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

Optimal. Leaf size=309 \[ \frac {a^2 \left (25 a^2 A+72 a b B+48 A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{120 d}-\frac {a \left (4 a^3 B+16 a^2 A b+27 a b^2 B+13 A b^3\right ) \sin ^3(c+d x)}{15 d}+\frac {\left (12 a^4 B+48 a^3 A b+87 a^2 b^2 B+53 a A b^3+15 b^4 B\right ) \sin (c+d x)}{15 d}+\frac {\left (5 a^4 A+24 a^3 b B+36 a^2 A b^2+32 a b^3 B+8 A b^4\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x \left (5 a^4 A+24 a^3 b B+36 a^2 A b^2+32 a b^3 B+8 A b^4\right )+\frac {a (2 a B+3 A b) \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{10 d}+\frac {a A \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^3}{6 d} \]

[Out]

1/16*(5*A*a^4+36*A*a^2*b^2+8*A*b^4+24*B*a^3*b+32*B*a*b^3)*x+1/15*(48*A*a^3*b+53*A*a*b^3+12*B*a^4+87*B*a^2*b^2+
15*B*b^4)*sin(d*x+c)/d+1/16*(5*A*a^4+36*A*a^2*b^2+8*A*b^4+24*B*a^3*b+32*B*a*b^3)*cos(d*x+c)*sin(d*x+c)/d+1/120
*a^2*(25*A*a^2+48*A*b^2+72*B*a*b)*cos(d*x+c)^3*sin(d*x+c)/d+1/10*a*(3*A*b+2*B*a)*cos(d*x+c)^4*(a+b*sec(d*x+c))
^2*sin(d*x+c)/d+1/6*a*A*cos(d*x+c)^5*(a+b*sec(d*x+c))^3*sin(d*x+c)/d-1/15*a*(16*A*a^2*b+13*A*b^3+4*B*a^3+27*B*
a*b^2)*sin(d*x+c)^3/d

________________________________________________________________________________________

Rubi [A]  time = 0.82, antiderivative size = 309, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {4025, 4094, 4074, 4047, 2635, 8, 4044, 3013} \[ -\frac {a \left (16 a^2 A b+4 a^3 B+27 a b^2 B+13 A b^3\right ) \sin ^3(c+d x)}{15 d}+\frac {\left (48 a^3 A b+87 a^2 b^2 B+12 a^4 B+53 a A b^3+15 b^4 B\right ) \sin (c+d x)}{15 d}+\frac {a^2 \left (25 a^2 A+72 a b B+48 A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{120 d}+\frac {\left (36 a^2 A b^2+5 a^4 A+24 a^3 b B+32 a b^3 B+8 A b^4\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} x \left (36 a^2 A b^2+5 a^4 A+24 a^3 b B+32 a b^3 B+8 A b^4\right )+\frac {a (2 a B+3 A b) \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{10 d}+\frac {a A \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^3}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((5*a^4*A + 36*a^2*A*b^2 + 8*A*b^4 + 24*a^3*b*B + 32*a*b^3*B)*x)/16 + ((48*a^3*A*b + 53*a*A*b^3 + 12*a^4*B + 8
7*a^2*b^2*B + 15*b^4*B)*Sin[c + d*x])/(15*d) + ((5*a^4*A + 36*a^2*A*b^2 + 8*A*b^4 + 24*a^3*b*B + 32*a*b^3*B)*C
os[c + d*x]*Sin[c + d*x])/(16*d) + (a^2*(25*a^2*A + 48*A*b^2 + 72*a*b*B)*Cos[c + d*x]^3*Sin[c + d*x])/(120*d)
+ (a*(3*A*b + 2*a*B)*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(10*d) + (a*A*Cos[c + d*x]^5*(a + b*S
ec[c + d*x])^3*Sin[c + d*x])/(6*d) - (a*(16*a^2*A*b + 13*A*b^3 + 4*a^3*B + 27*a*b^2*B)*Sin[c + d*x]^3)/(15*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 3013

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Dist[f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rule 4025

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[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[a*(a*B*n - A*b*(m - n - 1)) + (
2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; Free
Q[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LeQ[n, -1]

Rule 4044

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*
x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[{e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[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]

Rule 4074

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_)), 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) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rule 4094

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] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rubi steps

\begin {align*} \int \cos ^6(c+d x) (a+b \sec (c+d x))^4 (A+B \sec (c+d x)) \, dx &=\frac {a A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}-\frac {1}{6} \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (-3 a (3 A b+2 a B)-\left (5 a^2 A+6 A b^2+12 a b B\right ) \sec (c+d x)-2 b (a A+3 b B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a (3 A b+2 a B) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac {a A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}-\frac {1}{30} \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (-a \left (25 a^2 A+48 A b^2+72 a b B\right )-\left (71 a^2 A b+30 A b^3+24 a^3 B+90 a b^2 B\right ) \sec (c+d x)-2 b \left (14 a A b+6 a^2 B+15 b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac {a (3 A b+2 a B) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac {a A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {1}{120} \int \cos ^3(c+d x) \left (24 a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right )+15 \left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \sec (c+d x)+8 b^2 \left (14 a A b+6 a^2 B+15 b^2 B\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac {a (3 A b+2 a B) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac {a A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {1}{120} \int \cos ^3(c+d x) \left (24 a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right )+8 b^2 \left (14 a A b+6 a^2 B+15 b^2 B\right ) \sec ^2(c+d x)\right ) \, dx+\frac {1}{8} \left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {\left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac {a (3 A b+2 a B) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac {a A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {1}{120} \int \cos (c+d x) \left (8 b^2 \left (14 a A b+6 a^2 B+15 b^2 B\right )+24 a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right ) \cos ^2(c+d x)\right ) \, dx+\frac {1}{16} \left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \int 1 \, dx\\ &=\frac {1}{16} \left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) x+\frac {\left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac {a (3 A b+2 a B) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac {a A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}-\frac {\operatorname {Subst}\left (\int \left (8 b^2 \left (14 a A b+6 a^2 B+15 b^2 B\right )+24 a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right )-24 a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right ) x^2\right ) \, dx,x,-\sin (c+d x)\right )}{120 d}\\ &=\frac {1}{16} \left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) x+\frac {\left (48 a^3 A b+53 a A b^3+12 a^4 B+87 a^2 b^2 B+15 b^4 B\right ) \sin (c+d x)}{15 d}+\frac {\left (5 a^4 A+36 a^2 A b^2+8 A b^4+24 a^3 b B+32 a b^3 B\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^2 \left (25 a^2 A+48 A b^2+72 a b B\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac {a (3 A b+2 a B) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac {a A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}-\frac {a \left (16 a^2 A b+13 A b^3+4 a^3 B+27 a b^2 B\right ) \sin ^3(c+d x)}{15 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.24, size = 333, normalized size = 1.08 \[ \frac {45 a^4 A \sin (4 (c+d x))+5 a^4 A \sin (6 (c+d x))+300 a^4 A c+300 a^4 A d x+100 a^4 B \sin (3 (c+d x))+12 a^4 B \sin (5 (c+d x))+400 a^3 A b \sin (3 (c+d x))+48 a^3 A b \sin (5 (c+d x))+120 a^3 b B \sin (4 (c+d x))+1440 a^3 b B c+1440 a^3 b B d x+180 a^2 A b^2 \sin (4 (c+d x))+2160 a^2 A b^2 c+2160 a^2 A b^2 d x+480 a^2 b^2 B \sin (3 (c+d x))+120 \left (5 a^4 B+20 a^3 A b+36 a^2 b^2 B+24 a A b^3+8 b^4 B\right ) \sin (c+d x)+15 \left (15 a^4 A+64 a^3 b B+96 a^2 A b^2+64 a b^3 B+16 A b^4\right ) \sin (2 (c+d x))+320 a A b^3 \sin (3 (c+d x))+1920 a b^3 B c+1920 a b^3 B d x+480 A b^4 c+480 A b^4 d x}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(300*a^4*A*c + 2160*a^2*A*b^2*c + 480*A*b^4*c + 1440*a^3*b*B*c + 1920*a*b^3*B*c + 300*a^4*A*d*x + 2160*a^2*A*b
^2*d*x + 480*A*b^4*d*x + 1440*a^3*b*B*d*x + 1920*a*b^3*B*d*x + 120*(20*a^3*A*b + 24*a*A*b^3 + 5*a^4*B + 36*a^2
*b^2*B + 8*b^4*B)*Sin[c + d*x] + 15*(15*a^4*A + 96*a^2*A*b^2 + 16*A*b^4 + 64*a^3*b*B + 64*a*b^3*B)*Sin[2*(c +
d*x)] + 400*a^3*A*b*Sin[3*(c + d*x)] + 320*a*A*b^3*Sin[3*(c + d*x)] + 100*a^4*B*Sin[3*(c + d*x)] + 480*a^2*b^2
*B*Sin[3*(c + d*x)] + 45*a^4*A*Sin[4*(c + d*x)] + 180*a^2*A*b^2*Sin[4*(c + d*x)] + 120*a^3*b*B*Sin[4*(c + d*x)
] + 48*a^3*A*b*Sin[5*(c + d*x)] + 12*a^4*B*Sin[5*(c + d*x)] + 5*a^4*A*Sin[6*(c + d*x)])/(960*d)

________________________________________________________________________________________

fricas [A]  time = 0.47, size = 243, normalized size = 0.79 \[ \frac {15 \, {\left (5 \, A a^{4} + 24 \, B a^{3} b + 36 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} d x + {\left (40 \, A a^{4} \cos \left (d x + c\right )^{5} + 128 \, B a^{4} + 512 \, A a^{3} b + 960 \, B a^{2} b^{2} + 640 \, A a b^{3} + 240 \, B b^{4} + 48 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} \cos \left (d x + c\right )^{4} + 10 \, {\left (5 \, A a^{4} + 24 \, B a^{3} b + 36 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} + 32 \, {\left (2 \, B a^{4} + 8 \, A a^{3} b + 15 \, B a^{2} b^{2} + 10 \, A a b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (5 \, A a^{4} + 24 \, B a^{3} b + 36 \, A a^{2} b^{2} + 32 \, B a b^{3} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(5*A*a^4 + 24*B*a^3*b + 36*A*a^2*b^2 + 32*B*a*b^3 + 8*A*b^4)*d*x + (40*A*a^4*cos(d*x + c)^5 + 128*B*
a^4 + 512*A*a^3*b + 960*B*a^2*b^2 + 640*A*a*b^3 + 240*B*b^4 + 48*(B*a^4 + 4*A*a^3*b)*cos(d*x + c)^4 + 10*(5*A*
a^4 + 24*B*a^3*b + 36*A*a^2*b^2)*cos(d*x + c)^3 + 32*(2*B*a^4 + 8*A*a^3*b + 15*B*a^2*b^2 + 10*A*a*b^3)*cos(d*x
+ c)^2 + 15*(5*A*a^4 + 24*B*a^3*b + 36*A*a^2*b^2 + 32*B*a*b^3 + 8*A*b^4)*cos(d*x + c))*sin(d*x + c))/d

________________________________________________________________________________________

giac [B]  time = 3.05, size = 1127, normalized size = 3.65 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(5*A*a^4 + 24*B*a^3*b + 36*A*a^2*b^2 + 32*B*a*b^3 + 8*A*b^4)*(d*x + c) - 2*(165*A*a^4*tan(1/2*d*x +
1/2*c)^11 - 240*B*a^4*tan(1/2*d*x + 1/2*c)^11 - 960*A*a^3*b*tan(1/2*d*x + 1/2*c)^11 + 600*B*a^3*b*tan(1/2*d*x
+ 1/2*c)^11 + 900*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 - 1440*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 - 960*A*a*b^3*tan
(1/2*d*x + 1/2*c)^11 + 480*B*a*b^3*tan(1/2*d*x + 1/2*c)^11 + 120*A*b^4*tan(1/2*d*x + 1/2*c)^11 - 240*B*b^4*tan
(1/2*d*x + 1/2*c)^11 - 25*A*a^4*tan(1/2*d*x + 1/2*c)^9 - 560*B*a^4*tan(1/2*d*x + 1/2*c)^9 - 2240*A*a^3*b*tan(1
/2*d*x + 1/2*c)^9 + 840*B*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 1260*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 5280*B*a^2*b^
2*tan(1/2*d*x + 1/2*c)^9 - 3520*A*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 1440*B*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 360*A*b
^4*tan(1/2*d*x + 1/2*c)^9 - 1200*B*b^4*tan(1/2*d*x + 1/2*c)^9 + 450*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 1248*B*a^4*
tan(1/2*d*x + 1/2*c)^7 - 4992*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 240*B*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 360*A*a^2*
b^2*tan(1/2*d*x + 1/2*c)^7 - 8640*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 5760*A*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 960
*B*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 240*A*b^4*tan(1/2*d*x + 1/2*c)^7 - 2400*B*b^4*tan(1/2*d*x + 1/2*c)^7 - 450*A
*a^4*tan(1/2*d*x + 1/2*c)^5 - 1248*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 4992*A*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 240*B*
a^3*b*tan(1/2*d*x + 1/2*c)^5 - 360*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 8640*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 -
5760*A*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 960*B*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 240*A*b^4*tan(1/2*d*x + 1/2*c)^5 -
2400*B*b^4*tan(1/2*d*x + 1/2*c)^5 + 25*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 560*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 2240*
A*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 840*B*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 1260*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 -
5280*B*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 3520*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 1440*B*a*b^3*tan(1/2*d*x + 1/2*c
)^3 - 360*A*b^4*tan(1/2*d*x + 1/2*c)^3 - 1200*B*b^4*tan(1/2*d*x + 1/2*c)^3 - 165*A*a^4*tan(1/2*d*x + 1/2*c) -
240*B*a^4*tan(1/2*d*x + 1/2*c) - 960*A*a^3*b*tan(1/2*d*x + 1/2*c) - 600*B*a^3*b*tan(1/2*d*x + 1/2*c) - 900*A*a
^2*b^2*tan(1/2*d*x + 1/2*c) - 1440*B*a^2*b^2*tan(1/2*d*x + 1/2*c) - 960*A*a*b^3*tan(1/2*d*x + 1/2*c) - 480*B*a
*b^3*tan(1/2*d*x + 1/2*c) - 120*A*b^4*tan(1/2*d*x + 1/2*c) - 240*B*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/
2*c)^2 + 1)^6)/d

________________________________________________________________________________________

maple [A]  time = 1.85, size = 316, normalized size = 1.02 \[ \frac {A \,a^{4} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {a^{4} B \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {4 A \,a^{3} b \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+4 B \,a^{3} b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+6 A \,a^{2} b^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 a^{2} b^{2} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\frac {4 a A \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+4 B a \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,b^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,b^{4} \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

maxima [A]  time = 0.87, size = 307, normalized size = 0.99 \[ -\frac {5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{4} - 256 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{3} b - 120 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} b - 180 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b^{2} + 1920 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} b^{2} + 1280 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b^{3} - 960 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{3} - 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{4} - 960 \, B b^{4} \sin \left (d x + c\right )}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.18, size = 403, normalized size = 1.30 \[ \frac {5\,A\,a^4\,x}{16}+\frac {A\,b^4\,x}{2}+2\,B\,a\,b^3\,x+\frac {3\,B\,a^3\,b\,x}{2}+\frac {5\,B\,a^4\,\sin \left (c+d\,x\right )}{8\,d}+\frac {B\,b^4\,\sin \left (c+d\,x\right )}{d}+\frac {9\,A\,a^2\,b^2\,x}{4}+\frac {15\,A\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{64\,d}+\frac {3\,A\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{64\,d}+\frac {A\,a^4\,\sin \left (6\,c+6\,d\,x\right )}{192\,d}+\frac {A\,b^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {5\,B\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {B\,a^4\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {A\,a\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{3\,d}+\frac {5\,A\,a^3\,b\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {A\,a^3\,b\,\sin \left (5\,c+5\,d\,x\right )}{20\,d}+\frac {B\,a\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {B\,a^3\,b\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {B\,a^3\,b\,\sin \left (4\,c+4\,d\,x\right )}{8\,d}+\frac {9\,B\,a^2\,b^2\,\sin \left (c+d\,x\right )}{2\,d}+\frac {3\,A\,a^2\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {3\,A\,a^2\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{16\,d}+\frac {B\,a^2\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{2\,d}+\frac {3\,A\,a\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,A\,a^3\,b\,\sin \left (c+d\,x\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*A*a^4*x)/16 + (A*b^4*x)/2 + 2*B*a*b^3*x + (3*B*a^3*b*x)/2 + (5*B*a^4*sin(c + d*x))/(8*d) + (B*b^4*sin(c + d
*x))/d + (9*A*a^2*b^2*x)/4 + (15*A*a^4*sin(2*c + 2*d*x))/(64*d) + (3*A*a^4*sin(4*c + 4*d*x))/(64*d) + (A*a^4*s
in(6*c + 6*d*x))/(192*d) + (A*b^4*sin(2*c + 2*d*x))/(4*d) + (5*B*a^4*sin(3*c + 3*d*x))/(48*d) + (B*a^4*sin(5*c
+ 5*d*x))/(80*d) + (A*a*b^3*sin(3*c + 3*d*x))/(3*d) + (5*A*a^3*b*sin(3*c + 3*d*x))/(12*d) + (A*a^3*b*sin(5*c
+ 5*d*x))/(20*d) + (B*a*b^3*sin(2*c + 2*d*x))/d + (B*a^3*b*sin(2*c + 2*d*x))/d + (B*a^3*b*sin(4*c + 4*d*x))/(8
*d) + (9*B*a^2*b^2*sin(c + d*x))/(2*d) + (3*A*a^2*b^2*sin(2*c + 2*d*x))/(2*d) + (3*A*a^2*b^2*sin(4*c + 4*d*x))
/(16*d) + (B*a^2*b^2*sin(3*c + 3*d*x))/(2*d) + (3*A*a*b^3*sin(c + d*x))/d + (5*A*a^3*b*sin(c + d*x))/(2*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+b*sec(d*x+c))**4*(A+B*sec(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________