Integrand size = 38, antiderivative size = 288 \[ \int \sin (c+d x) \left (a \sin (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right )^2 \, dx=\frac {3 a b x}{4}+\frac {5 b c x}{8}-\frac {a^2 \cos (c+d x)}{d}-\frac {c^2 \cos (c+d x)}{d}-\frac {\left (b^2+2 a c\right ) \cos (c+d x)}{d}+\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {c^2 \cos ^3(c+d x)}{d}+\frac {2 \left (b^2+2 a c\right ) \cos ^3(c+d x)}{3 d}-\frac {3 c^2 \cos ^5(c+d x)}{5 d}-\frac {\left (b^2+2 a c\right ) \cos ^5(c+d x)}{5 d}+\frac {c^2 \cos ^7(c+d x)}{7 d}-\frac {3 a b \cos (c+d x) \sin (c+d x)}{4 d}-\frac {5 b c \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a b \cos (c+d x) \sin ^3(c+d x)}{2 d}-\frac {5 b c \cos (c+d x) \sin ^3(c+d x)}{12 d}-\frac {b c \cos (c+d x) \sin ^5(c+d x)}{3 d} \]
3/4*a*b*x+5/8*b*c*x-a^2*cos(d*x+c)/d-c^2*cos(d*x+c)/d-(2*a*c+b^2)*cos(d*x+ c)/d+1/3*a^2*cos(d*x+c)^3/d+c^2*cos(d*x+c)^3/d+2/3*(2*a*c+b^2)*cos(d*x+c)^ 3/d-3/5*c^2*cos(d*x+c)^5/d-1/5*(2*a*c+b^2)*cos(d*x+c)^5/d+1/7*c^2*cos(d*x+ c)^7/d-3/4*a*b*cos(d*x+c)*sin(d*x+c)/d-5/8*b*c*cos(d*x+c)*sin(d*x+c)/d-1/2 *a*b*cos(d*x+c)*sin(d*x+c)^3/d-5/12*b*c*cos(d*x+c)*sin(d*x+c)^3/d-1/3*b*c* cos(d*x+c)*sin(d*x+c)^5/d
Time = 1.66 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.58 \[ \int \sin (c+d x) \left (a \sin (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right )^2 \, dx=\frac {840 b (6 a+5 c) (c+d x)-105 \left (48 a^2+40 b^2+80 a c+35 c^2\right ) \cos (c+d x)+35 \left (16 a^2+20 b^2+40 a c+21 c^2\right ) \cos (3 (c+d x))-21 \left (4 b^2+c (8 a+7 c)\right ) \cos (5 (c+d x))+15 c^2 \cos (7 (c+d x))-210 b (16 a+15 c) \sin (2 (c+d x))+210 b (2 a+3 c) \sin (4 (c+d x))-70 b c \sin (6 (c+d x))}{6720 d} \]
(840*b*(6*a + 5*c)*(c + d*x) - 105*(48*a^2 + 40*b^2 + 80*a*c + 35*c^2)*Cos [c + d*x] + 35*(16*a^2 + 20*b^2 + 40*a*c + 21*c^2)*Cos[3*(c + d*x)] - 21*( 4*b^2 + c*(8*a + 7*c))*Cos[5*(c + d*x)] + 15*c^2*Cos[7*(c + d*x)] - 210*b* (16*a + 15*c)*Sin[2*(c + d*x)] + 210*b*(2*a + 3*c)*Sin[4*(c + d*x)] - 70*b *c*Sin[6*(c + d*x)])/(6720*d)
Time = 0.69 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3042, 4894, 3042, 3737, 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.
| \(\displaystyle \int \sin (c+d x) \left (a \sin (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x) \left (a \sin (c+d x)+b \sin (c+d x)^2+c \sin (c+d x)^3\right )^2dx\) |
| \(\Big \downarrow \) 4894 |
| \(\displaystyle \int \sin ^3(c+d x) \left (a+b \sin (c+d x)+c \sin ^2(c+d x)\right )^2dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x)^3 \left (a+b \sin (c+d x)+c \sin (c+d x)^2\right )^2dx\) |
| \(\Big \downarrow \) 3737 |
| \(\displaystyle \int \left (a^2 \sin ^3(c+d x)+\left (2 a c+b^2\right ) \sin ^5(c+d x)+2 a b \sin ^4(c+d x)+2 b c \sin ^6(c+d x)+c^2 \sin ^7(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^2 \cos ^3(c+d x)}{3 d}-\frac {a^2 \cos (c+d x)}{d}-\frac {\left (2 a c+b^2\right ) \cos ^5(c+d x)}{5 d}+\frac {2 \left (2 a c+b^2\right ) \cos ^3(c+d x)}{3 d}-\frac {\left (2 a c+b^2\right ) \cos (c+d x)}{d}-\frac {a b \sin ^3(c+d x) \cos (c+d x)}{2 d}-\frac {3 a b \sin (c+d x) \cos (c+d x)}{4 d}+\frac {3 a b x}{4}-\frac {b c \sin ^5(c+d x) \cos (c+d x)}{3 d}-\frac {5 b c \sin ^3(c+d x) \cos (c+d x)}{12 d}-\frac {5 b c \sin (c+d x) \cos (c+d x)}{8 d}+\frac {5 b c x}{8}+\frac {c^2 \cos ^7(c+d x)}{7 d}-\frac {3 c^2 \cos ^5(c+d x)}{5 d}+\frac {c^2 \cos ^3(c+d x)}{d}-\frac {c^2 \cos (c+d x)}{d}\) |
(3*a*b*x)/4 + (5*b*c*x)/8 - (a^2*Cos[c + d*x])/d - (c^2*Cos[c + d*x])/d - ((b^2 + 2*a*c)*Cos[c + d*x])/d + (a^2*Cos[c + d*x]^3)/(3*d) + (c^2*Cos[c + d*x]^3)/d + (2*(b^2 + 2*a*c)*Cos[c + d*x]^3)/(3*d) - (3*c^2*Cos[c + d*x]^ 5)/(5*d) - ((b^2 + 2*a*c)*Cos[c + d*x]^5)/(5*d) + (c^2*Cos[c + d*x]^7)/(7* d) - (3*a*b*Cos[c + d*x]*Sin[c + d*x])/(4*d) - (5*b*c*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (a*b*Cos[c + d*x]*Sin[c + d*x]^3)/(2*d) - (5*b*c*Cos[c + d* x]*Sin[c + d*x]^3)/(12*d) - (b*c*Cos[c + d*x]*Sin[c + d*x]^5)/(3*d)
3.10.37.3.1 Defintions of rubi rules used
Int[sin[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + (b_.)*sin[(d_.) + (e_.)*(x_)]^(n _.) + (c_.)*sin[(d_.) + (e_.)*(x_)]^(n2_.))^(p_), x_Symbol] :> Int[ExpandTr ig[sin[d + e*x]^m*(a + b*sin[d + e*x]^n + c*sin[d + e*x]^(2*n))^p, x], x] / ; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && Integ ersQ[m, n, p]
Int[(u_)*((a_)*(F_)[(d_.) + (e_.)*(x_)]^(p_.) + (b_.)*(F_)[(d_.) + (e_.)*(x _)]^(q_.) + (c_.)*(F_)[(d_.) + (e_.)*(x_)]^(r_.))^(n_.), x_Symbol] :> Int[A ctivateTrig[u*F[d + e*x]^(n*p)*(a + b*F[d + e*x]^(q - p) + c*F[d + e*x]^(r - p))^n], x] /; FreeQ[{a, b, c, d, e, p, q, r}, x] && InertTrigQ[F] && Inte gerQ[n] && PosQ[q - p] && PosQ[r - p]
Time = 3.72 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.63
| method | result | size |
| parallelrisch | \(\frac {\left (560 a^{2}+1400 a c +700 b^{2}+735 c^{2}\right ) \cos \left (3 d x +3 c \right )+\left (-168 a c -84 b^{2}-147 c^{2}\right ) \cos \left (5 d x +5 c \right )-3360 b \left (a +\frac {15 c}{16}\right ) \sin \left (2 d x +2 c \right )+420 b \left (a +\frac {3 c}{2}\right ) \sin \left (4 d x +4 c \right )+15 c^{2} \cos \left (7 d x +7 c \right )-70 b c \sin \left (6 d x +6 c \right )+\left (-5040 a^{2}-8400 a c -4200 b^{2}-3675 c^{2}\right ) \cos \left (d x +c \right )-3072 c^{2}+\left (4200 d x b -7168 a \right ) c +5040 x a b d -4480 a^{2}-3584 b^{2}}{6720 d}\) | \(182\) |
| parts | \(-\frac {a^{2} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3 d}-\frac {c^{2} \left (\frac {16}{5}+\sin \left (d x +c \right )^{6}+\frac {6 \sin \left (d x +c \right )^{4}}{5}+\frac {8 \sin \left (d x +c \right )^{2}}{5}\right ) \cos \left (d x +c \right )}{7 d}-\frac {\left (2 a c +b^{2}\right ) \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5 d}+\frac {2 a b \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {2 c b \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )}{d}\) | \(199\) |
| derivativedivides | \(\frac {-\frac {c^{2} \left (\frac {16}{5}+\sin \left (d x +c \right )^{6}+\frac {6 \sin \left (d x +c \right )^{4}}{5}+\frac {8 \sin \left (d x +c \right )^{2}}{5}\right ) \cos \left (d x +c \right )}{7}+2 b c \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )-\frac {2 a c \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}-\frac {b^{2} \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}+2 a b \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {a^{2} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}}{d}\) | \(213\) |
| default | \(\frac {-\frac {c^{2} \left (\frac {16}{5}+\sin \left (d x +c \right )^{6}+\frac {6 \sin \left (d x +c \right )^{4}}{5}+\frac {8 \sin \left (d x +c \right )^{2}}{5}\right ) \cos \left (d x +c \right )}{7}+2 b c \left (-\frac {\left (\sin \left (d x +c \right )^{5}+\frac {5 \sin \left (d x +c \right )^{3}}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )-\frac {2 a c \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}-\frac {b^{2} \left (\frac {8}{3}+\sin \left (d x +c \right )^{4}+\frac {4 \sin \left (d x +c \right )^{2}}{3}\right ) \cos \left (d x +c \right )}{5}+2 a b \left (-\frac {\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )-\frac {a^{2} \left (2+\sin \left (d x +c \right )^{2}\right ) \cos \left (d x +c \right )}{3}}{d}\) | \(213\) |
| risch | \(\frac {3 x a b}{4}+\frac {5 b c x}{8}-\frac {3 a^{2} \cos \left (d x +c \right )}{4 d}-\frac {5 \cos \left (d x +c \right ) a c}{4 d}-\frac {5 \cos \left (d x +c \right ) b^{2}}{8 d}-\frac {35 c^{2} \cos \left (d x +c \right )}{64 d}+\frac {c^{2} \cos \left (7 d x +7 c \right )}{448 d}-\frac {b c \sin \left (6 d x +6 c \right )}{96 d}-\frac {\cos \left (5 d x +5 c \right ) a c}{40 d}-\frac {\cos \left (5 d x +5 c \right ) b^{2}}{80 d}-\frac {7 \cos \left (5 d x +5 c \right ) c^{2}}{320 d}+\frac {a b \sin \left (4 d x +4 c \right )}{16 d}+\frac {3 b \sin \left (4 d x +4 c \right ) c}{32 d}+\frac {\cos \left (3 d x +3 c \right ) a^{2}}{12 d}+\frac {5 \cos \left (3 d x +3 c \right ) a c}{24 d}+\frac {5 \cos \left (3 d x +3 c \right ) b^{2}}{48 d}+\frac {7 \cos \left (3 d x +3 c \right ) c^{2}}{64 d}-\frac {a b \sin \left (2 d x +2 c \right )}{2 d}-\frac {15 b \sin \left (2 d x +2 c \right ) c}{32 d}\) | \(281\) |
| norman | \(\frac {-\frac {140 a^{2}+224 a c +112 b^{2}+96 c^{2}}{105 d}+\frac {b \left (6 a +5 c \right ) x}{8}-\frac {4 a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}-\frac {\left (52 a^{2}+64 a c +32 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{3 d}-\frac {\left (88 a^{2}+160 a c +80 b^{2}+96 c^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}-\frac {\left (120 a^{2}+224 a c +112 b^{2}+96 c^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5 d}-\frac {\left (140 a^{2}+224 a c +112 b^{2}+96 c^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{15 d}-\frac {b \left (6 a +5 c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {5 b \left (6 a +5 c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {5 b \left (6 a +5 c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {b \left (6 a +5 c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}+\frac {7 b \left (6 a +5 c \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {21 b \left (6 a +5 c \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8}+\frac {35 b \left (6 a +5 c \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8}+\frac {35 b \left (6 a +5 c \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {21 b \left (6 a +5 c \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {7 b \left (6 a +5 c \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{8}+\frac {b \left (6 a +5 c \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{8}-\frac {b \left (186 a +283 c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}+\frac {b \left (186 a +283 c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{7}}\) | \(508\) |
1/6720*((560*a^2+1400*a*c+700*b^2+735*c^2)*cos(3*d*x+3*c)+(-168*a*c-84*b^2 -147*c^2)*cos(5*d*x+5*c)-3360*b*(a+15/16*c)*sin(2*d*x+2*c)+420*b*(a+3/2*c) *sin(4*d*x+4*c)+15*c^2*cos(7*d*x+7*c)-70*b*c*sin(6*d*x+6*c)+(-5040*a^2-840 0*a*c-4200*b^2-3675*c^2)*cos(d*x+c)-3072*c^2+(4200*b*d*x-7168*a)*c+5040*x* a*b*d-4480*a^2-3584*b^2)/d
Time = 0.25 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.56 \[ \int \sin (c+d x) \left (a \sin (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right )^2 \, dx=\frac {120 \, c^{2} \cos \left (d x + c\right )^{7} - 168 \, {\left (b^{2} + 2 \, a c + 3 \, c^{2}\right )} \cos \left (d x + c\right )^{5} + 280 \, {\left (a^{2} + 2 \, b^{2} + 4 \, a c + 3 \, c^{2}\right )} \cos \left (d x + c\right )^{3} + 105 \, {\left (6 \, a b + 5 \, b c\right )} d x - 840 \, {\left (a^{2} + b^{2} + 2 \, a c + c^{2}\right )} \cos \left (d x + c\right ) - 35 \, {\left (8 \, b c \cos \left (d x + c\right )^{5} - 2 \, {\left (6 \, a b + 13 \, b c\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (10 \, a b + 11 \, b c\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{840 \, d} \]
1/840*(120*c^2*cos(d*x + c)^7 - 168*(b^2 + 2*a*c + 3*c^2)*cos(d*x + c)^5 + 280*(a^2 + 2*b^2 + 4*a*c + 3*c^2)*cos(d*x + c)^3 + 105*(6*a*b + 5*b*c)*d* x - 840*(a^2 + b^2 + 2*a*c + c^2)*cos(d*x + c) - 35*(8*b*c*cos(d*x + c)^5 - 2*(6*a*b + 13*b*c)*cos(d*x + c)^3 + 3*(10*a*b + 11*b*c)*cos(d*x + c))*si n(d*x + c))/d
Time = 0.55 (sec) , antiderivative size = 541, normalized size of antiderivative = 1.88 \[ \int \sin (c+d x) \left (a \sin (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right )^2 \, dx=\begin {cases} - \frac {a^{2} \sin ^{2}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} - \frac {2 a^{2} \cos ^{3}{\left (c + d x \right )}}{3 d} + \frac {3 a b x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {3 a b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 a b x \cos ^{4}{\left (c + d x \right )}}{4} - \frac {5 a b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} - \frac {3 a b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} - \frac {2 a c \sin ^{4}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} - \frac {8 a c \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {16 a c \cos ^{5}{\left (c + d x \right )}}{15 d} - \frac {b^{2} \sin ^{4}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} - \frac {4 b^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {8 b^{2} \cos ^{5}{\left (c + d x \right )}}{15 d} + \frac {5 b c x \sin ^{6}{\left (c + d x \right )}}{8} + \frac {15 b c x \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{8} + \frac {15 b c x \sin ^{2}{\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{8} + \frac {5 b c x \cos ^{6}{\left (c + d x \right )}}{8} - \frac {11 b c \sin ^{5}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} - \frac {5 b c \sin ^{3}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{3 d} - \frac {5 b c \sin {\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{8 d} - \frac {c^{2} \sin ^{6}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} - \frac {2 c^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{d} - \frac {8 c^{2} \sin ^{2}{\left (c + d x \right )} \cos ^{5}{\left (c + d x \right )}}{5 d} - \frac {16 c^{2} \cos ^{7}{\left (c + d x \right )}}{35 d} & \text {for}\: d \neq 0 \\x \left (a \sin {\left (c \right )} + b \sin ^{2}{\left (c \right )} + c \sin ^{3}{\left (c \right )}\right )^{2} \sin {\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((-a**2*sin(c + d*x)**2*cos(c + d*x)/d - 2*a**2*cos(c + d*x)**3/( 3*d) + 3*a*b*x*sin(c + d*x)**4/4 + 3*a*b*x*sin(c + d*x)**2*cos(c + d*x)**2 /2 + 3*a*b*x*cos(c + d*x)**4/4 - 5*a*b*sin(c + d*x)**3*cos(c + d*x)/(4*d) - 3*a*b*sin(c + d*x)*cos(c + d*x)**3/(4*d) - 2*a*c*sin(c + d*x)**4*cos(c + d*x)/d - 8*a*c*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - 16*a*c*cos(c + d*x )**5/(15*d) - b**2*sin(c + d*x)**4*cos(c + d*x)/d - 4*b**2*sin(c + d*x)**2 *cos(c + d*x)**3/(3*d) - 8*b**2*cos(c + d*x)**5/(15*d) + 5*b*c*x*sin(c + d *x)**6/8 + 15*b*c*x*sin(c + d*x)**4*cos(c + d*x)**2/8 + 15*b*c*x*sin(c + d *x)**2*cos(c + d*x)**4/8 + 5*b*c*x*cos(c + d*x)**6/8 - 11*b*c*sin(c + d*x) **5*cos(c + d*x)/(8*d) - 5*b*c*sin(c + d*x)**3*cos(c + d*x)**3/(3*d) - 5*b *c*sin(c + d*x)*cos(c + d*x)**5/(8*d) - c**2*sin(c + d*x)**6*cos(c + d*x)/ d - 2*c**2*sin(c + d*x)**4*cos(c + d*x)**3/d - 8*c**2*sin(c + d*x)**2*cos( c + d*x)**5/(5*d) - 16*c**2*cos(c + d*x)**7/(35*d), Ne(d, 0)), (x*(a*sin(c ) + b*sin(c)**2 + c*sin(c)**3)**2*sin(c), True))
Time = 0.22 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.76 \[ \int \sin (c+d x) \left (a \sin (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right )^2 \, dx=\frac {1120 \, {\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{2} + 210 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) - 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a b - 224 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} b^{2} - 448 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} a c + 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + 60 \, d x + 60 \, c + 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} b c + 96 \, {\left (5 \, \cos \left (d x + c\right )^{7} - 21 \, \cos \left (d x + c\right )^{5} + 35 \, \cos \left (d x + c\right )^{3} - 35 \, \cos \left (d x + c\right )\right )} c^{2}}{3360 \, d} \]
1/3360*(1120*(cos(d*x + c)^3 - 3*cos(d*x + c))*a^2 + 210*(12*d*x + 12*c + sin(4*d*x + 4*c) - 8*sin(2*d*x + 2*c))*a*b - 224*(3*cos(d*x + c)^5 - 10*co s(d*x + c)^3 + 15*cos(d*x + c))*b^2 - 448*(3*cos(d*x + c)^5 - 10*cos(d*x + c)^3 + 15*cos(d*x + c))*a*c + 35*(4*sin(2*d*x + 2*c)^3 + 60*d*x + 60*c + 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*b*c + 96*(5*cos(d*x + c)^7 - 21* cos(d*x + c)^5 + 35*cos(d*x + c)^3 - 35*cos(d*x + c))*c^2)/d
Time = 0.31 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.65 \[ \int \sin (c+d x) \left (a \sin (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right )^2 \, dx=\frac {1}{8} \, {\left (6 \, a b + 5 \, b c\right )} x + \frac {c^{2} \cos \left (7 \, d x + 7 \, c\right )}{448 \, d} - \frac {b c \sin \left (6 \, d x + 6 \, c\right )}{96 \, d} - \frac {{\left (4 \, b^{2} + 8 \, a c + 7 \, c^{2}\right )} \cos \left (5 \, d x + 5 \, c\right )}{320 \, d} + \frac {{\left (16 \, a^{2} + 20 \, b^{2} + 40 \, a c + 21 \, c^{2}\right )} \cos \left (3 \, d x + 3 \, c\right )}{192 \, d} - \frac {{\left (48 \, a^{2} + 40 \, b^{2} + 80 \, a c + 35 \, c^{2}\right )} \cos \left (d x + c\right )}{64 \, d} + \frac {{\left (2 \, a b + 3 \, b c\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {{\left (16 \, a b + 15 \, b c\right )} \sin \left (2 \, d x + 2 \, c\right )}{32 \, d} \]
1/8*(6*a*b + 5*b*c)*x + 1/448*c^2*cos(7*d*x + 7*c)/d - 1/96*b*c*sin(6*d*x + 6*c)/d - 1/320*(4*b^2 + 8*a*c + 7*c^2)*cos(5*d*x + 5*c)/d + 1/192*(16*a^ 2 + 20*b^2 + 40*a*c + 21*c^2)*cos(3*d*x + 3*c)/d - 1/64*(48*a^2 + 40*b^2 + 80*a*c + 35*c^2)*cos(d*x + c)/d + 1/32*(2*a*b + 3*b*c)*sin(4*d*x + 4*c)/d - 1/32*(16*a*b + 15*b*c)*sin(2*d*x + 2*c)/d
Time = 28.97 (sec) , antiderivative size = 456, normalized size of antiderivative = 1.58 \[ \int \sin (c+d x) \left (a \sin (c+d x)+b \sin ^2(c+d x)+c \sin ^3(c+d x)\right )^2 \, dx=\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,a+5\,c\right )}{4\,\left (\frac {3\,a\,b}{2}+\frac {5\,b\,c}{4}\right )}\right )\,\left (6\,a+5\,c\right )}{4\,d}-\frac {b\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )\,\left (6\,a+5\,c\right )}{4\,d}-\frac {\frac {32\,a\,c}{15}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}\,\left (\frac {3\,a\,b}{2}+\frac {5\,b\,c}{4}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (10\,a\,b+\frac {25\,b\,c}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}\,\left (10\,a\,b+\frac {25\,b\,c}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {31\,a\,b}{2}+\frac {283\,b\,c}{12}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {31\,a\,b}{2}+\frac {283\,b\,c}{12}\right )+4\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {52\,a^2}{3}+\frac {64\,c\,a}{3}+\frac {32\,b^2}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {88\,a^2}{3}+\frac {160\,a\,c}{3}+\frac {80\,b^2}{3}+32\,c^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {28\,a^2}{3}+\frac {224\,a\,c}{15}+\frac {112\,b^2}{15}+\frac {32\,c^2}{5}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (24\,a^2+\frac {224\,a\,c}{5}+\frac {112\,b^2}{5}+\frac {96\,c^2}{5}\right )+\frac {4\,a^2}{3}+\frac {16\,b^2}{15}+\frac {32\,c^2}{35}+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a\,b}{2}+\frac {5\,b\,c}{4}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
(b*atan((b*tan(c/2 + (d*x)/2)*(6*a + 5*c))/(4*((3*a*b)/2 + (5*b*c)/4)))*(6 *a + 5*c))/(4*d) - (b*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2)*(6*a + 5*c))/(4 *d) - ((32*a*c)/15 - tan(c/2 + (d*x)/2)^13*((3*a*b)/2 + (5*b*c)/4) + tan(c /2 + (d*x)/2)^3*(10*a*b + (25*b*c)/3) - tan(c/2 + (d*x)/2)^11*(10*a*b + (2 5*b*c)/3) + tan(c/2 + (d*x)/2)^5*((31*a*b)/2 + (283*b*c)/12) - tan(c/2 + ( d*x)/2)^9*((31*a*b)/2 + (283*b*c)/12) + 4*a^2*tan(c/2 + (d*x)/2)^10 + tan( c/2 + (d*x)/2)^8*((64*a*c)/3 + (52*a^2)/3 + (32*b^2)/3) + tan(c/2 + (d*x)/ 2)^6*((160*a*c)/3 + (88*a^2)/3 + (80*b^2)/3 + 32*c^2) + tan(c/2 + (d*x)/2) ^2*((224*a*c)/15 + (28*a^2)/3 + (112*b^2)/15 + (32*c^2)/5) + tan(c/2 + (d* x)/2)^4*((224*a*c)/5 + 24*a^2 + (112*b^2)/5 + (96*c^2)/5) + (4*a^2)/3 + (1 6*b^2)/15 + (32*c^2)/35 + tan(c/2 + (d*x)/2)*((3*a*b)/2 + (5*b*c)/4))/(d*( 7*tan(c/2 + (d*x)/2)^2 + 21*tan(c/2 + (d*x)/2)^4 + 35*tan(c/2 + (d*x)/2)^6 + 35*tan(c/2 + (d*x)/2)^8 + 21*tan(c/2 + (d*x)/2)^10 + 7*tan(c/2 + (d*x)/ 2)^12 + tan(c/2 + (d*x)/2)^14 + 1))