\(\int \cos ^6(c+d x) (a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\) [101]

Optimal result

Integrand size = 33, antiderivative size = 194 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{16} a^2 (11 A+14 C) x+\frac {2 a^2 (4 A+5 C) \sin (c+d x)}{5 d}+\frac {a^2 (11 A+14 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^2 (9 A+10 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {A \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}-\frac {2 a^2 (4 A+5 C) \sin ^3(c+d x)}{15 d} \]

[Out]

1/16*a^2*(11*A+14*C)*x+2/5*a^2*(4*A+5*C)*sin(d*x+c)/d+1/16*a^2*(11*A+14*C)*cos(d*x+c)*sin(d*x+c)/d+1/40*a^2*(9
*A+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*cos(d*x+c)^4*(a^2
+a^2*sec(d*x+c))*sin(d*x+c)/d-2/15*a^2*(4*A+5*C)*sin(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {4172, 4102, 4081, 3872, 2713, 2715, 8} \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 a^2 (4 A+5 C) \sin ^3(c+d x)}{15 d}+\frac {2 a^2 (4 A+5 C) \sin (c+d x)}{5 d}+\frac {a^2 (9 A+10 C) \sin (c+d x) \cos ^3(c+d x)}{40 d}+\frac {a^2 (11 A+14 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {A \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{15 d}+\frac {1}{16} a^2 x (11 A+14 C)+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d} \]

[In]

Int[Cos[c + d*x]^6*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^2*(11*A + 14*C)*x)/16 + (2*a^2*(4*A + 5*C)*Sin[c + d*x])/(5*d) + (a^2*(11*A + 14*C)*Cos[c + d*x]*Sin[c + d*
x])/(16*d) + (a^2*(9*A + 10*C)*Cos[c + d*x]^3*Sin[c + d*x])/(40*d) + (A*Cos[c + d*x]^5*(a + a*Sec[c + d*x])^2*
Sin[c + d*x])/(6*d) + (A*Cos[c + d*x]^4*(a^2 + a^2*Sec[c + d*x])*Sin[c + d*x])/(15*d) - (2*a^2*(4*A + 5*C)*Sin
[c + d*x]^3)/(15*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2715

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] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4081

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] + Dist[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]

Rule 4102

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]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[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]

Rule 4172

Int[((A_.) + 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] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2
^(-1)] || EqQ[m + n + 1, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {\int \cos ^5(c+d x) (a+a \sec (c+d x))^2 (2 a A+3 a (A+2 C) \sec (c+d x)) \, dx}{6 a} \\ & = \frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {A \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {\int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (3 a^2 (9 A+10 C)+3 a^2 (7 A+10 C) \sec (c+d x)\right ) \, dx}{30 a} \\ & = \frac {a^2 (9 A+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 \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}-\frac {\int \cos ^3(c+d x) \left (-48 a^3 (4 A+5 C)-15 a^3 (11 A+14 C) \sec (c+d x)\right ) \, dx}{120 a} \\ & = \frac {a^2 (9 A+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 \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{5} \left (2 a^2 (4 A+5 C)\right ) \int \cos ^3(c+d x) \, dx+\frac {1}{8} \left (a^2 (11 A+14 C)\right ) \int \cos ^2(c+d x) \, dx \\ & = \frac {a^2 (11 A+14 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^2 (9 A+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 \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{16} \left (a^2 (11 A+14 C)\right ) \int 1 \, dx-\frac {\left (2 a^2 (4 A+5 C)\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d} \\ & = \frac {1}{16} a^2 (11 A+14 C) x+\frac {2 a^2 (4 A+5 C) \sin (c+d x)}{5 d}+\frac {a^2 (11 A+14 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^2 (9 A+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 \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{15 d}-\frac {2 a^2 (4 A+5 C) \sin ^3(c+d x)}{15 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.00 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.63 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (240 A c+660 A d x+840 C d x+240 (5 A+6 C) \sin (c+d x)+15 (31 A+32 C) \sin (2 (c+d x))+200 A \sin (3 (c+d x))+160 C \sin (3 (c+d x))+75 A \sin (4 (c+d x))+30 C \sin (4 (c+d x))+24 A \sin (5 (c+d x))+5 A \sin (6 (c+d x)))}{960 d} \]

[In]

Integrate[Cos[c + d*x]^6*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^2*(240*A*c + 660*A*d*x + 840*C*d*x + 240*(5*A + 6*C)*Sin[c + d*x] + 15*(31*A + 32*C)*Sin[2*(c + d*x)] + 200
*A*Sin[3*(c + d*x)] + 160*C*Sin[3*(c + d*x)] + 75*A*Sin[4*(c + d*x)] + 30*C*Sin[4*(c + d*x)] + 24*A*Sin[5*(c +
d*x)] + 5*A*Sin[6*(c + d*x)]))/(960*d)

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.54

method	result	size
parallelrisch	\(\frac {5 a^{2} \left (\frac {\left (31 A +32 C \right ) \sin \left (2 d x +2 c \right )}{5}+\frac {8 \left (A +\frac {4 C}{5}\right ) \sin \left (3 d x +3 c \right )}{3}+\left (A +\frac {2 C}{5}\right ) \sin \left (4 d x +4 c \right )+\frac {8 A \sin \left (5 d x +5 c \right )}{25}+\frac {A \sin \left (6 d x +6 c \right )}{15}+16 \sin \left (d x +c \right ) \left (A +\frac {6 C}{5}\right )+\frac {44 x d \left (A +\frac {14 C}{11}\right )}{5}\right )}{64 d}\)	\(105\)
risch	\(\frac {11 a^{2} A x}{16}+\frac {7 a^{2} x C}{8}+\frac {5 \sin \left (d x +c \right ) a^{2} A}{4 d}+\frac {3 \sin \left (d x +c \right ) C \,a^{2}}{2 d}+\frac {a^{2} A \sin \left (6 d x +6 c \right )}{192 d}+\frac {a^{2} A \sin \left (5 d x +5 c \right )}{40 d}+\frac {5 a^{2} A \sin \left (4 d x +4 c \right )}{64 d}+\frac {\sin \left (4 d x +4 c \right ) C \,a^{2}}{32 d}+\frac {5 a^{2} A \sin \left (3 d x +3 c \right )}{24 d}+\frac {\sin \left (3 d x +3 c \right ) C \,a^{2}}{6 d}+\frac {31 a^{2} A \sin \left (2 d x +2 c \right )}{64 d}+\frac {\sin \left (2 d x +2 c \right ) C \,a^{2}}{2 d}\)	\(190\)
derivativedivides	\(\frac {a^{2} A \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+C \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 a^{2} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {2 C \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{2} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+C \,a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(211\)
default	\(\frac {a^{2} A \left (\frac {\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+C \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 a^{2} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {2 C \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{2} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+C \,a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(211\)

[In]

int(cos(d*x+c)^6*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

5/64*a^2*(1/5*(31*A+32*C)*sin(2*d*x+2*c)+8/3*(A+4/5*C)*sin(3*d*x+3*c)+(A+2/5*C)*sin(4*d*x+4*c)+8/25*A*sin(5*d*
x+5*c)+1/15*A*sin(6*d*x+6*c)+16*sin(d*x+c)*(A+6/5*C)+44/5*x*d*(A+14/11*C))/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.65 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (11 \, A + 14 \, C\right )} a^{2} d x + {\left (40 \, A a^{2} \cos \left (d x + c\right )^{5} + 96 \, A a^{2} \cos \left (d x + c\right )^{4} + 10 \, {\left (11 \, A + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 32 \, {\left (4 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (11 \, A + 14 \, C\right )} a^{2} \cos \left (d x + c\right ) + 64 \, {\left (4 \, A + 5 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{240 \, d} \]

[In]

integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(15*(11*A + 14*C)*a^2*d*x + (40*A*a^2*cos(d*x + c)^5 + 96*A*a^2*cos(d*x + c)^4 + 10*(11*A + 6*C)*a^2*cos
(d*x + c)^3 + 32*(4*A + 5*C)*a^2*cos(d*x + c)^2 + 15*(11*A + 14*C)*a^2*cos(d*x + c) + 64*(4*A + 5*C)*a^2)*sin(
d*x + c))/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**6*(a+a*sec(d*x+c))**2*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.05 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {128 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{2} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 640 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2}}{960 \, d} \]

[In]

integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

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 - 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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.26 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (11 \, A a^{2} + 14 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (165 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 210 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 935 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1190 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1986 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2580 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3006 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3180 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1305 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2330 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 795 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 750 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]

[In]

integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/240*(15*(11*A*a^2 + 14*C*a^2)*(d*x + c) + 2*(165*A*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 + 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 + 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 + 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 + 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) + 7
50*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d

Mupad [B] (verification not implemented)

Time = 18.01 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.47 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (\frac {11\,A\,a^2}{8}+\frac {7\,C\,a^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {187\,A\,a^2}{24}+\frac {119\,C\,a^2}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {331\,A\,a^2}{20}+\frac {43\,C\,a^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {501\,A\,a^2}{20}+\frac {53\,C\,a^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {87\,A\,a^2}{8}+\frac {233\,C\,a^2}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {53\,A\,a^2}{8}+\frac {25\,C\,a^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (11\,A+14\,C\right )}{8\,\left (\frac {11\,A\,a^2}{8}+\frac {7\,C\,a^2}{4}\right )}\right )\,\left (11\,A+14\,C\right )}{8\,d} \]

[In]

int(cos(c + d*x)^6*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^2,x)

[Out]

(tan(c/2 + (d*x)/2)*((53*A*a^2)/8 + (25*C*a^2)/4) + tan(c/2 + (d*x)/2)^11*((11*A*a^2)/8 + (7*C*a^2)/4) + tan(c
/2 + (d*x)/2)^3*((87*A*a^2)/8 + (233*C*a^2)/12) + tan(c/2 + (d*x)/2)^9*((187*A*a^2)/24 + (119*C*a^2)/12) + tan
(c/2 + (d*x)/2)^7*((331*A*a^2)/20 + (43*C*a^2)/2) + tan(c/2 + (d*x)/2)^5*((501*A*a^2)/20 + (53*C*a^2)/2))/(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 + 14*C))/(8*((11*A*a^
2)/8 + (7*C*a^2)/4)))*(11*A + 14*C))/(8*d)