Integrand size = 40, antiderivative size = 198 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {a^3 (7 A-B) \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{105 f \sqrt {a+a \sin (e+f x)}}-\frac {2 a^2 (7 A-B) \cos (e+f x) \sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))^{7/2}}{105 f}-\frac {a (7 A-B) \cos (e+f x) (a+a \sin (e+f x))^{3/2} (c-c \sin (e+f x))^{7/2}}{42 f}-\frac {B \cos (e+f x) (a+a \sin (e+f x))^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f} \]
-1/42*a*(7*A-B)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)*(c-c*sin(f*x+e))^(7/2)/f -1/7*B*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)*(c-c*sin(f*x+e))^(7/2)/f-1/105*a^ 3*(7*A-B)*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)/f/(a+a*sin(f*x+e))^(1/2)-2/105 *a^2*(7*A-B)*cos(f*x+e)*(c-c*sin(f*x+e))^(7/2)*(a+a*sin(f*x+e))^(1/2)/f
Time = 7.51 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.13 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {c^3 (-1+\sin (e+f x))^3 (a (1+\sin (e+f x)))^{5/2} \sqrt {c-c \sin (e+f x)} (525 (A-B) \cos (2 (e+f x))+210 (A-B) \cos (4 (e+f x))+35 A \cos (6 (e+f x))-35 B \cos (6 (e+f x))+4200 A \sin (e+f x)-525 B \sin (e+f x)+700 A \sin (3 (e+f x))+35 B \sin (3 (e+f x))+84 A \sin (5 (e+f x))+63 B \sin (5 (e+f x))+15 B \sin (7 (e+f x)))}{6720 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5} \]
-1/6720*(c^3*(-1 + Sin[e + f*x])^3*(a*(1 + Sin[e + f*x]))^(5/2)*Sqrt[c - c *Sin[e + f*x]]*(525*(A - B)*Cos[2*(e + f*x)] + 210*(A - B)*Cos[4*(e + f*x) ] + 35*A*Cos[6*(e + f*x)] - 35*B*Cos[6*(e + f*x)] + 4200*A*Sin[e + f*x] - 525*B*Sin[e + f*x] + 700*A*Sin[3*(e + f*x)] + 35*B*Sin[3*(e + f*x)] + 84*A *Sin[5*(e + f*x)] + 63*B*Sin[5*(e + f*x)] + 15*B*Sin[7*(e + f*x)]))/(f*(Co s[(e + f*x)/2] - Sin[(e + f*x)/2])^7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) ^5)
Time = 0.95 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3452, 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.
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2} (A+B \sin (e+f x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2} (A+B \sin (e+f x))dx\) |
| \(\Big \downarrow \) 3452 |
| \(\displaystyle \frac {1}{7} (7 A-B) \int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{7/2}dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} (7 A-B) \int (\sin (e+f x) a+a)^{5/2} (c-c \sin (e+f x))^{7/2}dx-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {1}{7} (7 A-B) \left (\frac {2}{3} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} (7 A-B) \left (\frac {2}{3} a \int (\sin (e+f x) a+a)^{3/2} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}\) |
| \(\Big \downarrow \) 3219 |
| \(\displaystyle \frac {1}{7} (7 A-B) \left (\frac {2}{3} a \left (\frac {2}{5} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{7} (7 A-B) \left (\frac {2}{3} a \left (\frac {2}{5} a \int \sqrt {\sin (e+f x) a+a} (c-c \sin (e+f x))^{7/2}dx-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}\) |
| \(\Big \downarrow \) 3217 |
| \(\displaystyle \frac {1}{7} (7 A-B) \left (\frac {2}{3} a \left (-\frac {a^2 \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{10 f \sqrt {a \sin (e+f x)+a}}-\frac {a \cos (e+f x) \sqrt {a \sin (e+f x)+a} (c-c \sin (e+f x))^{7/2}}{5 f}\right )-\frac {a \cos (e+f x) (a \sin (e+f x)+a)^{3/2} (c-c \sin (e+f x))^{7/2}}{6 f}\right )-\frac {B \cos (e+f x) (a \sin (e+f x)+a)^{5/2} (c-c \sin (e+f x))^{7/2}}{7 f}\) |
-1/7*(B*Cos[e + f*x]*(a + a*Sin[e + f*x])^(5/2)*(c - c*Sin[e + f*x])^(7/2) )/f + ((7*A - B)*(-1/6*(a*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2)*(c - c*S in[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
3.2.49.3.1 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 = 73.46 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.01
| method | result | size |
| default | \(-\frac {a^{2} c^{3} \tan \left (f x +e \right ) \left (30 B \left (\cos ^{4}\left (f x +e \right )\right ) \left (\sin ^{2}\left (f x +e \right )\right )+35 A \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )+35 B \left (\cos ^{2}\left (f x +e \right )\right ) \left (\sin ^{3}\left (f x +e \right )\right )-42 A \left (\cos ^{4}\left (f x +e \right )\right )+24 B \left (\sin ^{2}\left (f x +e \right )\right ) \left (\cos ^{2}\left (f x +e \right )\right )+35 A \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+70 B \left (\sin ^{3}\left (f x +e \right )\right )-56 A \left (\cos ^{2}\left (f x +e \right )\right )+16 B \left (\sin ^{2}\left (f x +e \right )\right )+35 A \sin \left (f x +e \right )-105 B \sin \left (f x +e \right )-112 A \right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}}{210 f}\) | \(200\) |
| parts | \(\frac {A \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, a^{2} c^{3} \left (5 \left (\cos ^{5}\left (f x +e \right )\right )+6 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right )+8 \cos \left (f x +e \right ) \sin \left (f x +e \right )+16 \tan \left (f x +e \right )-5 \sec \left (f x +e \right )\right )}{30 f}+\frac {B \sec \left (f x +e \right ) \left (30 \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )-35 \left (\cos ^{4}\left (f x +e \right )\right )+24 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-35 \left (\cos ^{2}\left (f x +e \right )\right )+16 \sin \left (f x +e \right )-35\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, c^{3} a^{2} \left (\cos ^{2}\left (f x +e \right )-1\right )}{210 f}\) | \(211\) |
-1/210*a^2*c^3/f*tan(f*x+e)*(30*B*cos(f*x+e)^4*sin(f*x+e)^2+35*A*cos(f*x+e )^4*sin(f*x+e)+35*B*cos(f*x+e)^2*sin(f*x+e)^3-42*A*cos(f*x+e)^4+24*B*sin(f *x+e)^2*cos(f*x+e)^2+35*A*sin(f*x+e)*cos(f*x+e)^2+70*B*sin(f*x+e)^3-56*A*c os(f*x+e)^2+16*B*sin(f*x+e)^2+35*A*sin(f*x+e)-105*B*sin(f*x+e)-112*A)*(-c* (sin(f*x+e)-1))^(1/2)*(a*(1+sin(f*x+e)))^(1/2)
Time = 0.31 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.81 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\frac {{\left (35 \, {\left (A - B\right )} a^{2} c^{3} \cos \left (f x + e\right )^{6} - 35 \, {\left (A - B\right )} a^{2} c^{3} + 2 \, {\left (15 \, B a^{2} c^{3} \cos \left (f x + e\right )^{6} + 3 \, {\left (7 \, A - B\right )} a^{2} c^{3} \cos \left (f x + e\right )^{4} + 4 \, {\left (7 \, A - B\right )} a^{2} c^{3} \cos \left (f x + e\right )^{2} + 8 \, {\left (7 \, A - B\right )} a^{2} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{210 \, f \cos \left (f x + e\right )} \]
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x , algorithm="fricas")
1/210*(35*(A - B)*a^2*c^3*cos(f*x + e)^6 - 35*(A - B)*a^2*c^3 + 2*(15*B*a^ 2*c^3*cos(f*x + e)^6 + 3*(7*A - B)*a^2*c^3*cos(f*x + e)^4 + 4*(7*A - B)*a^ 2*c^3*cos(f*x + e)^2 + 8*(7*A - B)*a^2*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. \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\text {Timed out} \]
\[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\int { {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}} \,d x } \]
integrate((a+a*sin(f*x+e))^(5/2)*(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. 355 vs. \(2 (174) = 348\).
Time = 0.49 (sec) , antiderivative size = 355, normalized size of antiderivative = 1.79 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=-\frac {16 \, {\left (120 \, B a^{2} 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^{2} 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} - 350 \, B a^{2} 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} + 168 \, A a^{2} 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} + 336 \, B a^{2} 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} - 105 \, A a^{2} 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} - 105 \, B a^{2} 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}}{105 \, f} \]
integrate((a+a*sin(f*x+e))^(5/2)*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(7/2),x , algorithm="giac")
-16/105*(120*B*a^2*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^2*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 - 350*B*a^2*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)^1 2 + 168*A*a^2*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 + 336*B*a^2*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*p i + 1/2*f*x + 1/2*e)^10 - 105*A*a^2*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 - 1 05*B*a^2*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 = 18.72 (sec) , antiderivative size = 383, normalized size of antiderivative = 1.93 \[ \int (a+a \sin (e+f x))^{5/2} (A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2} \, dx=\frac {{\mathrm {e}}^{-e\,7{}\mathrm {i}-f\,x\,7{}\mathrm {i}}\,\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (-\frac {a^2\,c^3\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (2\,e+2\,f\,x\right )\,\left (A\,1{}\mathrm {i}-B\,1{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,5{}\mathrm {i}}{32\,f}-\frac {a^2\,c^3\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (4\,e+4\,f\,x\right )\,\left (A\,1{}\mathrm {i}-B\,1{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,1{}\mathrm {i}}{16\,f}-\frac {a^2\,c^3\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\cos \left (6\,e+6\,f\,x\right )\,\left (A\,1{}\mathrm {i}-B\,1{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,1{}\mathrm {i}}{96\,f}+\frac {a^2\,c^3\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (5\,e+5\,f\,x\right )\,\left (4\,A+3\,B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{160\,f}+\frac {5\,a^2\,c^3\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (e+f\,x\right )\,\left (8\,A-B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{32\,f}+\frac {a^2\,c^3\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (3\,e+3\,f\,x\right )\,\left (20\,A+B\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{96\,f}+\frac {B\,a^2\,c^3\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,\sin \left (7\,e+7\,f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}{224\,f}\right )}{2\,\cos \left (e+f\,x\right )} \]
(exp(- e*7i - f*x*7i)*(c - c*sin(e + f*x))^(1/2)*((a^2*c^3*exp(e*7i + f*x* 7i)*sin(5*e + 5*f*x)*(4*A + 3*B)*(a + a*sin(e + f*x))^(1/2))/(160*f) - (a^ 2*c^3*exp(e*7i + f*x*7i)*cos(4*e + 4*f*x)*(A*1i - B*1i)*(a + a*sin(e + f*x ))^(1/2)*1i)/(16*f) - (a^2*c^3*exp(e*7i + f*x*7i)*cos(6*e + 6*f*x)*(A*1i - B*1i)*(a + a*sin(e + f*x))^(1/2)*1i)/(96*f) - (a^2*c^3*exp(e*7i + f*x*7i) *cos(2*e + 2*f*x)*(A*1i - B*1i)*(a + a*sin(e + f*x))^(1/2)*5i)/(32*f) + (5 *a^2*c^3*exp(e*7i + f*x*7i)*sin(e + f*x)*(8*A - B)*(a + a*sin(e + f*x))^(1 /2))/(32*f) + (a^2*c^3*exp(e*7i + f*x*7i)*sin(3*e + 3*f*x)*(20*A + B)*(a + a*sin(e + f*x))^(1/2))/(96*f) + (B*a^2*c^3*exp(e*7i + f*x*7i)*sin(7*e + 7 *f*x)*(a + a*sin(e + f*x))^(1/2))/(224*f)))/(2*cos(e + f*x))