Integrand size = 33, antiderivative size = 138 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {1}{8} a^2 (12 A+8 B+7 C) x+\frac {a^2 (12 A+8 B+7 C) \sin (c+d x)}{6 d}+\frac {a^2 (12 A+8 B+7 C) \cos (c+d x) \sin (c+d x)}{24 d}+\frac {(4 B-C) (a+a \cos (c+d x))^2 \sin (c+d x)}{12 d}+\frac {C (a+a \cos (c+d x))^3 \sin (c+d x)}{4 a d} \] Output:
1/8*a^2*(12*A+8*B+7*C)*x+1/6*a^2*(12*A+8*B+7*C)*sin(d*x+c)/d+1/24*a^2*(12* A+8*B+7*C)*cos(d*x+c)*sin(d*x+c)/d+1/12*(4*B-C)*(a+a*cos(d*x+c))^2*sin(d*x +c)/d+1/4*C*(a+a*cos(d*x+c))^3*sin(d*x+c)/a/d
Time = 0.72 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.93 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {a^2 \sin (c+d x) \left (6 (12 A+8 B+7 C) \arcsin \left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right )+\left (8 (6 A+5 B+4 C)+3 (4 A+8 B+7 C) \cos (c+d x)+8 (B+2 C) \cos ^2(c+d x)+6 C \cos ^3(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )}{24 d \sqrt {\sin ^2(c+d x)}} \] Input:
Integrate[(a + a*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x ]
Output:
(a^2*Sin[c + d*x]*(6*(12*A + 8*B + 7*C)*ArcSin[Sqrt[Sin[(c + d*x)/2]^2]] + (8*(6*A + 5*B + 4*C) + 3*(4*A + 8*B + 7*C)*Cos[c + d*x] + 8*(B + 2*C)*Cos [c + d*x]^2 + 6*C*Cos[c + d*x]^3)*Sqrt[Sin[c + d*x]^2]))/(24*d*Sqrt[Sin[c + d*x]^2])
Time = 0.49 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 3502, 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.
| \(\displaystyle \int (a \cos (c+d x)+a)^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^2 (a (4 A+3 C)+a (4 B-C) \cos (c+d x))dx}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (a (4 A+3 C)+a (4 B-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 a d}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\frac {1}{3} a (12 A+8 B+7 C) \int (\cos (c+d x) a+a)^2dx+\frac {a (4 B-C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{3} a (12 A+8 B+7 C) \int \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2dx+\frac {a (4 B-C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 a d}\) |
| \(\Big \downarrow \) 3123 |
| \(\displaystyle \frac {\frac {1}{3} a (12 A+8 B+7 C) \left (\frac {2 a^2 \sin (c+d x)}{d}+\frac {a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac {3 a^2 x}{2}\right )+\frac {a (4 B-C) \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}}{4 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^3}{4 a d}\) |
Input:
Int[(a + a*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]
Output:
(C*(a + a*Cos[c + d*x])^3*Sin[c + d*x])/(4*a*d) + ((a*(4*B - C)*(a + a*Cos [c + d*x])^2*Sin[c + d*x])/(3*d) + (a*(12*A + 8*B + 7*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)/(4*a)
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_.)*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]
Time = 12.56 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.59
| method | result | size |
| parallelrisch | \(\frac {3 a^{2} \left (\frac {\left (\frac {A}{2}+B +C \right ) \sin \left (2 d x +2 c \right )}{3}+\frac {\left (\frac {B}{2}+C \right ) \sin \left (3 d x +3 c \right )}{9}+\frac {\sin \left (4 d x +4 c \right ) C}{48}+\left (\frac {4 A}{3}+\frac {7 B}{6}+C \right ) \sin \left (d x +c \right )+d x \left (A +\frac {2 B}{3}+\frac {7 C}{12}\right )\right )}{2 d}\) | \(81\) |
| parts | \(a^{2} x A +\frac {\left (2 a^{2} A +B \,a^{2}\right ) \sin \left (d x +c \right )}{d}+\frac {\left (B \,a^{2}+2 a^{2} C \right ) \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3 d}+\frac {\left (a^{2} A +2 B \,a^{2}+a^{2} C \right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{2} C \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(147\) |
| risch | \(\frac {3 a^{2} x A}{2}+a^{2} x B +\frac {7 a^{2} C x}{8}+\frac {2 \sin \left (d x +c \right ) a^{2} A}{d}+\frac {7 \sin \left (d x +c \right ) B \,a^{2}}{4 d}+\frac {3 \sin \left (d x +c \right ) a^{2} C}{2 d}+\frac {\sin \left (4 d x +4 c \right ) a^{2} C}{32 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{2}}{12 d}+\frac {\sin \left (3 d x +3 c \right ) a^{2} C}{6 d}+\frac {\sin \left (2 d x +2 c \right ) a^{2} A}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{2}}{2 d}+\frac {\sin \left (2 d x +2 c \right ) a^{2} C}{2 d}\) | \(175\) |
| derivativedivides | \(\frac {a^{2} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {B \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{2} C \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a^{2} A \sin \left (d x +c \right )+2 B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 a^{2} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{2} A \left (d x +c \right )+B \,a^{2} \sin \left (d x +c \right )+a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(203\) |
| default | \(\frac {a^{2} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {B \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{2} C \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a^{2} A \sin \left (d x +c \right )+2 B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 a^{2} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{2} A \left (d x +c \right )+B \,a^{2} \sin \left (d x +c \right )+a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(203\) |
| norman | \(\frac {\frac {a^{2} \left (12 A +8 B +7 C \right ) x}{8}+\frac {11 a^{2} \left (12 A +8 B +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}+\frac {a^{2} \left (12 A +8 B +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {a^{2} \left (12 A +8 B +7 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {3 a^{2} \left (12 A +8 B +7 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4}+\frac {a^{2} \left (12 A +8 B +7 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}+\frac {a^{2} \left (12 A +8 B +7 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {a^{2} \left (20 A +24 B +25 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{2} \left (156 A +136 B +83 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\) | \(256\) |
| orering | \(\text {Expression too large to display}\) | \(4114\) |
Input:
int((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x,method=_RETURNVER BOSE)
Output:
3/2*a^2*(1/3*(1/2*A+B+C)*sin(2*d*x+2*c)+1/9*(1/2*B+C)*sin(3*d*x+3*c)+1/48* sin(4*d*x+4*c)*C+(4/3*A+7/6*B+C)*sin(d*x+c)+d*x*(A+2/3*B+7/12*C))/d
Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.72 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (12 \, A + 8 \, B + 7 \, C\right )} a^{2} d x + {\left (6 \, C a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (4 \, A + 8 \, B + 7 \, C\right )} a^{2} \cos \left (d x + c\right ) + 8 \, {\left (6 \, A + 5 \, B + 4 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{24 \, d} \] Input:
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm= "fricas")
Output:
1/24*(3*(12*A + 8*B + 7*C)*a^2*d*x + (6*C*a^2*cos(d*x + c)^3 + 8*(B + 2*C) *a^2*cos(d*x + c)^2 + 3*(4*A + 8*B + 7*C)*a^2*cos(d*x + c) + 8*(6*A + 5*B + 4*C)*a^2)*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 420 vs. \(2 (122) = 244\).
Time = 0.20 (sec) , antiderivative size = 420, normalized size of antiderivative = 3.04 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\begin {cases} \frac {A a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {A a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + A a^{2} x + \frac {A a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} + \frac {2 A a^{2} \sin {\left (c + d x \right )}}{d} + B a^{2} x \sin ^{2}{\left (c + d x \right )} + B a^{2} x \cos ^{2}{\left (c + d x \right )} + \frac {2 B a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {B a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {B a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{d} + \frac {B a^{2} \sin {\left (c + d x \right )}}{d} + \frac {3 C a^{2} x \sin ^{4}{\left (c + d x \right )}}{8} + \frac {3 C a^{2} x \sin ^{2}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {C a^{2} x \sin ^{2}{\left (c + d x \right )}}{2} + \frac {3 C a^{2} x \cos ^{4}{\left (c + d x \right )}}{8} + \frac {C a^{2} x \cos ^{2}{\left (c + d x \right )}}{2} + \frac {3 C a^{2} \sin ^{3}{\left (c + d x \right )} \cos {\left (c + d x \right )}}{8 d} + \frac {4 C a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {5 C a^{2} \sin {\left (c + d x \right )} \cos ^{3}{\left (c + d x \right )}}{8 d} + \frac {2 C a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} + \frac {C a^{2} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a \cos {\left (c \right )} + a\right )^{2} \left (A + B \cos {\left (c \right )} + C \cos ^{2}{\left (c \right )}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((a+a*cos(d*x+c))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2),x)
Output:
Piecewise((A*a**2*x*sin(c + d*x)**2/2 + A*a**2*x*cos(c + d*x)**2/2 + A*a** 2*x + A*a**2*sin(c + d*x)*cos(c + d*x)/(2*d) + 2*A*a**2*sin(c + d*x)/d + B *a**2*x*sin(c + d*x)**2 + B*a**2*x*cos(c + d*x)**2 + 2*B*a**2*sin(c + d*x) **3/(3*d) + B*a**2*sin(c + d*x)*cos(c + d*x)**2/d + B*a**2*sin(c + d*x)*co s(c + d*x)/d + B*a**2*sin(c + d*x)/d + 3*C*a**2*x*sin(c + d*x)**4/8 + 3*C* a**2*x*sin(c + d*x)**2*cos(c + d*x)**2/4 + C*a**2*x*sin(c + d*x)**2/2 + 3* C*a**2*x*cos(c + d*x)**4/8 + C*a**2*x*cos(c + d*x)**2/2 + 3*C*a**2*sin(c + d*x)**3*cos(c + d*x)/(8*d) + 4*C*a**2*sin(c + d*x)**3/(3*d) + 5*C*a**2*si n(c + d*x)*cos(c + d*x)**3/(8*d) + 2*C*a**2*sin(c + d*x)*cos(c + d*x)**2/d + C*a**2*sin(c + d*x)*cos(c + d*x)/(2*d), Ne(d, 0)), (x*(a*cos(c) + a)**2 *(A + B*cos(c) + C*cos(c)**2), True))
Time = 0.05 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.38 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 96 \, {\left (d x + c\right )} A a^{2} - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} + 48 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 64 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 192 \, A a^{2} \sin \left (d x + c\right ) + 96 \, B a^{2} \sin \left (d x + c\right )}{96 \, d} \] Input:
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm= "maxima")
Output:
1/96*(24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^2 + 96*(d*x + c)*A*a^2 - 32* (sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^2 + 48*(2*d*x + 2*c + sin(2*d*x + 2* c))*B*a^2 - 64*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^2 + 3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^2 + 24*(2*d*x + 2*c + sin(2*d *x + 2*c))*C*a^2 + 192*A*a^2*sin(d*x + c) + 96*B*a^2*sin(d*x + c))/d
Time = 0.13 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.93 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {C a^{2} \sin \left (4 \, d x + 4 \, c\right )}{32 \, d} + \frac {1}{8} \, {\left (12 \, A a^{2} + 8 \, B a^{2} + 7 \, C a^{2}\right )} x + \frac {{\left (B a^{2} + 2 \, C a^{2}\right )} \sin \left (3 \, d x + 3 \, c\right )}{12 \, d} + \frac {{\left (A a^{2} + 2 \, B a^{2} + 2 \, C a^{2}\right )} \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac {{\left (8 \, A a^{2} + 7 \, B a^{2} + 6 \, C a^{2}\right )} \sin \left (d x + c\right )}{4 \, d} \] Input:
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm= "giac")
Output:
1/32*C*a^2*sin(4*d*x + 4*c)/d + 1/8*(12*A*a^2 + 8*B*a^2 + 7*C*a^2)*x + 1/1 2*(B*a^2 + 2*C*a^2)*sin(3*d*x + 3*c)/d + 1/4*(A*a^2 + 2*B*a^2 + 2*C*a^2)*s in(2*d*x + 2*c)/d + 1/4*(8*A*a^2 + 7*B*a^2 + 6*C*a^2)*sin(d*x + c)/d
Time = 0.35 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.26 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3\,A\,a^2\,x}{2}+B\,a^2\,x+\frac {7\,C\,a^2\,x}{8}+\frac {2\,A\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {7\,B\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,C\,a^2\,\sin \left (c+d\,x\right )}{2\,d}+\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {B\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {C\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{6\,d}+\frac {C\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{32\,d} \] Input:
int((a + a*cos(c + d*x))^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2),x)
Output:
(3*A*a^2*x)/2 + B*a^2*x + (7*C*a^2*x)/8 + (2*A*a^2*sin(c + d*x))/d + (7*B* a^2*sin(c + d*x))/(4*d) + (3*C*a^2*sin(c + d*x))/(2*d) + (A*a^2*sin(2*c + 2*d*x))/(4*d) + (B*a^2*sin(2*c + 2*d*x))/(2*d) + (B*a^2*sin(3*c + 3*d*x))/ (12*d) + (C*a^2*sin(2*c + 2*d*x))/(2*d) + (C*a^2*sin(3*c + 3*d*x))/(6*d) + (C*a^2*sin(4*c + 4*d*x))/(32*d)
Time = 0.17 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.98 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {a^{2} \left (-6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} c +12 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +24 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b +27 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c -8 \sin \left (d x +c \right )^{3} b -16 \sin \left (d x +c \right )^{3} c +48 \sin \left (d x +c \right ) a +48 \sin \left (d x +c \right ) b +48 \sin \left (d x +c \right ) c +36 a d x +24 b d x +21 c d x \right )}{24 d} \] Input:
int((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x)
Output:
(a**2*( - 6*cos(c + d*x)*sin(c + d*x)**3*c + 12*cos(c + d*x)*sin(c + d*x)* a + 24*cos(c + d*x)*sin(c + d*x)*b + 27*cos(c + d*x)*sin(c + d*x)*c - 8*si n(c + d*x)**3*b - 16*sin(c + d*x)**3*c + 48*sin(c + d*x)*a + 48*sin(c + d* x)*b + 48*sin(c + d*x)*c + 36*a*d*x + 24*b*d*x + 21*c*d*x))/(24*d)