Integrand size = 41, antiderivative size = 232 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (20 a b B+5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac {\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a (2 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d} \] Output:
1/8*(6*A*a*b+3*B*a^2+4*B*b^2+8*C*a*b)*arctanh(sin(d*x+c))/d+1/15*(20*B*a*b +5*b^2*(2*A+3*C)+2*a^2*(4*A+5*C))*tan(d*x+c)/d+1/8*(6*A*a*b+3*B*a^2+4*B*b^ 2+8*C*a*b)*sec(d*x+c)*tan(d*x+c)/d+1/15*(2*A*b^2+10*B*a*b+a^2*(4*A+5*C))*s ec(d*x+c)^2*tan(d*x+c)/d+1/20*a*(2*A*b+5*B*a)*sec(d*x+c)^3*tan(d*x+c)/d+1/ 5*A*(a+b*cos(d*x+c))^2*sec(d*x+c)^4*tan(d*x+c)/d
Time = 2.71 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.72 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (15 \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \sec (c+d x)+30 a (2 A b+a B) \sec ^3(c+d x)+8 \left (15 \left (2 a b B+a^2 (A+C)+b^2 (A+C)\right )+5 \left (A b^2+2 a b B+a^2 (2 A+C)\right ) \tan ^2(c+d x)+3 a^2 A \tan ^4(c+d x)\right )\right )}{120 d} \] Input:
Integrate[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*S ec[c + d*x]^6,x]
Output:
(15*(6*a*A*b + 3*a^2*B + 4*b^2*B + 8*a*b*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(6*a*A*b + 3*a^2*B + 4*b^2*B + 8*a*b*C)*Sec[c + d*x] + 30*a*(2* A*b + a*B)*Sec[c + d*x]^3 + 8*(15*(2*a*b*B + a^2*(A + C) + b^2*(A + C)) + 5*(A*b^2 + 2*a*b*B + a^2*(2*A + C))*Tan[c + d*x]^2 + 3*a^2*A*Tan[c + d*x]^ 4)))/(120*d)
Time = 1.41 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.97, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.366, Rules used = {3042, 3526, 3042, 3510, 25, 3042, 3500, 3042, 3227, 3042, 4254, 24, 4255, 3042, 4257}
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.
| \(\displaystyle \int \sec ^6(c+d x) (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 3526 |
| \(\displaystyle \frac {1}{5} \int (a+b \cos (c+d x)) \left (b (2 A+5 C) \cos ^2(c+d x)+(4 a A+5 b B+5 a C) \cos (c+d x)+2 A b+5 a B\right ) \sec ^5(c+d x)dx+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b (2 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+(4 a A+5 b B+5 a C) \sin \left (c+d x+\frac {\pi }{2}\right )+2 A b+5 a B\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {1}{5} \left (\frac {a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}-\frac {1}{4} \int -\left (\left (4 b^2 (2 A+5 C) \cos ^2(c+d x)+5 \left (3 B a^2+6 A b a+8 b C a+4 b^2 B\right ) \cos (c+d x)+4 \left ((4 A+5 C) a^2+10 b B a+2 A b^2\right )\right ) \sec ^4(c+d x)\right )dx\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \left (4 b^2 (2 A+5 C) \cos ^2(c+d x)+5 \left (3 B a^2+6 A b a+8 b C a+4 b^2 B\right ) \cos (c+d x)+4 \left ((4 A+5 C) a^2+10 b B a+2 A b^2\right )\right ) \sec ^4(c+d x)dx+\frac {a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \frac {4 b^2 (2 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2+5 \left (3 B a^2+6 A b a+8 b C a+4 b^2 B\right ) \sin \left (c+d x+\frac {\pi }{2}\right )+4 \left ((4 A+5 C) a^2+10 b B a+2 A b^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \int \left (15 \left (3 B a^2+6 A b a+8 b C a+4 b^2 B\right )+4 \left (2 (4 A+5 C) a^2+20 b B a+5 b^2 (2 A+3 C)\right ) \cos (c+d x)\right ) \sec ^3(c+d x)dx+\frac {4 \tan (c+d x) \sec ^2(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{3 d}\right )+\frac {a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \int \frac {15 \left (3 B a^2+6 A b a+8 b C a+4 b^2 B\right )+4 \left (2 (4 A+5 C) a^2+20 b B a+5 b^2 (2 A+3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {4 \tan (c+d x) \sec ^2(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{3 d}\right )+\frac {a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (15 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \int \sec ^3(c+d x)dx+4 \left (2 a^2 (4 A+5 C)+20 a b B+5 b^2 (2 A+3 C)\right ) \int \sec ^2(c+d x)dx\right )+\frac {4 \tan (c+d x) \sec ^2(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{3 d}\right )+\frac {a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (4 \left (2 a^2 (4 A+5 C)+20 a b B+5 b^2 (2 A+3 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+15 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {4 \tan (c+d x) \sec ^2(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{3 d}\right )+\frac {a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (15 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-\frac {4 \left (2 a^2 (4 A+5 C)+20 a b B+5 b^2 (2 A+3 C)\right ) \int 1d(-\tan (c+d x))}{d}\right )+\frac {4 \tan (c+d x) \sec ^2(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{3 d}\right )+\frac {a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (15 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {4 \tan (c+d x) \left (2 a^2 (4 A+5 C)+20 a b B+5 b^2 (2 A+3 C)\right )}{d}\right )+\frac {4 \tan (c+d x) \sec ^2(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{3 d}\right )+\frac {a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (15 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {4 \tan (c+d x) \left (2 a^2 (4 A+5 C)+20 a b B+5 b^2 (2 A+3 C)\right )}{d}\right )+\frac {4 \tan (c+d x) \sec ^2(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{3 d}\right )+\frac {a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (15 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {4 \tan (c+d x) \left (2 a^2 (4 A+5 C)+20 a b B+5 b^2 (2 A+3 C)\right )}{d}\right )+\frac {4 \tan (c+d x) \sec ^2(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{3 d}\right )+\frac {a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\frac {1}{3} \left (15 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {4 \tan (c+d x) \left (2 a^2 (4 A+5 C)+20 a b B+5 b^2 (2 A+3 C)\right )}{d}\right )+\frac {4 \tan (c+d x) \sec ^2(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )}{3 d}\right )+\frac {a (5 a B+2 A b) \tan (c+d x) \sec ^3(c+d x)}{4 d}\right )+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d}\) |
Input:
Int[(a + b*Cos[c + d*x])^2*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]
Output:
(A*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + ((a*(2*A*b + 5*a*B)*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((4*(2*A*b^2 + 10*a*b*B + a^ 2*(4*A + 5*C))*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((4*(20*a*b*B + 5*b^2* (2*A + 3*C) + 2*a^2*(4*A + 5*C))*Tan[c + d*x])/d + 15*(6*a*A*b + 3*a^2*B + 4*b^2*B + 8*a*b*C)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d *x])/(2*d)))/3)/4)/5
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_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
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[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S imp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) ))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F reeQ[{a, b, c, d, e, f, A, B, C}, 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_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c*C - B* d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x ] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f *x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d , 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.12 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.85
| method | result | size |
| parts | \(-\frac {A \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (2 a A b +B \,a^{2}\right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {\left (B \,b^{2}+2 a b C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (A \,b^{2}+2 B a b +a^{2} C \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {C \,b^{2} \tan \left (d x +c \right )}{d}\) | \(197\) |
| derivativedivides | \(\frac {-A \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+B \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{2} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 a A b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-2 B a b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 a b C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-A \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \tan \left (d x +c \right ) b^{2}}{d}\) | \(292\) |
| default | \(\frac {-A \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+B \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{2} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 a A b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-2 B a b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 a b C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-A \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \tan \left (d x +c \right ) b^{2}}{d}\) | \(292\) |
| parallelrisch | \(\frac {-450 \left (\frac {\cos \left (5 d x +5 c \right )}{5}+\cos \left (3 d x +3 c \right )+2 \cos \left (d x +c \right )\right ) \left (\frac {B \,a^{2}}{2}+b \left (A +\frac {4 C}{3}\right ) a +\frac {2 B \,b^{2}}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+450 \left (\frac {\cos \left (5 d x +5 c \right )}{5}+\cos \left (3 d x +3 c \right )+2 \cos \left (d x +c \right )\right ) \left (\frac {B \,a^{2}}{2}+b \left (A +\frac {4 C}{3}\right ) a +\frac {2 B \,b^{2}}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (\left (320 A +400 C \right ) a^{2}+800 B a b +400 b^{2} \left (A +\frac {9 C}{10}\right )\right ) \sin \left (3 d x +3 c \right )+\left (\left (64 A +80 C \right ) a^{2}+160 B a b +80 b^{2} \left (A +\frac {3 C}{2}\right )\right ) \sin \left (5 d x +5 c \right )+\left (420 B \,a^{2}+840 b \left (A +\frac {4 C}{7}\right ) a +240 B \,b^{2}\right ) \sin \left (2 d x +2 c \right )+\left (90 B \,a^{2}+180 b \left (A +\frac {4 C}{3}\right ) a +120 B \,b^{2}\right ) \sin \left (4 d x +4 c \right )+640 \sin \left (d x +c \right ) \left (\frac {\left (A +\frac {3 C}{4}\right ) b^{2}}{2}+B a b +a^{2} \left (A +\frac {C}{2}\right )\right )}{600 \left (\frac {\cos \left (5 d x +5 c \right )}{5}+\cos \left (3 d x +3 c \right )+2 \cos \left (d x +c \right )\right ) d}\) | \(336\) |
| risch | \(-\frac {i \left (-160 B a b -80 A \,b^{2}-120 C \,b^{2}-80 a^{2} C -64 A \,a^{2}+60 B \,b^{2} {\mathrm e}^{9 i \left (d x +c \right )}+210 B \,a^{2} {\mathrm e}^{7 i \left (d x +c \right )}-120 C \,b^{2} {\mathrm e}^{8 i \left (d x +c \right )}-240 C \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}-480 C \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}+45 B \,a^{2} {\mathrm e}^{9 i \left (d x +c \right )}+120 B \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-640 A \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-240 A \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-480 C \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-720 C \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-210 B \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-120 B \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-60 B \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-560 A \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-560 C \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-320 A \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-400 A \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-400 C \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-45 B \,a^{2} {\mathrm e}^{i \left (d x +c \right )}-240 C a b \,{\mathrm e}^{3 i \left (d x +c \right )}+90 A a b \,{\mathrm e}^{9 i \left (d x +c \right )}+120 C a b \,{\mathrm e}^{9 i \left (d x +c \right )}+420 A a b \,{\mathrm e}^{7 i \left (d x +c \right )}+240 C a b \,{\mathrm e}^{7 i \left (d x +c \right )}-480 B a b \,{\mathrm e}^{6 i \left (d x +c \right )}-420 A a b \,{\mathrm e}^{3 i \left (d x +c \right )}-120 C a b \,{\mathrm e}^{i \left (d x +c \right )}-1120 B a b \,{\mathrm e}^{4 i \left (d x +c \right )}-800 B a b \,{\mathrm e}^{2 i \left (d x +c \right )}-90 A a b \,{\mathrm e}^{i \left (d x +c \right )}\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A b}{4 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B \,b^{2}}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b C}{d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A b}{4 d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B \,b^{2}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b C}{d}\) | \(676\) |
Input:
int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x,meth od=_RETURNVERBOSE)
Output:
-A*a^2/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+(2*A*a*b+B* a^2)/d*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+t an(d*x+c)))+(B*b^2+2*C*a*b)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c) +tan(d*x+c)))-(A*b^2+2*B*a*b+C*a^2)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+C *b^2/d*tan(d*x+c)
Time = 0.12 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.05 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {15 \, {\left (3 \, B a^{2} + 2 \, {\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, B a^{2} + 2 \, {\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (2 \, {\left (4 \, A + 5 \, C\right )} a^{2} + 20 \, B a b + 5 \, {\left (2 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left (3 \, B a^{2} + 2 \, {\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 24 \, A a^{2} + 8 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{2} + 10 \, B a b + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6, x, algorithm="fricas")
Output:
1/240*(15*(3*B*a^2 + 2*(3*A + 4*C)*a*b + 4*B*b^2)*cos(d*x + c)^5*log(sin(d *x + c) + 1) - 15*(3*B*a^2 + 2*(3*A + 4*C)*a*b + 4*B*b^2)*cos(d*x + c)^5*l og(-sin(d*x + c) + 1) + 2*(8*(2*(4*A + 5*C)*a^2 + 20*B*a*b + 5*(2*A + 3*C) *b^2)*cos(d*x + c)^4 + 15*(3*B*a^2 + 2*(3*A + 4*C)*a*b + 4*B*b^2)*cos(d*x + c)^3 + 24*A*a^2 + 8*((4*A + 5*C)*a^2 + 10*B*a*b + 5*A*b^2)*cos(d*x + c)^ 2 + 30*(B*a^2 + 2*A*a*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)
Timed out. \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Timed out} \] Input:
integrate((a+b*cos(d*x+c))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)* *6,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 357, normalized size of antiderivative = 1.54 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{2} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{2} - 15 \, B a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 30 \, A a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, C a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, B b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, C b^{2} \tan \left (d x + c\right )}{240 \, d} \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6, x, algorithm="maxima")
Output:
1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^2 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 + 160*(tan(d*x + c)^3 + 3*tan( d*x + c))*B*a*b + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b^2 - 15*B*a^2*(2 *(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 30*A*a*b*(2*(3*s in(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 120*C*a*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 60*B*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 240*C*b^2*tan(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 766 vs. \(2 (220) = 440\).
Time = 0.17 (sec) , antiderivative size = 766, normalized size of antiderivative = 3.30 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\text {Too large to display} \] Input:
integrate((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6, x, algorithm="giac")
Output:
1/120*(15*(3*B*a^2 + 6*A*a*b + 8*C*a*b + 4*B*b^2)*log(abs(tan(1/2*d*x + 1/ 2*c) + 1)) - 15*(3*B*a^2 + 6*A*a*b + 8*C*a*b + 4*B*b^2)*log(abs(tan(1/2*d* x + 1/2*c) - 1)) - 2*(120*A*a^2*tan(1/2*d*x + 1/2*c)^9 - 75*B*a^2*tan(1/2* d*x + 1/2*c)^9 + 120*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 150*A*a*b*tan(1/2*d*x + 1/2*c)^9 + 240*B*a*b*tan(1/2*d*x + 1/2*c)^9 - 120*C*a*b*tan(1/2*d*x + 1/ 2*c)^9 + 120*A*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*B*b^2*tan(1/2*d*x + 1/2*c)^ 9 + 120*C*b^2*tan(1/2*d*x + 1/2*c)^9 - 160*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 30*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 320*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 60*A* a*b*tan(1/2*d*x + 1/2*c)^7 - 640*B*a*b*tan(1/2*d*x + 1/2*c)^7 + 240*C*a*b* tan(1/2*d*x + 1/2*c)^7 - 320*A*b^2*tan(1/2*d*x + 1/2*c)^7 + 120*B*b^2*tan( 1/2*d*x + 1/2*c)^7 - 480*C*b^2*tan(1/2*d*x + 1/2*c)^7 + 464*A*a^2*tan(1/2* d*x + 1/2*c)^5 + 400*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 800*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 400*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 720*C*b^2*tan(1/2*d*x + 1/ 2*c)^5 - 160*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 30*B*a^2*tan(1/2*d*x + 1/2*c)^ 3 - 320*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 60*A*a*b*tan(1/2*d*x + 1/2*c)^3 - 6 40*B*a*b*tan(1/2*d*x + 1/2*c)^3 - 240*C*a*b*tan(1/2*d*x + 1/2*c)^3 - 320*A *b^2*tan(1/2*d*x + 1/2*c)^3 - 120*B*b^2*tan(1/2*d*x + 1/2*c)^3 - 480*C*b^2 *tan(1/2*d*x + 1/2*c)^3 + 120*A*a^2*tan(1/2*d*x + 1/2*c) + 75*B*a^2*tan(1/ 2*d*x + 1/2*c) + 120*C*a^2*tan(1/2*d*x + 1/2*c) + 150*A*a*b*tan(1/2*d*x + 1/2*c) + 240*B*a*b*tan(1/2*d*x + 1/2*c) + 120*C*a*b*tan(1/2*d*x + 1/2*c...
Time = 2.79 (sec) , antiderivative size = 455, normalized size of antiderivative = 1.96 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,B\,a^2}{8}+\frac {B\,b^2}{2}+\frac {3\,A\,a\,b}{4}+C\,a\,b\right )}{\frac {3\,B\,a^2}{2}+2\,B\,b^2+3\,A\,a\,b+4\,C\,a\,b}\right )\,\left (\frac {3\,B\,a^2}{4}+B\,b^2+\frac {3\,A\,a\,b}{2}+2\,C\,a\,b\right )}{d}-\frac {\left (2\,A\,a^2+2\,A\,b^2-\frac {5\,B\,a^2}{4}-B\,b^2+2\,C\,a^2+2\,C\,b^2-\frac {5\,A\,a\,b}{2}+4\,B\,a\,b-2\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {B\,a^2}{2}-\frac {16\,A\,b^2}{3}-\frac {8\,A\,a^2}{3}+2\,B\,b^2-\frac {16\,C\,a^2}{3}-8\,C\,b^2+A\,a\,b-\frac {32\,B\,a\,b}{3}+4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^2}{15}+\frac {20\,A\,b^2}{3}+\frac {20\,C\,a^2}{3}+12\,C\,b^2+\frac {40\,B\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a^2}{3}-\frac {16\,A\,b^2}{3}-\frac {B\,a^2}{2}-2\,B\,b^2-\frac {16\,C\,a^2}{3}-8\,C\,b^2-A\,a\,b-\frac {32\,B\,a\,b}{3}-4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^2+2\,A\,b^2+\frac {5\,B\,a^2}{4}+B\,b^2+2\,C\,a^2+2\,C\,b^2+\frac {5\,A\,a\,b}{2}+4\,B\,a\,b+2\,C\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(((a + b*cos(c + d*x))^2*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^6,x)
Output:
(atanh((4*tan(c/2 + (d*x)/2)*((3*B*a^2)/8 + (B*b^2)/2 + (3*A*a*b)/4 + C*a* b))/((3*B*a^2)/2 + 2*B*b^2 + 3*A*a*b + 4*C*a*b))*((3*B*a^2)/4 + B*b^2 + (3 *A*a*b)/2 + 2*C*a*b))/d - (tan(c/2 + (d*x)/2)*(2*A*a^2 + 2*A*b^2 + (5*B*a^ 2)/4 + B*b^2 + 2*C*a^2 + 2*C*b^2 + (5*A*a*b)/2 + 4*B*a*b + 2*C*a*b) + tan( c/2 + (d*x)/2)^9*(2*A*a^2 + 2*A*b^2 - (5*B*a^2)/4 - B*b^2 + 2*C*a^2 + 2*C* b^2 - (5*A*a*b)/2 + 4*B*a*b - 2*C*a*b) - tan(c/2 + (d*x)/2)^3*((8*A*a^2)/3 + (16*A*b^2)/3 + (B*a^2)/2 + 2*B*b^2 + (16*C*a^2)/3 + 8*C*b^2 + A*a*b + ( 32*B*a*b)/3 + 4*C*a*b) - tan(c/2 + (d*x)/2)^7*((8*A*a^2)/3 + (16*A*b^2)/3 - (B*a^2)/2 - 2*B*b^2 + (16*C*a^2)/3 + 8*C*b^2 - A*a*b + (32*B*a*b)/3 - 4* C*a*b) + tan(c/2 + (d*x)/2)^5*((116*A*a^2)/15 + (20*A*b^2)/3 + (20*C*a^2)/ 3 + 12*C*b^2 + (40*B*a*b)/3))/(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))
Time = 0.21 (sec) , antiderivative size = 819, normalized size of antiderivative = 3.53 \[ \int (a+b \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx =\text {Too large to display} \] Input:
int((a+b*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)
Output:
( - 135*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b - 12 0*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b*c - 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b**3 + 270*cos(c + d*x)*l og(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b + 240*cos(c + d*x)*log(tan ((c + d*x)/2) - 1)*sin(c + d*x)**2*a*b*c + 120*cos(c + d*x)*log(tan((c + d *x)/2) - 1)*sin(c + d*x)**2*b**3 - 135*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*b - 120*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b*c - 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*b**3 + 135*cos(c + d*x)*log(tan((c + d*x)/ 2) + 1)*sin(c + d*x)**4*a**2*b + 120*cos(c + d*x)*log(tan((c + d*x)/2) + 1 )*sin(c + d*x)**4*a*b*c + 60*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*b**3 - 270*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)** 2*a**2*b - 240*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b* c - 120*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b**3 + 135* cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2*b + 120*cos(c + d*x)*log(tan(( c + d*x)/2) + 1)*a*b*c + 60*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b**3 - 135*cos(c + d*x)*sin(c + d*x)**3*a**2*b - 120*cos(c + d*x)*sin(c + d*x)**3 *a*b*c - 60*cos(c + d*x)*sin(c + d*x)**3*b**3 + 225*cos(c + d*x)*sin(c + d *x)*a**2*b + 120*cos(c + d*x)*sin(c + d*x)*a*b*c + 60*cos(c + d*x)*sin(c + d*x)*b**3 + 64*sin(c + d*x)**5*a**3 + 80*sin(c + d*x)**5*a**2*c + 240*sin (c + d*x)**5*a*b**2 + 120*sin(c + d*x)**5*b**2*c - 160*sin(c + d*x)**3*...