3.80 \(\int \cos ^6(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=220 \[ \frac {a^4 (72 A+83 B) \sin (c+d x)}{15 d}+\frac {a^4 (159 A+176 B) \sin (c+d x) \cos ^2(c+d x)}{120 d}+\frac {7 a^4 (7 A+8 B) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {(73 A+72 B) \sin (c+d x) \cos ^3(c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{120 d}+\frac {7}{16} a^4 x (7 A+8 B)+\frac {(3 A+2 B) \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{10 d}+\frac {a A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^3}{6 d} \]

[Out]

7/16*a^4*(7*A+8*B)*x+1/15*a^4*(72*A+83*B)*sin(d*x+c)/d+7/16*a^4*(7*A+8*B)*cos(d*x+c)*sin(d*x+c)/d+1/120*a^4*(1
59*A+176*B)*cos(d*x+c)^2*sin(d*x+c)/d+1/6*a*A*cos(d*x+c)^5*(a+a*sec(d*x+c))^3*sin(d*x+c)/d+1/10*(3*A+2*B)*cos(
d*x+c)^4*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/d+1/120*(73*A+72*B)*cos(d*x+c)^3*(a^4+a^4*sec(d*x+c))*sin(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.53, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4017, 3996, 3787, 2635, 8, 2637} \[ \frac {a^4 (72 A+83 B) \sin (c+d x)}{15 d}+\frac {a^4 (159 A+176 B) \sin (c+d x) \cos ^2(c+d x)}{120 d}+\frac {7 a^4 (7 A+8 B) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {(3 A+2 B) \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{10 d}+\frac {(73 A+72 B) \sin (c+d x) \cos ^3(c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{120 d}+\frac {7}{16} a^4 x (7 A+8 B)+\frac {a A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^3}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

(7*a^4*(7*A + 8*B)*x)/16 + (a^4*(72*A + 83*B)*Sin[c + d*x])/(15*d) + (7*a^4*(7*A + 8*B)*Cos[c + d*x]*Sin[c + d
*x])/(16*d) + (a^4*(159*A + 176*B)*Cos[c + d*x]^2*Sin[c + d*x])/(120*d) + (a*A*Cos[c + d*x]^5*(a + a*Sec[c + d
*x])^3*Sin[c + d*x])/(6*d) + ((3*A + 2*B)*Cos[c + d*x]^4*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(10*d) + ((7
3*A + 72*B)*Cos[c + d*x]^3*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(120*d)

Rule 8

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

Rule 2635

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] && Integer
Q[2*n]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3787

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 3996

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 4017

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]

Rubi steps

\begin {align*} \int \cos ^6(c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac {a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {1}{6} \int \cos ^5(c+d x) (a+a \sec (c+d x))^3 (3 a (3 A+2 B)+2 a (A+3 B) \sec (c+d x)) \, dx\\ &=\frac {a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {(3 A+2 B) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{10 d}+\frac {1}{30} \int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (a^2 (73 A+72 B)+14 a^2 (2 A+3 B) \sec (c+d x)\right ) \, dx\\ &=\frac {a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {(3 A+2 B) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{10 d}+\frac {(73 A+72 B) \cos ^3(c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{120 d}+\frac {1}{120} \int \cos ^3(c+d x) (a+a \sec (c+d x)) \left (3 a^3 (159 A+176 B)+6 a^3 (43 A+52 B) \sec (c+d x)\right ) \, dx\\ &=\frac {a^4 (159 A+176 B) \cos ^2(c+d x) \sin (c+d x)}{120 d}+\frac {a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {(3 A+2 B) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{10 d}+\frac {(73 A+72 B) \cos ^3(c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{120 d}-\frac {1}{360} \int \cos ^2(c+d x) \left (-315 a^4 (7 A+8 B)-24 a^4 (72 A+83 B) \sec (c+d x)\right ) \, dx\\ &=\frac {a^4 (159 A+176 B) \cos ^2(c+d x) \sin (c+d x)}{120 d}+\frac {a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {(3 A+2 B) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{10 d}+\frac {(73 A+72 B) \cos ^3(c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{120 d}+\frac {1}{8} \left (7 a^4 (7 A+8 B)\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{15} \left (a^4 (72 A+83 B)\right ) \int \cos (c+d x) \, dx\\ &=\frac {a^4 (72 A+83 B) \sin (c+d x)}{15 d}+\frac {7 a^4 (7 A+8 B) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^4 (159 A+176 B) \cos ^2(c+d x) \sin (c+d x)}{120 d}+\frac {a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {(3 A+2 B) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{10 d}+\frac {(73 A+72 B) \cos ^3(c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{120 d}+\frac {1}{16} \left (7 a^4 (7 A+8 B)\right ) \int 1 \, dx\\ &=\frac {7}{16} a^4 (7 A+8 B) x+\frac {a^4 (72 A+83 B) \sin (c+d x)}{15 d}+\frac {7 a^4 (7 A+8 B) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^4 (159 A+176 B) \cos ^2(c+d x) \sin (c+d x)}{120 d}+\frac {a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {(3 A+2 B) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{10 d}+\frac {(73 A+72 B) \cos ^3(c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{120 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.61, size = 134, normalized size = 0.61 \[ \frac {a^4 (120 (44 A+49 B) \sin (c+d x)+15 (127 A+128 B) \sin (2 (c+d x))+720 A \sin (3 (c+d x))+225 A \sin (4 (c+d x))+48 A \sin (5 (c+d x))+5 A \sin (6 (c+d x))+2940 A c+2940 A d x+580 B \sin (3 (c+d x))+120 B \sin (4 (c+d x))+12 B \sin (5 (c+d x))+3360 B d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + a*Sec[c + d*x])^4*(A + B*Sec[c + d*x]),x]

[Out]

(a^4*(2940*A*c + 2940*A*d*x + 3360*B*d*x + 120*(44*A + 49*B)*Sin[c + d*x] + 15*(127*A + 128*B)*Sin[2*(c + d*x)
] + 720*A*Sin[3*(c + d*x)] + 580*B*Sin[3*(c + d*x)] + 225*A*Sin[4*(c + d*x)] + 120*B*Sin[4*(c + d*x)] + 48*A*S
in[5*(c + d*x)] + 12*B*Sin[5*(c + d*x)] + 5*A*Sin[6*(c + d*x)]))/(960*d)

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 130, normalized size = 0.59 \[ \frac {105 \, {\left (7 \, A + 8 \, B\right )} a^{4} d x + {\left (40 \, A a^{4} \cos \left (d x + c\right )^{5} + 48 \, {\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{4} + 10 \, {\left (41 \, A + 24 \, B\right )} a^{4} \cos \left (d x + c\right )^{3} + 32 \, {\left (18 \, A + 17 \, B\right )} a^{4} \cos \left (d x + c\right )^{2} + 105 \, {\left (7 \, A + 8 \, B\right )} a^{4} \cos \left (d x + c\right ) + 16 \, {\left (72 \, A + 83 \, B\right )} a^{4}\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/240*(105*(7*A + 8*B)*a^4*d*x + (40*A*a^4*cos(d*x + c)^5 + 48*(4*A + B)*a^4*cos(d*x + c)^4 + 10*(41*A + 24*B)
*a^4*cos(d*x + c)^3 + 32*(18*A + 17*B)*a^4*cos(d*x + c)^2 + 105*(7*A + 8*B)*a^4*cos(d*x + c) + 16*(72*A + 83*B
)*a^4)*sin(d*x + c))/d

________________________________________________________________________________________

giac [A]  time = 0.93, size = 244, normalized size = 1.11 \[ \frac {105 \, {\left (7 \, A a^{4} + 8 \, B a^{4}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (735 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 840 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 4165 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 4760 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 9702 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 11088 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 11802 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 13488 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 7355 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9320 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3105 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3000 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/240*(105*(7*A*a^4 + 8*B*a^4)*(d*x + c) + 2*(735*A*a^4*tan(1/2*d*x + 1/2*c)^11 + 840*B*a^4*tan(1/2*d*x + 1/2*
c)^11 + 4165*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 4760*B*a^4*tan(1/2*d*x + 1/2*c)^9 + 9702*A*a^4*tan(1/2*d*x + 1/2*c
)^7 + 11088*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 11802*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 13488*B*a^4*tan(1/2*d*x + 1/2*
c)^5 + 7355*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 9320*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 3105*A*a^4*tan(1/2*d*x + 1/2*c)
+ 3000*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d

________________________________________________________________________________________

maple [A]  time = 2.00, size = 306, normalized size = 1.39 \[ \frac {A \,a^{4} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {a^{4} B \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {4 A \,a^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+4 a^{4} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+6 A \,a^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a^{4} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\frac {4 A \,a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+4 a^{4} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} B \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x)

[Out]

1/d*(A*a^4*(1/6*(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c)+1/5*a^4*B*(8/3+cos
(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+4/5*A*a^4*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+4*a^4*B*(1/4*(
cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+6*A*a^4*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3
/8*d*x+3/8*c)+2*a^4*B*(2+cos(d*x+c)^2)*sin(d*x+c)+4/3*A*a^4*(2+cos(d*x+c)^2)*sin(d*x+c)+4*a^4*B*(1/2*cos(d*x+c
)*sin(d*x+c)+1/2*d*x+1/2*c)+A*a^4*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+a^4*B*sin(d*x+c))

________________________________________________________________________________________

maxima [A]  time = 0.35, size = 297, normalized size = 1.35 \[ \frac {256 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{4} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 1280 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} + 180 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{4} - 1920 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{4} + 120 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 960 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{4} + 960 \, B a^{4} \sin \left (d x + c\right )}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/960*(256*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^4 - 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^4 - 1280*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 + 180*(
12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^4 + 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 + 64
*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*B*a^4 - 1920*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^4
+ 120*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^4 + 960*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*
a^4 + 960*B*a^4*sin(d*x + c))/d

________________________________________________________________________________________

mupad [B]  time = 4.62, size = 286, normalized size = 1.30 \[ \frac {\left (\frac {49\,A\,a^4}{8}+7\,B\,a^4\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {833\,A\,a^4}{24}+\frac {119\,B\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {1617\,A\,a^4}{20}+\frac {462\,B\,a^4}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {1967\,A\,a^4}{20}+\frac {562\,B\,a^4}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {1471\,A\,a^4}{24}+\frac {233\,B\,a^4}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {207\,A\,a^4}{8}+25\,B\,a^4\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {7\,a^4\,\mathrm {atan}\left (\frac {7\,a^4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (7\,A+8\,B\right )}{8\,\left (\frac {49\,A\,a^4}{8}+7\,B\,a^4\right )}\right )\,\left (7\,A+8\,B\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6*(A + B/cos(c + d*x))*(a + a/cos(c + d*x))^4,x)

[Out]

(tan(c/2 + (d*x)/2)*((207*A*a^4)/8 + 25*B*a^4) + tan(c/2 + (d*x)/2)^11*((49*A*a^4)/8 + 7*B*a^4) + tan(c/2 + (d
*x)/2)^9*((833*A*a^4)/24 + (119*B*a^4)/3) + tan(c/2 + (d*x)/2)^3*((1471*A*a^4)/24 + (233*B*a^4)/3) + tan(c/2 +
(d*x)/2)^7*((1617*A*a^4)/20 + (462*B*a^4)/5) + tan(c/2 + (d*x)/2)^5*((1967*A*a^4)/20 + (562*B*a^4)/5))/(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)) + (7*a^4*atan((7*a^4*tan(c/2 + (d*x)/2)*(7*A + 8*B))/(8*((49*A*a^
4)/8 + 7*B*a^4)))*(7*A + 8*B))/(8*d)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________