Integrand size = 40, antiderivative size = 180 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} \left (6 a b B+3 a^2 C+4 b^2 C\right ) x+\frac {\left (4 a^2 B+5 b^2 B+10 a b C\right ) \sin (c+d x)}{5 d}+\frac {\left (6 a b B+3 a^2 C+4 b^2 C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (2 b B+a C) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a^2 B \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {\left (4 a^2 B+5 b^2 B+10 a b C\right ) \sin ^3(c+d x)}{15 d} \]
1/8*(6*B*a*b+3*C*a^2+4*C*b^2)*x+1/5*(4*B*a^2+5*B*b^2+10*C*a*b)*sin(d*x+c)/ d+1/8*(6*B*a*b+3*C*a^2+4*C*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/4*a*(2*B*b+C*a)* cos(d*x+c)^3*sin(d*x+c)/d+1/5*a^2*B*cos(d*x+c)^4*sin(d*x+c)/d-1/15*(4*B*a^ 2+5*B*b^2+10*C*a*b)*sin(d*x+c)^3/d
Time = 1.11 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.81 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {60 \left (6 a b B+3 a^2 C+4 b^2 C\right ) (c+d x)+60 \left (5 a^2 B+6 b^2 B+12 a b C\right ) \sin (c+d x)+120 \left (2 a b B+a^2 C+b^2 C\right ) \sin (2 (c+d x))+10 \left (5 a^2 B+4 b^2 B+8 a b C\right ) \sin (3 (c+d x))+15 a (2 b B+a C) \sin (4 (c+d x))+6 a^2 B \sin (5 (c+d x))}{480 d} \]
(60*(6*a*b*B + 3*a^2*C + 4*b^2*C)*(c + d*x) + 60*(5*a^2*B + 6*b^2*B + 12*a *b*C)*Sin[c + d*x] + 120*(2*a*b*B + a^2*C + b^2*C)*Sin[2*(c + d*x)] + 10*( 5*a^2*B + 4*b^2*B + 8*a*b*C)*Sin[3*(c + d*x)] + 15*a*(2*b*B + a*C)*Sin[4*( c + d*x)] + 6*a^2*B*Sin[5*(c + d*x)])/(480*d)
Time = 0.88 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.84, 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, 4512, 25, 3042, 4535, 3042, 3113, 2009, 4533, 3042, 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.
| \(\displaystyle \int \cos ^6(c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 4560 |
| \(\displaystyle \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 (B+C \sec (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 4512 |
| \(\displaystyle \frac {a^2 B \sin (c+d x) \cos ^4(c+d x)}{5 d}-\frac {1}{5} \int -\cos ^4(c+d x) \left (5 b^2 C \sec ^2(c+d x)+\left (4 B a^2+10 b C a+5 b^2 B\right ) \sec (c+d x)+5 a (2 b B+a C)\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \int \cos ^4(c+d x) \left (5 b^2 C \sec ^2(c+d x)+\left (4 B a^2+10 b C a+5 b^2 B\right ) \sec (c+d x)+5 a (2 b B+a C)\right )dx+\frac {a^2 B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {5 b^2 C \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (4 B a^2+10 b C a+5 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+5 a (2 b B+a C)}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a^2 B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{5} \left (\left (4 a^2 B+10 a b C+5 b^2 B\right ) \int \cos ^3(c+d x)dx+\int \cos ^4(c+d x) \left (5 b^2 C \sec ^2(c+d x)+5 a (2 b B+a C)\right )dx\right )+\frac {a^2 B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\left (4 a^2 B+10 a b C+5 b^2 B\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx+\int \frac {5 b^2 C \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a (2 b B+a C)}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\right )+\frac {a^2 B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 b^2 C \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a (2 b B+a C)}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx-\frac {\left (4 a^2 B+10 a b C+5 b^2 B\right ) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a^2 B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 b^2 C \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a (2 b B+a C)}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx-\frac {\left (4 a^2 B+10 a b C+5 b^2 B\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a^2 B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4533 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} \left (3 a^2 C+6 a b B+4 b^2 C\right ) \int \cos ^2(c+d x)dx-\frac {\left (4 a^2 B+10 a b C+5 b^2 B\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5 a (a C+2 b B) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a^2 B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} \left (3 a^2 C+6 a b B+4 b^2 C\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {\left (4 a^2 B+10 a b C+5 b^2 B\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5 a (a C+2 b B) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a^2 B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} \left (3 a^2 C+6 a b B+4 b^2 C\right ) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {\left (4 a^2 B+10 a b C+5 b^2 B\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5 a (a C+2 b B) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a^2 B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} \left (-\frac {\left (4 a^2 B+10 a b C+5 b^2 B\right ) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5}{4} \left (3 a^2 C+6 a b B+4 b^2 C\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )+\frac {5 a (a C+2 b B) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a^2 B \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
(a^2*B*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) + ((5*a*(2*b*B + a*C)*Cos[c + d* x]^3*Sin[c + d*x])/(4*d) + (5*(6*a*b*B + 3*a^2*C + 4*b^2*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/4 - ((4*a^2*B + 5*b^2*B + 10*a*b*C)*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d)/5
3.8.83.3.1 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_))^2*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a^2*A*Cos[ e + f*x]*((d*Csc[e + f*x])^(n + 1)/(d*f*n)), x] + Simp[1/(d*n) Int[(d*Csc [e + f*x])^(n + 1)*(a*(2*A*b + a*B)*n + (2*a*b*B*n + A*(b^2*n + a^2*(n + 1) ))*Csc[e + f*x] + b^2*B*n*Csc[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[A*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*m)), x] + Simp[(C*m + A*(m + 1))/(b^2*m) Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
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 = 0.72 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.82
| method | result | size |
| parallelrisch | \(\frac {120 \left (2 B a b +C \,a^{2}+C \,b^{2}\right ) \sin \left (2 d x +2 c \right )+10 \left (5 B \,a^{2}+4 B \,b^{2}+8 C a b \right ) \sin \left (3 d x +3 c \right )+15 \left (2 B a b +C \,a^{2}\right ) \sin \left (4 d x +4 c \right )+6 B \,a^{2} \sin \left (5 d x +5 c \right )+60 \left (5 B \,a^{2}+6 B \,b^{2}+12 C a b \right ) \sin \left (d x +c \right )+360 x d \left (B a b +\frac {1}{2} C \,a^{2}+\frac {2}{3} C \,b^{2}\right )}{480 d}\) | \(147\) |
| derivativedivides | \(\frac {\frac {B \,a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+2 B a b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+C \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {2 C a b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \,b^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(184\) |
| default | \(\frac {\frac {B \,a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+2 B a b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+C \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {2 C a b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \,b^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(184\) |
| risch | \(\frac {3 B a b x}{4}+\frac {3 a^{2} x C}{8}+\frac {x C \,b^{2}}{2}+\frac {5 a^{2} B \sin \left (d x +c \right )}{8 d}+\frac {3 \sin \left (d x +c \right ) B \,b^{2}}{4 d}+\frac {3 \sin \left (d x +c \right ) C a b}{2 d}+\frac {B \,a^{2} \sin \left (5 d x +5 c \right )}{80 d}+\frac {\sin \left (4 d x +4 c \right ) B a b}{16 d}+\frac {\sin \left (4 d x +4 c \right ) C \,a^{2}}{32 d}+\frac {5 B \,a^{2} \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) B \,b^{2}}{12 d}+\frac {\sin \left (3 d x +3 c \right ) C a b}{6 d}+\frac {\sin \left (2 d x +2 c \right ) B a b}{2 d}+\frac {\sin \left (2 d x +2 c \right ) C \,a^{2}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C \,b^{2}}{4 d}\) | \(225\) |
1/480*(120*(2*B*a*b+C*a^2+C*b^2)*sin(2*d*x+2*c)+10*(5*B*a^2+4*B*b^2+8*C*a* b)*sin(3*d*x+3*c)+15*(2*B*a*b+C*a^2)*sin(4*d*x+4*c)+6*B*a^2*sin(5*d*x+5*c) +60*(5*B*a^2+6*B*b^2+12*C*a*b)*sin(d*x+c)+360*x*d*(B*a*b+1/2*C*a^2+2/3*C*b ^2))/d
Time = 0.27 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.79 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, C a^{2} + 6 \, B a b + 4 \, C b^{2}\right )} d x + {\left (24 \, B a^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (C a^{2} + 2 \, B a b\right )} \cos \left (d x + c\right )^{3} + 64 \, B a^{2} + 160 \, C a b + 80 \, B b^{2} + 8 \, {\left (4 \, B a^{2} + 10 \, C a b + 5 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, C a^{2} + 6 \, B a b + 4 \, C b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]
integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
1/120*(15*(3*C*a^2 + 6*B*a*b + 4*C*b^2)*d*x + (24*B*a^2*cos(d*x + c)^4 + 3 0*(C*a^2 + 2*B*a*b)*cos(d*x + c)^3 + 64*B*a^2 + 160*C*a*b + 80*B*b^2 + 8*( 4*B*a^2 + 10*C*a*b + 5*B*b^2)*cos(d*x + c)^2 + 15*(3*C*a^2 + 6*B*a*b + 4*C *b^2)*cos(d*x + c))*sin(d*x + c))/d
Timed out. \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Time = 0.26 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.98 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a b - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a b - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{2}}{480 \, d} \]
integrate(cos(d*x+c)^6*(a+b*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 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^2 + 30*(12 *d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a*b - 320*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a*b - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*b ^2 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*b^2)/d
Leaf count of result is larger than twice the leaf count of optimal. 487 vs. \(2 (168) = 336\).
Time = 0.32 (sec) , antiderivative size = 487, normalized size of antiderivative = 2.71 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, C a^{2} + 6 \, B a b + 4 \, C b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (120 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 150 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 240 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 60 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 640 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 120 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 800 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 640 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 150 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 240 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C b^{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*(3*C*a^2 + 6*B*a*b + 4*C*b^2)*(d*x + c) + 2*(120*B*a^2*tan(1/2*d *x + 1/2*c)^9 - 75*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 150*B*a*b*tan(1/2*d*x + 1/2*c)^9 + 240*C*a*b*tan(1/2*d*x + 1/2*c)^9 + 120*B*b^2*tan(1/2*d*x + 1/2* c)^9 - 60*C*b^2*tan(1/2*d*x + 1/2*c)^9 + 160*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 30*C*a^2*tan(1/2*d*x + 1/2*c)^7 - 60*B*a*b*tan(1/2*d*x + 1/2*c)^7 + 640* C*a*b*tan(1/2*d*x + 1/2*c)^7 + 320*B*b^2*tan(1/2*d*x + 1/2*c)^7 - 120*C*b^ 2*tan(1/2*d*x + 1/2*c)^7 + 464*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 800*C*a*b*ta n(1/2*d*x + 1/2*c)^5 + 400*B*b^2*tan(1/2*d*x + 1/2*c)^5 + 160*B*a^2*tan(1/ 2*d*x + 1/2*c)^3 + 30*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 60*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 640*C*a*b*tan(1/2*d*x + 1/2*c)^3 + 320*B*b^2*tan(1/2*d*x + 1/ 2*c)^3 + 120*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*B*a^2*tan(1/2*d*x + 1/2*c) + 75*C*a^2*tan(1/2*d*x + 1/2*c) + 150*B*a*b*tan(1/2*d*x + 1/2*c) + 240*C* a*b*tan(1/2*d*x + 1/2*c) + 120*B*b^2*tan(1/2*d*x + 1/2*c) + 60*C*b^2*tan(1 /2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d
Time = 20.66 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.71 \[ \int \cos ^6(c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {x\,\left (\frac {3\,C\,a^2}{4}+\frac {3\,B\,a\,b}{2}+C\,b^2\right )}{2}+\frac {\left (2\,B\,a^2+2\,B\,b^2-\frac {5\,C\,a^2}{4}-C\,b^2-\frac {5\,B\,a\,b}{2}+4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {8\,B\,a^2}{3}+\frac {16\,B\,b^2}{3}-\frac {C\,a^2}{2}-2\,C\,b^2-B\,a\,b+\frac {32\,C\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,B\,a^2}{15}+\frac {40\,C\,a\,b}{3}+\frac {20\,B\,b^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {8\,B\,a^2}{3}+\frac {16\,B\,b^2}{3}+\frac {C\,a^2}{2}+2\,C\,b^2+B\,a\,b+\frac {32\,C\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,B\,a^2+2\,B\,b^2+\frac {5\,C\,a^2}{4}+C\,b^2+\frac {5\,B\,a\,b}{2}+4\,C\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]
(x*((3*C*a^2)/4 + C*b^2 + (3*B*a*b)/2))/2 + (tan(c/2 + (d*x)/2)^5*((116*B* a^2)/15 + (20*B*b^2)/3 + (40*C*a*b)/3) + tan(c/2 + (d*x)/2)^9*(2*B*a^2 + 2 *B*b^2 - (5*C*a^2)/4 - C*b^2 - (5*B*a*b)/2 + 4*C*a*b) + tan(c/2 + (d*x)/2) ^3*((8*B*a^2)/3 + (16*B*b^2)/3 + (C*a^2)/2 + 2*C*b^2 + B*a*b + (32*C*a*b)/ 3) + tan(c/2 + (d*x)/2)^7*((8*B*a^2)/3 + (16*B*b^2)/3 - (C*a^2)/2 - 2*C*b^ 2 - B*a*b + (32*C*a*b)/3) + tan(c/2 + (d*x)/2)*(2*B*a^2 + 2*B*b^2 + (5*C*a ^2)/4 + C*b^2 + (5*B*a*b)/2 + 4*C*a*b))/(d*(5*tan(c/2 + (d*x)/2)^2 + 10*ta n(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + ta n(c/2 + (d*x)/2)^10 + 1))