Integrand size = 31, antiderivative size = 134 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {8 (A-B) (a+a \sin (c+d x))^7}{7 a^4 d}-\frac {(3 A-5 B) (a+a \sin (c+d x))^8}{2 a^5 d}+\frac {2 (A-3 B) (a+a \sin (c+d x))^9}{3 a^6 d}-\frac {(A-7 B) (a+a \sin (c+d x))^{10}}{10 a^7 d}-\frac {B (a+a \sin (c+d x))^{11}}{11 a^8 d} \]
8/7*(A-B)*(a+a*sin(d*x+c))^7/a^4/d-1/2*(3*A-5*B)*(a+a*sin(d*x+c))^8/a^5/d+ 2/3*(A-3*B)*(a+a*sin(d*x+c))^9/a^6/d-1/10*(A-7*B)*(a+a*sin(d*x+c))^10/a^7/ d-1/11*B*(a+a*sin(d*x+c))^11/a^8/d
Time = 1.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.64 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {a^3 (1+\sin (c+d x))^7 \left (-484 A+78 B+14 (77 A-39 B) \sin (c+d x)+(-847 A+1029 B) \sin ^2(c+d x)+21 (11 A-37 B) \sin ^3(c+d x)+210 B \sin ^4(c+d x)\right )}{2310 d} \]
-1/2310*(a^3*(1 + Sin[c + d*x])^7*(-484*A + 78*B + 14*(77*A - 39*B)*Sin[c + d*x] + (-847*A + 1029*B)*Sin[c + d*x]^2 + 21*(11*A - 37*B)*Sin[c + d*x]^ 3 + 210*B*Sin[c + d*x]^4))/d
Time = 0.38 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.90, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3315, 27, 86, 2009}
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.
| \(\displaystyle \int \cos ^7(c+d x) (a \sin (c+d x)+a)^3 (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^7 (a \sin (c+d x)+a)^3 (A+B \sin (c+d x))dx\) |
| \(\Big \downarrow \) 3315 |
| \(\displaystyle \frac {\int \frac {(a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^6 (a A+a B \sin (c+d x))}{a}d(a \sin (c+d x))}{a^7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (a-a \sin (c+d x))^3 (\sin (c+d x) a+a)^6 (a A+a B \sin (c+d x))d(a \sin (c+d x))}{a^8 d}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {\int \left (-B (\sin (c+d x) a+a)^{10}-a (A-7 B) (\sin (c+d x) a+a)^9+6 a^2 (A-3 B) (\sin (c+d x) a+a)^8-4 a^3 (3 A-5 B) (\sin (c+d x) a+a)^7+8 a^4 (A-B) (\sin (c+d x) a+a)^6\right )d(a \sin (c+d x))}{a^8 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {8}{7} a^4 (A-B) (a \sin (c+d x)+a)^7-\frac {1}{2} a^3 (3 A-5 B) (a \sin (c+d x)+a)^8+\frac {2}{3} a^2 (A-3 B) (a \sin (c+d x)+a)^9-\frac {1}{10} a (A-7 B) (a \sin (c+d x)+a)^{10}-\frac {1}{11} B (a \sin (c+d x)+a)^{11}}{a^8 d}\) |
((8*a^4*(A - B)*(a + a*Sin[c + d*x])^7)/7 - (a^3*(3*A - 5*B)*(a + a*Sin[c + d*x])^8)/2 + (2*a^2*(A - 3*B)*(a + a*Sin[c + d*x])^9)/3 - (a*(A - 7*B)*( a + a*Sin[c + d*x])^10)/10 - (B*(a + a*Sin[c + d*x])^11)/11)/(a^8*d)
3.10.86.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]
Time = 1.53 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.17
| method | result | size |
| derivativedivides | \(-\frac {a^{3} \left (\frac {B \left (\sin ^{11}\left (d x +c \right )\right )}{11}+\frac {\left (A +3 B \right ) \left (\sin ^{10}\left (d x +c \right )\right )}{10}+\frac {\left (\sin ^{9}\left (d x +c \right )\right ) A}{3}-B \left (\sin ^{8}\left (d x +c \right )\right )+\frac {\left (-6 B -8 A \right ) \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (6 B -6 A \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (8 B +6 A \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+2 \left (\sin ^{4}\left (d x +c \right )\right ) A -B \left (\sin ^{3}\left (d x +c \right )\right )+\frac {\left (-3 A -B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right )\right )}{d}\) | \(157\) |
| default | \(-\frac {a^{3} \left (\frac {B \left (\sin ^{11}\left (d x +c \right )\right )}{11}+\frac {\left (A +3 B \right ) \left (\sin ^{10}\left (d x +c \right )\right )}{10}+\frac {\left (\sin ^{9}\left (d x +c \right )\right ) A}{3}-B \left (\sin ^{8}\left (d x +c \right )\right )+\frac {\left (-6 B -8 A \right ) \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (6 B -6 A \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (8 B +6 A \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+2 \left (\sin ^{4}\left (d x +c \right )\right ) A -B \left (\sin ^{3}\left (d x +c \right )\right )+\frac {\left (-3 A -B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right )\right )}{d}\) | \(157\) |
| parallelrisch | \(-\frac {a^{3} \left (\left (-\frac {A}{10}-\frac {3 B}{10}\right ) \cos \left (10 d x +10 c \right )+\left (91 A +49 B \right ) \cos \left (2 d x +2 c \right )+\left (44 A +20 B \right ) \cos \left (4 d x +4 c \right )+\left (\frac {23 A}{2}+\frac {5 B}{2}\right ) \cos \left (6 d x +6 c \right )+\left (A -B \right ) \cos \left (8 d x +8 c \right )+\left (-56 A +B \right ) \sin \left (3 d x +3 c \right )+\left (-\frac {8 A}{5}+\frac {107 B}{10}\right ) \sin \left (5 d x +5 c \right )+\left (\frac {22 A}{7}+\frac {61 B}{14}\right ) \sin \left (7 d x +7 c \right )+\left (\frac {2 A}{3}+\frac {B}{2}\right ) \sin \left (9 d x +9 c \right )-\frac {B \sin \left (11 d x +11 c \right )}{22}+\left (-364 A -91 B \right ) \sin \left (d x +c \right )-\frac {737 A}{5}-\frac {351 B}{5}\right )}{512 d}\) | \(191\) |
| risch | \(\frac {91 \sin \left (d x +c \right ) A \,a^{3}}{128 d}+\frac {91 a^{3} B \sin \left (d x +c \right )}{512 d}+\frac {B \,a^{3} \sin \left (11 d x +11 c \right )}{11264 d}+\frac {a^{3} \cos \left (10 d x +10 c \right ) A}{5120 d}+\frac {3 a^{3} \cos \left (10 d x +10 c \right ) B}{5120 d}-\frac {\sin \left (9 d x +9 c \right ) A \,a^{3}}{768 d}-\frac {\sin \left (9 d x +9 c \right ) B \,a^{3}}{1024 d}-\frac {a^{3} \cos \left (8 d x +8 c \right ) A}{512 d}+\frac {a^{3} \cos \left (8 d x +8 c \right ) B}{512 d}-\frac {11 \sin \left (7 d x +7 c \right ) A \,a^{3}}{1792 d}-\frac {61 \sin \left (7 d x +7 c \right ) B \,a^{3}}{7168 d}-\frac {23 a^{3} \cos \left (6 d x +6 c \right ) A}{1024 d}-\frac {5 a^{3} \cos \left (6 d x +6 c \right ) B}{1024 d}+\frac {\sin \left (5 d x +5 c \right ) A \,a^{3}}{320 d}-\frac {107 \sin \left (5 d x +5 c \right ) B \,a^{3}}{5120 d}-\frac {11 a^{3} \cos \left (4 d x +4 c \right ) A}{128 d}-\frac {5 a^{3} \cos \left (4 d x +4 c \right ) B}{128 d}+\frac {7 \sin \left (3 d x +3 c \right ) A \,a^{3}}{64 d}-\frac {\sin \left (3 d x +3 c \right ) B \,a^{3}}{512 d}-\frac {91 a^{3} \cos \left (2 d x +2 c \right ) A}{512 d}-\frac {49 a^{3} \cos \left (2 d x +2 c \right ) B}{512 d}\) | \(374\) |
-a^3/d*(1/11*B*sin(d*x+c)^11+1/10*(A+3*B)*sin(d*x+c)^10+1/3*sin(d*x+c)^9*A -B*sin(d*x+c)^8+1/7*(-6*B-8*A)*sin(d*x+c)^7+1/6*(6*B-6*A)*sin(d*x+c)^6+1/5 *(8*B+6*A)*sin(d*x+c)^5+2*sin(d*x+c)^4*A-B*sin(d*x+c)^3+1/2*(-3*A-B)*sin(d *x+c)^2-A*sin(d*x+c))
Time = 0.32 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.16 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {231 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{10} - 1155 \, {\left (A + B\right )} a^{3} \cos \left (d x + c\right )^{8} + 2 \, {\left (105 \, B a^{3} \cos \left (d x + c\right )^{10} - 35 \, {\left (11 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right )^{8} + 20 \, {\left (11 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{6} + 24 \, {\left (11 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} + 32 \, {\left (11 \, A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 64 \, {\left (11 \, A + 3 \, B\right )} a^{3}\right )} \sin \left (d x + c\right )}{2310 \, d} \]
1/2310*(231*(A + 3*B)*a^3*cos(d*x + c)^10 - 1155*(A + B)*a^3*cos(d*x + c)^ 8 + 2*(105*B*a^3*cos(d*x + c)^10 - 35*(11*A + 15*B)*a^3*cos(d*x + c)^8 + 2 0*(11*A + 3*B)*a^3*cos(d*x + c)^6 + 24*(11*A + 3*B)*a^3*cos(d*x + c)^4 + 3 2*(11*A + 3*B)*a^3*cos(d*x + c)^2 + 64*(11*A + 3*B)*a^3)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 530 vs. \(2 (124) = 248\).
Time = 1.82 (sec) , antiderivative size = 530, normalized size of antiderivative = 3.96 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {16 A a^{3} \sin ^{9}{\left (c + d x \right )}}{105 d} + \frac {24 A a^{3} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {16 A a^{3} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {6 A a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {8 A a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} + \frac {2 A a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} - \frac {A a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{8 d} + \frac {A a^{3} \sin {\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} - \frac {A a^{3} \cos ^{10}{\left (c + d x \right )}}{40 d} - \frac {3 A a^{3} \cos ^{8}{\left (c + d x \right )}}{8 d} + \frac {16 B a^{3} \sin ^{11}{\left (c + d x \right )}}{1155 d} + \frac {8 B a^{3} \sin ^{9}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{105 d} + \frac {16 B a^{3} \sin ^{9}{\left (c + d x \right )}}{105 d} + \frac {6 B a^{3} \sin ^{7}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{35 d} + \frac {24 B a^{3} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{5 d} + \frac {6 B a^{3} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {B a^{3} \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} - \frac {3 B a^{3} \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{8 d} - \frac {3 B a^{3} \cos ^{10}{\left (c + d x \right )}}{40 d} - \frac {B a^{3} \cos ^{8}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a \sin {\left (c \right )} + a\right )^{3} \cos ^{7}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((16*A*a**3*sin(c + d*x)**9/(105*d) + 24*A*a**3*sin(c + d*x)**7*c os(c + d*x)**2/(35*d) + 16*A*a**3*sin(c + d*x)**7/(35*d) + 6*A*a**3*sin(c + d*x)**5*cos(c + d*x)**4/(5*d) + 8*A*a**3*sin(c + d*x)**5*cos(c + d*x)**2 /(5*d) + A*a**3*sin(c + d*x)**3*cos(c + d*x)**6/d + 2*A*a**3*sin(c + d*x)* *3*cos(c + d*x)**4/d - A*a**3*sin(c + d*x)**2*cos(c + d*x)**8/(8*d) + A*a* *3*sin(c + d*x)*cos(c + d*x)**6/d - A*a**3*cos(c + d*x)**10/(40*d) - 3*A*a **3*cos(c + d*x)**8/(8*d) + 16*B*a**3*sin(c + d*x)**11/(1155*d) + 8*B*a**3 *sin(c + d*x)**9*cos(c + d*x)**2/(105*d) + 16*B*a**3*sin(c + d*x)**9/(105* d) + 6*B*a**3*sin(c + d*x)**7*cos(c + d*x)**4/(35*d) + 24*B*a**3*sin(c + d *x)**7*cos(c + d*x)**2/(35*d) + B*a**3*sin(c + d*x)**5*cos(c + d*x)**6/(5* d) + 6*B*a**3*sin(c + d*x)**5*cos(c + d*x)**4/(5*d) + B*a**3*sin(c + d*x)* *3*cos(c + d*x)**6/d - 3*B*a**3*sin(c + d*x)**2*cos(c + d*x)**8/(8*d) - 3* B*a**3*cos(c + d*x)**10/(40*d) - B*a**3*cos(c + d*x)**8/(8*d), Ne(d, 0)), (x*(A + B*sin(c))*(a*sin(c) + a)**3*cos(c)**7, True))
Time = 0.30 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.36 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=-\frac {210 \, B a^{3} \sin \left (d x + c\right )^{11} + 231 \, {\left (A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{10} + 770 \, A a^{3} \sin \left (d x + c\right )^{9} - 2310 \, B a^{3} \sin \left (d x + c\right )^{8} - 660 \, {\left (4 \, A + 3 \, B\right )} a^{3} \sin \left (d x + c\right )^{7} - 2310 \, {\left (A - B\right )} a^{3} \sin \left (d x + c\right )^{6} + 924 \, {\left (3 \, A + 4 \, B\right )} a^{3} \sin \left (d x + c\right )^{5} + 4620 \, A a^{3} \sin \left (d x + c\right )^{4} - 2310 \, B a^{3} \sin \left (d x + c\right )^{3} - 1155 \, {\left (3 \, A + B\right )} a^{3} \sin \left (d x + c\right )^{2} - 2310 \, A a^{3} \sin \left (d x + c\right )}{2310 \, d} \]
-1/2310*(210*B*a^3*sin(d*x + c)^11 + 231*(A + 3*B)*a^3*sin(d*x + c)^10 + 7 70*A*a^3*sin(d*x + c)^9 - 2310*B*a^3*sin(d*x + c)^8 - 660*(4*A + 3*B)*a^3* sin(d*x + c)^7 - 2310*(A - B)*a^3*sin(d*x + c)^6 + 924*(3*A + 4*B)*a^3*sin (d*x + c)^5 + 4620*A*a^3*sin(d*x + c)^4 - 2310*B*a^3*sin(d*x + c)^3 - 1155 *(3*A + B)*a^3*sin(d*x + c)^2 - 2310*A*a^3*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (124) = 248\).
Time = 0.44 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.11 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {B a^{3} \sin \left (11 \, d x + 11 \, c\right )}{11264 \, d} + \frac {{\left (A a^{3} + 3 \, B a^{3}\right )} \cos \left (10 \, d x + 10 \, c\right )}{5120 \, d} - \frac {{\left (A a^{3} - B a^{3}\right )} \cos \left (8 \, d x + 8 \, c\right )}{512 \, d} - \frac {{\left (23 \, A a^{3} + 5 \, B a^{3}\right )} \cos \left (6 \, d x + 6 \, c\right )}{1024 \, d} - \frac {{\left (11 \, A a^{3} + 5 \, B a^{3}\right )} \cos \left (4 \, d x + 4 \, c\right )}{128 \, d} - \frac {7 \, {\left (13 \, A a^{3} + 7 \, B a^{3}\right )} \cos \left (2 \, d x + 2 \, c\right )}{512 \, d} - \frac {{\left (4 \, A a^{3} + 3 \, B a^{3}\right )} \sin \left (9 \, d x + 9 \, c\right )}{3072 \, d} - \frac {{\left (44 \, A a^{3} + 61 \, B a^{3}\right )} \sin \left (7 \, d x + 7 \, c\right )}{7168 \, d} + \frac {{\left (16 \, A a^{3} - 107 \, B a^{3}\right )} \sin \left (5 \, d x + 5 \, c\right )}{5120 \, d} + \frac {{\left (56 \, A a^{3} - B a^{3}\right )} \sin \left (3 \, d x + 3 \, c\right )}{512 \, d} + \frac {91 \, {\left (4 \, A a^{3} + B a^{3}\right )} \sin \left (d x + c\right )}{512 \, d} \]
1/11264*B*a^3*sin(11*d*x + 11*c)/d + 1/5120*(A*a^3 + 3*B*a^3)*cos(10*d*x + 10*c)/d - 1/512*(A*a^3 - B*a^3)*cos(8*d*x + 8*c)/d - 1/1024*(23*A*a^3 + 5 *B*a^3)*cos(6*d*x + 6*c)/d - 1/128*(11*A*a^3 + 5*B*a^3)*cos(4*d*x + 4*c)/d - 7/512*(13*A*a^3 + 7*B*a^3)*cos(2*d*x + 2*c)/d - 1/3072*(4*A*a^3 + 3*B*a ^3)*sin(9*d*x + 9*c)/d - 1/7168*(44*A*a^3 + 61*B*a^3)*sin(7*d*x + 7*c)/d + 1/5120*(16*A*a^3 - 107*B*a^3)*sin(5*d*x + 5*c)/d + 1/512*(56*A*a^3 - B*a^ 3)*sin(3*d*x + 3*c)/d + 91/512*(4*A*a^3 + B*a^3)*sin(d*x + c)/d
Time = 0.22 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.32 \[ \int \cos ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx=\frac {\frac {a^3\,{\sin \left (c+d\,x\right )}^2\,\left (3\,A+B\right )}{2}-\frac {A\,a^3\,{\sin \left (c+d\,x\right )}^9}{3}-2\,A\,a^3\,{\sin \left (c+d\,x\right )}^4+a^3\,{\sin \left (c+d\,x\right )}^6\,\left (A-B\right )-\frac {a^3\,{\sin \left (c+d\,x\right )}^{10}\,\left (A+3\,B\right )}{10}+B\,a^3\,{\sin \left (c+d\,x\right )}^3+B\,a^3\,{\sin \left (c+d\,x\right )}^8-\frac {B\,a^3\,{\sin \left (c+d\,x\right )}^{11}}{11}-\frac {2\,a^3\,{\sin \left (c+d\,x\right )}^5\,\left (3\,A+4\,B\right )}{5}+\frac {2\,a^3\,{\sin \left (c+d\,x\right )}^7\,\left (4\,A+3\,B\right )}{7}+A\,a^3\,\sin \left (c+d\,x\right )}{d} \]
((a^3*sin(c + d*x)^2*(3*A + B))/2 - (A*a^3*sin(c + d*x)^9)/3 - 2*A*a^3*sin (c + d*x)^4 + a^3*sin(c + d*x)^6*(A - B) - (a^3*sin(c + d*x)^10*(A + 3*B)) /10 + B*a^3*sin(c + d*x)^3 + B*a^3*sin(c + d*x)^8 - (B*a^3*sin(c + d*x)^11 )/11 - (2*a^3*sin(c + d*x)^5*(3*A + 4*B))/5 + (2*a^3*sin(c + d*x)^7*(4*A + 3*B))/7 + A*a^3*sin(c + d*x))/d