Integrand size = 40, antiderivative size = 226 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {2 a^4 A \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{35 f \sqrt {a+a \sin (e+f x)}}-\frac {4 a^3 A \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{35 f}-\frac {a^2 A \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{7 f}-\frac {a A \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{7/2} (c-c \sin (e+f x))^{7/2}}{8 f} \]
-1/7*a^2*A*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(7/2)/f-1/7* a*A*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(7/2)/f-1/8*B*cos(f *x+e)*(a+a*sin(f*x+e))^(7/2)*(c-c*sin(f*x+e))^(7/2)/f-2/35*a^4*A*cos(f*x+e )*(c-c*sin(f*x+e))^(7/2)/f/(a+a*sin(f*x+e))^(1/2)-4/35*a^3*A*cos(f*x+e)*(c -c*sin(f*x+e))^(7/2)*(a+a*sin(f*x+e))^(1/2)/f
Time = 4.39 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.61 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {a^3 c^3 \sec (e+f x) \sqrt {a (1+\sin (e+f x))} \sqrt {c-c \sin (e+f x)} (1225 B+1960 B \cos (2 (e+f x))+980 B \cos (4 (e+f x))+280 B \cos (6 (e+f x))+35 B \cos (8 (e+f x))-19600 A \sin (e+f x)-3920 A \sin (3 (e+f x))-784 A \sin (5 (e+f x))-80 A \sin (7 (e+f x)))}{35840 f} \]
-1/35840*(a^3*c^3*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[c - c*Sin[e + f*x]]*(1225*B + 1960*B*Cos[2*(e + f*x)] + 980*B*Cos[4*(e + f*x)] + 280* B*Cos[6*(e + f*x)] + 35*B*Cos[8*(e + f*x)] - 19600*A*Sin[e + f*x] - 3920*A *Sin[3*(e + f*x)] - 784*A*Sin[5*(e + f*x)] - 80*A*Sin[7*(e + f*x)]))/f
Time = 1.18 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3452, 3042, 3219, 3042, 3219, 3042, 3219, 3042, 3217}
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.
| \(\displaystyle \int (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{7/2} (A+B \sin (e+f x))dx\) |
| \(\Big \downarrow \) 3452 |
| \(\displaystyle A \int (\sin (e+f x) a+a)^{7/2} (c-c \sin (e+f x))^{7/2}dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{7/2}}{8 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle A \int (\sin (e+f x) a+a)^{7/2} (c-c \sin (e+f x))^{7/2}dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{7/2}}{8 f}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle A \left (\frac {6}{7} a \int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{7/2}}{8 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle A \left (\frac {6}{7} a \int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{7/2}}{8 f}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle A \left (\frac {6}{7} a \left (\frac {2}{3} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{7/2}}{8 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle A \left (\frac {6}{7} a \left (\frac {2}{3} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{7/2}}{8 f}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle A \left (\frac {6}{7} a \left (\frac {2}{3} a \left (\frac {2}{5} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{7/2}}{8 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle A \left (\frac {6}{7} a \left (\frac {2}{3} a \left (\frac {2}{5} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{7/2}}{8 f}\) |
| \(\Big \downarrow \) 3217 |
| \(\displaystyle A \left (\frac {6}{7} a \left (\frac {2}{3} a \left (-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{10 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{7/2} (c-c \sin (e+f x))^{7/2}}{8 f}\) |
-1/8*(B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(7/2)*(c - c*Sin[e + f*x])^(7/2) )/f + A*(-1/7*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f* x])^(7/2))/f + (6*a*(-1/6*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*Sin[e + f*x])^(7/2))/f + (2*a*(-1/10*(a^2*Cos[e + f*x]*(c - c*Sin[e + f* x])^(7/2))/(f*Sqrt[a + a*Sin[e + f*x]]) - (a*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]]*(c - c*Sin[e + f*x])^(7/2))/(5*f)))/3))/7)
3.2.61.3.1 Defintions of rubi rules used
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f _.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x])^ n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] /; FreeQ[{a, b, c, d, e, f, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*(a + b*Sin[e + f*x])^ (m - 1)*((c + d*Sin[e + f*x])^n/(f*(m + n))), x] + Simp[a*((2*m - 1)/(m + n )) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; Fre eQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I GtQ[m - 1/2, 0] && !LtQ[n, -1] && !(IGtQ[n - 1/2, 0] && LtQ[n, m]) && !( ILtQ[m + n, 0] && GtQ[2*m + n + 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] :> Sim p[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + n + 1))), x] - Simp[(B*c*(m - n) - A*d*(m + n + 1))/(d*(m + n + 1)) Int[( a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && NeQ[m + n + 1, 0]
Time = 74.58 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.62
| method | result | size |
| default | \(\frac {a^{3} c^{3} \tan \left (f x +e \right ) \left (35 B \sin \left (f x +e \right ) \left (\cos ^{6}\left (f x +e \right )\right )+40 A \left (\cos ^{6}\left (f x +e \right )\right )+35 B \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )+48 A \left (\cos ^{4}\left (f x +e \right )\right )+35 B \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+64 A \left (\cos ^{2}\left (f x +e \right )\right )+35 B \sin \left (f x +e \right )+128 A \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{280 f}\) | \(140\) |
| parts | \(\frac {A \tan \left (f x +e \right ) a^{3} c^{3} \left (5 \left (\cos ^{6}\left (f x +e \right )\right )+6 \left (\cos ^{4}\left (f x +e \right )\right )+8 \left (\cos ^{2}\left (f x +e \right )\right )+16\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}}{35 f}-\frac {B \sec \left (f x +e \right ) a^{3} c^{3} \left (\cos ^{6}\left (f x +e \right )+\cos ^{4}\left (f x +e \right )+\cos ^{2}\left (f x +e \right )+1\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \left (\cos ^{2}\left (f x +e \right )-1\right )}{8 f}\) | \(156\) |
1/280*a^3*c^3/f*tan(f*x+e)*(35*B*sin(f*x+e)*cos(f*x+e)^6+40*A*cos(f*x+e)^6 +35*B*cos(f*x+e)^4*sin(f*x+e)+48*A*cos(f*x+e)^4+35*B*cos(f*x+e)^2*sin(f*x+ e)+64*A*cos(f*x+e)^2+35*B*sin(f*x+e)+128*A)*(-c*(sin(f*x+e)-1))^(1/2)*(a*( 1+sin(f*x+e)))^(1/2)
Time = 0.29 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.59 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {{\left (35 \, B a^{3} c^{3} \cos \left (f x + e\right )^{8} - 35 \, B a^{3} c^{3} - 8 \, {\left (5 \, A a^{3} c^{3} \cos \left (f x + e\right )^{6} + 6 \, A a^{3} c^{3} \cos \left (f x + e\right )^{4} + 8 \, A a^{3} c^{3} \cos \left (f x + e\right )^{2} + 16 \, A a^{3} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{280 \, f \cos \left (f x + e\right )} \]
integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x , algorithm="fricas")
-1/280*(35*B*a^3*c^3*cos(f*x + e)^8 - 35*B*a^3*c^3 - 8*(5*A*a^3*c^3*cos(f* x + e)^6 + 6*A*a^3*c^3*cos(f*x + e)^4 + 8*A*a^3*c^3*cos(f*x + e)^2 + 16*A* a^3*c^3)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/ (f*cos(f*x + e))
Timed out. \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\text {Timed out} \]
\[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {7}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}} \,d x } \]
integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x , algorithm="maxima")
Leaf count of result is larger than twice the leaf count of optimal. 453 vs. \(2 (196) = 392\).
Time = 0.54 (sec) , antiderivative size = 453, normalized size of antiderivative = 2.00 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\frac {32 \, {\left (35 \, B a^{3} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{16} - 20 \, A a^{3} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{14} - 140 \, B a^{3} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{14} + 70 \, A a^{3} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} + 210 \, B a^{3} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{12} - 84 \, A a^{3} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} - 140 \, B a^{3} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{10} + 35 \, A a^{3} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8} + 35 \, B a^{3} c^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{8}\right )} \sqrt {a} \sqrt {c}}{35 \, f} \]
integrate((a+a*sin(f*x+e))^(7/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x , algorithm="giac")
32/35*(35*B*a^3*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^16 - 20*A*a^3*c^3*sgn(cos (-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4* pi + 1/2*f*x + 1/2*e)^14 - 140*B*a^3*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e ))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^14 + 70*A*a^3*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f* x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^12 + 210*B*a^3*c^3*sgn(cos(-1/4 *pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^12 - 84*A*a^3*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn (sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^10 - 140*B *a^3*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1 /2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^10 + 35*A*a^3*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f* x + 1/2*e)^8 + 35*B*a^3*c^3*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1 /4*pi + 1/2*f*x + 1/2*e))*sin(-1/4*pi + 1/2*f*x + 1/2*e)^8)*sqrt(a)*sqrt(c )/f
Time = 17.25 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.70 \[ \int (a+a \sin (e+f x))^{7/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\frac {{\mathrm {e}}^{-e\,8{}\mathrm {i}-f\,x\,8{}\mathrm {i}}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {35\,A\,a^3\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}-\frac {7\,B\,a^3\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{64\,f}-\frac {7\,B\,a^3\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{128\,f}-\frac {B\,a^3\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{64\,f}-\frac {B\,a^3\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\cos \left (8\,e+8\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{512\,f}+\frac {7\,A\,a^3\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}+\frac {7\,A\,a^3\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (5\,e+5\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{160\,f}+\frac {A\,a^3\,c^3\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,\sin \left (7\,e+7\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{224\,f}\right )}{2\,\cos \left (e+f\,x\right )} \]
(exp(- e*8i - f*x*8i)*(c - c*sin(e + f*x))^(1/2)*((35*A*a^3*c^3*exp(e*8i + f*x*8i)*sin(e + f*x)*(a + a*sin(e + f*x))^(1/2))/(32*f) - (7*B*a^3*c^3*ex p(e*8i + f*x*8i)*cos(2*e + 2*f*x)*(a + a*sin(e + f*x))^(1/2))/(64*f) - (7* B*a^3*c^3*exp(e*8i + f*x*8i)*cos(4*e + 4*f*x)*(a + a*sin(e + f*x))^(1/2))/ (128*f) - (B*a^3*c^3*exp(e*8i + f*x*8i)*cos(6*e + 6*f*x)*(a + a*sin(e + f* x))^(1/2))/(64*f) - (B*a^3*c^3*exp(e*8i + f*x*8i)*cos(8*e + 8*f*x)*(a + a* sin(e + f*x))^(1/2))/(512*f) + (7*A*a^3*c^3*exp(e*8i + f*x*8i)*sin(3*e + 3 *f*x)*(a + a*sin(e + f*x))^(1/2))/(32*f) + (7*A*a^3*c^3*exp(e*8i + f*x*8i) *sin(5*e + 5*f*x)*(a + a*sin(e + f*x))^(1/2))/(160*f) + (A*a^3*c^3*exp(e*8 i + f*x*8i)*sin(7*e + 7*f*x)*(a + a*sin(e + f*x))^(1/2))/(224*f)))/(2*cos( e + f*x))