\right )\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle a c \left (\frac {1}{5} (5 A-2 B) \left (\frac {\cos ^3(e+f x) \left (c^2-c^2 \sin (e+f x)\right )}{4 f}+\frac {5}{4} c \left (\frac {c \cos ^3(e+f x)}{3 f}+c \left (\frac {\sin (e+f x) \cos (e+f x)}{2 f}+\frac {x}{2}\right )\right )\right )-\frac {B \cos ^3(e+f x) (c-c \sin (e+f x))^2}{5 f}\right )\) |
Input:
Int[(a + a*Sin[e + f*x])*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^3,x]
Output:
a*c*(-1/5*(B*Cos[e + f*x]^3*(c - c*Sin[e + f*x])^2)/f + ((5*A - 2*B)*((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)
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]
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])
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]
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]
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]))
Time = 176.26 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.69
| method | result | size |
| parallelrisch | \(\frac {c^{3} a \left (\left (\frac {2 A}{3}-\frac {5 B}{12}\right ) \cos \left (3 f x +3 e \right )+\left (-\frac {A}{8}+\frac {B}{4}\right ) \sin \left (4 f x +4 e \right )+\frac {\cos \left (5 f x +5 e \right ) B}{20}+\sin \left (2 f x +2 e \right ) A +\left (2 A -\frac {3 B}{2}\right ) \cos \left (f x +e \right )+\frac {5 f x A}{2}-f x B +\frac {8 A}{3}-\frac {28 B}{15}\right )}{4 f}\) | \(98\) |
| risch | \(\frac {5 a \,c^{3} x A}{8}-\frac {a \,c^{3} x B}{4}+\frac {a \,c^{3} \cos \left (f x +e \right ) A}{2 f}-\frac {3 a \,c^{3} \cos \left (f x +e \right ) B}{8 f}+\frac {B a \,c^{3} \cos \left (5 f x +5 e \right )}{80 f}-\frac {\sin \left (4 f x +4 e \right ) A a \,c^{3}}{32 f}+\frac {\sin \left (4 f x +4 e \right ) B a \,c^{3}}{16 f}+\frac {a \,c^{3} \cos \left (3 f x +3 e \right ) A}{6 f}-\frac {5 a \,c^{3} \cos \left (3 f x +3 e \right ) B}{48 f}+\frac {A a \,c^{3} \sin \left (2 f x +2 e \right )}{4 f}\) | \(164\) |
| parts | \(-\frac {\left (-2 A a \,c^{3}+B a \,c^{3}\right ) \cos \left (f x +e \right )}{f}+\frac {\left (-A a \,c^{3}+2 B a \,c^{3}\right ) \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f}+a \,c^{3} x A -\frac {2 A a \,c^{3} \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3 f}-\frac {2 B a \,c^{3} \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {B a \,c^{3} \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5 f}\) | \(180\) |
| derivativedivides | \(\frac {-\frac {2 A a \,c^{3} \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}+2 \cos \left (f x +e \right ) A a \,c^{3}+A a \,c^{3} \left (f x +e \right )+2 B a \,c^{3} \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-2 B a \,c^{3} \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\cos \left (f x +e \right ) B a \,c^{3}-A a \,c^{3} \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+\frac {B a \,c^{3} \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5}}{f}\) | \(208\) |
| default | \(\frac {-\frac {2 A a \,c^{3} \left (2+\sin \left (f x +e \right )^{2}\right ) \cos \left (f x +e \right )}{3}+2 \cos \left (f x +e \right ) A a \,c^{3}+A a \,c^{3} \left (f x +e \right )+2 B a \,c^{3} \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )-2 B a \,c^{3} \left (-\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )-\cos \left (f x +e \right ) B a \,c^{3}-A a \,c^{3} \left (-\frac {\left (\sin \left (f x +e \right )^{3}+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+\frac {B a \,c^{3} \left (\frac {8}{3}+\sin \left (f x +e \right )^{4}+\frac {4 \sin \left (f x +e \right )^{2}}{3}\right ) \cos \left (f x +e \right )}{5}}{f}\) | \(208\) |
| norman | \(\frac {\left (\frac {5}{8} A a \,c^{3}-\frac {1}{4} B a \,c^{3}\right ) x +\left (\frac {5}{8} A a \,c^{3}-\frac {1}{4} B a \,c^{3}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{10}+\left (\frac {25}{4} A a \,c^{3}-\frac {5}{2} B a \,c^{3}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\left (\frac {25}{4} A a \,c^{3}-\frac {5}{2} B a \,c^{3}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}+\left (\frac {25}{8} A a \,c^{3}-\frac {5}{4} B a \,c^{3}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\left (\frac {25}{8} A a \,c^{3}-\frac {5}{4} B a \,c^{3}\right ) x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}+\frac {\left (4 A a \,c^{3}-2 B a \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{8}}{f}+\frac {20 A a \,c^{3}-14 B a \,c^{3}}{15 f}+\frac {2 \left (4 A a \,c^{3}-4 B a \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{f}+\frac {\left (8 A a \,c^{3}-8 B a \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3 f}+\frac {2 \left (8 A a \,c^{3}-2 B a \,c^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{3 f}+\frac {a \,c^{3} \left (3 A +2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f}-\frac {a \,c^{3} \left (3 A +2 B \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{4 f}+\frac {a \,c^{3} \left (-6 B +7 A \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2 f}-\frac {a \,c^{3} \left (-6 B +7 A \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{2 f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{5}}\) | \(424\) |
| orering | \(\text {Expression too large to display}\) | \(7002\) |
Input:
int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^3,x,method=_RETURNV ERBOSE)
Output:
1/4*c^3*a*((2/3*A-5/12*B)*cos(3*f*x+3*e)+(-1/8*A+1/4*B)*sin(4*f*x+4*e)+1/2 0*cos(5*f*x+5*e)*B+sin(2*f*x+2*e)*A+(2*A-3/2*B)*cos(f*x+e)+5/2*f*x*A-f*x*B +8/3*A-28/15*B)/f
Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.72 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^3 \, dx=\frac {24 \, B a c^{3} \cos \left (f x + e\right )^{5} + 80 \, {\left (A - B\right )} a c^{3} \cos \left (f x + e\right )^{3} + 15 \, {\left (5 \, A - 2 \, B\right )} a c^{3} f x - 15 \, {\left (2 \, {\left (A - 2 \, B\right )} a c^{3} \cos \left (f x + e\right )^{3} - {\left (5 \, A - 2 \, B\right )} a c^{3} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{120 \, f} \] Input:
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^3,x, algorith m="fricas")
Output:
1/120*(24*B*a*c^3*cos(f*x + e)^5 + 80*(A - B)*a*c^3*cos(f*x + e)^3 + 15*(5 *A - 2*B)*a*c^3*f*x - 15*(2*(A - 2*B)*a*c^3*cos(f*x + e)^3 - (5*A - 2*B)*a *c^3*cos(f*x + e))*sin(f*x + e))/f
Leaf count of result is larger than twice the leaf count of optimal. 486 vs. \(2 (129) = 258\).
Time = 0.31 (sec) , antiderivative size = 486, normalized size of antiderivative = 3.42 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^3 \, dx=\begin {cases} - \frac {3 A a c^{3} x \sin ^{4}{\left (e + f x \right )}}{8} - \frac {3 A a c^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{4} - \frac {3 A a c^{3} x \cos ^{4}{\left (e + f x \right )}}{8} + A a c^{3} x + \frac {5 A a c^{3} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{8 f} - \frac {2 A a c^{3} \sin ^{2}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {3 A a c^{3} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{8 f} - \frac {4 A a c^{3} \cos ^{3}{\left (e + f x \right )}}{3 f} + \frac {2 A a c^{3} \cos {\left (e + f x \right )}}{f} + \frac {3 B a c^{3} x \sin ^{4}{\left (e + f x \right )}}{4} + \frac {3 B a c^{3} x \sin ^{2}{\left (e + f x \right )} \cos ^{2}{\left (e + f x \right )}}{2} - B a c^{3} x \sin ^{2}{\left (e + f x \right )} + \frac {3 B a c^{3} x \cos ^{4}{\left (e + f x \right )}}{4} - B a c^{3} x \cos ^{2}{\left (e + f x \right )} + \frac {B a c^{3} \sin ^{4}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} - \frac {5 B a c^{3} \sin ^{3}{\left (e + f x \right )} \cos {\left (e + f x \right )}}{4 f} + \frac {4 B a c^{3} \sin ^{2}{\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{3 f} - \frac {3 B a c^{3} \sin {\left (e + f x \right )} \cos ^{3}{\left (e + f x \right )}}{4 f} + \frac {B a c^{3} \sin {\left (e + f x \right )} \cos {\left (e + f x \right )}}{f} + \frac {8 B a c^{3} \cos ^{5}{\left (e + f x \right )}}{15 f} - \frac {B a c^{3} \cos {\left (e + f x \right )}}{f} & \text {for}\: f \neq 0 \\x \left (A + B \sin {\left (e \right )}\right ) \left (a \sin {\left (e \right )} + a\right ) \left (- c \sin {\left (e \right )} + c\right )^{3} & \text {otherwise} \end {cases} \] Input:
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**3,x)
Output:
Piecewise((-3*A*a*c**3*x*sin(e + f*x)**4/8 - 3*A*a*c**3*x*sin(e + f*x)**2* cos(e + f*x)**2/4 - 3*A*a*c**3*x*cos(e + f*x)**4/8 + A*a*c**3*x + 5*A*a*c* *3*sin(e + f*x)**3*cos(e + f*x)/(8*f) - 2*A*a*c**3*sin(e + f*x)**2*cos(e + f*x)/f + 3*A*a*c**3*sin(e + f*x)*cos(e + f*x)**3/(8*f) - 4*A*a*c**3*cos(e + f*x)**3/(3*f) + 2*A*a*c**3*cos(e + f*x)/f + 3*B*a*c**3*x*sin(e + f*x)** 4/4 + 3*B*a*c**3*x*sin(e + f*x)**2*cos(e + f*x)**2/2 - B*a*c**3*x*sin(e + f*x)**2 + 3*B*a*c**3*x*cos(e + f*x)**4/4 - B*a*c**3*x*cos(e + f*x)**2 + B* a*c**3*sin(e + f*x)**4*cos(e + f*x)/f - 5*B*a*c**3*sin(e + f*x)**3*cos(e + f*x)/(4*f) + 4*B*a*c**3*sin(e + f*x)**2*cos(e + f*x)**3/(3*f) - 3*B*a*c** 3*sin(e + f*x)*cos(e + f*x)**3/(4*f) + B*a*c**3*sin(e + f*x)*cos(e + f*x)/ f + 8*B*a*c**3*cos(e + f*x)**5/(15*f) - B*a*c**3*cos(e + f*x)/f, Ne(f, 0)) , (x*(A + B*sin(e))*(a*sin(e) + a)*(-c*sin(e) + c)**3, True))
Time = 0.04 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.41 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^3 \, dx=\frac {320 \, {\left (\cos \left (f x + e\right )^{3} - 3 \, \cos \left (f x + e\right )\right )} A a c^{3} - 15 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} A a c^{3} + 480 \, {\left (f x + e\right )} A a c^{3} + 32 \, {\left (3 \, \cos \left (f x + e\right )^{5} - 10 \, \cos \left (f x + e\right )^{3} + 15 \, \cos \left (f x + e\right )\right )} B a c^{3} + 30 \, {\left (12 \, f x + 12 \, e + \sin \left (4 \, f x + 4 \, e\right ) - 8 \, \sin \left (2 \, f x + 2 \, e\right )\right )} B a c^{3} - 240 \, {\left (2 \, f x + 2 \, e - \sin \left (2 \, f x + 2 \, e\right )\right )} B a c^{3} + 960 \, A a c^{3} \cos \left (f x + e\right ) - 480 \, B a c^{3} \cos \left (f x + e\right )}{480 \, f} \] Input:
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^3,x, algorith m="maxima")
Output:
1/480*(320*(cos(f*x + e)^3 - 3*cos(f*x + e))*A*a*c^3 - 15*(12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*A*a*c^3 + 480*(f*x + e)*A*a*c^3 + 32*(3*cos(f*x + e)^5 - 10*cos(f*x + e)^3 + 15*cos(f*x + e))*B*a*c^3 + 30*( 12*f*x + 12*e + sin(4*f*x + 4*e) - 8*sin(2*f*x + 2*e))*B*a*c^3 - 240*(2*f* x + 2*e - sin(2*f*x + 2*e))*B*a*c^3 + 960*A*a*c^3*cos(f*x + e) - 480*B*a*c ^3*cos(f*x + e))/f
Time = 0.26 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.99 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^3 \, dx=\frac {B a c^{3} \cos \left (5 \, f x + 5 \, e\right )}{80 \, f} + \frac {A a c^{3} \sin \left (2 \, f x + 2 \, e\right )}{4 \, f} + \frac {1}{8} \, {\left (5 \, A a c^{3} - 2 \, B a c^{3}\right )} x + \frac {{\left (8 \, A a c^{3} - 5 \, B a c^{3}\right )} \cos \left (3 \, f x + 3 \, e\right )}{48 \, f} + \frac {{\left (4 \, A a c^{3} - 3 \, B a c^{3}\right )} \cos \left (f x + e\right )}{8 \, f} - \frac {{\left (A a c^{3} - 2 \, B a c^{3}\right )} \sin \left (4 \, f x + 4 \, e\right )}{32 \, f} \] Input:
integrate((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^3,x, algorith m="giac")
Output:
1/80*B*a*c^3*cos(5*f*x + 5*e)/f + 1/4*A*a*c^3*sin(2*f*x + 2*e)/f + 1/8*(5* A*a*c^3 - 2*B*a*c^3)*x + 1/48*(8*A*a*c^3 - 5*B*a*c^3)*cos(3*f*x + 3*e)/f + 1/8*(4*A*a*c^3 - 3*B*a*c^3)*cos(f*x + e)/f - 1/32*(A*a*c^3 - 2*B*a*c^3)*s in(4*f*x + 4*e)/f
Time = 36.13 (sec) , antiderivative size = 389, normalized size of antiderivative = 2.74 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^3 \, dx=\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {3\,A\,a\,c^3}{4}+\frac {B\,a\,c^3}{2}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8\,\left (4\,A\,a\,c^3-2\,B\,a\,c^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {7\,A\,a\,c^3}{2}-3\,B\,a\,c^3\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7\,\left (\frac {7\,A\,a\,c^3}{2}-3\,B\,a\,c^3\right )-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^9\,\left (\frac {3\,A\,a\,c^3}{4}+\frac {B\,a\,c^3}{2}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6\,\left (8\,A\,a\,c^3-8\,B\,a\,c^3\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {8\,A\,a\,c^3}{3}-\frac {8\,B\,a\,c^3}{3}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {16\,A\,a\,c^3}{3}-\frac {4\,B\,a\,c^3}{3}\right )+\frac {4\,A\,a\,c^3}{3}-\frac {14\,B\,a\,c^3}{15}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}+\frac {a\,c^3\,\mathrm {atan}\left (\frac {a\,c^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (5\,A-2\,B\right )}{4\,\left (\frac {5\,A\,a\,c^3}{4}-\frac {B\,a\,c^3}{2}\right )}\right )\,\left (5\,A-2\,B\right )}{4\,f}-\frac {a\,c^3\,\left (5\,A-2\,B\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )-\frac {f\,x}{2}\right )}{4\,f} \] Input:
int((A + B*sin(e + f*x))*(a + a*sin(e + f*x))*(c - c*sin(e + f*x))^3,x)
Output:
(tan(e/2 + (f*x)/2)*((3*A*a*c^3)/4 + (B*a*c^3)/2) + tan(e/2 + (f*x)/2)^8*( 4*A*a*c^3 - 2*B*a*c^3) + tan(e/2 + (f*x)/2)^3*((7*A*a*c^3)/2 - 3*B*a*c^3) - tan(e/2 + (f*x)/2)^7*((7*A*a*c^3)/2 - 3*B*a*c^3) - tan(e/2 + (f*x)/2)^9* ((3*A*a*c^3)/4 + (B*a*c^3)/2) + tan(e/2 + (f*x)/2)^6*(8*A*a*c^3 - 8*B*a*c^ 3) + tan(e/2 + (f*x)/2)^2*((8*A*a*c^3)/3 - (8*B*a*c^3)/3) + tan(e/2 + (f*x )/2)^4*((16*A*a*c^3)/3 - (4*B*a*c^3)/3) + (4*A*a*c^3)/3 - (14*B*a*c^3)/15) /(f*(5*tan(e/2 + (f*x)/2)^2 + 10*tan(e/2 + (f*x)/2)^4 + 10*tan(e/2 + (f*x) /2)^6 + 5*tan(e/2 + (f*x)/2)^8 + tan(e/2 + (f*x)/2)^10 + 1)) + (a*c^3*atan ((a*c^3*tan(e/2 + (f*x)/2)*(5*A - 2*B))/(4*((5*A*a*c^3)/4 - (B*a*c^3)/2))) *(5*A - 2*B))/(4*f) - (a*c^3*(5*A - 2*B)*(atan(tan(e/2 + (f*x)/2)) - (f*x) /2))/(4*f)
Time = 0.18 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.12 \[ \int (a+a \sin (e+f x)) (A+B \sin (e+f x)) (c-c \sin (e+f x))^3 \, dx=\frac {a \,c^{3} \left (24 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} b +30 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} a -60 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{3} b -80 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a +32 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b +45 \cos \left (f x +e \right ) \sin \left (f x +e \right ) a +30 \cos \left (f x +e \right ) \sin \left (f x +e \right ) b +80 \cos \left (f x +e \right ) a -56 \cos \left (f x +e \right ) b +75 a f x -80 a -30 b f x +56 b \right )}{120 f} \] Input:
int((a+a*sin(f*x+e))*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^3,x)
Output:
(a*c**3*(24*cos(e + f*x)*sin(e + f*x)**4*b + 30*cos(e + f*x)*sin(e + f*x)* *3*a - 60*cos(e + f*x)*sin(e + f*x)**3*b - 80*cos(e + f*x)*sin(e + f*x)**2 *a + 32*cos(e + f*x)*sin(e + f*x)**2*b + 45*cos(e + f*x)*sin(e + f*x)*a + 30*cos(e + f*x)*sin(e + f*x)*b + 80*cos(e + f*x)*a - 56*cos(e + f*x)*b + 7 5*a*f*x - 80*a - 30*b*f*x + 56*b))/(120*f)