3.662 \(\int \cos ^6(c+d x) (a+b \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=257 \[ -\frac{b \left (3 a^2 (4 A+5 C)+A b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{b \left (9 a^2 (4 A+5 C)+b^2 (11 A+15 C)\right ) \sin (c+d x)}{15 d}+\frac{a \left (5 a^2 (5 A+6 C)+6 A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{120 d}+\frac{a \left (a^2 (5 A+6 C)+6 b^2 (3 A+4 C)\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} a x \left (a^2 (5 A+6 C)+6 b^2 (3 A+4 C)\right )+\frac{A \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^3}{6 d}+\frac{A b \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{10 d} \]

[Out]

(a*(6*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*x)/16 + (b*(9*a^2*(4*A + 5*C) + b^2*(11*A + 15*C))*Sin[c + d*x])/(15*
d) + (a*(6*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (a*(6*A*b^2 + 5*a^2*(5*A + 6
*C))*Cos[c + d*x]^3*Sin[c + d*x])/(120*d) + (A*b*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(10*d) +
(A*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(6*d) - (b*(A*b^2 + 3*a^2*(4*A + 5*C))*Sin[c + d*x]^3)/
(15*d)

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Rubi [A]  time = 0.752351, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {4095, 4094, 4074, 4047, 2635, 8, 4044, 3013} \[ -\frac{b \left (3 a^2 (4 A+5 C)+A b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{b \left (9 a^2 (4 A+5 C)+b^2 (11 A+15 C)\right ) \sin (c+d x)}{15 d}+\frac{a \left (5 a^2 (5 A+6 C)+6 A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{120 d}+\frac{a \left (a^2 (5 A+6 C)+6 b^2 (3 A+4 C)\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} a x \left (a^2 (5 A+6 C)+6 b^2 (3 A+4 C)\right )+\frac{A \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^3}{6 d}+\frac{A b \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{10 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]

[Out]

(a*(6*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*x)/16 + (b*(9*a^2*(4*A + 5*C) + b^2*(11*A + 15*C))*Sin[c + d*x])/(15*
d) + (a*(6*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (a*(6*A*b^2 + 5*a^2*(5*A + 6
*C))*Cos[c + d*x]^3*Sin[c + d*x])/(120*d) + (A*b*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(10*d) +
(A*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(6*d) - (b*(A*b^2 + 3*a^2*(4*A + 5*C))*Sin[c + d*x]^3)/
(15*d)

Rule 4095

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

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 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 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 8

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

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 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]

Rubi steps

\begin{align*} \int \cos ^6(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{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 b+a (5 A+6 C) \sec (c+d x)+2 b (A+3 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{A b \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac{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 (6 A b^2+5 a^2 (5 A+6 C)+a b (47 A+60 C) \sec (c+d x)+2 b^2 (8 A+15 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a \left (6 A b^2+5 a^2 (5 A+6 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac{A b \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac{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 b \left (A b^2+3 a^2 (4 A+5 C)\right )-15 a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \sec (c+d x)-8 b^3 (8 A+15 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a \left (6 A b^2+5 a^2 (5 A+6 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac{A b \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac{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 b \left (A b^2+3 a^2 (4 A+5 C)\right )-8 b^3 (8 A+15 C) \sec ^2(c+d x)\right ) \, dx+\frac{1}{8} \left (a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right )\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a \left (6 A b^2+5 a^2 (5 A+6 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac{A b \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac{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^3 (8 A+15 C)-24 b \left (A b^2+3 a^2 (4 A+5 C)\right ) \cos ^2(c+d x)\right ) \, dx+\frac{1}{16} \left (a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right )\right ) \int 1 \, dx\\ &=\frac{1}{16} a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) x+\frac{a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a \left (6 A b^2+5 a^2 (5 A+6 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac{A b \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac{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^3 (8 A+15 C)-24 b \left (A b^2+3 a^2 (4 A+5 C)\right )+24 b \left (A b^2+3 a^2 (4 A+5 C)\right ) x^2\right ) \, dx,x,-\sin (c+d x)\right )}{120 d}\\ &=\frac{1}{16} a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) x+\frac{b \left (9 a^2 (4 A+5 C)+b^2 (11 A+15 C)\right ) \sin (c+d x)}{15 d}+\frac{a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a \left (6 A b^2+5 a^2 (5 A+6 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac{A b \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac{A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}-\frac{b \left (A b^2+3 a^2 (4 A+5 C)\right ) \sin ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 1.12658, size = 253, normalized size = 0.98 \[ \frac{15 a \left (a^2 (15 A+16 C)+48 b^2 (A+C)\right ) \sin (2 (c+d x))+120 b \left (3 a^2 (5 A+6 C)+2 b^2 (3 A+4 C)\right ) \sin (c+d x)+300 a^2 A b \sin (3 (c+d x))+36 a^2 A b \sin (5 (c+d x))+45 a^3 A \sin (4 (c+d x))+5 a^3 A \sin (6 (c+d x))+300 a^3 A c+300 a^3 A d x+240 a^2 b C \sin (3 (c+d x))+30 a^3 C \sin (4 (c+d x))+360 a^3 c C+360 a^3 C d x+90 a A b^2 \sin (4 (c+d x))+1080 a A b^2 c+1080 a A b^2 d x+1440 a b^2 c C+1440 a b^2 C d x+80 A b^3 \sin (3 (c+d x))}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + b*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]

[Out]

(300*a^3*A*c + 1080*a*A*b^2*c + 360*a^3*c*C + 1440*a*b^2*c*C + 300*a^3*A*d*x + 1080*a*A*b^2*d*x + 360*a^3*C*d*
x + 1440*a*b^2*C*d*x + 120*b*(2*b^2*(3*A + 4*C) + 3*a^2*(5*A + 6*C))*Sin[c + d*x] + 15*a*(48*b^2*(A + C) + a^2
*(15*A + 16*C))*Sin[2*(c + d*x)] + 300*a^2*A*b*Sin[3*(c + d*x)] + 80*A*b^3*Sin[3*(c + d*x)] + 240*a^2*b*C*Sin[
3*(c + d*x)] + 45*a^3*A*Sin[4*(c + d*x)] + 90*a*A*b^2*Sin[4*(c + d*x)] + 30*a^3*C*Sin[4*(c + d*x)] + 36*a^2*A*
b*Sin[5*(c + d*x)] + 5*a^3*A*Sin[6*(c + d*x)])/(960*d)

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Maple [A]  time = 0.084, size = 249, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( A{a}^{3} \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) +{a}^{3}C \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +{\frac{3\,A{a}^{2}b\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+{a}^{2}bC \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +3\,Aa{b}^{2} \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +3\,Ca{b}^{2} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{\frac{A{b}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+C{b}^{3}\sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x)

[Out]

1/d*(A*a^3*(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)+a^3*C*(1/4*(cos(d*
x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+3/5*A*a^2*b*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+a^
2*b*C*(2+cos(d*x+c)^2)*sin(d*x+c)+3*A*a*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)+3*C*a
*b^2*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+1/3*A*b^3*(2+cos(d*x+c)^2)*sin(d*x+c)+C*b^3*sin(d*x+c))

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Maxima [A]  time = 1.02243, size = 328, normalized size = 1.28 \begin{align*} -\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^{3} - 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^{3} - 192 \,{\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} b + 960 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} b - 90 \,{\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 b^{2} - 720 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{2} + 320 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{3} - 960 \, C b^{3} \sin \left (d x + c\right )}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),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^3 - 30*(12*d*x
+ 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^3 - 192*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*
x + c))*A*a^2*b + 960*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^2*b - 90*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin
(2*d*x + 2*c))*A*a*b^2 - 720*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a*b^2 + 320*(sin(d*x + c)^3 - 3*sin(d*x + c))*
A*b^3 - 960*C*b^3*sin(d*x + c))/d

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Fricas [A]  time = 0.553126, size = 450, normalized size = 1.75 \begin{align*} \frac{15 \,{\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 6 \,{\left (3 \, A + 4 \, C\right )} a b^{2}\right )} d x +{\left (40 \, A a^{3} \cos \left (d x + c\right )^{5} + 144 \, A a^{2} b \cos \left (d x + c\right )^{4} + 96 \,{\left (4 \, A + 5 \, C\right )} a^{2} b + 80 \,{\left (2 \, A + 3 \, C\right )} b^{3} + 10 \,{\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 18 \, A a b^{2}\right )} \cos \left (d x + c\right )^{3} + 16 \,{\left (3 \,{\left (4 \, A + 5 \, C\right )} a^{2} b + 5 \, A b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 6 \,{\left (3 \, A + 4 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(15*((5*A + 6*C)*a^3 + 6*(3*A + 4*C)*a*b^2)*d*x + (40*A*a^3*cos(d*x + c)^5 + 144*A*a^2*b*cos(d*x + c)^4
+ 96*(4*A + 5*C)*a^2*b + 80*(2*A + 3*C)*b^3 + 10*((5*A + 6*C)*a^3 + 18*A*a*b^2)*cos(d*x + c)^3 + 16*(3*(4*A +
5*C)*a^2*b + 5*A*b^3)*cos(d*x + c)^2 + 15*((5*A + 6*C)*a^3 + 6*(3*A + 4*C)*a*b^2)*cos(d*x + c))*sin(d*x + c))/
d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*(a+b*sec(d*x+c))**3*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.26488, size = 1191, normalized size = 4.63 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/240*(15*(5*A*a^3 + 6*C*a^3 + 18*A*a*b^2 + 24*C*a*b^2)*(d*x + c) - 2*(165*A*a^3*tan(1/2*d*x + 1/2*c)^11 + 150
*C*a^3*tan(1/2*d*x + 1/2*c)^11 - 720*A*a^2*b*tan(1/2*d*x + 1/2*c)^11 - 720*C*a^2*b*tan(1/2*d*x + 1/2*c)^11 + 4
50*A*a*b^2*tan(1/2*d*x + 1/2*c)^11 + 360*C*a*b^2*tan(1/2*d*x + 1/2*c)^11 - 240*A*b^3*tan(1/2*d*x + 1/2*c)^11 -
240*C*b^3*tan(1/2*d*x + 1/2*c)^11 - 25*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 210*C*a^3*tan(1/2*d*x + 1/2*c)^9 - 1680
*A*a^2*b*tan(1/2*d*x + 1/2*c)^9 - 2640*C*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 630*A*a*b^2*tan(1/2*d*x + 1/2*c)^9 + 1
080*C*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 880*A*b^3*tan(1/2*d*x + 1/2*c)^9 - 1200*C*b^3*tan(1/2*d*x + 1/2*c)^9 + 45
0*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 60*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 3744*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 4320*
C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 180*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 720*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 144
0*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 2400*C*b^3*tan(1/2*d*x + 1/2*c)^7 - 450*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 60*C*a
^3*tan(1/2*d*x + 1/2*c)^5 - 3744*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 4320*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 180*A*
a*b^2*tan(1/2*d*x + 1/2*c)^5 - 720*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 1440*A*b^3*tan(1/2*d*x + 1/2*c)^5 - 2400*C
*b^3*tan(1/2*d*x + 1/2*c)^5 + 25*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 210*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 1680*A*a^2*
b*tan(1/2*d*x + 1/2*c)^3 - 2640*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 630*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 1080*C*a
*b^2*tan(1/2*d*x + 1/2*c)^3 - 880*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 1200*C*b^3*tan(1/2*d*x + 1/2*c)^3 - 165*A*a^3
*tan(1/2*d*x + 1/2*c) - 150*C*a^3*tan(1/2*d*x + 1/2*c) - 720*A*a^2*b*tan(1/2*d*x + 1/2*c) - 720*C*a^2*b*tan(1/
2*d*x + 1/2*c) - 450*A*a*b^2*tan(1/2*d*x + 1/2*c) - 360*C*a*b^2*tan(1/2*d*x + 1/2*c) - 240*A*b^3*tan(1/2*d*x +
1/2*c) - 240*C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d