\(\int \cos ^6(c+d x) (a+a \sec (c+d x))^2 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [322]

Optimal result

Integrand size = 40, antiderivative size = 160 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} a^2 (6 B+7 C) x+\frac {a^2 (9 B+10 C) \sin (c+d x)}{5 d}+\frac {a^2 (6 B+7 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (6 B+5 C) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {B \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}-\frac {a^2 (9 B+10 C) \sin ^3(c+d x)}{15 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {4157, 4102, 4081, 3872, 2713, 2715, 8} \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {a^2 (9 B+10 C) \sin ^3(c+d x)}{15 d}+\frac {a^2 (9 B+10 C) \sin (c+d x)}{5 d}+\frac {a^2 (6 B+5 C) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {a^2 (6 B+7 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {B \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d}+\frac {1}{8} a^2 x (6 B+7 C) \]

[In]

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

[Out]

(a^2*(6*B + 7*C)*x)/8 + (a^2*(9*B + 10*C)*Sin[c + d*x])/(5*d) + (a^2*(6*B + 7*C)*Cos[c + d*x]*Sin[c + d*x])/(8
*d) + (a^2*(6*B + 5*C)*Cos[c + d*x]^3*Sin[c + d*x])/(20*d) + (B*Cos[c + d*x]^4*(a^2 + a^2*Sec[c + d*x])*Sin[c
+ d*x])/(5*d) - (a^2*(9*B + 10*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 4157

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] :> Dist[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]

Rubi steps \begin{align*} \text {integral}& = \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 (B+C \sec (c+d x)) \, dx \\ & = \frac {B \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{5 d}+\frac {1}{5} \int \cos ^4(c+d x) (a+a \sec (c+d x)) (a (6 B+5 C)+a (3 B+5 C) \sec (c+d x)) \, dx \\ & = \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 {1}{20} \int \cos ^3(c+d x) \left (-4 a^2 (9 B+10 C)-5 a^2 (6 B+7 C) \sec (c+d x)\right ) \, dx \\ & = \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 {1}{4} \left (a^2 (6 B+7 C)\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{5} \left (a^2 (9 B+10 C)\right ) \int \cos ^3(c+d x) \, dx \\ & = \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 {1}{8} \left (a^2 (6 B+7 C)\right ) \int 1 \, dx-\frac {\left (a^2 (9 B+10 C)\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.68 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (360 B c+360 B d x+420 C d x+60 (11 B+12 C) \sin (c+d x)+240 (B+C) \sin (2 (c+d x))+90 B \sin (3 (c+d x))+80 C \sin (3 (c+d x))+30 B \sin (4 (c+d x))+15 C \sin (4 (c+d x))+6 B \sin (5 (c+d x)))}{480 d} \]

[In]

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

[Out]

(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*Si
n[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)

Maple [A] (verified)

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	\(\frac {\frac {B \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \,a^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 B \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {2 C \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {B \,a^{2} \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+C \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\)	\(186\)

[In]

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

[Out]

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)*sin(4*d*x+4*c)+1/5*B*sin(5*d*x+5*c)+(2
2*B+24*C)*sin(d*x+c)+12*(B+7/6*C)*x*d)/d

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.69 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (6 \, B + 7 \, C\right )} a^{2} d x + {\left (24 \, B a^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (2 \, B + C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (9 \, B + 10 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (6 \, B + 7 \, C\right )} a^{2} \cos \left (d x + c\right ) + 16 \, {\left (9 \, B + 10 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{120 \, d} \]

[In]

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")

[Out]

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

Sympy [F(-1)]

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

[In]

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)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.11 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{2} - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2}}{480 \, d} \]

[In]

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")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.31 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (6 \, B a^{2} + 7 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (90 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 105 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 420 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 490 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 864 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 800 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 540 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 790 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 390 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 375 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]

[In]

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="giac")

[Out]

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 + 8
00*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

Mupad [B] (verification not implemented)

Time = 18.85 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.54 \[ \int \cos ^6(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (\frac {3\,B\,a^2}{2}+\frac {7\,C\,a^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (7\,B\,a^2+\frac {49\,C\,a^2}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {72\,B\,a^2}{5}+\frac {40\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (9\,B\,a^2+\frac {79\,C\,a^2}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,B\,a^2}{2}+\frac {25\,C\,a^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (6\,B+7\,C\right )}{4\,\left (\frac {3\,B\,a^2}{2}+\frac {7\,C\,a^2}{4}\right )}\right )\,\left (6\,B+7\,C\right )}{4\,d} \]

[In]

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

[Out]

(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)