Integrand size = 41, antiderivative size = 213 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{16} a^2 (11 A+12 B+14 C) x+\frac {a^2 (8 A+9 B+10 C) \sin (c+d x)}{5 d}+\frac {a^2 (11 A+12 B+14 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^2 (9 A+12 B+10 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(A+3 B) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}-\frac {a^2 (8 A+9 B+10 C) \sin ^3(c+d x)}{15 d} \]
1/16*a^2*(11*A+12*B+14*C)*x+1/5*a^2*(8*A+9*B+10*C)*sin(d*x+c)/d+1/16*a^2*( 11*A+12*B+14*C)*cos(d*x+c)*sin(d*x+c)/d+1/40*a^2*(9*A+12*B+10*C)*cos(d*x+c )^3*sin(d*x+c)/d+1/6*A*cos(d*x+c)^5*(a+a*sec(d*x+c))^2*sin(d*x+c)/d+1/15*( A+3*B)*cos(d*x+c)^4*(a^2+a^2*sec(d*x+c))*sin(d*x+c)/d-1/15*a^2*(8*A+9*B+10 *C)*sin(d*x+c)^3/d
Time = 0.57 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.80 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (240 A c+720 B c+660 A d x+720 B d x+840 C d x+120 (10 A+11 B+12 C) \sin (c+d x)+15 (31 A+32 (B+C)) \sin (2 (c+d x))+200 A \sin (3 (c+d x))+180 B \sin (3 (c+d x))+160 C \sin (3 (c+d x))+75 A \sin (4 (c+d x))+60 B \sin (4 (c+d x))+30 C \sin (4 (c+d x))+24 A \sin (5 (c+d x))+12 B \sin (5 (c+d x))+5 A \sin (6 (c+d x)))}{960 d} \]
(a^2*(240*A*c + 720*B*c + 660*A*d*x + 720*B*d*x + 840*C*d*x + 120*(10*A + 11*B + 12*C)*Sin[c + d*x] + 15*(31*A + 32*(B + C))*Sin[2*(c + d*x)] + 200* A*Sin[3*(c + d*x)] + 180*B*Sin[3*(c + d*x)] + 160*C*Sin[3*(c + d*x)] + 75* A*Sin[4*(c + d*x)] + 60*B*Sin[4*(c + d*x)] + 30*C*Sin[4*(c + d*x)] + 24*A* Sin[5*(c + d*x)] + 12*B*Sin[5*(c + d*x)] + 5*A*Sin[6*(c + d*x)]))/(960*d)
Time = 1.14 (sec) , antiderivative size = 206, 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, 4574, 3042, 4505, 27, 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.
| \(\displaystyle \int \cos ^6(c+d x) (a \sec (c+d x)+a)^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^6}dx\) |
| \(\Big \downarrow \) 4574 |
| \(\displaystyle \frac {\int \cos ^5(c+d x) (\sec (c+d x) a+a)^2 (2 a (A+3 B)+3 a (A+2 C) \sec (c+d x))dx}{6 a}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 a (A+3 B)+3 a (A+2 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx}{6 a}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {1}{5} \int 3 \cos ^4(c+d x) (\sec (c+d x) a+a) \left ((9 A+12 B+10 C) a^2+(7 A+6 B+10 C) \sec (c+d x) a^2\right )dx+\frac {2 (A+3 B) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3}{5} \int \cos ^4(c+d x) (\sec (c+d x) a+a) \left ((9 A+12 B+10 C) a^2+(7 A+6 B+10 C) \sec (c+d x) a^2\right )dx+\frac {2 (A+3 B) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{5} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((9 A+12 B+10 C) a^2+(7 A+6 B+10 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {2 (A+3 B) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 4484 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {a^3 (9 A+12 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {1}{4} \int -\cos ^3(c+d x) \left (8 (8 A+9 B+10 C) a^3+5 (11 A+12 B+14 C) \sec (c+d x) a^3\right )dx\right )+\frac {2 (A+3 B) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {1}{4} \int \cos ^3(c+d x) \left (8 (8 A+9 B+10 C) a^3+5 (11 A+12 B+14 C) \sec (c+d x) a^3\right )dx+\frac {a^3 (9 A+12 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {2 (A+3 B) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {1}{4} \int \frac {8 (8 A+9 B+10 C) a^3+5 (11 A+12 B+14 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a^3 (9 A+12 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {2 (A+3 B) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {1}{4} \left (8 a^3 (8 A+9 B+10 C) \int \cos ^3(c+d x)dx+5 a^3 (11 A+12 B+14 C) \int \cos ^2(c+d x)dx\right )+\frac {a^3 (9 A+12 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {2 (A+3 B) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {1}{4} \left (5 a^3 (11 A+12 B+14 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+8 a^3 (8 A+9 B+10 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx\right )+\frac {a^3 (9 A+12 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {2 (A+3 B) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {1}{4} \left (5 a^3 (11 A+12 B+14 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {8 a^3 (8 A+9 B+10 C) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a^3 (9 A+12 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {2 (A+3 B) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {1}{4} \left (5 a^3 (11 A+12 B+14 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {8 a^3 (8 A+9 B+10 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a^3 (9 A+12 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {2 (A+3 B) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {1}{4} \left (5 a^3 (11 A+12 B+14 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {8 a^3 (8 A+9 B+10 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a^3 (9 A+12 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {2 (A+3 B) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {3}{5} \left (\frac {a^3 (9 A+12 B+10 C) \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {1}{4} \left (5 a^3 (11 A+12 B+14 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {8 a^3 (8 A+9 B+10 C) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )\right )+\frac {2 (A+3 B) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{5 d}}{6 a}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d}\) |
(A*Cos[c + d*x]^5*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(6*d) + ((2*(A + 3* B)*Cos[c + d*x]^4*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(5*d) + (3*((a^3* (9*A + 12*B + 10*C)*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (5*a^3*(11*A + 12 *B + 14*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)) - (8*a^3*(8*A + 9*B + 10*C)*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d)/4))/5)/(6*a)
3.5.26.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[ e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x] , x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Time = 0.52 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.59
| method | result | size |
| parallelrisch | \(\frac {5 a^{2} \left (\left (\frac {93 A}{40}+\frac {12 B}{5}+\frac {12 C}{5}\right ) \sin \left (2 d x +2 c \right )+\left (A +\frac {9 B}{10}+\frac {4 C}{5}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {3 A}{8}+\frac {3 B}{10}+\frac {3 C}{20}\right ) \sin \left (4 d x +4 c \right )+\left (\frac {3 A}{25}+\frac {3 B}{50}\right ) \sin \left (5 d x +5 c \right )+\frac {A \sin \left (6 d x +6 c \right )}{40}+\left (6 A +\frac {33 B}{5}+\frac {36 C}{5}\right ) \sin \left (d x +c \right )+\frac {33 x d \left (A +\frac {12 B}{11}+\frac {14 C}{11}\right )}{10}\right )}{24 d}\) | \(126\) |
| risch | \(\frac {11 a^{2} A x}{16}+\frac {3 a^{2} B x}{4}+\frac {7 a^{2} x C}{8}+\frac {5 \sin \left (d x +c \right ) a^{2} A}{4 d}+\frac {11 a^{2} B \sin \left (d x +c \right )}{8 d}+\frac {3 \sin \left (d x +c \right ) C \,a^{2}}{2 d}+\frac {a^{2} A \sin \left (6 d x +6 c \right )}{192 d}+\frac {a^{2} A \sin \left (5 d x +5 c \right )}{40 d}+\frac {\sin \left (5 d x +5 c \right ) B \,a^{2}}{80 d}+\frac {5 a^{2} A \sin \left (4 d x +4 c \right )}{64 d}+\frac {\sin \left (4 d x +4 c \right ) B \,a^{2}}{16 d}+\frac {\sin \left (4 d x +4 c \right ) C \,a^{2}}{32 d}+\frac {5 a^{2} A \sin \left (3 d x +3 c \right )}{24 d}+\frac {3 \sin \left (3 d x +3 c \right ) B \,a^{2}}{16 d}+\frac {\sin \left (3 d x +3 c \right ) C \,a^{2}}{6 d}+\frac {31 a^{2} A \sin \left (2 d x +2 c \right )}{64 d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{2}}{2 d}+\frac {\sin \left (2 d x +2 c \right ) C \,a^{2}}{2 d}\) | \(284\) |
| derivativedivides | \(\frac {a^{2} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \,a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 a^{2} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+2 B \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 C \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{2} A \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {B \,a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+C \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(304\) |
| default | \(\frac {a^{2} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {B \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \,a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {2 a^{2} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+2 B \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 C \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{2} A \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {B \,a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+C \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(304\) |
5/24*a^2*((93/40*A+12/5*B+12/5*C)*sin(2*d*x+2*c)+(A+9/10*B+4/5*C)*sin(3*d* x+3*c)+(3/8*A+3/10*B+3/20*C)*sin(4*d*x+4*c)+(3/25*A+3/50*B)*sin(5*d*x+5*c) +1/40*A*sin(6*d*x+6*c)+(6*A+33/5*B+36/5*C)*sin(d*x+c)+33/10*x*d*(A+12/11*B +14/11*C))/d
Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.68 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (11 \, A + 12 \, B + 14 \, C\right )} a^{2} d x + {\left (40 \, A a^{2} \cos \left (d x + c\right )^{5} + 48 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right )^{4} + 10 \, {\left (11 \, A + 12 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 16 \, {\left (8 \, A + 9 \, B + 10 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (11 \, A + 12 \, B + 14 \, C\right )} a^{2} \cos \left (d x + c\right ) + 32 \, {\left (8 \, A + 9 \, B + 10 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{240 \, d} \]
integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="fricas")
1/240*(15*(11*A + 12*B + 14*C)*a^2*d*x + (40*A*a^2*cos(d*x + c)^5 + 48*(2* A + B)*a^2*cos(d*x + c)^4 + 10*(11*A + 12*B + 6*C)*a^2*cos(d*x + c)^3 + 16 *(8*A + 9*B + 10*C)*a^2*cos(d*x + c)^2 + 15*(11*A + 12*B + 14*C)*a^2*cos(d *x + c) + 32*(8*A + 9*B + 10*C)*a^2)*sin(d*x + c))/d
Timed out. \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Time = 0.23 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.39 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {128 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{2} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{2} - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} + 60 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 640 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2}}{960 \, d} \]
integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="maxima")
1/960*(128*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^2 - 5*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 48*sin(2* d*x + 2*c))*A*a^2 + 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2 *c))*A*a^2 + 64*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*B *a^2 - 320*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^2 + 60*(12*d*x + 12*c + s in(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^2 - 640*(sin(d*x + c)^3 - 3*sin( d*x + c))*C*a^2 + 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c ))*C*a^2 + 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2)/d
Time = 0.36 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.64 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (11 \, A a^{2} + 12 \, B a^{2} + 14 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (165 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 180 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 210 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 935 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1020 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1190 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1986 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2568 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2580 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3006 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2808 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3180 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1305 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1860 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2330 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 795 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 780 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 750 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]
integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="giac")
1/240*(15*(11*A*a^2 + 12*B*a^2 + 14*C*a^2)*(d*x + c) + 2*(165*A*a^2*tan(1/ 2*d*x + 1/2*c)^11 + 180*B*a^2*tan(1/2*d*x + 1/2*c)^11 + 210*C*a^2*tan(1/2* d*x + 1/2*c)^11 + 935*A*a^2*tan(1/2*d*x + 1/2*c)^9 + 1020*B*a^2*tan(1/2*d* x + 1/2*c)^9 + 1190*C*a^2*tan(1/2*d*x + 1/2*c)^9 + 1986*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 2568*B*a^2*tan(1/2*d*x + 1/2*c)^7 + 2580*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 3006*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 2808*B*a^2*tan(1/2*d*x + 1/ 2*c)^5 + 3180*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 1305*A*a^2*tan(1/2*d*x + 1/2* c)^3 + 1860*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 2330*C*a^2*tan(1/2*d*x + 1/2*c) ^3 + 795*A*a^2*tan(1/2*d*x + 1/2*c) + 780*B*a^2*tan(1/2*d*x + 1/2*c) + 750 *C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d
Time = 18.68 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.56 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (\frac {11\,A\,a^2}{8}+\frac {3\,B\,a^2}{2}+\frac {7\,C\,a^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {187\,A\,a^2}{24}+\frac {17\,B\,a^2}{2}+\frac {119\,C\,a^2}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {331\,A\,a^2}{20}+\frac {107\,B\,a^2}{5}+\frac {43\,C\,a^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {501\,A\,a^2}{20}+\frac {117\,B\,a^2}{5}+\frac {53\,C\,a^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {87\,A\,a^2}{8}+\frac {31\,B\,a^2}{2}+\frac {233\,C\,a^2}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {53\,A\,a^2}{8}+\frac {13\,B\,a^2}{2}+\frac {25\,C\,a^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (11\,A+12\,B+14\,C\right )}{8\,\left (\frac {11\,A\,a^2}{8}+\frac {3\,B\,a^2}{2}+\frac {7\,C\,a^2}{4}\right )}\right )\,\left (11\,A+12\,B+14\,C\right )}{8\,d} \]
(tan(c/2 + (d*x)/2)^11*((11*A*a^2)/8 + (3*B*a^2)/2 + (7*C*a^2)/4) + tan(c/ 2 + (d*x)/2)^9*((187*A*a^2)/24 + (17*B*a^2)/2 + (119*C*a^2)/12) + tan(c/2 + (d*x)/2)^3*((87*A*a^2)/8 + (31*B*a^2)/2 + (233*C*a^2)/12) + tan(c/2 + (d *x)/2)^7*((331*A*a^2)/20 + (107*B*a^2)/5 + (43*C*a^2)/2) + tan(c/2 + (d*x) /2)^5*((501*A*a^2)/20 + (117*B*a^2)/5 + (53*C*a^2)/2) + tan(c/2 + (d*x)/2) *((53*A*a^2)/8 + (13*B*a^2)/2 + (25*C*a^2)/4))/(d*(6*tan(c/2 + (d*x)/2)^2 + 15*tan(c/2 + (d*x)/2)^4 + 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2 )^8 + 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1)) + (a^2*atan((a ^2*tan(c/2 + (d*x)/2)*(11*A + 12*B + 14*C))/(8*((11*A*a^2)/8 + (3*B*a^2)/2 + (7*C*a^2)/4)))*(11*A + 12*B + 14*C))/(8*d)