Integrand size = 31, antiderivative size = 178 \[ \int \cos (c+d x) (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{4} a b (4 A+3 C) x+\frac {\left (5 a^2 (3 A+2 C)+2 b^2 (5 A+4 C)\right ) \sin (c+d x)}{15 d}+\frac {a b (4 A+3 C) \cos (c+d x) \sin (c+d x)}{4 d}+\frac {\left (2 a^2 C+b^2 (5 A+4 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{15 d}+\frac {a b C \cos ^3(c+d x) \sin (c+d x)}{10 d}+\frac {C \cos ^2(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{5 d} \]
1/4*a*b*(4*A+3*C)*x+1/15*(5*a^2*(3*A+2*C)+2*b^2*(5*A+4*C))*sin(d*x+c)/d+1/ 4*a*b*(4*A+3*C)*cos(d*x+c)*sin(d*x+c)/d+1/15*(2*a^2*C+b^2*(5*A+4*C))*cos(d *x+c)^2*sin(d*x+c)/d+1/10*a*b*C*cos(d*x+c)^3*sin(d*x+c)/d+1/5*C*cos(d*x+c) ^2*(a+b*cos(d*x+c))^2*sin(d*x+c)/d
Time = 1.66 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.71 \[ \int \cos (c+d x) (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {60 a b (4 A+3 C) (c+d x)+30 \left (b^2 (6 A+5 C)+a^2 (8 A+6 C)\right ) \sin (c+d x)+120 a b (A+C) \sin (2 (c+d x))+5 \left (4 A b^2+4 a^2 C+5 b^2 C\right ) \sin (3 (c+d x))+15 a b C \sin (4 (c+d x))+3 b^2 C \sin (5 (c+d x))}{240 d} \]
(60*a*b*(4*A + 3*C)*(c + d*x) + 30*(b^2*(6*A + 5*C) + a^2*(8*A + 6*C))*Sin [c + d*x] + 120*a*b*(A + C)*Sin[2*(c + d*x)] + 5*(4*A*b^2 + 4*a^2*C + 5*b^ 2*C)*Sin[3*(c + d*x)] + 15*a*b*C*Sin[4*(c + d*x)] + 3*b^2*C*Sin[5*(c + d*x )])/(240*d)
Time = 0.77 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 3529, 3042, 3512, 27, 3042, 3502, 3042, 3213}
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.
| \(\displaystyle \int \cos (c+d x) (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3529 |
| \(\displaystyle \frac {1}{5} \int \cos (c+d x) (a+b \cos (c+d x)) \left (2 a C \cos ^2(c+d x)+b (5 A+4 C) \cos (c+d x)+a (5 A+2 C)\right )dx+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (2 a C \sin \left (c+d x+\frac {\pi }{2}\right )^2+b (5 A+4 C) \sin \left (c+d x+\frac {\pi }{2}\right )+a (5 A+2 C)\right )dx+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3512 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int 2 \cos (c+d x) \left (2 (5 A+2 C) a^2+5 b (4 A+3 C) \cos (c+d x) a+2 \left (2 C a^2+b^2 (5 A+4 C)\right ) \cos ^2(c+d x)\right )dx+\frac {a b C \sin (c+d x) \cos ^3(c+d x)}{2 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \int \cos (c+d x) \left (2 (5 A+2 C) a^2+5 b (4 A+3 C) \cos (c+d x) a+2 \left (2 C a^2+b^2 (5 A+4 C)\right ) \cos ^2(c+d x)\right )dx+\frac {a b C \sin (c+d x) \cos ^3(c+d x)}{2 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (2 (5 A+2 C) a^2+5 b (4 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a+2 \left (2 C a^2+b^2 (5 A+4 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+\frac {a b C \sin (c+d x) \cos ^3(c+d x)}{2 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (\frac {1}{3} \int \cos (c+d x) \left (2 \left (5 (3 A+2 C) a^2+2 b^2 (5 A+4 C)\right )+15 a b (4 A+3 C) \cos (c+d x)\right )dx+\frac {2 \left (2 a^2 C+b^2 (5 A+4 C)\right ) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {a b C \sin (c+d x) \cos ^3(c+d x)}{2 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (\frac {1}{3} \int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (2 \left (5 (3 A+2 C) a^2+2 b^2 (5 A+4 C)\right )+15 a b (4 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 \left (2 a^2 C+b^2 (5 A+4 C)\right ) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {a b C \sin (c+d x) \cos ^3(c+d x)}{2 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3213 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{2} \left (\frac {2 \left (2 a^2 C+b^2 (5 A+4 C)\right ) \sin (c+d x) \cos ^2(c+d x)}{3 d}+\frac {1}{3} \left (\frac {2 \left (5 a^2 (3 A+2 C)+2 b^2 (5 A+4 C)\right ) \sin (c+d x)}{d}+\frac {15 a b (4 A+3 C) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {15}{2} a b x (4 A+3 C)\right )\right )+\frac {a b C \sin (c+d x) \cos ^3(c+d x)}{2 d}\right )+\frac {C \sin (c+d x) \cos ^2(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
(C*Cos[c + d*x]^2*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(5*d) + ((a*b*C*Cos [c + d*x]^3*Sin[c + d*x])/(2*d) + ((2*(2*a^2*C + b^2*(5*A + 4*C))*Cos[c + d*x]^2*Sin[c + d*x])/(3*d) + ((15*a*b*(4*A + 3*C)*x)/2 + (2*(5*a^2*(3*A + 2*C) + 2*b^2*(5*A + 4*C))*Sin[c + d*x])/d + (15*a*b*(4*A + 3*C)*Cos[c + d* x]*Sin[c + d*x])/(2*d))/3)/2)/5
3.6.32.3.1 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[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(2*a*c + b*d)*(x/2), x] + (-Simp[(b*c + a*d)*(Co s[e + f*x]/f), x] - Simp[b*d*Cos[e + f*x]*(Sin[e + f*x]/(2*f)), x]) /; Free Q[{a, b, c, d, e, f}, 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_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si n[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[(a + b*Si n[e + f*x])^m*Simp[a*C*d + A*b*c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e + f*x]^2 , x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && !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/(d*(m + n + 2)) Int[(a + b*Sin[e + f*x ])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*( n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + C* (a*d*m - b*c*(m + 1))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f , A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))
Time = 5.94 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.67
| method | result | size |
| parallelrisch | \(\frac {\left (\left (20 A +25 C \right ) b^{2}+20 a^{2} C \right ) \sin \left (3 d x +3 c \right )+120 b a \left (A +C \right ) \sin \left (2 d x +2 c \right )+15 C a b \sin \left (4 d x +4 c \right )+3 b^{2} C \sin \left (5 d x +5 c \right )+\left (\left (180 A +150 C \right ) b^{2}+240 a^{2} \left (A +\frac {3 C}{4}\right )\right ) \sin \left (d x +c \right )+240 x b \left (A +\frac {3 C}{4}\right ) d a}{240 d}\) | \(120\) |
| parts | \(\frac {\left (A \,b^{2}+a^{2} C \right ) \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {\sin \left (d x +c \right ) A \,a^{2}}{d}+\frac {b^{2} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5 d}+\frac {2 A a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 C a b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(154\) |
| derivativedivides | \(\frac {\frac {b^{2} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+2 C a b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 A a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,a^{2} \sin \left (d x +c \right )}{d}\) | \(158\) |
| default | \(\frac {\frac {b^{2} C \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+2 C a b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 A a b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,a^{2} \sin \left (d x +c \right )}{d}\) | \(158\) |
| risch | \(a b x A +\frac {3 a b C x}{4}+\frac {\sin \left (d x +c \right ) A \,a^{2}}{d}+\frac {3 \sin \left (d x +c \right ) A \,b^{2}}{4 d}+\frac {3 \sin \left (d x +c \right ) a^{2} C}{4 d}+\frac {5 b^{2} C \sin \left (d x +c \right )}{8 d}+\frac {b^{2} C \sin \left (5 d x +5 c \right )}{80 d}+\frac {C a b \sin \left (4 d x +4 c \right )}{16 d}+\frac {\sin \left (3 d x +3 c \right ) A \,b^{2}}{12 d}+\frac {\sin \left (3 d x +3 c \right ) a^{2} C}{12 d}+\frac {5 \sin \left (3 d x +3 c \right ) b^{2} C}{48 d}+\frac {\sin \left (2 d x +2 c \right ) A a b}{2 d}+\frac {\sin \left (2 d x +2 c \right ) C a b}{2 d}\) | \(195\) |
| norman | \(\frac {\left (A a b +\frac {3}{4} C a b \right ) x +\left (A a b +\frac {3}{4} C a b \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (5 A a b +\frac {15}{4} C a b \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (5 A a b +\frac {15}{4} C a b \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (10 A a b +\frac {15}{2} C a b \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (10 A a b +\frac {15}{2} C a b \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {4 \left (45 A \,a^{2}+25 A \,b^{2}+25 a^{2} C +29 b^{2} C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{15 d}+\frac {\left (4 A \,a^{2}-4 A a b +4 A \,b^{2}+4 a^{2} C -5 C a b +4 b^{2} C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d}+\frac {\left (4 A \,a^{2}+4 A a b +4 A \,b^{2}+4 a^{2} C +5 C a b +4 b^{2} C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {\left (24 A \,a^{2}-12 A a b +16 A \,b^{2}+16 a^{2} C -3 C a b +8 b^{2} C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {\left (24 A \,a^{2}+12 A a b +16 A \,b^{2}+16 a^{2} C +3 C a b +8 b^{2} C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}\) | \(392\) |
1/240*(((20*A+25*C)*b^2+20*a^2*C)*sin(3*d*x+3*c)+120*b*a*(A+C)*sin(2*d*x+2 *c)+15*C*a*b*sin(4*d*x+4*c)+3*b^2*C*sin(5*d*x+5*c)+((180*A+150*C)*b^2+240* a^2*(A+3/4*C))*sin(d*x+c)+240*x*b*(A+3/4*C)*d*a)/d
Time = 0.29 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.69 \[ \int \cos (c+d x) (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, A + 3 \, C\right )} a b d x + {\left (12 \, C b^{2} \cos \left (d x + c\right )^{4} + 30 \, C a b \cos \left (d x + c\right )^{3} + 15 \, {\left (4 \, A + 3 \, C\right )} a b \cos \left (d x + c\right ) + 20 \, {\left (3 \, A + 2 \, C\right )} a^{2} + 8 \, {\left (5 \, A + 4 \, C\right )} b^{2} + 4 \, {\left (5 \, C a^{2} + {\left (5 \, A + 4 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, d} \]
1/60*(15*(4*A + 3*C)*a*b*d*x + (12*C*b^2*cos(d*x + c)^4 + 30*C*a*b*cos(d*x + c)^3 + 15*(4*A + 3*C)*a*b*cos(d*x + c) + 20*(3*A + 2*C)*a^2 + 8*(5*A + 4*C)*b^2 + 4*(5*C*a^2 + (5*A + 4*C)*b^2)*cos(d*x + c)^2)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (167) = 334\).
Time = 0.26 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.97 \[ \int \cos (c+d x) (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx=\begin {cases} \frac {A a^{2} \sin {\left (c + d x \right )}}{d} + A a b x \sin ^{2}{\left (c + d x \right )} + A a b x \cos ^{2}{\left (c + d x \right )} + \frac {A a b \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {2 A b^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A b^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {2 C a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {C a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 C a b x \sin ^{4}{\left (c + d x \right )}}{4} + \frac {3 C a b x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 C a b x \cos ^{4}{\left (c + d x \right )}}{4} + \frac {3 C a b \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} + \frac {5 C a b \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{4 d} + \frac {8 C b^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {4 C b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {C b^{2} \sin {\left (c + d x \right )} \cos ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (A + C \cos ^{2}{\left (c \right )}\right ) \left (a + b \cos {\left (c \right )}\right )^{2} \cos {\left (c \right )} & \text {otherwise} \end {cases} \]
Piecewise((A*a**2*sin(c + d*x)/d + A*a*b*x*sin(c + d*x)**2 + A*a*b*x*cos(c + d*x)**2 + A*a*b*sin(c + d*x)*cos(c + d*x)/d + 2*A*b**2*sin(c + d*x)**3/ (3*d) + A*b**2*sin(c + d*x)*cos(c + d*x)**2/d + 2*C*a**2*sin(c + d*x)**3/( 3*d) + C*a**2*sin(c + d*x)*cos(c + d*x)**2/d + 3*C*a*b*x*sin(c + d*x)**4/4 + 3*C*a*b*x*sin(c + d*x)**2*cos(c + d*x)**2/2 + 3*C*a*b*x*cos(c + d*x)**4 /4 + 3*C*a*b*sin(c + d*x)**3*cos(c + d*x)/(4*d) + 5*C*a*b*sin(c + d*x)*cos (c + d*x)**3/(4*d) + 8*C*b**2*sin(c + d*x)**5/(15*d) + 4*C*b**2*sin(c + d* x)**3*cos(c + d*x)**2/(3*d) + C*b**2*sin(c + d*x)*cos(c + d*x)**4/d, Ne(d, 0)), (x*(A + C*cos(c)**2)*(a + b*cos(c))**2*cos(c), True))
Time = 0.21 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.87 \[ \int \cos (c+d x) (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {80 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a b - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a b + 80 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{2} - 16 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C b^{2} - 240 \, A a^{2} \sin \left (d x + c\right )}{240 \, d} \]
-1/240*(80*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^2 - 120*(2*d*x + 2*c + si n(2*d*x + 2*c))*A*a*b - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a*b + 80*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b^2 - 16*(3*sin(d* x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*b^2 - 240*A*a^2*sin(d*x + c))/d
Time = 0.31 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.80 \[ \int \cos (c+d x) (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {C b^{2} \sin \left (5 \, d x + 5 \, c\right )}{80 \, d} + \frac {C a b \sin \left (4 \, d x + 4 \, c\right )}{16 \, d} + \frac {1}{4} \, {\left (4 \, A a b + 3 \, C a b\right )} x + \frac {{\left (4 \, C a^{2} + 4 \, A b^{2} + 5 \, C b^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{48 \, d} + \frac {{\left (A a b + C a b\right )} \sin \left (2 \, d x + 2 \, c\right )}{2 \, d} + \frac {{\left (8 \, A a^{2} + 6 \, C a^{2} + 6 \, A b^{2} + 5 \, C b^{2}\right )} \sin \left (d x + c\right )}{8 \, d} \]
1/80*C*b^2*sin(5*d*x + 5*c)/d + 1/16*C*a*b*sin(4*d*x + 4*c)/d + 1/4*(4*A*a *b + 3*C*a*b)*x + 1/48*(4*C*a^2 + 4*A*b^2 + 5*C*b^2)*sin(3*d*x + 3*c)/d + 1/2*(A*a*b + C*a*b)*sin(2*d*x + 2*c)/d + 1/8*(8*A*a^2 + 6*C*a^2 + 6*A*b^2 + 5*C*b^2)*sin(d*x + c)/d
Time = 2.82 (sec) , antiderivative size = 371, normalized size of antiderivative = 2.08 \[ \int \cos (c+d x) (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {\left (2\,A\,a^2+2\,A\,b^2+2\,C\,a^2+2\,C\,b^2-2\,A\,a\,b-\frac {5\,C\,a\,b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (8\,A\,a^2+\frac {16\,A\,b^2}{3}+\frac {16\,C\,a^2}{3}+\frac {8\,C\,b^2}{3}-4\,A\,a\,b-C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (12\,A\,a^2+\frac {20\,A\,b^2}{3}+\frac {20\,C\,a^2}{3}+\frac {116\,C\,b^2}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (8\,A\,a^2+\frac {16\,A\,b^2}{3}+\frac {16\,C\,a^2}{3}+\frac {8\,C\,b^2}{3}+4\,A\,a\,b+C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^2+2\,A\,b^2+2\,C\,a^2+2\,C\,b^2+2\,A\,a\,b+\frac {5\,C\,a\,b}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,b\,\mathrm {atan}\left (\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (4\,A+3\,C\right )}{2\,\left (2\,A\,a\,b+\frac {3\,C\,a\,b}{2}\right )}\right )\,\left (4\,A+3\,C\right )}{2\,d}-\frac {a\,b\,\left (4\,A+3\,C\right )\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{2\,d} \]
(tan(c/2 + (d*x)/2)^9*(2*A*a^2 + 2*A*b^2 + 2*C*a^2 + 2*C*b^2 - 2*A*a*b - ( 5*C*a*b)/2) + tan(c/2 + (d*x)/2)^3*(8*A*a^2 + (16*A*b^2)/3 + (16*C*a^2)/3 + (8*C*b^2)/3 + 4*A*a*b + C*a*b) + tan(c/2 + (d*x)/2)^7*(8*A*a^2 + (16*A*b ^2)/3 + (16*C*a^2)/3 + (8*C*b^2)/3 - 4*A*a*b - C*a*b) + tan(c/2 + (d*x)/2) ^5*(12*A*a^2 + (20*A*b^2)/3 + (20*C*a^2)/3 + (116*C*b^2)/15) + tan(c/2 + ( d*x)/2)*(2*A*a^2 + 2*A*b^2 + 2*C*a^2 + 2*C*b^2 + 2*A*a*b + (5*C*a*b)/2))/( 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*b*atan((a* b*tan(c/2 + (d*x)/2)*(4*A + 3*C))/(2*(2*A*a*b + (3*C*a*b)/2)))*(4*A + 3*C) )/(2*d) - (a*b*(4*A + 3*C)*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(2*d)