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

Optimal. Leaf size=339 \[ -\frac{\left (3 a^2 b^2 (50 A+63 C)+4 a^4 (6 A+7 C)+4 A b^4\right ) \sin ^3(c+d x)}{105 d}+\frac{\left (3 a^2 b^2 (162 A+203 C)+12 a^4 (6 A+7 C)+b^4 (74 A+105 C)\right ) \sin (c+d x)}{105 d}+\frac{a b \left (a^2 (103 A+126 C)+6 A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{210 d}+\frac{a b \left (a^2 (5 A+6 C)+2 b^2 (3 A+4 C)\right ) \sin (c+d x) \cos (c+d x)}{4 d}+\frac{\left (a^2 (6 A+7 C)+2 A b^2\right ) \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{35 d}+\frac{1}{4} a b x \left (a^2 (5 A+6 C)+2 b^2 (3 A+4 C)\right )+\frac{A \sin (c+d x) \cos ^6(c+d x) (a+b \sec (c+d x))^4}{7 d}+\frac{2 A b \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^3}{21 d} \]

[Out]

(a*b*(2*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*x)/4 + ((12*a^4*(6*A + 7*C) + b^4*(74*A + 105*C) + 3*a^2*b^2*(162*A
+ 203*C))*Sin[c + d*x])/(105*d) + (a*b*(2*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*Cos[c + d*x]*Sin[c + d*x])/(4*d)
+ (a*b*(6*A*b^2 + a^2*(103*A + 126*C))*Cos[c + d*x]^3*Sin[c + d*x])/(210*d) + ((2*A*b^2 + a^2*(6*A + 7*C))*Co
s[c + d*x]^4*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(35*d) + (2*A*b*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*Sin[c
+ d*x])/(21*d) + (A*Cos[c + d*x]^6*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(7*d) - ((4*A*b^4 + 4*a^4*(6*A + 7*C)
+ 3*a^2*b^2*(50*A + 63*C))*Sin[c + d*x]^3)/(105*d)

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Rubi [A]  time = 1.15132, antiderivative size = 339, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ -\frac{\left (3 a^2 b^2 (50 A+63 C)+4 a^4 (6 A+7 C)+4 A b^4\right ) \sin ^3(c+d x)}{105 d}+\frac{\left (3 a^2 b^2 (162 A+203 C)+12 a^4 (6 A+7 C)+b^4 (74 A+105 C)\right ) \sin (c+d x)}{105 d}+\frac{a b \left (a^2 (103 A+126 C)+6 A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{210 d}+\frac{a b \left (a^2 (5 A+6 C)+2 b^2 (3 A+4 C)\right ) \sin (c+d x) \cos (c+d x)}{4 d}+\frac{\left (a^2 (6 A+7 C)+2 A b^2\right ) \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{35 d}+\frac{1}{4} a b x \left (a^2 (5 A+6 C)+2 b^2 (3 A+4 C)\right )+\frac{A \sin (c+d x) \cos ^6(c+d x) (a+b \sec (c+d x))^4}{7 d}+\frac{2 A b \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^3}{21 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*b*(2*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*x)/4 + ((12*a^4*(6*A + 7*C) + b^4*(74*A + 105*C) + 3*a^2*b^2*(162*A
+ 203*C))*Sin[c + d*x])/(105*d) + (a*b*(2*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*Cos[c + d*x]*Sin[c + d*x])/(4*d)
+ (a*b*(6*A*b^2 + a^2*(103*A + 126*C))*Cos[c + d*x]^3*Sin[c + d*x])/(210*d) + ((2*A*b^2 + a^2*(6*A + 7*C))*Co
s[c + d*x]^4*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(35*d) + (2*A*b*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*Sin[c
+ d*x])/(21*d) + (A*Cos[c + d*x]^6*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(7*d) - ((4*A*b^4 + 4*a^4*(6*A + 7*C)
+ 3*a^2*b^2*(50*A + 63*C))*Sin[c + d*x]^3)/(105*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 ^7(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^6(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac{1}{7} \int \cos ^6(c+d x) (a+b \sec (c+d x))^3 \left (4 A b+a (6 A+7 C) \sec (c+d x)+b (2 A+7 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{2 A b \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{A \cos ^6(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac{1}{42} \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (6 \left (2 A b^2+a^2 (6 A+7 C)\right )+4 a b (17 A+21 C) \sec (c+d x)+2 b^2 (10 A+21 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{\left (2 A b^2+a^2 (6 A+7 C)\right ) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d}+\frac{2 A b \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{A \cos ^6(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac{1}{210} \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (4 b \left (6 A b^2+a^2 (103 A+126 C)\right )+2 a \left (12 a^2 (6 A+7 C)+b^2 (244 A+315 C)\right ) \sec (c+d x)+2 b \left (6 a^2 (6 A+7 C)+b^2 (62 A+105 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a b \left (6 A b^2+a^2 (103 A+126 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{210 d}+\frac{\left (2 A b^2+a^2 (6 A+7 C)\right ) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d}+\frac{2 A b \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{A \cos ^6(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d}-\frac{1}{840} \int \cos ^3(c+d x) \left (-24 \left (4 A b^4+4 a^4 (6 A+7 C)+3 a^2 b^2 (50 A+63 C)\right )-420 a b \left (2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \sec (c+d x)-8 b^2 \left (6 a^2 (6 A+7 C)+b^2 (62 A+105 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a b \left (6 A b^2+a^2 (103 A+126 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{210 d}+\frac{\left (2 A b^2+a^2 (6 A+7 C)\right ) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d}+\frac{2 A b \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{A \cos ^6(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d}-\frac{1}{840} \int \cos ^3(c+d x) \left (-24 \left (4 A b^4+4 a^4 (6 A+7 C)+3 a^2 b^2 (50 A+63 C)\right )-8 b^2 \left (6 a^2 (6 A+7 C)+b^2 (62 A+105 C)\right ) \sec ^2(c+d x)\right ) \, dx+\frac{1}{2} \left (a b \left (2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right )\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{a b \left (2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{4 d}+\frac{a b \left (6 A b^2+a^2 (103 A+126 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{210 d}+\frac{\left (2 A b^2+a^2 (6 A+7 C)\right ) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d}+\frac{2 A b \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{A \cos ^6(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d}-\frac{1}{840} \int \cos (c+d x) \left (-8 b^2 \left (6 a^2 (6 A+7 C)+b^2 (62 A+105 C)\right )-24 \left (4 A b^4+4 a^4 (6 A+7 C)+3 a^2 b^2 (50 A+63 C)\right ) \cos ^2(c+d x)\right ) \, dx+\frac{1}{4} \left (a b \left (2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right )\right ) \int 1 \, dx\\ &=\frac{1}{4} a b \left (2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) x+\frac{a b \left (2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{4 d}+\frac{a b \left (6 A b^2+a^2 (103 A+126 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{210 d}+\frac{\left (2 A b^2+a^2 (6 A+7 C)\right ) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d}+\frac{2 A b \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{A \cos ^6(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d}+\frac{\operatorname{Subst}\left (\int \left (-24 \left (4 A b^4+4 a^4 (6 A+7 C)+3 a^2 b^2 (50 A+63 C)\right )-8 b^2 \left (6 a^2 (6 A+7 C)+b^2 (62 A+105 C)\right )+24 \left (4 A b^4+4 a^4 (6 A+7 C)+3 a^2 b^2 (50 A+63 C)\right ) x^2\right ) \, dx,x,-\sin (c+d x)\right )}{840 d}\\ &=\frac{1}{4} a b \left (2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) x+\frac{\left (12 a^4 (6 A+7 C)+b^4 (74 A+105 C)+3 a^2 b^2 (162 A+203 C)\right ) \sin (c+d x)}{105 d}+\frac{a b \left (2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{4 d}+\frac{a b \left (6 A b^2+a^2 (103 A+126 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{210 d}+\frac{\left (2 A b^2+a^2 (6 A+7 C)\right ) \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{35 d}+\frac{2 A b \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{21 d}+\frac{A \cos ^6(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{7 d}-\frac{\left (4 A b^4+4 a^4 (6 A+7 C)+3 a^2 b^2 (50 A+63 C)\right ) \sin ^3(c+d x)}{105 d}\\ \end{align*}

Mathematica [A]  time = 0.858602, size = 351, normalized size = 1.04 \[ \frac{420 a b \left (a^2 (15 A+16 C)+16 b^2 (A+C)\right ) \sin (2 (c+d x))+105 \left (48 a^2 b^2 (5 A+6 C)+5 a^4 (7 A+8 C)+16 b^4 (3 A+4 C)\right ) \sin (c+d x)+4200 a^2 A b^2 \sin (3 (c+d x))+504 a^2 A b^2 \sin (5 (c+d x))+1260 a^3 A b \sin (4 (c+d x))+140 a^3 A b \sin (6 (c+d x))+8400 a^3 A b c+8400 a^3 A b d x+735 a^4 A \sin (3 (c+d x))+147 a^4 A \sin (5 (c+d x))+15 a^4 A \sin (7 (c+d x))+3360 a^2 b^2 C \sin (3 (c+d x))+840 a^3 b C \sin (4 (c+d x))+10080 a^3 b c C+10080 a^3 b C d x+700 a^4 C \sin (3 (c+d x))+84 a^4 C \sin (5 (c+d x))+840 a A b^3 \sin (4 (c+d x))+10080 a A b^3 c+10080 a A b^3 d x+13440 a b^3 c C+13440 a b^3 C d x+560 A b^4 \sin (3 (c+d x))}{6720 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8400*a^3*A*b*c + 10080*a*A*b^3*c + 10080*a^3*b*c*C + 13440*a*b^3*c*C + 8400*a^3*A*b*d*x + 10080*a*A*b^3*d*x +
10080*a^3*b*C*d*x + 13440*a*b^3*C*d*x + 105*(16*b^4*(3*A + 4*C) + 48*a^2*b^2*(5*A + 6*C) + 5*a^4*(7*A + 8*C))
*Sin[c + d*x] + 420*a*b*(16*b^2*(A + C) + a^2*(15*A + 16*C))*Sin[2*(c + d*x)] + 735*a^4*A*Sin[3*(c + d*x)] + 4
200*a^2*A*b^2*Sin[3*(c + d*x)] + 560*A*b^4*Sin[3*(c + d*x)] + 700*a^4*C*Sin[3*(c + d*x)] + 3360*a^2*b^2*C*Sin[
3*(c + d*x)] + 1260*a^3*A*b*Sin[4*(c + d*x)] + 840*a*A*b^3*Sin[4*(c + d*x)] + 840*a^3*b*C*Sin[4*(c + d*x)] + 1
47*a^4*A*Sin[5*(c + d*x)] + 504*a^2*A*b^2*Sin[5*(c + d*x)] + 84*a^4*C*Sin[5*(c + d*x)] + 140*a^3*A*b*Sin[6*(c
+ d*x)] + 15*a^4*A*Sin[7*(c + d*x)])/(6720*d)

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Maple [A]  time = 0.1, size = 332, normalized size = 1. \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}^{3}b \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}^{3}bC \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}^{2}{b}^{2}\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\,C{a}^{2}{b}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +4\,Aa{b}^{3} \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\,Ca{b}^{3} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +{\frac{A{b}^{4} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+C{b}^{4}\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+b*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^3*b*(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^3*b*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^2*b^2*(8/3+cos(d*x+c)^4
+4/3*cos(d*x+c)^2)*sin(d*x+c)+2*C*a^2*b^2*(2+cos(d*x+c)^2)*sin(d*x+c)+4*A*a*b^3*(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*C*a*b^3*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+1/3*A*b^4*(2+cos(d*x+c)^2)*
sin(d*x+c)+C*b^4*sin(d*x+c))

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Maxima [A]  time = 0.991636, size = 444, normalized size = 1.31 \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} - 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} + 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^{3} b - 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^{3} b - 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^{2} b^{2} + 3360 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} b^{2} - 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 b^{3} - 1680 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{3} + 560 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{4} - 1680 \, C b^{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+b*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 - 112*(3*sin(d*
x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*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^3*b - 210*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^3*b -
672*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^2*b^2 + 3360*(sin(d*x + c)^3 - 3*sin(d*x + c
))*C*a^2*b^2 - 210*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a*b^3 - 1680*(2*d*x + 2*c + sin(2
*d*x + 2*c))*C*a*b^3 + 560*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b^4 - 1680*C*b^4*sin(d*x + c))/d

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Fricas [A]  time = 0.581904, size = 595, normalized size = 1.76 \begin{align*} \frac{105 \,{\left ({\left (5 \, A + 6 \, C\right )} a^{3} b + 2 \,{\left (3 \, A + 4 \, C\right )} a b^{3}\right )} d x +{\left (60 \, A a^{4} \cos \left (d x + c\right )^{6} + 280 \, A a^{3} b \cos \left (d x + c\right )^{5} + 32 \,{\left (6 \, A + 7 \, C\right )} a^{4} + 336 \,{\left (4 \, A + 5 \, C\right )} a^{2} b^{2} + 140 \,{\left (2 \, A + 3 \, C\right )} b^{4} + 12 \,{\left ({\left (6 \, A + 7 \, C\right )} a^{4} + 42 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{4} + 70 \,{\left ({\left (5 \, A + 6 \, C\right )} a^{3} b + 6 \, A a b^{3}\right )} \cos \left (d x + c\right )^{3} + 4 \,{\left (4 \,{\left (6 \, A + 7 \, C\right )} a^{4} + 42 \,{\left (4 \, A + 5 \, C\right )} a^{2} b^{2} + 35 \, A b^{4}\right )} \cos \left (d x + c\right )^{2} + 105 \,{\left ({\left (5 \, A + 6 \, C\right )} a^{3} b + 2 \,{\left (3 \, A + 4 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )\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+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/420*(105*((5*A + 6*C)*a^3*b + 2*(3*A + 4*C)*a*b^3)*d*x + (60*A*a^4*cos(d*x + c)^6 + 280*A*a^3*b*cos(d*x + c)
^5 + 32*(6*A + 7*C)*a^4 + 336*(4*A + 5*C)*a^2*b^2 + 140*(2*A + 3*C)*b^4 + 12*((6*A + 7*C)*a^4 + 42*A*a^2*b^2)*
cos(d*x + c)^4 + 70*((5*A + 6*C)*a^3*b + 6*A*a*b^3)*cos(d*x + c)^3 + 4*(4*(6*A + 7*C)*a^4 + 42*(4*A + 5*C)*a^2
*b^2 + 35*A*b^4)*cos(d*x + c)^2 + 105*((5*A + 6*C)*a^3*b + 2*(3*A + 4*C)*a*b^3)*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)**7*(a+b*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.25626, size = 1658, normalized size = 4.89 \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)^7*(a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/420*(105*(5*A*a^3*b + 6*C*a^3*b + 6*A*a*b^3 + 8*C*a*b^3)*(d*x + c) + 2*(420*A*a^4*tan(1/2*d*x + 1/2*c)^13 +
420*C*a^4*tan(1/2*d*x + 1/2*c)^13 - 1155*A*a^3*b*tan(1/2*d*x + 1/2*c)^13 - 1050*C*a^3*b*tan(1/2*d*x + 1/2*c)^1
3 + 2520*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^13 + 2520*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^13 - 1050*A*a*b^3*tan(1/2*d*x
+ 1/2*c)^13 - 840*C*a*b^3*tan(1/2*d*x + 1/2*c)^13 + 420*A*b^4*tan(1/2*d*x + 1/2*c)^13 + 420*C*b^4*tan(1/2*d*x
+ 1/2*c)^13 + 840*A*a^4*tan(1/2*d*x + 1/2*c)^11 + 1400*C*a^4*tan(1/2*d*x + 1/2*c)^11 - 980*A*a^3*b*tan(1/2*d*
x + 1/2*c)^11 - 2520*C*a^3*b*tan(1/2*d*x + 1/2*c)^11 + 8400*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 + 11760*C*a^2*b^
2*tan(1/2*d*x + 1/2*c)^11 - 2520*A*a*b^3*tan(1/2*d*x + 1/2*c)^11 - 3360*C*a*b^3*tan(1/2*d*x + 1/2*c)^11 + 1960
*A*b^4*tan(1/2*d*x + 1/2*c)^11 + 2520*C*b^4*tan(1/2*d*x + 1/2*c)^11 + 3612*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 3164
*C*a^4*tan(1/2*d*x + 1/2*c)^9 - 2975*A*a^3*b*tan(1/2*d*x + 1/2*c)^9 - 1890*C*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 18
984*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 24360*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 1890*A*a*b^3*tan(1/2*d*x + 1/2
*c)^9 - 4200*C*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 4060*A*b^4*tan(1/2*d*x + 1/2*c)^9 + 6300*C*b^4*tan(1/2*d*x + 1/2
*c)^9 + 2544*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 4368*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 26208*A*a^2*b^2*tan(1/2*d*x +
1/2*c)^7 + 30240*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 5040*A*b^4*tan(1/2*d*x + 1/2*c)^7 + 8400*C*b^4*tan(1/2*d*x
+ 1/2*c)^7 + 3612*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 3164*C*a^4*tan(1/2*d*x + 1/2*c)^5 + 2975*A*a^3*b*tan(1/2*d*x
+ 1/2*c)^5 + 1890*C*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 18984*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 24360*C*a^2*b^2*t
an(1/2*d*x + 1/2*c)^5 + 1890*A*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 4200*C*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 4060*A*b^4
*tan(1/2*d*x + 1/2*c)^5 + 6300*C*b^4*tan(1/2*d*x + 1/2*c)^5 + 840*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 1400*C*a^4*ta
n(1/2*d*x + 1/2*c)^3 + 980*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 2520*C*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 8400*A*a^2*b
^2*tan(1/2*d*x + 1/2*c)^3 + 11760*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 + 2520*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 336
0*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 1960*A*b^4*tan(1/2*d*x + 1/2*c)^3 + 2520*C*b^4*tan(1/2*d*x + 1/2*c)^3 + 420
*A*a^4*tan(1/2*d*x + 1/2*c) + 420*C*a^4*tan(1/2*d*x + 1/2*c) + 1155*A*a^3*b*tan(1/2*d*x + 1/2*c) + 1050*C*a^3*
b*tan(1/2*d*x + 1/2*c) + 2520*A*a^2*b^2*tan(1/2*d*x + 1/2*c) + 2520*C*a^2*b^2*tan(1/2*d*x + 1/2*c) + 1050*A*a*
b^3*tan(1/2*d*x + 1/2*c) + 840*C*a*b^3*tan(1/2*d*x + 1/2*c) + 420*A*b^4*tan(1/2*d*x + 1/2*c) + 420*C*b^4*tan(1
/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^7)/d