Integrand size = 38, antiderivative size = 160 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{8} a^2 (7 B+6 C) x+\frac {a^2 (10 B+9 C) \sin (c+d x)}{5 d}+\frac {a^2 (7 B+6 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (5 B+6 C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {C \cos ^3(c+d x) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{5 d}-\frac {a^2 (10 B+9 C) \sin ^3(c+d x)}{15 d} \]
1/8*a^2*(7*B+6*C)*x+1/5*a^2*(10*B+9*C)*sin(d*x+c)/d+1/8*a^2*(7*B+6*C)*cos( d*x+c)*sin(d*x+c)/d+1/20*a^2*(5*B+6*C)*cos(d*x+c)^3*sin(d*x+c)/d+1/5*C*cos (d*x+c)^3*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d-1/15*a^2*(10*B+9*C)*sin(d*x+c) ^3/d
Time = 0.31 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.65 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {a^2 (420 B d x+360 C d x+60 (12 B+11 C) \sin (c+d x)+240 (B+C) \sin (2 (c+d x))+80 B \sin (3 (c+d x))+90 C \sin (3 (c+d x))+15 B \sin (4 (c+d x))+30 C \sin (4 (c+d x))+6 C \sin (5 (c+d x)))}{480 d} \]
(a^2*(420*B*d*x + 360*C*d*x + 60*(12*B + 11*C)*Sin[c + d*x] + 240*(B + C)* Sin[2*(c + d*x)] + 80*B*Sin[3*(c + d*x)] + 90*C*Sin[3*(c + d*x)] + 15*B*Si n[4*(c + d*x)] + 30*C*Sin[4*(c + d*x)] + 6*C*Sin[5*(c + d*x)]))/(480*d)
Time = 0.96 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.94, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.395, Rules used = {3042, 3508, 3042, 3455, 3042, 3447, 3042, 3502, 3042, 3227, 3042, 3113, 2009, 3115, 24}
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.
| \(\displaystyle \int \cos (c+d x) (a \cos (c+d x)+a)^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \cos ^2(c+d x) (a \cos (c+d x)+a)^2 (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {1}{5} \int \cos ^2(c+d x) (\cos (c+d x) a+a) (a (5 B+3 C)+a (5 B+6 C) \cos (c+d x))dx+\frac {C \sin (c+d x) \cos ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a (5 B+3 C)+a (5 B+6 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {C \sin (c+d x) \cos ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {1}{5} \int \cos ^2(c+d x) \left ((5 B+6 C) \cos ^2(c+d x) a^2+(5 B+3 C) a^2+\left ((5 B+3 C) a^2+(5 B+6 C) a^2\right ) \cos (c+d x)\right )dx+\frac {C \sin (c+d x) \cos ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left ((5 B+6 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^2+(5 B+3 C) a^2+\left ((5 B+3 C) a^2+(5 B+6 C) a^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {C \sin (c+d x) \cos ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \cos ^2(c+d x) \left (5 (7 B+6 C) a^2+4 (10 B+9 C) \cos (c+d x) a^2\right )dx+\frac {a^2 (5 B+6 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (5 (7 B+6 C) a^2+4 (10 B+9 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {a^2 (5 B+6 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (4 a^2 (10 B+9 C) \int \cos ^3(c+d x)dx+5 a^2 (7 B+6 C) \int \cos ^2(c+d x)dx\right )+\frac {a^2 (5 B+6 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (7 B+6 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+4 a^2 (10 B+9 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {a^2 (5 B+6 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (7 B+6 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {4 a^2 (10 B+9 C) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a^2 (5 B+6 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (7 B+6 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {4 a^2 (10 B+9 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a^2 (5 B+6 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (7 B+6 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {4 a^2 (10 B+9 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a^2 (5 B+6 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {C \sin (c+d x) \cos ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} \left (\frac {a^2 (5 B+6 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {1}{4} \left (5 a^2 (7 B+6 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {4 a^2 (10 B+9 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )\right )+\frac {C \sin (c+d x) \cos ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{5 d}\) |
(C*Cos[c + d*x]^3*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(5*d) + ((a^2*(5* B + 6*C)*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (5*a^2*(7*B + 6*C)*(x/2 + (C os[c + d*x]*Sin[c + d*x])/(2*d)) - (4*a^2*(10*B + 9*C)*(-Sin[c + d*x] + Si n[c + d*x]^3/3))/d)/4)/5
3.3.35.3.1 Defintions of rubi rules used
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && !LtQ[n, -1 ] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 5.90 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.58
| method | result | size |
| parallelrisch | \(\frac {\left (16 \left (B +C \right ) \sin \left (2 d x +2 c \right )+2 \left (\frac {8 B}{3}+3 C \right ) \sin \left (3 d x +3 c \right )+\left (B +2 C \right ) \sin \left (4 d x +4 c \right )+\frac {2 \sin \left (5 d x +5 c \right ) C}{5}+4 \left (12 B +11 C \right ) \sin \left (d x +c \right )+28 \left (B +\frac {6 C}{7}\right ) x d \right ) a^{2}}{32 d}\) | \(93\) |
| parts | \(\frac {\left (B \,a^{2}+2 a^{2} C \right ) \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {\left (2 B \,a^{2}+a^{2} C \right ) \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{2} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5 d}\) | \(149\) |
| risch | \(\frac {7 a^{2} B x}{8}+\frac {3 a^{2} C x}{4}+\frac {3 \sin \left (d x +c \right ) B \,a^{2}}{2 d}+\frac {11 \sin \left (d x +c \right ) a^{2} C}{8 d}+\frac {a^{2} C \sin \left (5 d x +5 c \right )}{80 d}+\frac {\sin \left (4 d x +4 c \right ) B \,a^{2}}{32 d}+\frac {\sin \left (4 d x +4 c \right ) a^{2} C}{16 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{2}}{6 d}+\frac {3 \sin \left (3 d x +3 c \right ) a^{2} C}{16 d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{2}}{2 d}+\frac {\sin \left (2 d x +2 c \right ) a^{2} C}{2 d}\) | \(172\) |
| derivativedivides | \(\frac {\frac {a^{2} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+B \,a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a^{2} C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 B \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(186\) |
| default | \(\frac {\frac {a^{2} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+B \,a^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a^{2} C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 B \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(186\) |
| norman | \(\frac {\frac {a^{2} \left (7 B +6 C \right ) x}{8}+\frac {7 a^{2} \left (7 B +6 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {a^{2} \left (7 B +6 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {5 a^{2} \left (7 B +6 C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {5 a^{2} \left (7 B +6 C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 a^{2} \left (7 B +6 C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 a^{2} \left (7 B +6 C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {a^{2} \left (7 B +6 C \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8}+\frac {a^{2} \left (25 B +26 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {8 a^{2} \left (25 B +27 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {a^{2} \left (79 B +54 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\) | \(279\) |
1/32*(16*(B+C)*sin(2*d*x+2*c)+2*(8/3*B+3*C)*sin(3*d*x+3*c)+(B+2*C)*sin(4*d *x+4*c)+2/5*sin(5*d*x+5*c)*C+4*(12*B+11*C)*sin(d*x+c)+28*(B+6/7*C)*x*d)*a^ 2/d
Time = 0.25 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.69 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (7 \, B + 6 \, C\right )} a^{2} d x + {\left (24 \, C a^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (10 \, B + 9 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right ) + 16 \, {\left (10 \, B + 9 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{120 \, d} \]
1/120*(15*(7*B + 6*C)*a^2*d*x + (24*C*a^2*cos(d*x + c)^4 + 30*(B + 2*C)*a^ 2*cos(d*x + c)^3 + 8*(10*B + 9*C)*a^2*cos(d*x + c)^2 + 15*(7*B + 6*C)*a^2* cos(d*x + c) + 16*(10*B + 9*C)*a^2)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 462 vs. \(2 (144) = 288\).
Time = 0.29 (sec) , antiderivative size = 462, normalized size of antiderivative = 2.89 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\begin {cases} \frac {3 B a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 B a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {B a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {B a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {4 B a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 B a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {2 B a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {3 C a^{2} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {3 C a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 C a^{2} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {8 C a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {3 C a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {2 C a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 C a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {C a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (B \cos {\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) \left (a \cos {\left (c \right )} + a\right )^{2} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((3*B*a**2*x*sin(c + d*x)**4/8 + 3*B*a**2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + B*a**2*x*sin(c + d*x)**2/2 + 3*B*a**2*x*cos(c + d*x)**4/8 + B*a**2*x*cos(c + d*x)**2/2 + 3*B*a**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 4*B*a**2*sin(c + d*x)**3/(3*d) + 5*B*a**2*sin(c + d*x)*cos(c + d*x)**3/( 8*d) + 2*B*a**2*sin(c + d*x)*cos(c + d*x)**2/d + B*a**2*sin(c + d*x)*cos(c + d*x)/(2*d) + 3*C*a**2*x*sin(c + d*x)**4/4 + 3*C*a**2*x*sin(c + d*x)**2* cos(c + d*x)**2/2 + 3*C*a**2*x*cos(c + d*x)**4/4 + 8*C*a**2*sin(c + d*x)** 5/(15*d) + 4*C*a**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) + 3*C*a**2*sin(c + d*x)**3*cos(c + d*x)/(4*d) + 2*C*a**2*sin(c + d*x)**3/(3*d) + C*a**2*si n(c + d*x)*cos(c + d*x)**4/d + 5*C*a**2*sin(c + d*x)*cos(c + d*x)**3/(4*d) + C*a**2*sin(c + d*x)*cos(c + d*x)**2/d, Ne(d, 0)), (x*(B*cos(c) + C*cos( c)**2)*(a*cos(c) + a)**2*cos(c), True))
Time = 0.22 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.11 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=-\frac {320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a^{2} + 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} - 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2}}{480 \, d} \]
-1/480*(320*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^2 - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^2 - 120*(2*d*x + 2*c + sin(2*d* x + 2*c))*B*a^2 - 32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*a^2 + 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^2 - 30*(12*d*x + 12* c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^2)/d
Time = 0.33 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.86 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {C a^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {1}{8} \, {\left (7 \, B a^{2} + 6 \, C a^{2}\right )} x + \frac {{\left (B a^{2} + 2 \, C a^{2}\right )} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {{\left (8 \, B a^{2} + 9 \, C a^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (B a^{2} + C a^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac {{\left (12 \, B a^{2} + 11 \, C a^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \]
1/80*C*a^2*sin(5*d*x + 5*c)/d + 1/8*(7*B*a^2 + 6*C*a^2)*x + 1/32*(B*a^2 + 2*C*a^2)*sin(4*d*x + 4*c)/d + 1/48*(8*B*a^2 + 9*C*a^2)*sin(3*d*x + 3*c)/d + 1/2*(B*a^2 + C*a^2)*sin(2*d*x + 2*c)/d + 1/8*(12*B*a^2 + 11*C*a^2)*sin(d *x + c)/d
Time = 2.72 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.73 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {\left (\frac {7\,B\,a^2}{4}+\frac {3\,C\,a^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {49\,B\,a^2}{6}+7\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {40\,B\,a^2}{3}+\frac {72\,C\,a^2}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {79\,B\,a^2}{6}+9\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {25\,B\,a^2}{4}+\frac {13\,C\,a^2}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (7\,B+6\,C\right )}{4\,\left (\frac {7\,B\,a^2}{4}+\frac {3\,C\,a^2}{2}\right )}\right )\,\left (7\,B+6\,C\right )}{4\,d}-\frac {a^2\,\left (7\,B+6\,C\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{4\,d} \]
(tan(c/2 + (d*x)/2)*((25*B*a^2)/4 + (13*C*a^2)/2) + tan(c/2 + (d*x)/2)^9*( (7*B*a^2)/4 + (3*C*a^2)/2) + tan(c/2 + (d*x)/2)^7*((49*B*a^2)/6 + 7*C*a^2) + tan(c/2 + (d*x)/2)^3*((79*B*a^2)/6 + 9*C*a^2) + tan(c/2 + (d*x)/2)^5*(( 40*B*a^2)/3 + (72*C*a^2)/5))/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d* x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d* x)/2)^10 + 1)) + (a^2*atan((a^2*tan(c/2 + (d*x)/2)*(7*B + 6*C))/(4*((7*B*a ^2)/4 + (3*C*a^2)/2)))*(7*B + 6*C))/(4*d) - (a^2*(7*B + 6*C)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(4*d)