Integrand size = 40, antiderivative size = 160 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} a^2 (6 B+7 C) x+\frac {a^2 (9 B+10 C) \sin (c+d x)}{5 d}+\frac {a^2 (6 B+7 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (6 B+5 C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {B \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}-\frac {a^2 (9 B+10 C) \sin ^3(c+d x)}{15 d} \]
1/8*a^2*(6*B+7*C)*x+1/5*a^2*(9*B+10*C)*sin(d*x+c)/d+1/8*a^2*(6*B+7*C)*cos( d*x+c)*sin(d*x+c)/d+1/20*a^2*(6*B+5*C)*cos(d*x+c)^3*sin(d*x+c)/d+1/5*B*cos (d*x+c)^4*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d-1/15*a^2*(9*B+10*C)*sin(d*x+c) ^3/d
Time = 0.23 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.68 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (360 B c+360 B d x+420 C d x+60 (11 B+12 C) \sin (c+d x)+240 (B+C) \sin (2 (c+d x))+90 B \sin (3 (c+d x))+80 C \sin (3 (c+d x))+30 B \sin (4 (c+d x))+15 C \sin (4 (c+d x))+6 B \sin (5 (c+d x)))}{480 d} \]
(a^2*(360*B*c + 360*B*d*x + 420*C*d*x + 60*(11*B + 12*C)*Sin[c + d*x] + 24 0*(B + C)*Sin[2*(c + d*x)] + 90*B*Sin[3*(c + d*x)] + 80*C*Sin[3*(c + d*x)] + 30*B*Sin[4*(c + d*x)] + 15*C*Sin[4*(c + d*x)] + 6*B*Sin[5*(c + d*x)]))/ (480*d)
Time = 0.89 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.94, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 4560, 3042, 4505, 3042, 4484, 25, 3042, 4274, 3042, 3113, 2009, 3115, 24}
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.
| \(\displaystyle \int \cos ^6(c+d x) (a \sec (c+d x)+a)^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \cos ^5(c+d x) (a \sec (c+d x)+a)^2 (B+C \sec (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {1}{5} \int \cos ^4(c+d x) (\sec (c+d x) a+a) (a (6 B+5 C)+a (3 B+5 C) \sec (c+d x))dx+\frac {B \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a (6 B+5 C)+a (3 B+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {B \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 4484 |
| \(\displaystyle \frac {1}{5} \left (\frac {a^2 (6 B+5 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {1}{4} \int -\cos ^3(c+d x) \left (4 (9 B+10 C) a^2+5 (6 B+7 C) \sec (c+d x) a^2\right )dx\right )+\frac {B \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \cos ^3(c+d x) \left (4 (9 B+10 C) a^2+5 (6 B+7 C) \sec (c+d x) a^2\right )dx+\frac {a^2 (6 B+5 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {B \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \frac {4 (9 B+10 C) a^2+5 (6 B+7 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^2 (6 B+5 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {B \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (4 a^2 (9 B+10 C) \int \cos ^3(c+d x)dx+5 a^2 (6 B+7 C) \int \cos ^2(c+d x)dx\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {B \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (6 B+7 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+4 a^2 (9 B+10 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {B \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (6 B+7 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {4 a^2 (9 B+10 C) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {B \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (6 B+7 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {4 a^2 (9 B+10 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {B \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 a^2 (6 B+7 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {4 a^2 (9 B+10 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a^2 (6 B+5 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {B \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} \left (\frac {a^2 (6 B+5 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {1}{4} \left (5 a^2 (6 B+7 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {4 a^2 (9 B+10 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )\right )+\frac {B \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}\) |
(B*Cos[c + d*x]^4*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(5*d) + ((a^2*(6* B + 5*C)*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (5*a^2*(6*B + 7*C)*(x/2 + (C os[c + d*x]*Sin[c + d*x])/(2*d)) - (4*a^2*(9*B + 10*C)*(-Sin[c + d*x] + Si n[c + d*x]^3/3))/d)/4)/5
3.4.22.3.1 Defintions of rubi rules used
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Simp[1/(d*n) Int[(d*Csc[e + f*x])^( n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 ] && GtQ[m, 1/2] && LtQ[n, -1]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Time = 0.33 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.59
| method | result | size |
| parallelrisch | \(\frac {a^{2} \left (\left (8 B +8 C \right ) \sin \left (2 d x +2 c \right )+\left (3 B +\frac {8 C}{3}\right ) \sin \left (3 d x +3 c \right )+\left (B +\frac {C}{2}\right ) \sin \left (4 d x +4 c \right )+\frac {B \sin \left (5 d x +5 c \right )}{5}+\left (22 B +24 C \right ) \sin \left (d x +c \right )+12 \left (B +\frac {7 C}{6}\right ) x d \right )}{16 d}\) | \(94\) |
| risch | \(\frac {3 a^{2} B x}{4}+\frac {7 a^{2} x C}{8}+\frac {11 a^{2} B \sin \left (d x +c \right )}{8 d}+\frac {3 \sin \left (d x +c \right ) C \,a^{2}}{2 d}+\frac {B \,a^{2} \sin \left (5 d x +5 c \right )}{80 d}+\frac {B \,a^{2} \sin \left (4 d x +4 c \right )}{16 d}+\frac {\sin \left (4 d x +4 c \right ) C \,a^{2}}{32 d}+\frac {3 B \,a^{2} \sin \left (3 d x +3 c \right )}{16 d}+\frac {\sin \left (3 d x +3 c \right ) C \,a^{2}}{6 d}+\frac {B \,a^{2} \sin \left (2 d x +2 c \right )}{2 d}+\frac {\sin \left (2 d x +2 c \right ) C \,a^{2}}{2 d}\) | \(172\) |
| derivativedivides | \(\frac {\frac {B \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \,a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 B \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 C \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {B \,a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+C \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(186\) |
| default | \(\frac {\frac {B \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \,a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 B \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 C \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {B \,a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+C \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(186\) |
1/16*a^2*((8*B+8*C)*sin(2*d*x+2*c)+(3*B+8/3*C)*sin(3*d*x+3*c)+(B+1/2*C)*si n(4*d*x+4*c)+1/5*B*sin(5*d*x+5*c)+(22*B+24*C)*sin(d*x+c)+12*(B+7/6*C)*x*d) /d
Time = 0.25 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.69 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (6 \, B + 7 \, C\right )} a^{2} d x + {\left (24 \, B a^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (9 \, B + 10 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (6 \, B + 7 \, C\right )} a^{2} \cos \left (d x + c\right ) + 16 \, {\left (9 \, B + 10 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{120 \, d} \]
integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
1/120*(15*(6*B + 7*C)*a^2*d*x + (24*B*a^2*cos(d*x + c)^4 + 30*(2*B + C)*a^ 2*cos(d*x + c)^3 + 8*(9*B + 10*C)*a^2*cos(d*x + c)^2 + 15*(6*B + 7*C)*a^2* cos(d*x + c) + 16*(9*B + 10*C)*a^2)*sin(d*x + c))/d
Timed out. \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Time = 0.26 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.11 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{2} - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2}}{480 \, d} \]
integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*B*a^2 - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^2 + 30*(12*d*x + 12*c + sin(4*d *x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^2 - 320*(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 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2)/d
Time = 0.33 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.31 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (6 \, B a^{2} + 7 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (90 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 105 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 420 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 490 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 864 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 800 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 540 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 790 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 390 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 375 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]
1/120*(15*(6*B*a^2 + 7*C*a^2)*(d*x + c) + 2*(90*B*a^2*tan(1/2*d*x + 1/2*c) ^9 + 105*C*a^2*tan(1/2*d*x + 1/2*c)^9 + 420*B*a^2*tan(1/2*d*x + 1/2*c)^7 + 490*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 864*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 800 *C*a^2*tan(1/2*d*x + 1/2*c)^5 + 540*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 790*C*a ^2*tan(1/2*d*x + 1/2*c)^3 + 390*B*a^2*tan(1/2*d*x + 1/2*c) + 375*C*a^2*tan (1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d
Time = 18.85 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.54 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (\frac {3\,B\,a^2}{2}+\frac {7\,C\,a^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (7\,B\,a^2+\frac {49\,C\,a^2}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {72\,B\,a^2}{5}+\frac {40\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (9\,B\,a^2+\frac {79\,C\,a^2}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,B\,a^2}{2}+\frac {25\,C\,a^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,B+7\,C\right )}{4\,\left (\frac {3\,B\,a^2}{2}+\frac {7\,C\,a^2}{4}\right )}\right )\,\left (6\,B+7\,C\right )}{4\,d} \]
(tan(c/2 + (d*x)/2)*((13*B*a^2)/2 + (25*C*a^2)/4) + tan(c/2 + (d*x)/2)^9*( (3*B*a^2)/2 + (7*C*a^2)/4) + tan(c/2 + (d*x)/2)^7*(7*B*a^2 + (49*C*a^2)/6) + tan(c/2 + (d*x)/2)^3*(9*B*a^2 + (79*C*a^2)/6) + tan(c/2 + (d*x)/2)^5*(( 72*B*a^2)/5 + (40*C*a^2)/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)) + (a^2*atan((a^2*tan(c/2 + (d*x)/2)*(6*B + 7*C))/(4*((3*B*a ^2)/2 + (7*C*a^2)/4)))*(6*B + 7*C))/(4*d)