Integrand size = 48, antiderivative size = 176 \[ \int (a+b \cos (c+d x))^2 \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx=\frac {1}{8} \left (8 a^3 b B+12 a b^3 B-8 a^4 C+3 b^4 C\right ) x+\frac {b \left (16 a^2 b B+4 b^3 B-13 a^3 C+8 a b^2 C\right ) \sin (c+d x)}{6 d}+\frac {b^2 \left (20 a b B-14 a^2 C+9 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {b (4 b B-a C) (a+b \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac {b C (a+b \cos (c+d x))^3 \sin (c+d x)}{4 d} \]
1/8*(8*B*a^3*b+12*B*a*b^3-8*C*a^4+3*C*b^4)*x+1/6*b*(16*B*a^2*b+4*B*b^3-13* C*a^3+8*C*a*b^2)*sin(d*x+c)/d+1/24*b^2*(20*B*a*b-14*C*a^2+9*C*b^2)*cos(d*x +c)*sin(d*x+c)/d+1/12*b*(4*B*b-C*a)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d+1/4*b* C*(a+b*cos(d*x+c))^3*sin(d*x+c)/d
Time = 1.81 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.76 \[ \int (a+b \cos (c+d x))^2 \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx=\frac {-12 \left (-8 a^3 b B-12 a b^3 B+8 a^4 C-3 b^4 C\right ) (c+d x)+24 b \left (12 a^2 b B+3 b^3 B-8 a^3 C+6 a b^2 C\right ) \sin (c+d x)+24 b^3 (3 a B+b C) \sin (2 (c+d x))+8 b^3 (b B+2 a C) \sin (3 (c+d x))+3 b^4 C \sin (4 (c+d x))}{96 d} \]
Integrate[(a + b*Cos[c + d*x])^2*(a*b*B - a^2*C + b^2*B*Cos[c + d*x] + b^2 *C*Cos[c + d*x]^2),x]
(-12*(-8*a^3*b*B - 12*a*b^3*B + 8*a^4*C - 3*b^4*C)*(c + d*x) + 24*b*(12*a^ 2*b*B + 3*b^3*B - 8*a^3*C + 6*a*b^2*C)*Sin[c + d*x] + 24*b^3*(3*a*B + b*C) *Sin[2*(c + d*x)] + 8*b^3*(b*B + 2*a*C)*Sin[3*(c + d*x)] + 3*b^4*C*Sin[4*( c + d*x)])/(96*d)
Time = 0.74 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3042, 3494, 3042, 3232, 3042, 3232, 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.
| \(\displaystyle \int (a+b \cos (c+d x))^2 \left (a^2 (-C)+a b B+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (a^2 (-C)+a b B+b^2 B \sin \left (c+d x+\frac {\pi }{2}\right )+b^2 C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3494 |
| \(\displaystyle \frac {\int (a+b \cos (c+d x))^3 \left (C \cos (c+d x) b^3+(b B-a C) b^2\right )dx}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (C \sin \left (c+d x+\frac {\pi }{2}\right ) b^3+(b B-a C) b^2\right )dx}{b^2}\) |
| \(\Big \downarrow \) 3232 |
| \(\displaystyle \frac {\frac {1}{4} \int (a+b \cos (c+d x))^2 \left ((4 b B-a C) \cos (c+d x) b^3+\left (3 C b^2+4 a (b B-a C)\right ) b^2\right )dx+\frac {b^3 C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left ((4 b B-a C) \sin \left (c+d x+\frac {\pi }{2}\right ) b^3+\left (3 C b^2+4 a (b B-a C)\right ) b^2\right )dx+\frac {b^3 C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}}{b^2}\) |
| \(\Big \downarrow \) 3232 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int (a+b \cos (c+d x)) \left (\left (-14 C a^2+20 b B a+9 b^2 C\right ) \cos (c+d x) b^3+\left (-12 C a^3+12 b B a^2+7 b^2 C a+8 b^3 B\right ) b^2\right )dx+\frac {b^3 (4 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\right )+\frac {b^3 C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}}{b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (\left (-14 C a^2+20 b B a+9 b^2 C\right ) \sin \left (c+d x+\frac {\pi }{2}\right ) b^3+\left (-12 C a^3+12 b B a^2+7 b^2 C a+8 b^3 B\right ) b^2\right )dx+\frac {b^3 (4 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\right )+\frac {b^3 C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}}{b^2}\) |
| \(\Big \downarrow \) 3213 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {b^4 \left (-14 a^2 C+20 a b B+9 b^2 C\right ) \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3}{2} b^2 x \left (-8 a^4 C+8 a^3 b B+12 a b^3 B+3 b^4 C\right )+\frac {2 b^3 \left (-13 a^3 C+16 a^2 b B+8 a b^2 C+4 b^3 B\right ) \sin (c+d x)}{d}\right )+\frac {b^3 (4 b B-a C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}\right )+\frac {b^3 C \sin (c+d x) (a+b \cos (c+d x))^3}{4 d}}{b^2}\) |
((b^3*C*(a + b*Cos[c + d*x])^3*Sin[c + d*x])/(4*d) + ((b^3*(4*b*B - a*C)*( a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d) + ((3*b^2*(8*a^3*b*B + 12*a*b^3* B - 8*a^4*C + 3*b^4*C)*x)/2 + (2*b^3*(16*a^2*b*B + 4*b^3*B - 13*a^3*C + 8* a*b^2*C)*Sin[c + d*x])/d + (b^4*(20*a*b*B - 14*a^2*C + 9*b^2*C)*Cos[c + d* x]*Sin[c + d*x])/(2*d))/3)/4)/b^2
3.10.75.3.1 Defintions of rubi rules used
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_)*((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[1/(m + 1) Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[b* d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ [{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((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)*Simp[b*B - a*C + b*C*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 0.27 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.78
| method | result | size |
| parallelrisch | \(\frac {24 \left (3 B a \,b^{3}+C \,b^{4}\right ) \sin \left (2 d x +2 c \right )+8 \left (B \,b^{4}+2 C a \,b^{3}\right ) \sin \left (3 d x +3 c \right )+3 C \sin \left (4 d x +4 c \right ) b^{4}+24 \left (12 B \,a^{2} b^{2}+3 B \,b^{4}-8 a^{3} b C +6 C a \,b^{3}\right ) \sin \left (d x +c \right )+96 \left (B \,a^{3} b +\frac {3}{2} B a \,b^{3}-a^{4} C +\frac {3}{8} C \,b^{4}\right ) x d}{96 d}\) | \(138\) |
| parts | \(a^{3} \left (B b -C a \right ) x +\frac {\left (B \,b^{4}+2 C a \,b^{3}\right ) \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {\left (3 B \,a^{2} b^{2}-2 a^{3} b C \right ) \sin \left (d x +c \right )}{d}+\frac {C \,b^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {3 B a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(149\) |
| derivativedivides | \(\frac {C \,b^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,b^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {2 C a \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 B a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 B \sin \left (d x +c \right ) a^{2} b^{2}-2 C \sin \left (d x +c \right ) a^{3} b +B \,a^{3} b \left (d x +c \right )-a^{4} C \left (d x +c \right )}{d}\) | \(168\) |
| default | \(\frac {C \,b^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,b^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {2 C a \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+3 B a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 B \sin \left (d x +c \right ) a^{2} b^{2}-2 C \sin \left (d x +c \right ) a^{3} b +B \,a^{3} b \left (d x +c \right )-a^{4} C \left (d x +c \right )}{d}\) | \(168\) |
| risch | \(x B \,a^{3} b +\frac {3 x B a \,b^{3}}{2}-a^{4} C x +\frac {3 b^{4} C x}{8}+\frac {3 \sin \left (d x +c \right ) B \,a^{2} b^{2}}{d}+\frac {3 \sin \left (d x +c \right ) B \,b^{4}}{4 d}-\frac {2 \sin \left (d x +c \right ) a^{3} b C}{d}+\frac {3 \sin \left (d x +c \right ) C a \,b^{3}}{2 d}+\frac {C \,b^{4} \sin \left (4 d x +4 c \right )}{32 d}+\frac {\sin \left (3 d x +3 c \right ) B \,b^{4}}{12 d}+\frac {\sin \left (3 d x +3 c \right ) C a \,b^{3}}{6 d}+\frac {3 \sin \left (2 d x +2 c \right ) B a \,b^{3}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C \,b^{4}}{4 d}\) | \(188\) |
| norman | \(\frac {\left (B \,a^{3} b +\frac {3}{2} B a \,b^{3}-a^{4} C +\frac {3}{8} C \,b^{4}\right ) x +\left (B \,a^{3} b +\frac {3}{2} B a \,b^{3}-a^{4} C +\frac {3}{8} C \,b^{4}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 B \,a^{3} b +6 B a \,b^{3}-4 a^{4} C +\frac {3}{2} C \,b^{4}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (4 B \,a^{3} b +6 B a \,b^{3}-4 a^{4} C +\frac {3}{2} C \,b^{4}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (6 B \,a^{3} b +9 B a \,b^{3}-6 a^{4} C +\frac {9}{4} C \,b^{4}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {b \left (24 B \,a^{2} b -12 B a \,b^{2}+8 B \,b^{3}-16 a^{3} C +16 C a \,b^{2}-5 C \,b^{3}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {b \left (24 B \,a^{2} b +12 B a \,b^{2}+8 B \,b^{3}-16 a^{3} C +16 C a \,b^{2}+5 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {b \left (216 B \,a^{2} b -36 B a \,b^{2}+40 B \,b^{3}-144 a^{3} C +80 C a \,b^{2}+9 C \,b^{3}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {b \left (216 B \,a^{2} b +36 B a \,b^{2}+40 B \,b^{3}-144 a^{3} C +80 C a \,b^{2}-9 C \,b^{3}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}\) | \(431\) |
int((a+b*cos(d*x+c))^2*(B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2),x ,method=_RETURNVERBOSE)
1/96*(24*(3*B*a*b^3+C*b^4)*sin(2*d*x+2*c)+8*(B*b^4+2*C*a*b^3)*sin(3*d*x+3* c)+3*C*sin(4*d*x+4*c)*b^4+24*(12*B*a^2*b^2+3*B*b^4-8*C*a^3*b+6*C*a*b^3)*si n(d*x+c)+96*(B*a^3*b+3/2*B*a*b^3-a^4*C+3/8*C*b^4)*x*d)/d
Time = 0.28 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.76 \[ \int (a+b \cos (c+d x))^2 \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx=-\frac {3 \, {\left (8 \, C a^{4} - 8 \, B a^{3} b - 12 \, B a b^{3} - 3 \, C b^{4}\right )} d x - {\left (6 \, C b^{4} \cos \left (d x + c\right )^{3} - 48 \, C a^{3} b + 72 \, B a^{2} b^{2} + 32 \, C a b^{3} + 16 \, B b^{4} + 8 \, {\left (2 \, C a b^{3} + B b^{4}\right )} \cos \left (d x + c\right )^{2} + 9 \, {\left (4 \, B a b^{3} + C b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]
integrate((a+b*cos(d*x+c))^2*(B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c )^2),x, algorithm="fricas")
-1/24*(3*(8*C*a^4 - 8*B*a^3*b - 12*B*a*b^3 - 3*C*b^4)*d*x - (6*C*b^4*cos(d *x + c)^3 - 48*C*a^3*b + 72*B*a^2*b^2 + 32*C*a*b^3 + 16*B*b^4 + 8*(2*C*a*b ^3 + B*b^4)*cos(d*x + c)^2 + 9*(4*B*a*b^3 + C*b^4)*cos(d*x + c))*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 357 vs. \(2 (172) = 344\).
Time = 0.20 (sec) , antiderivative size = 357, normalized size of antiderivative = 2.03 \[ \int (a+b \cos (c+d x))^2 \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx=\begin {cases} B a^{3} b x + \frac {3 B a^{2} b^{2} \sin {\left (c + d x \right )}}{d} + \frac {3 B a b^{3} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a b^{3} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 B a b^{3} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 B b^{4} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B b^{4} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - C a^{4} x - \frac {2 C a^{3} b \sin {\left (c + d x \right )}}{d} + \frac {4 C a b^{3} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {2 C a b^{3} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {3 C b^{4} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C b^{4} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {3 C b^{4} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {3 C b^{4} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {5 C b^{4} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} & \text {for}\: d \neq 0 \\x \left (a + b \cos {\left (c \right )}\right )^{2} \left (B a b + B b^{2} \cos {\left (c \right )} - C a^{2} + C b^{2} \cos ^{2}{\left (c \right )}\right ) & \text {otherwise} \end {cases} \]
Piecewise((B*a**3*b*x + 3*B*a**2*b**2*sin(c + d*x)/d + 3*B*a*b**3*x*sin(c + d*x)**2/2 + 3*B*a*b**3*x*cos(c + d*x)**2/2 + 3*B*a*b**3*sin(c + d*x)*cos (c + d*x)/(2*d) + 2*B*b**4*sin(c + d*x)**3/(3*d) + B*b**4*sin(c + d*x)*cos (c + d*x)**2/d - C*a**4*x - 2*C*a**3*b*sin(c + d*x)/d + 4*C*a*b**3*sin(c + d*x)**3/(3*d) + 2*C*a*b**3*sin(c + d*x)*cos(c + d*x)**2/d + 3*C*b**4*x*si n(c + d*x)**4/8 + 3*C*b**4*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + 3*C*b**4* x*cos(c + d*x)**4/8 + 3*C*b**4*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 5*C*b* *4*sin(c + d*x)*cos(c + d*x)**3/(8*d), Ne(d, 0)), (x*(a + b*cos(c))**2*(B* a*b + B*b**2*cos(c) - C*a**2 + C*b**2*cos(c)**2), True))
Time = 0.20 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.92 \[ \int (a+b \cos (c+d x))^2 \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx=-\frac {96 \, {\left (d x + c\right )} C a^{4} - 96 \, {\left (d x + c\right )} B a^{3} b - 72 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{3} + 64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a b^{3} + 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b^{4} - 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{4} + 192 \, C a^{3} b \sin \left (d x + c\right ) - 288 \, B a^{2} b^{2} \sin \left (d x + c\right )}{96 \, d} \]
integrate((a+b*cos(d*x+c))^2*(B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c )^2),x, algorithm="maxima")
-1/96*(96*(d*x + c)*C*a^4 - 96*(d*x + c)*B*a^3*b - 72*(2*d*x + 2*c + sin(2 *d*x + 2*c))*B*a*b^3 + 64*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a*b^3 + 32*( sin(d*x + c)^3 - 3*sin(d*x + c))*B*b^4 - 3*(12*d*x + 12*c + sin(4*d*x + 4* c) + 8*sin(2*d*x + 2*c))*C*b^4 + 192*C*a^3*b*sin(d*x + c) - 288*B*a^2*b^2* sin(d*x + c))/d
Time = 0.32 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.82 \[ \int (a+b \cos (c+d x))^2 \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx=\frac {C b^{4} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} - \frac {1}{8} \, {\left (8 \, C a^{4} - 8 \, B a^{3} b - 12 \, B a b^{3} - 3 \, C b^{4}\right )} x + \frac {{\left (2 \, C a b^{3} + B b^{4}\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (3 \, B a b^{3} + C b^{4}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} - \frac {{\left (8 \, C a^{3} b - 12 \, B a^{2} b^{2} - 6 \, C a b^{3} - 3 \, B b^{4}\right )} \sin \left (d x + c\right )}{4 \, d} \]
integrate((a+b*cos(d*x+c))^2*(B*a*b-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c )^2),x, algorithm="giac")
1/32*C*b^4*sin(4*d*x + 4*c)/d - 1/8*(8*C*a^4 - 8*B*a^3*b - 12*B*a*b^3 - 3* C*b^4)*x + 1/12*(2*C*a*b^3 + B*b^4)*sin(3*d*x + 3*c)/d + 1/4*(3*B*a*b^3 + C*b^4)*sin(2*d*x + 2*c)/d - 1/4*(8*C*a^3*b - 12*B*a^2*b^2 - 6*C*a*b^3 - 3* B*b^4)*sin(d*x + c)/d
Time = 2.19 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.06 \[ \int (a+b \cos (c+d x))^2 \left (a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)\right ) \, dx=\frac {3\,C\,b^4\,x}{8}-C\,a^4\,x+\frac {3\,B\,a\,b^3\,x}{2}+B\,a^3\,b\,x+\frac {3\,B\,b^4\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,b^4\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,b^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,b^4\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,B\,a\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {3\,B\,a^2\,b^2\,\sin \left (c+d\,x\right )}{d}+\frac {C\,a\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{6\,d}+\frac {3\,C\,a\,b^3\,\sin \left (c+d\,x\right )}{2\,d}-\frac {2\,C\,a^3\,b\,\sin \left (c+d\,x\right )}{d} \]
(3*C*b^4*x)/8 - C*a^4*x + (3*B*a*b^3*x)/2 + B*a^3*b*x + (3*B*b^4*sin(c + d *x))/(4*d) + (B*b^4*sin(3*c + 3*d*x))/(12*d) + (C*b^4*sin(2*c + 2*d*x))/(4 *d) + (C*b^4*sin(4*c + 4*d*x))/(32*d) + (3*B*a*b^3*sin(2*c + 2*d*x))/(4*d) + (3*B*a^2*b^2*sin(c + d*x))/d + (C*a*b^3*sin(3*c + 3*d*x))/(6*d) + (3*C* a*b^3*sin(c + d*x))/(2*d) - (2*C*a^3*b*sin(c + d*x))/d