Integrand size = 31, antiderivative size = 349 \[ \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {\left (a^2-b^2\right )^3 (A b-a B) (a+b \sin (c+d x))^3}{3 b^8 d}+\frac {\left (a^2-b^2\right )^2 \left (6 a A b-7 a^2 B+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^8 d}-\frac {3 \left (a^2-b^2\right ) \left (5 a^2 A b-A b^3-7 a^3 B+3 a b^2 B\right ) (a+b \sin (c+d x))^5}{5 b^8 d}+\frac {\left (20 a^3 A b-12 a A b^3-35 a^4 B+30 a^2 b^2 B-3 b^4 B\right ) (a+b \sin (c+d x))^6}{6 b^8 d}-\frac {\left (15 a^2 A b-3 A b^3-35 a^3 B+15 a b^2 B\right ) (a+b \sin (c+d x))^7}{7 b^8 d}+\frac {3 \left (2 a A b-7 a^2 B+b^2 B\right ) (a+b \sin (c+d x))^8}{8 b^8 d}-\frac {(A b-7 a B) (a+b \sin (c+d x))^9}{9 b^8 d}-\frac {B (a+b \sin (c+d x))^{10}}{10 b^8 d} \]
-1/3*(a^2-b^2)^3*(A*b-B*a)*(a+b*sin(d*x+c))^3/b^8/d+1/4*(a^2-b^2)^2*(6*A*a *b-7*B*a^2+B*b^2)*(a+b*sin(d*x+c))^4/b^8/d-3/5*(a^2-b^2)*(5*A*a^2*b-A*b^3- 7*B*a^3+3*B*a*b^2)*(a+b*sin(d*x+c))^5/b^8/d+1/6*(20*A*a^3*b-12*A*a*b^3-35* B*a^4+30*B*a^2*b^2-3*B*b^4)*(a+b*sin(d*x+c))^6/b^8/d-1/7*(15*A*a^2*b-3*A*b ^3-35*B*a^3+15*B*a*b^2)*(a+b*sin(d*x+c))^7/b^8/d+3/8*(2*A*a*b-7*B*a^2+B*b^ 2)*(a+b*sin(d*x+c))^8/b^8/d-1/9*(A*b-7*B*a)*(a+b*sin(d*x+c))^9/b^8/d-1/10* B*(a+b*sin(d*x+c))^10/b^8/d
Time = 0.88 (sec) , antiderivative size = 295, normalized size of antiderivative = 0.85 \[ \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {-3 a^4 \left (a^6-9 a^4 b^2+42 a^2 b^4-210 b^6\right ) B+2520 a^2 A b^8 \sin (c+d x)+1260 a b^8 (2 A b+a B) \sin ^2(c+d x)+840 b^8 \left (-3 a^2 A+A b^2+2 a b B\right ) \sin ^3(c+d x)+630 b^8 \left (-6 a A b-3 a^2 B+b^2 B\right ) \sin ^4(c+d x)-1512 b^8 \left (-a^2 A+A b^2+2 a b B\right ) \sin ^5(c+d x)+1260 b^8 \left (2 a A b+a^2 B-b^2 B\right ) \sin ^6(c+d x)+360 b^8 \left (-a^2 A+3 A b^2+6 a b B\right ) \sin ^7(c+d x)-315 b^8 \left (2 a A b+a^2 B-3 b^2 B\right ) \sin ^8(c+d x)-280 b^9 (A b+2 a B) \sin ^9(c+d x)-252 b^{10} B \sin ^{10}(c+d x)}{2520 b^8 d} \]
(-3*a^4*(a^6 - 9*a^4*b^2 + 42*a^2*b^4 - 210*b^6)*B + 2520*a^2*A*b^8*Sin[c + d*x] + 1260*a*b^8*(2*A*b + a*B)*Sin[c + d*x]^2 + 840*b^8*(-3*a^2*A + A*b ^2 + 2*a*b*B)*Sin[c + d*x]^3 + 630*b^8*(-6*a*A*b - 3*a^2*B + b^2*B)*Sin[c + d*x]^4 - 1512*b^8*(-(a^2*A) + A*b^2 + 2*a*b*B)*Sin[c + d*x]^5 + 1260*b^8 *(2*a*A*b + a^2*B - b^2*B)*Sin[c + d*x]^6 + 360*b^8*(-(a^2*A) + 3*A*b^2 + 6*a*b*B)*Sin[c + d*x]^7 - 315*b^8*(2*a*A*b + a^2*B - 3*b^2*B)*Sin[c + d*x] ^8 - 280*b^9*(A*b + 2*a*B)*Sin[c + d*x]^9 - 252*b^10*B*Sin[c + d*x]^10)/(2 520*b^8*d)
Time = 0.62 (sec) , antiderivative size = 308, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3316, 27, 652, 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.
| \(\displaystyle \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^7 (a+b \sin (c+d x))^2 (A+B \sin (c+d x))dx\) |
| \(\Big \downarrow \) 3316 |
| \(\displaystyle \frac {\int \frac {(a+b \sin (c+d x))^2 (A b+B \sin (c+d x) b) \left (b^2-b^2 \sin ^2(c+d x)\right )^3}{b}d(b \sin (c+d x))}{b^7 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int (a+b \sin (c+d x))^2 (A b+B \sin (c+d x) b) \left (b^2-b^2 \sin ^2(c+d x)\right )^3d(b \sin (c+d x))}{b^8 d}\) |
| \(\Big \downarrow \) 652 |
| \(\displaystyle \frac {\int \left (-B (a+b \sin (c+d x))^9+(7 a B-A b) (a+b \sin (c+d x))^8-3 \left (7 B a^2-2 A b a-b^2 B\right ) (a+b \sin (c+d x))^7+\left (35 B a^3-15 A b a^2-15 b^2 B a+3 A b^3\right ) (a+b \sin (c+d x))^6+\left (-35 B a^4+20 A b a^3+30 b^2 B a^2-12 A b^3 a-3 b^4 B\right ) (a+b \sin (c+d x))^5+3 \left (a^2-b^2\right ) \left (7 B a^3-5 A b a^2-3 b^2 B a+A b^3\right ) (a+b \sin (c+d x))^4-\left (a^2-b^2\right )^2 \left (7 B a^2-6 A b a-b^2 B\right ) (a+b \sin (c+d x))^3+\left (a^2-b^2\right )^3 (a B-A b) (a+b \sin (c+d x))^2\right )d(b \sin (c+d x))}{b^8 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {3}{8} \left (-7 a^2 B+2 a A b+b^2 B\right ) (a+b \sin (c+d x))^8+\frac {1}{4} \left (a^2-b^2\right )^2 \left (-7 a^2 B+6 a A b+b^2 B\right ) (a+b \sin (c+d x))^4-\frac {1}{3} \left (a^2-b^2\right )^3 (A b-a B) (a+b \sin (c+d x))^3-\frac {1}{7} \left (-35 a^3 B+15 a^2 A b+15 a b^2 B-3 A b^3\right ) (a+b \sin (c+d x))^7-\frac {3}{5} \left (a^2-b^2\right ) \left (-7 a^3 B+5 a^2 A b+3 a b^2 B-A b^3\right ) (a+b \sin (c+d x))^5+\frac {1}{6} \left (-35 a^4 B+20 a^3 A b+30 a^2 b^2 B-12 a A b^3-3 b^4 B\right ) (a+b \sin (c+d x))^6-\frac {1}{9} (A b-7 a B) (a+b \sin (c+d x))^9-\frac {1}{10} B (a+b \sin (c+d x))^{10}}{b^8 d}\) |
(-1/3*((a^2 - b^2)^3*(A*b - a*B)*(a + b*Sin[c + d*x])^3) + ((a^2 - b^2)^2* (6*a*A*b - 7*a^2*B + b^2*B)*(a + b*Sin[c + d*x])^4)/4 - (3*(a^2 - b^2)*(5* a^2*A*b - A*b^3 - 7*a^3*B + 3*a*b^2*B)*(a + b*Sin[c + d*x])^5)/5 + ((20*a^ 3*A*b - 12*a*A*b^3 - 35*a^4*B + 30*a^2*b^2*B - 3*b^4*B)*(a + b*Sin[c + d*x ])^6)/6 - ((15*a^2*A*b - 3*A*b^3 - 35*a^3*B + 15*a*b^2*B)*(a + b*Sin[c + d *x])^7)/7 + (3*(2*a*A*b - 7*a^2*B + b^2*B)*(a + b*Sin[c + d*x])^8)/8 - ((A *b - 7*a*B)*(a + b*Sin[c + d*x])^9)/9 - (B*(a + b*Sin[c + d*x])^10)/10)/(b ^8*d)
3.16.36.3.1 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[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ )^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + c *x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && IGtQ[p, 0]
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*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) /2] && NeQ[a^2 - b^2, 0]
Time = 1.94 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.70
| method | result | size |
| derivativedivides | \(-\frac {\frac {B \,b^{2} \left (\sin ^{10}\left (d x +c \right )\right )}{10}+\frac {\left (A \,b^{2}+2 B a b \right ) \left (\sin ^{9}\left (d x +c \right )\right )}{9}+\frac {\left (\left (a^{2}-3 b^{2}\right ) B +2 A a b \right ) \left (\sin ^{8}\left (d x +c \right )\right )}{8}+\frac {\left (-6 B a b +\left (a^{2}-3 b^{2}\right ) A \right ) \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (\left (-3 a^{2}+3 b^{2}\right ) B -6 A a b \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (6 B a b +\left (-3 a^{2}+3 b^{2}\right ) A \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\left (3 a^{2}-b^{2}\right ) B +6 A a b \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (-2 B a b +\left (3 a^{2}-b^{2}\right ) A \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-2 A a b -B \,a^{2}\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right ) a^{2}}{d}\) | \(246\) |
| default | \(-\frac {\frac {B \,b^{2} \left (\sin ^{10}\left (d x +c \right )\right )}{10}+\frac {\left (A \,b^{2}+2 B a b \right ) \left (\sin ^{9}\left (d x +c \right )\right )}{9}+\frac {\left (\left (a^{2}-3 b^{2}\right ) B +2 A a b \right ) \left (\sin ^{8}\left (d x +c \right )\right )}{8}+\frac {\left (-6 B a b +\left (a^{2}-3 b^{2}\right ) A \right ) \left (\sin ^{7}\left (d x +c \right )\right )}{7}+\frac {\left (\left (-3 a^{2}+3 b^{2}\right ) B -6 A a b \right ) \left (\sin ^{6}\left (d x +c \right )\right )}{6}+\frac {\left (6 B a b +\left (-3 a^{2}+3 b^{2}\right ) A \right ) \left (\sin ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\left (3 a^{2}-b^{2}\right ) B +6 A a b \right ) \left (\sin ^{4}\left (d x +c \right )\right )}{4}+\frac {\left (-2 B a b +\left (3 a^{2}-b^{2}\right ) A \right ) \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (-2 A a b -B \,a^{2}\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right ) a^{2}}{d}\) | \(246\) |
| parallelrisch | \(\frac {\left (-35280 A a b -17640 B \,a^{2}-4410 B \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (-17640 A a b -8820 B \,a^{2}-1260 B \,b^{2}\right ) \cos \left (4 d x +4 c \right )+\left (-5040 A a b -2520 B \,a^{2}+315 B \,b^{2}\right ) \cos \left (6 d x +6 c \right )+\left (-630 A a b -315 B \,a^{2}+315 B \,b^{2}\right ) \cos \left (8 d x +8 c \right )+\left (7056 A \,a^{2}-2016 A \,b^{2}-4032 B a b \right ) \sin \left (5 d x +5 c \right )+\left (720 A \,a^{2}-900 A \,b^{2}-1800 B a b \right ) \sin \left (7 d x +7 c \right )+\left (-140 A \,b^{2}-280 B a b \right ) \sin \left (9 d x +9 c \right )+63 B \,b^{2} \cos \left (10 d x +10 c \right )+35280 A \,a^{2} \sin \left (3 d x +3 c \right )+\left (176400 A \,a^{2}+17640 A \,b^{2}+35280 B a b \right ) \sin \left (d x +c \right )+58590 A a b +29295 B \,a^{2}+4977 B \,b^{2}}{322560 d}\) | \(269\) |
| risch | \(\frac {7 A \,a^{2} \sin \left (3 d x +3 c \right )}{64 d}+\frac {7 \sin \left (d x +c \right ) B a b}{64 d}-\frac {\sin \left (9 d x +9 c \right ) B a b}{1152 d}-\frac {\cos \left (8 d x +8 c \right ) A a b}{512 d}-\frac {5 \sin \left (7 d x +7 c \right ) B a b}{896 d}-\frac {\cos \left (6 d x +6 c \right ) A a b}{64 d}-\frac {\sin \left (5 d x +5 c \right ) B a b}{80 d}-\frac {7 \cos \left (4 d x +4 c \right ) A a b}{128 d}+\frac {35 \sin \left (d x +c \right ) A \,a^{2}}{64 d}+\frac {7 \sin \left (d x +c \right ) A \,b^{2}}{128 d}-\frac {\sin \left (9 d x +9 c \right ) A \,b^{2}}{2304 d}-\frac {\cos \left (8 d x +8 c \right ) B \,a^{2}}{1024 d}+\frac {\cos \left (8 d x +8 c \right ) B \,b^{2}}{1024 d}+\frac {\sin \left (7 d x +7 c \right ) A \,a^{2}}{448 d}-\frac {5 \sin \left (7 d x +7 c \right ) A \,b^{2}}{1792 d}-\frac {\cos \left (6 d x +6 c \right ) B \,a^{2}}{128 d}+\frac {\cos \left (6 d x +6 c \right ) B \,b^{2}}{1024 d}+\frac {7 \sin \left (5 d x +5 c \right ) A \,a^{2}}{320 d}-\frac {\sin \left (5 d x +5 c \right ) A \,b^{2}}{160 d}-\frac {7 \cos \left (4 d x +4 c \right ) B \,a^{2}}{256 d}-\frac {\cos \left (4 d x +4 c \right ) B \,b^{2}}{256 d}-\frac {7 \cos \left (2 d x +2 c \right ) B \,a^{2}}{128 d}-\frac {7 \cos \left (2 d x +2 c \right ) B \,b^{2}}{512 d}-\frac {7 \cos \left (2 d x +2 c \right ) A a b}{64 d}+\frac {B \,b^{2} \cos \left (10 d x +10 c \right )}{5120 d}\) | \(435\) |
-1/d*(1/10*B*b^2*sin(d*x+c)^10+1/9*(A*b^2+2*B*a*b)*sin(d*x+c)^9+1/8*((a^2- 3*b^2)*B+2*A*a*b)*sin(d*x+c)^8+1/7*(-6*B*a*b+(a^2-3*b^2)*A)*sin(d*x+c)^7+1 /6*((-3*a^2+3*b^2)*B-6*A*a*b)*sin(d*x+c)^6+1/5*(6*B*a*b+(-3*a^2+3*b^2)*A)* sin(d*x+c)^5+1/4*((3*a^2-b^2)*B+6*A*a*b)*sin(d*x+c)^4+1/3*(-2*B*a*b+(3*a^2 -b^2)*A)*sin(d*x+c)^3+1/2*(-2*A*a*b-B*a^2)*sin(d*x+c)^2-A*sin(d*x+c)*a^2)
Time = 0.30 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.50 \[ \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {252 \, B b^{2} \cos \left (d x + c\right )^{10} - 315 \, {\left (B a^{2} + 2 \, A a b + B b^{2}\right )} \cos \left (d x + c\right )^{8} - 8 \, {\left (35 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{8} - 5 \, {\left (9 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{6} - 6 \, {\left (9 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{4} - 144 \, A a^{2} - 32 \, B a b - 16 \, A b^{2} - 8 \, {\left (9 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2520 \, d} \]
1/2520*(252*B*b^2*cos(d*x + c)^10 - 315*(B*a^2 + 2*A*a*b + B*b^2)*cos(d*x + c)^8 - 8*(35*(2*B*a*b + A*b^2)*cos(d*x + c)^8 - 5*(9*A*a^2 + 2*B*a*b + A *b^2)*cos(d*x + c)^6 - 6*(9*A*a^2 + 2*B*a*b + A*b^2)*cos(d*x + c)^4 - 144* A*a^2 - 32*B*a*b - 16*A*b^2 - 8*(9*A*a^2 + 2*B*a*b + A*b^2)*cos(d*x + c)^2 )*sin(d*x + c))/d
Time = 1.31 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.11 \[ \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {16 A a^{2} \sin ^{7}{\left (c + d x \right )}}{35 d} + \frac {8 A a^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{5 d} + \frac {2 A a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{d} - \frac {A a b \cos ^{8}{\left (c + d x \right )}}{4 d} + \frac {16 A b^{2} \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {8 A b^{2} \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {2 A b^{2} \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {A b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} - \frac {B a^{2} \cos ^{8}{\left (c + d x \right )}}{8 d} + \frac {32 B a b \sin ^{9}{\left (c + d x \right )}}{315 d} + \frac {16 B a b \sin ^{7}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{35 d} + \frac {4 B a b \sin ^{5}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{5 d} + \frac {2 B a b \sin ^{3}{\left (c + d x \right )} \cos ^{6}{\left (c + d x \right )}}{3 d} - \frac {B b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{8}{\left (c + d x \right )}}{8 d} - \frac {B b^{2} \cos ^{10}{\left (c + d x \right )}}{40 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a + b \sin {\left (c \right )}\right )^{2} \cos ^{7}{\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((16*A*a**2*sin(c + d*x)**7/(35*d) + 8*A*a**2*sin(c + d*x)**5*cos (c + d*x)**2/(5*d) + 2*A*a**2*sin(c + d*x)**3*cos(c + d*x)**4/d + A*a**2*s in(c + d*x)*cos(c + d*x)**6/d - A*a*b*cos(c + d*x)**8/(4*d) + 16*A*b**2*si n(c + d*x)**9/(315*d) + 8*A*b**2*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 2*A*b**2*sin(c + d*x)**5*cos(c + d*x)**4/(5*d) + A*b**2*sin(c + d*x)**3*co s(c + d*x)**6/(3*d) - B*a**2*cos(c + d*x)**8/(8*d) + 32*B*a*b*sin(c + d*x) **9/(315*d) + 16*B*a*b*sin(c + d*x)**7*cos(c + d*x)**2/(35*d) + 4*B*a*b*si n(c + d*x)**5*cos(c + d*x)**4/(5*d) + 2*B*a*b*sin(c + d*x)**3*cos(c + d*x) **6/(3*d) - B*b**2*sin(c + d*x)**2*cos(c + d*x)**8/(8*d) - B*b**2*cos(c + d*x)**10/(40*d), Ne(d, 0)), (x*(A + B*sin(c))*(a + b*sin(c))**2*cos(c)**7, True))
Time = 0.21 (sec) , antiderivative size = 238, normalized size of antiderivative = 0.68 \[ \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {252 \, B b^{2} \sin \left (d x + c\right )^{10} + 280 \, {\left (2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )^{9} + 315 \, {\left (B a^{2} + 2 \, A a b - 3 \, B b^{2}\right )} \sin \left (d x + c\right )^{8} + 360 \, {\left (A a^{2} - 6 \, B a b - 3 \, A b^{2}\right )} \sin \left (d x + c\right )^{7} - 1260 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \sin \left (d x + c\right )^{6} - 1512 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{5} + 630 \, {\left (3 \, B a^{2} + 6 \, A a b - B b^{2}\right )} \sin \left (d x + c\right )^{4} - 2520 \, A a^{2} \sin \left (d x + c\right ) + 840 \, {\left (3 \, A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{3} - 1260 \, {\left (B a^{2} + 2 \, A a b\right )} \sin \left (d x + c\right )^{2}}{2520 \, d} \]
-1/2520*(252*B*b^2*sin(d*x + c)^10 + 280*(2*B*a*b + A*b^2)*sin(d*x + c)^9 + 315*(B*a^2 + 2*A*a*b - 3*B*b^2)*sin(d*x + c)^8 + 360*(A*a^2 - 6*B*a*b - 3*A*b^2)*sin(d*x + c)^7 - 1260*(B*a^2 + 2*A*a*b - B*b^2)*sin(d*x + c)^6 - 1512*(A*a^2 - 2*B*a*b - A*b^2)*sin(d*x + c)^5 + 630*(3*B*a^2 + 6*A*a*b - B *b^2)*sin(d*x + c)^4 - 2520*A*a^2*sin(d*x + c) + 840*(3*A*a^2 - 2*B*a*b - A*b^2)*sin(d*x + c)^3 - 1260*(B*a^2 + 2*A*a*b)*sin(d*x + c)^2)/d
Time = 0.50 (sec) , antiderivative size = 279, normalized size of antiderivative = 0.80 \[ \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {B b^{2} \cos \left (10 \, d x + 10 \, c\right )}{5120 \, d} + \frac {7 \, A a^{2} \sin \left (3 \, d x + 3 \, c\right )}{64 \, d} - \frac {{\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \cos \left (8 \, d x + 8 \, c\right )}{1024 \, d} - \frac {{\left (8 \, B a^{2} + 16 \, A a b - B b^{2}\right )} \cos \left (6 \, d x + 6 \, c\right )}{1024 \, d} - \frac {{\left (7 \, B a^{2} + 14 \, A a b + B b^{2}\right )} \cos \left (4 \, d x + 4 \, c\right )}{256 \, d} - \frac {7 \, {\left (4 \, B a^{2} + 8 \, A a b + B b^{2}\right )} \cos \left (2 \, d x + 2 \, c\right )}{512 \, d} - \frac {{\left (2 \, B a b + A b^{2}\right )} \sin \left (9 \, d x + 9 \, c\right )}{2304 \, d} + \frac {{\left (4 \, A a^{2} - 10 \, B a b - 5 \, A b^{2}\right )} \sin \left (7 \, d x + 7 \, c\right )}{1792 \, d} + \frac {{\left (7 \, A a^{2} - 4 \, B a b - 2 \, A b^{2}\right )} \sin \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {7 \, {\left (10 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )}{128 \, d} \]
1/5120*B*b^2*cos(10*d*x + 10*c)/d + 7/64*A*a^2*sin(3*d*x + 3*c)/d - 1/1024 *(B*a^2 + 2*A*a*b - B*b^2)*cos(8*d*x + 8*c)/d - 1/1024*(8*B*a^2 + 16*A*a*b - B*b^2)*cos(6*d*x + 6*c)/d - 1/256*(7*B*a^2 + 14*A*a*b + B*b^2)*cos(4*d* x + 4*c)/d - 7/512*(4*B*a^2 + 8*A*a*b + B*b^2)*cos(2*d*x + 2*c)/d - 1/2304 *(2*B*a*b + A*b^2)*sin(9*d*x + 9*c)/d + 1/1792*(4*A*a^2 - 10*B*a*b - 5*A*b ^2)*sin(7*d*x + 7*c)/d + 1/320*(7*A*a^2 - 4*B*a*b - 2*A*b^2)*sin(5*d*x + 5 *c)/d + 7/128*(10*A*a^2 + 2*B*a*b + A*b^2)*sin(d*x + c)/d
Time = 0.19 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.68 \[ \int \cos ^7(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {B\,a^2}{2}+A\,b\,a\right )-{\sin \left (c+d\,x\right )}^9\,\left (\frac {A\,b^2}{9}+\frac {2\,B\,a\,b}{9}\right )+{\sin \left (c+d\,x\right )}^3\,\left (-A\,a^2+\frac {2\,B\,a\,b}{3}+\frac {A\,b^2}{3}\right )-{\sin \left (c+d\,x\right )}^5\,\left (-\frac {3\,A\,a^2}{5}+\frac {6\,B\,a\,b}{5}+\frac {3\,A\,b^2}{5}\right )+{\sin \left (c+d\,x\right )}^7\,\left (-\frac {A\,a^2}{7}+\frac {6\,B\,a\,b}{7}+\frac {3\,A\,b^2}{7}\right )+{\sin \left (c+d\,x\right )}^6\,\left (\frac {B\,a^2}{2}+A\,a\,b-\frac {B\,b^2}{2}\right )-{\sin \left (c+d\,x\right )}^4\,\left (\frac {3\,B\,a^2}{4}+\frac {3\,A\,a\,b}{2}-\frac {B\,b^2}{4}\right )-{\sin \left (c+d\,x\right )}^8\,\left (\frac {B\,a^2}{8}+\frac {A\,a\,b}{4}-\frac {3\,B\,b^2}{8}\right )-\frac {B\,b^2\,{\sin \left (c+d\,x\right )}^{10}}{10}+A\,a^2\,\sin \left (c+d\,x\right )}{d} \]
(sin(c + d*x)^2*((B*a^2)/2 + A*a*b) - sin(c + d*x)^9*((A*b^2)/9 + (2*B*a*b )/9) + sin(c + d*x)^3*((A*b^2)/3 - A*a^2 + (2*B*a*b)/3) - sin(c + d*x)^5*( (3*A*b^2)/5 - (3*A*a^2)/5 + (6*B*a*b)/5) + sin(c + d*x)^7*((3*A*b^2)/7 - ( A*a^2)/7 + (6*B*a*b)/7) + sin(c + d*x)^6*((B*a^2)/2 - (B*b^2)/2 + A*a*b) - sin(c + d*x)^4*((3*B*a^2)/4 - (B*b^2)/4 + (3*A*a*b)/2) - sin(c + d*x)^8*( (B*a^2)/8 - (3*B*b^2)/8 + (A*a*b)/4) - (B*b^2*sin(c + d*x)^10)/10 + A*a^2* sin(c + d*x))/d