Integrand size = 31, antiderivative size = 163 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{4} a^2 (4 A+3 C) x+\frac {a^2 (4 A+3 C) \sin (c+d x)}{3 d}+\frac {a^2 (4 A+3 C) \cos (c+d x) \sin (c+d x)}{12 d}+\frac {(10 A+3 C) (a+a \cos (c+d x))^2 \sin (c+d x)}{30 d}+\frac {C \cos ^2(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d}+\frac {C (a+a \cos (c+d x))^3 \sin (c+d x)}{10 a d} \]
1/4*a^2*(4*A+3*C)*x+1/3*a^2*(4*A+3*C)*sin(d*x+c)/d+1/12*a^2*(4*A+3*C)*cos( d*x+c)*sin(d*x+c)/d+1/30*(10*A+3*C)*(a+a*cos(d*x+c))^2*sin(d*x+c)/d+1/5*C* cos(d*x+c)^2*(a+a*cos(d*x+c))^2*sin(d*x+c)/d+1/10*C*(a+a*cos(d*x+c))^3*sin (d*x+c)/a/d
Time = 0.20 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.60 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {a^2 (120 c C+240 A d x+180 C d x+30 (14 A+11 C) \sin (c+d x)+120 (A+C) \sin (2 (c+d x))+20 A \sin (3 (c+d x))+45 C \sin (3 (c+d x))+15 C \sin (4 (c+d x))+3 C \sin (5 (c+d x)))}{240 d} \]
(a^2*(120*c*C + 240*A*d*x + 180*C*d*x + 30*(14*A + 11*C)*Sin[c + d*x] + 12 0*(A + C)*Sin[2*(c + d*x)] + 20*A*Sin[3*(c + d*x)] + 45*C*Sin[3*(c + d*x)] + 15*C*Sin[4*(c + d*x)] + 3*C*Sin[5*(c + d*x)]))/(240*d)
Time = 0.78 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 3525, 3042, 3447, 3042, 3502, 27, 3042, 3230, 3042, 3123}
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.
| \(\displaystyle \int \cos (c+d x) (a \cos (c+d x)+a)^2 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3525 |
| \(\displaystyle \frac {\int \cos (c+d x) (\cos (c+d x) a+a)^2 (a (5 A+2 C)+2 a C \cos (c+d x))dx}{5 a}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a (5 A+2 C)+2 a C \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 a}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^2 \left (2 a C \cos ^2(c+d x)+a (5 A+2 C) \cos (c+d x)\right )dx}{5 a}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 a C \sin \left (c+d x+\frac {\pi }{2}\right )^2+a (5 A+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 a}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {\int 2 (\cos (c+d x) a+a)^2 \left (3 C a^2+(10 A+3 C) \cos (c+d x) a^2\right )dx}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{5 a}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int (\cos (c+d x) a+a)^2 \left (3 C a^2+(10 A+3 C) \cos (c+d x) a^2\right )dx}{2 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{5 a}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (3 C a^2+(10 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx}{2 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{5 a}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\frac {\frac {5}{3} a^2 (4 A+3 C) \int (\cos (c+d x) a+a)^2dx+\frac {(10 A+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{2 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{5 a}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\frac {5}{3} a^2 (4 A+3 C) \int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2dx+\frac {(10 A+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{2 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{5 a}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3123 |
| \(\displaystyle \frac {\frac {\frac {5}{3} a^2 (4 A+3 C) \left (\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3 a^2 x}{2}\right )+\frac {(10 A+3 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{3 d}}{2 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{2 d}}{5 a}+\frac {C \sin (c+d x) \cos ^2(c+d x) (a \cos (c+d x)+a)^2}{5 d}\) |
(C*Cos[c + d*x]^2*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(5*d) + ((C*(a + a* Cos[c + d*x])^3*Sin[c + d*x])/(2*d) + (((10*A + 3*C)*(a^2 + a^2*Cos[c + d* x])^2*Sin[c + d*x])/(3*d) + (5*a^2*(4*A + 3*C)*((3*a^2*x)/2 + (2*a^2*Sin[c + d*x])/d + (a^2*Cos[c + d*x]*Sin[c + d*x])/(2*d)))/3)/(2*a))/(5*a)
3.1.10.3.1 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_.)*sin[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(2*a^2 + b^ 2)*(x/2), x] + (-Simp[2*a*b*(Cos[c + d*x]/d), x] - Simp[b^2*Cos[c + d*x]*(S in[c + d*x]/(2*d)), x]) /; FreeQ[{a, b, c, d}, x]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
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_.)*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_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1 )/(d*f*(m + n + 2))), x] + Simp[1/(b*d*(m + n + 2)) Int[(a + b*Sin[e + f* x])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1 )) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && NeQ[m + n + 2, 0]
Time = 6.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.54
| method | result | size |
| parallelrisch | \(\frac {\left (\frac {\left (A +C \right ) \sin \left (2 d x +2 c \right )}{2}+\frac {\left (\frac {A}{3}+\frac {3 C}{4}\right ) \sin \left (3 d x +3 c \right )}{4}+\frac {\sin \left (4 d x +4 c \right ) C}{16}+\frac {\sin \left (5 d x +5 c \right ) C}{80}+\frac {\left (7 A +\frac {11 C}{2}\right ) \sin \left (d x +c \right )}{4}+d x \left (A +\frac {3 C}{4}\right )\right ) a^{2}}{d}\) | \(88\) |
| risch | \(a^{2} x A +\frac {3 a^{2} C x}{4}+\frac {7 \sin \left (d x +c \right ) A \,a^{2}}{4 d}+\frac {11 \sin \left (d x +c \right ) a^{2} C}{8 d}+\frac {a^{2} C \sin \left (5 d x +5 c \right )}{80 d}+\frac {\sin \left (4 d x +4 c \right ) a^{2} C}{16 d}+\frac {\sin \left (3 d x +3 c \right ) A \,a^{2}}{12 d}+\frac {3 \sin \left (3 d x +3 c \right ) a^{2} C}{16 d}+\frac {\sin \left (2 d x +2 c \right ) A \,a^{2}}{2 d}+\frac {\sin \left (2 d x +2 c \right ) a^{2} C}{2 d}\) | \(153\) |
| parts | \(\frac {\left (A \,a^{2}+a^{2} C \right ) \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {\sin \left (d x +c \right ) A \,a^{2}}{d}+\frac {a^{2} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5 d}+\frac {2 A \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{2} C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(156\) |
| derivativedivides | \(\frac {\frac {A \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a^{2} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+2 A \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a^{2} C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+A \,a^{2} \sin \left (d x +c \right )+\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(160\) |
| default | \(\frac {\frac {A \,a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a^{2} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+2 A \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a^{2} C \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+A \,a^{2} \sin \left (d x +c \right )+\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\) | \(160\) |
| norman | \(\frac {\frac {a^{2} \left (4 A +3 C \right ) x}{4}+\frac {7 a^{2} \left (4 A +3 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {a^{2} \left (4 A +3 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {5 a^{2} \left (4 A +3 C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {5 a^{2} \left (4 A +3 C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a^{2} \left (4 A +3 C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2}+\frac {5 a^{2} \left (4 A +3 C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {a^{2} \left (4 A +3 C \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}+\frac {a^{2} \left (12 A +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {8 a^{2} \left (35 A +27 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {a^{2} \left (52 A +27 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\) | \(279\) |
(1/2*(A+C)*sin(2*d*x+2*c)+1/4*(1/3*A+3/4*C)*sin(3*d*x+3*c)+1/16*sin(4*d*x+ 4*c)*C+1/80*sin(5*d*x+5*c)*C+1/4*(7*A+11/2*C)*sin(d*x+c)+d*x*(A+3/4*C))*a^ 2/d
Time = 0.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.65 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, A + 3 \, C\right )} a^{2} d x + {\left (12 \, C a^{2} \cos \left (d x + c\right )^{4} + 30 \, C a^{2} \cos \left (d x + c\right )^{3} + 4 \, {\left (5 \, A + 9 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (4 \, A + 3 \, C\right )} a^{2} \cos \left (d x + c\right ) + 4 \, {\left (25 \, A + 18 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{60 \, d} \]
1/60*(15*(4*A + 3*C)*a^2*d*x + (12*C*a^2*cos(d*x + c)^4 + 30*C*a^2*cos(d*x + c)^3 + 4*(5*A + 9*C)*a^2*cos(d*x + c)^2 + 15*(4*A + 3*C)*a^2*cos(d*x + c) + 4*(25*A + 18*C)*a^2)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (144) = 288\).
Time = 0.27 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.15 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx=\begin {cases} A a^{2} x \sin ^{2}{\left (c + d x \right )} + A a^{2} x \cos ^{2}{\left (c + d x \right )} + \frac {2 A a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {A a^{2} \sin {\left (c + d x \right )}}{d} + \frac {3 C a^{2} x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {3 C a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 C a^{2} x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {8 C a^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C a^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {3 C a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {2 C a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C a^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} + \frac {5 C a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {C a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\left (c \right )}\right ) \left (a \cos {\left (c \right )} + a\right )^{2} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((A*a**2*x*sin(c + d*x)**2 + A*a**2*x*cos(c + d*x)**2 + 2*A*a**2* sin(c + d*x)**3/(3*d) + A*a**2*sin(c + d*x)*cos(c + d*x)**2/d + A*a**2*sin (c + d*x)*cos(c + d*x)/d + A*a**2*sin(c + d*x)/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*sin(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*(A + C*cos(c)**2)*(a*cos(c) + a)**2*cos(c), True))
Time = 0.25 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.96 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {80 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 16 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a^{2} + 80 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} - 240 \, A a^{2} \sin \left (d x + c\right )}{240 \, d} \]
-1/240*(80*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^2 - 120*(2*d*x + 2*c + si n(2*d*x + 2*c))*A*a^2 - 16*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin( d*x + c))*C*a^2 + 80*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^2 - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^2 - 240*A*a^2*sin(d*x + c))/d
Time = 0.27 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.79 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {C a^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {C a^{2} \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} + \frac {1}{4} \, {\left (4 \, A a^{2} + 3 \, C a^{2}\right )} x + \frac {{\left (4 \, A a^{2} + 9 \, C a^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (A a^{2} + C a^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac {{\left (14 \, A a^{2} + 11 \, C a^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \]
1/80*C*a^2*sin(5*d*x + 5*c)/d + 1/16*C*a^2*sin(4*d*x + 4*c)/d + 1/4*(4*A*a ^2 + 3*C*a^2)*x + 1/48*(4*A*a^2 + 9*C*a^2)*sin(3*d*x + 3*c)/d + 1/2*(A*a^2 + C*a^2)*sin(2*d*x + 2*c)/d + 1/8*(14*A*a^2 + 11*C*a^2)*sin(d*x + c)/d
Time = 2.35 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.70 \[ \int \cos (c+d x) (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\left (2\,A\,a^2+\frac {3\,C\,a^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {28\,A\,a^2}{3}+7\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {56\,A\,a^2}{3}+\frac {72\,C\,a^2}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {52\,A\,a^2}{3}+9\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (6\,A\,a^2+\frac {13\,C\,a^2}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {a^2\,\left (4\,A+3\,C\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{2\,d}+\frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,A+3\,C\right )}{2\,\left (2\,A\,a^2+\frac {3\,C\,a^2}{2}\right )}\right )\,\left (4\,A+3\,C\right )}{2\,d} \]
(tan(c/2 + (d*x)/2)*(6*A*a^2 + (13*C*a^2)/2) + tan(c/2 + (d*x)/2)^9*(2*A*a ^2 + (3*C*a^2)/2) + tan(c/2 + (d*x)/2)^7*((28*A*a^2)/3 + 7*C*a^2) + tan(c/ 2 + (d*x)/2)^3*((52*A*a^2)/3 + 9*C*a^2) + tan(c/2 + (d*x)/2)^5*((56*A*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*(4*A + 3*C)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(2*d) + (a^ 2*atan((a^2*tan(c/2 + (d*x)/2)*(4*A + 3*C))/(2*(2*A*a^2 + (3*C*a^2)/2)))*( 4*A + 3*C))/(2*d)