3.120 \(\int \cos ^7(c+d x) (a+a \sec (c+d x))^4 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=254 \[ \frac{a^4 (454 A+581 C) \sin (c+d x)}{105 d}+\frac{a^4 (247 A+308 C) \sin (c+d x) \cos ^2(c+d x)}{210 d}+\frac{a^4 (11 A+14 C) \sin (c+d x) \cos (c+d x)}{4 d}+\frac{(8 A+7 C) \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{35 d}+\frac{(109 A+126 C) \sin (c+d x) \cos ^3(c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{210 d}+\frac{1}{4} a^4 x (11 A+14 C)+\frac{A \sin (c+d x) \cos ^6(c+d x) (a \sec (c+d x)+a)^4}{7 d}+\frac{2 a A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^3}{21 d} \]

[Out]

(a^4*(11*A + 14*C)*x)/4 + (a^4*(454*A + 581*C)*Sin[c + d*x])/(105*d) + (a^4*(11*A + 14*C)*Cos[c + d*x]*Sin[c +
d*x])/(4*d) + (a^4*(247*A + 308*C)*Cos[c + d*x]^2*Sin[c + d*x])/(210*d) + (2*a*A*Cos[c + d*x]^5*(a + a*Sec[c
+ d*x])^3*Sin[c + d*x])/(21*d) + (A*Cos[c + d*x]^6*(a + a*Sec[c + d*x])^4*Sin[c + d*x])/(7*d) + ((8*A + 7*C)*C
os[c + d*x]^4*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(35*d) + ((109*A + 126*C)*Cos[c + d*x]^3*(a^4 + a^4*Sec
[c + d*x])*Sin[c + d*x])/(210*d)

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Rubi [A]  time = 0.718647, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4087, 4017, 3996, 3787, 2635, 8, 2637} \[ \frac{a^4 (454 A+581 C) \sin (c+d x)}{105 d}+\frac{a^4 (247 A+308 C) \sin (c+d x) \cos ^2(c+d x)}{210 d}+\frac{a^4 (11 A+14 C) \sin (c+d x) \cos (c+d x)}{4 d}+\frac{(8 A+7 C) \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{35 d}+\frac{(109 A+126 C) \sin (c+d x) \cos ^3(c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{210 d}+\frac{1}{4} a^4 x (11 A+14 C)+\frac{A \sin (c+d x) \cos ^6(c+d x) (a \sec (c+d x)+a)^4}{7 d}+\frac{2 a A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^3}{21 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^4*(11*A + 14*C)*x)/4 + (a^4*(454*A + 581*C)*Sin[c + d*x])/(105*d) + (a^4*(11*A + 14*C)*Cos[c + d*x]*Sin[c +
d*x])/(4*d) + (a^4*(247*A + 308*C)*Cos[c + d*x]^2*Sin[c + d*x])/(210*d) + (2*a*A*Cos[c + d*x]^5*(a + a*Sec[c
+ d*x])^3*Sin[c + d*x])/(21*d) + (A*Cos[c + d*x]^6*(a + a*Sec[c + d*x])^4*Sin[c + d*x])/(7*d) + ((8*A + 7*C)*C
os[c + d*x]^4*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(35*d) + ((109*A + 126*C)*Cos[c + d*x]^3*(a^4 + a^4*Sec
[c + d*x])*Sin[c + d*x])/(210*d)

Rule 4087

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])

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]

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 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 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 2637

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

Rubi steps

\begin{align*} \int \cos ^7(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^6(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac{\int \cos ^6(c+d x) (a+a \sec (c+d x))^4 (4 a A+a (2 A+7 C) \sec (c+d x)) \, dx}{7 a}\\ &=\frac{2 a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{A \cos ^6(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac{\int \cos ^5(c+d x) (a+a \sec (c+d x))^3 \left (6 a^2 (8 A+7 C)+2 a^2 (10 A+21 C) \sec (c+d x)\right ) \, dx}{42 a}\\ &=\frac{2 a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{A \cos ^6(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac{(8 A+7 C) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 d}+\frac{\int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (4 a^3 (109 A+126 C)+98 a^3 (2 A+3 C) \sec (c+d x)\right ) \, dx}{210 a}\\ &=\frac{2 a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{A \cos ^6(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac{(8 A+7 C) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 d}+\frac{(109 A+126 C) \cos ^3(c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{210 d}+\frac{\int \cos ^3(c+d x) (a+a \sec (c+d x)) \left (12 a^4 (247 A+308 C)+24 a^4 (69 A+91 C) \sec (c+d x)\right ) \, dx}{840 a}\\ &=\frac{a^4 (247 A+308 C) \cos ^2(c+d x) \sin (c+d x)}{210 d}+\frac{2 a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{A \cos ^6(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac{(8 A+7 C) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 d}+\frac{(109 A+126 C) \cos ^3(c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{210 d}-\frac{\int \cos ^2(c+d x) \left (-1260 a^5 (11 A+14 C)-24 a^5 (454 A+581 C) \sec (c+d x)\right ) \, dx}{2520 a}\\ &=\frac{a^4 (247 A+308 C) \cos ^2(c+d x) \sin (c+d x)}{210 d}+\frac{2 a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{A \cos ^6(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac{(8 A+7 C) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 d}+\frac{(109 A+126 C) \cos ^3(c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{210 d}+\frac{1}{2} \left (a^4 (11 A+14 C)\right ) \int \cos ^2(c+d x) \, dx+\frac{1}{105} \left (a^4 (454 A+581 C)\right ) \int \cos (c+d x) \, dx\\ &=\frac{a^4 (454 A+581 C) \sin (c+d x)}{105 d}+\frac{a^4 (11 A+14 C) \cos (c+d x) \sin (c+d x)}{4 d}+\frac{a^4 (247 A+308 C) \cos ^2(c+d x) \sin (c+d x)}{210 d}+\frac{2 a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{A \cos ^6(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac{(8 A+7 C) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 d}+\frac{(109 A+126 C) \cos ^3(c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{210 d}+\frac{1}{4} \left (a^4 (11 A+14 C)\right ) \int 1 \, dx\\ &=\frac{1}{4} a^4 (11 A+14 C) x+\frac{a^4 (454 A+581 C) \sin (c+d x)}{105 d}+\frac{a^4 (11 A+14 C) \cos (c+d x) \sin (c+d x)}{4 d}+\frac{a^4 (247 A+308 C) \cos ^2(c+d x) \sin (c+d x)}{210 d}+\frac{2 a A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{A \cos ^6(c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac{(8 A+7 C) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{35 d}+\frac{(109 A+126 C) \cos ^3(c+d x) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{210 d}\\ \end{align*}

Mathematica [A]  time = 0.637744, size = 145, normalized size = 0.57 \[ \frac{a^4 (105 (323 A+392 C) \sin (c+d x)+420 (31 A+32 C) \sin (2 (c+d x))+5495 A \sin (3 (c+d x))+2100 A \sin (4 (c+d x))+651 A \sin (5 (c+d x))+140 A \sin (6 (c+d x))+15 A \sin (7 (c+d x))+11760 A c+18480 A d x+4060 C \sin (3 (c+d x))+840 C \sin (4 (c+d x))+84 C \sin (5 (c+d x))+23520 C d x)}{6720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^4*(11760*A*c + 18480*A*d*x + 23520*C*d*x + 105*(323*A + 392*C)*Sin[c + d*x] + 420*(31*A + 32*C)*Sin[2*(c +
d*x)] + 5495*A*Sin[3*(c + d*x)] + 4060*C*Sin[3*(c + d*x)] + 2100*A*Sin[4*(c + d*x)] + 840*C*Sin[4*(c + d*x)] +
651*A*Sin[5*(c + d*x)] + 84*C*Sin[5*(c + d*x)] + 140*A*Sin[6*(c + d*x)] + 15*A*Sin[7*(c + d*x)]))/(6720*d)

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Maple [A]  time = 0.119, size = 322, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ({\frac{A{a}^{4}\sin \left ( dx+c \right ) }{7} \left ({\frac{16}{5}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{6}+{\frac{6\, \left ( \cos \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{5}} \right ) }+{\frac{{a}^{4}C\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 ) }+4\,A{a}^{4} \left ( 1/6\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+5/4\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) \sin \left ( dx+c \right ) +{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) +4\,{a}^{4}C \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 ) +{\frac{6\,A{a}^{4}\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 ) }+2\,{a}^{4}C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +4\,A{a}^{4} \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 ) +4\,{a}^{4}C \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{\frac{A{a}^{4} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{a}^{4}C\sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x)

[Out]

1/d*(1/7*A*a^4*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x+c)^2)*sin(d*x+c)+1/5*a^4*C*(8/3+cos(d*x+c)^4+4/
3*cos(d*x+c)^2)*sin(d*x+c)+4*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)+4*a^4*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)+6/5*A*a^4*(8/3+cos(d*x+c)^4+4/3*cos
(d*x+c)^2)*sin(d*x+c)+2*a^4*C*(2+cos(d*x+c)^2)*sin(d*x+c)+4*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)+4*a^4*C*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+1/3*A*a^4*(2+cos(d*x+c)^2)*sin(d*x+c)+a^4*C
*sin(d*x+c))

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Maxima [A]  time = 0.956173, size = 431, normalized size = 1.7 \begin{align*} -\frac{48 \,{\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} A a^{4} - 672 \,{\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} + 35 \,{\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} + 560 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 210 \,{\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} - 112 \,{\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 )} C a^{4} + 3360 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} - 210 \,{\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^{4} - 1680 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} - 1680 \, C a^{4} \sin \left (d x + c\right )}{1680 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/1680*(48*(5*sin(d*x + c)^7 - 21*sin(d*x + c)^5 + 35*sin(d*x + c)^3 - 35*sin(d*x + c))*A*a^4 - 672*(3*sin(d*
x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^4 + 35*(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 + 560*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 210*(12*d*x + 12*c + sin(
4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^4 - 112*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*a^4
+ 3360*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^4 - 210*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C
*a^4 - 1680*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^4 - 1680*C*a^4*sin(d*x + c))/d

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Fricas [A]  time = 0.525151, size = 377, normalized size = 1.48 \begin{align*} \frac{105 \,{\left (11 \, A + 14 \, C\right )} a^{4} d x +{\left (60 \, A a^{4} \cos \left (d x + c\right )^{6} + 280 \, A a^{4} \cos \left (d x + c\right )^{5} + 12 \,{\left (48 \, A + 7 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 70 \,{\left (11 \, A + 6 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 4 \,{\left (227 \, A + 238 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 105 \,{\left (11 \, A + 14 \, C\right )} a^{4} \cos \left (d x + c\right ) + 4 \,{\left (454 \, A + 581 \, C\right )} a^{4}\right )} \sin \left (d x + c\right )}{420 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/420*(105*(11*A + 14*C)*a^4*d*x + (60*A*a^4*cos(d*x + c)^6 + 280*A*a^4*cos(d*x + c)^5 + 12*(48*A + 7*C)*a^4*c
os(d*x + c)^4 + 70*(11*A + 6*C)*a^4*cos(d*x + c)^3 + 4*(227*A + 238*C)*a^4*cos(d*x + c)^2 + 105*(11*A + 14*C)*
a^4*cos(d*x + c) + 4*(454*A + 581*C)*a^4)*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)**7*(a+a*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.25544, size = 375, normalized size = 1.48 \begin{align*} \frac{105 \,{\left (11 \, A a^{4} + 14 \, C a^{4}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (1155 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{13} + 1470 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{13} + 7700 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 9800 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 21791 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 27734 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 33792 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 43008 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 31521 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 39914 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 14700 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 21560 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5565 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5250 \, C 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 )}^{7}}}{420 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/420*(105*(11*A*a^4 + 14*C*a^4)*(d*x + c) + 2*(1155*A*a^4*tan(1/2*d*x + 1/2*c)^13 + 1470*C*a^4*tan(1/2*d*x +
1/2*c)^13 + 7700*A*a^4*tan(1/2*d*x + 1/2*c)^11 + 9800*C*a^4*tan(1/2*d*x + 1/2*c)^11 + 21791*A*a^4*tan(1/2*d*x
+ 1/2*c)^9 + 27734*C*a^4*tan(1/2*d*x + 1/2*c)^9 + 33792*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 43008*C*a^4*tan(1/2*d*x
+ 1/2*c)^7 + 31521*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 39914*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 14700*A*a^4*tan(1/2*d*
x + 1/2*c)^3 + 21560*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 5565*A*a^4*tan(1/2*d*x + 1/2*c) + 5250*C*a^4*tan(1/2*d*x +
1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^7)/d