3.437 \(\int \cos ^7(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=265 \[ -\frac {a^3 (108 A+119 B+133 C) \sin ^3(c+d x)}{105 d}+\frac {a^3 (108 A+119 B+133 C) \sin (c+d x)}{35 d}+\frac {a^3 (129 A+147 B+154 C) \sin (c+d x) \cos ^3(c+d x)}{280 d}+\frac {a^3 (21 A+23 B+26 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {(3 A+4 B+3 C) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac {1}{16} a^3 x (21 A+23 B+26 C)+\frac {(3 A+7 B) \sin (c+d x) \cos ^5(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{42 a d}+\frac {A \sin (c+d x) \cos ^6(c+d x) (a \sec (c+d x)+a)^3}{7 d} \]

[Out]

1/16*a^3*(21*A+23*B+26*C)*x+1/35*a^3*(108*A+119*B+133*C)*sin(d*x+c)/d+1/16*a^3*(21*A+23*B+26*C)*cos(d*x+c)*sin
(d*x+c)/d+1/280*a^3*(129*A+147*B+154*C)*cos(d*x+c)^3*sin(d*x+c)/d+1/7*A*cos(d*x+c)^6*(a+a*sec(d*x+c))^3*sin(d*
x+c)/d+1/42*(3*A+7*B)*cos(d*x+c)^5*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/a/d+1/15*(3*A+4*B+3*C)*cos(d*x+c)^4*(a^3+
a^3*sec(d*x+c))*sin(d*x+c)/d-1/105*a^3*(108*A+119*B+133*C)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.61, antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4086, 4017, 3996, 3787, 2633, 2635, 8} \[ -\frac {a^3 (108 A+119 B+133 C) \sin ^3(c+d x)}{105 d}+\frac {a^3 (108 A+119 B+133 C) \sin (c+d x)}{35 d}+\frac {a^3 (129 A+147 B+154 C) \sin (c+d x) \cos ^3(c+d x)}{280 d}+\frac {a^3 (21 A+23 B+26 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {(3 A+4 B+3 C) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac {(3 A+7 B) \sin (c+d x) \cos ^5(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{42 a d}+\frac {1}{16} a^3 x (21 A+23 B+26 C)+\frac {A \sin (c+d x) \cos ^6(c+d x) (a \sec (c+d x)+a)^3}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(21*A + 23*B + 26*C)*x)/16 + (a^3*(108*A + 119*B + 133*C)*Sin[c + d*x])/(35*d) + (a^3*(21*A + 23*B + 26*C
)*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (a^3*(129*A + 147*B + 154*C)*Cos[c + d*x]^3*Sin[c + d*x])/(280*d) + (A*C
os[c + d*x]^6*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(7*d) + ((3*A + 7*B)*Cos[c + d*x]^5*(a^2 + a^2*Sec[c + d*x]
)^2*Sin[c + d*x])/(42*a*d) + ((3*A + 4*B + 3*C)*Cos[c + d*x]^4*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(15*d) -
(a^3*(108*A + 119*B + 133*C)*Sin[c + d*x]^3)/(105*d)

Rule 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 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]

Rule 4086

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

Rubi steps

\begin {align*} \int \cos ^7(c+d x) (a+a \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^6(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {\int \cos ^6(c+d x) (a+a \sec (c+d x))^3 (a (3 A+7 B)+a (3 A+7 C) \sec (c+d x)) \, dx}{7 a}\\ &=\frac {A \cos ^6(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {(3 A+7 B) \cos ^5(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{42 a d}+\frac {\int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (14 a^2 (3 A+4 B+3 C)+3 a^2 (9 A+7 B+14 C) \sec (c+d x)\right ) \, dx}{42 a}\\ &=\frac {A \cos ^6(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {(3 A+7 B) \cos ^5(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{42 a d}+\frac {(3 A+4 B+3 C) \cos ^4(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {\int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (3 a^3 (129 A+147 B+154 C)+3 a^3 (87 A+91 B+112 C) \sec (c+d x)\right ) \, dx}{210 a}\\ &=\frac {a^3 (129 A+147 B+154 C) \cos ^3(c+d x) \sin (c+d x)}{280 d}+\frac {A \cos ^6(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {(3 A+7 B) \cos ^5(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{42 a d}+\frac {(3 A+4 B+3 C) \cos ^4(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}-\frac {\int \cos ^3(c+d x) \left (-24 a^4 (108 A+119 B+133 C)-105 a^4 (21 A+23 B+26 C) \sec (c+d x)\right ) \, dx}{840 a}\\ &=\frac {a^3 (129 A+147 B+154 C) \cos ^3(c+d x) \sin (c+d x)}{280 d}+\frac {A \cos ^6(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {(3 A+7 B) \cos ^5(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{42 a d}+\frac {(3 A+4 B+3 C) \cos ^4(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{8} \left (a^3 (21 A+23 B+26 C)\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{35} \left (a^3 (108 A+119 B+133 C)\right ) \int \cos ^3(c+d x) \, dx\\ &=\frac {a^3 (21 A+23 B+26 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^3 (129 A+147 B+154 C) \cos ^3(c+d x) \sin (c+d x)}{280 d}+\frac {A \cos ^6(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {(3 A+7 B) \cos ^5(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{42 a d}+\frac {(3 A+4 B+3 C) \cos ^4(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}+\frac {1}{16} \left (a^3 (21 A+23 B+26 C)\right ) \int 1 \, dx-\frac {\left (a^3 (108 A+119 B+133 C)\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{35 d}\\ &=\frac {1}{16} a^3 (21 A+23 B+26 C) x+\frac {a^3 (108 A+119 B+133 C) \sin (c+d x)}{35 d}+\frac {a^3 (21 A+23 B+26 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^3 (129 A+147 B+154 C) \cos ^3(c+d x) \sin (c+d x)}{280 d}+\frac {A \cos ^6(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{7 d}+\frac {(3 A+7 B) \cos ^5(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{42 a d}+\frac {(3 A+4 B+3 C) \cos ^4(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{15 d}-\frac {a^3 (108 A+119 B+133 C) \sin ^3(c+d x)}{105 d}\\ \end {align*}

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Mathematica [A]  time = 1.31, size = 204, normalized size = 0.77 \[ \frac {a^3 (105 (155 A+168 B+184 C) \sin (c+d x)+105 (61 A+63 B+64 C) \sin (2 (c+d x))+2835 A \sin (3 (c+d x))+1155 A \sin (4 (c+d x))+399 A \sin (5 (c+d x))+105 A \sin (6 (c+d x))+15 A \sin (7 (c+d x))+3360 A c+8820 A d x+2660 B \sin (3 (c+d x))+945 B \sin (4 (c+d x))+252 B \sin (5 (c+d x))+35 B \sin (6 (c+d x))+9660 B c+9660 B d x+2380 C \sin (3 (c+d x))+630 C \sin (4 (c+d x))+84 C \sin (5 (c+d x))+10920 C d x)}{6720 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(3360*A*c + 9660*B*c + 8820*A*d*x + 9660*B*d*x + 10920*C*d*x + 105*(155*A + 168*B + 184*C)*Sin[c + d*x] +
105*(61*A + 63*B + 64*C)*Sin[2*(c + d*x)] + 2835*A*Sin[3*(c + d*x)] + 2660*B*Sin[3*(c + d*x)] + 2380*C*Sin[3*
(c + d*x)] + 1155*A*Sin[4*(c + d*x)] + 945*B*Sin[4*(c + d*x)] + 630*C*Sin[4*(c + d*x)] + 399*A*Sin[5*(c + d*x)
] + 252*B*Sin[5*(c + d*x)] + 84*C*Sin[5*(c + d*x)] + 105*A*Sin[6*(c + d*x)] + 35*B*Sin[6*(c + d*x)] + 15*A*Sin
[7*(c + d*x)]))/(6720*d)

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fricas [A]  time = 0.46, size = 168, normalized size = 0.63 \[ \frac {105 \, {\left (21 \, A + 23 \, B + 26 \, C\right )} a^{3} d x + {\left (240 \, A a^{3} \cos \left (d x + c\right )^{6} + 280 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{5} + 48 \, {\left (27 \, A + 21 \, B + 7 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 70 \, {\left (21 \, A + 23 \, B + 18 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 16 \, {\left (108 \, A + 119 \, B + 133 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 105 \, {\left (21 \, A + 23 \, B + 26 \, C\right )} a^{3} \cos \left (d x + c\right ) + 32 \, {\left (108 \, A + 119 \, B + 133 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{1680 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/1680*(105*(21*A + 23*B + 26*C)*a^3*d*x + (240*A*a^3*cos(d*x + c)^6 + 280*(3*A + B)*a^3*cos(d*x + c)^5 + 48*(
27*A + 21*B + 7*C)*a^3*cos(d*x + c)^4 + 70*(21*A + 23*B + 18*C)*a^3*cos(d*x + c)^3 + 16*(108*A + 119*B + 133*C
)*a^3*cos(d*x + c)^2 + 105*(21*A + 23*B + 26*C)*a^3*cos(d*x + c) + 32*(108*A + 119*B + 133*C)*a^3)*sin(d*x + c
))/d

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giac [A]  time = 0.34, size = 401, normalized size = 1.51 \[ \frac {105 \, {\left (21 \, A a^{3} + 23 \, B a^{3} + 26 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (2205 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 2415 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 2730 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{13} + 14700 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 16100 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 18200 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 41601 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45563 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 51506 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 62592 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 72576 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 77952 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 63231 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 62853 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 71246 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 25620 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33180 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40040 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 11235 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11025 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 10710 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{7}}}{1680 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/1680*(105*(21*A*a^3 + 23*B*a^3 + 26*C*a^3)*(d*x + c) + 2*(2205*A*a^3*tan(1/2*d*x + 1/2*c)^13 + 2415*B*a^3*ta
n(1/2*d*x + 1/2*c)^13 + 2730*C*a^3*tan(1/2*d*x + 1/2*c)^13 + 14700*A*a^3*tan(1/2*d*x + 1/2*c)^11 + 16100*B*a^3
*tan(1/2*d*x + 1/2*c)^11 + 18200*C*a^3*tan(1/2*d*x + 1/2*c)^11 + 41601*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 45563*B*
a^3*tan(1/2*d*x + 1/2*c)^9 + 51506*C*a^3*tan(1/2*d*x + 1/2*c)^9 + 62592*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 72576*B
*a^3*tan(1/2*d*x + 1/2*c)^7 + 77952*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 63231*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 62853*
B*a^3*tan(1/2*d*x + 1/2*c)^5 + 71246*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 25620*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 33180
*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 40040*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 11235*A*a^3*tan(1/2*d*x + 1/2*c) + 11025*
B*a^3*tan(1/2*d*x + 1/2*c) + 10710*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^7)/d

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maple [A]  time = 2.63, size = 427, normalized size = 1.61 \[ \frac {\frac {A \,a^{3} \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}+a^{3} B \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 {C \,a^{3} \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}+3 A \,a^{3} \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 {3 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}+3 C \,a^{3} \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 )+\frac {3 A \,a^{3} \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}+3 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 )+C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+A \,a^{3} \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 )+\frac {a^{3} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/d*(1/7*A*a^3*(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)+a^3*B*(1/6*(cos(d*x+c)^5+5/4*c
os(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)+5/16*d*x+5/16*c)+1/5*C*a^3*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x
+c)+3*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)+3/5*a^3*B*(8/3+co
s(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+3*C*a^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)+3
/5*A*a^3*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+3*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)+C*a^3*(2+cos(d*x+c)^2)*sin(d*x+c)+A*a^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)+1/3*a^3*B*(2+cos(d*x+c)^2)*sin(d*x+c)+C*a^3*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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maxima [A]  time = 0.36, size = 425, normalized size = 1.60 \[ -\frac {192 \, {\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} A a^{3} - 1344 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{3} + 105 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 210 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 1344 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{3} + 35 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 2240 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} - 630 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 448 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a^{3} + 6720 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} - 630 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 1680 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3}}{6720 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/6720*(192*(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^3 - 1344*(3*sin(
d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^3 + 105*(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 - 210*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^3 -
1344*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*B*a^3 + 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))*B*a^3 + 2240*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 - 630*(12*d*
x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^3 - 448*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d
*x + c))*C*a^3 + 6720*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3 - 630*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(
2*d*x + 2*c))*C*a^3 - 1680*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3)/d

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mupad [B]  time = 6.07, size = 377, normalized size = 1.42 \[ \frac {\left (\frac {21\,A\,a^3}{8}+\frac {23\,B\,a^3}{8}+\frac {13\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{13}+\left (\frac {35\,A\,a^3}{2}+\frac {115\,B\,a^3}{6}+\frac {65\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {1981\,A\,a^3}{40}+\frac {6509\,B\,a^3}{120}+\frac {3679\,C\,a^3}{60}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {2608\,A\,a^3}{35}+\frac {432\,B\,a^3}{5}+\frac {464\,C\,a^3}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3011\,A\,a^3}{40}+\frac {2993\,B\,a^3}{40}+\frac {5089\,C\,a^3}{60}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {61\,A\,a^3}{2}+\frac {79\,B\,a^3}{2}+\frac {143\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {107\,A\,a^3}{8}+\frac {105\,B\,a^3}{8}+\frac {51\,C\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{14}+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^3\,\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (21\,A+23\,B+26\,C\right )}{8\,\left (\frac {21\,A\,a^3}{8}+\frac {23\,B\,a^3}{8}+\frac {13\,C\,a^3}{4}\right )}\right )\,\left (21\,A+23\,B+26\,C\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^7*(a + a/cos(c + d*x))^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

(tan(c/2 + (d*x)/2)^13*((21*A*a^3)/8 + (23*B*a^3)/8 + (13*C*a^3)/4) + tan(c/2 + (d*x)/2)^11*((35*A*a^3)/2 + (1
15*B*a^3)/6 + (65*C*a^3)/3) + tan(c/2 + (d*x)/2)^3*((61*A*a^3)/2 + (79*B*a^3)/2 + (143*C*a^3)/3) + tan(c/2 + (
d*x)/2)^7*((2608*A*a^3)/35 + (432*B*a^3)/5 + (464*C*a^3)/5) + tan(c/2 + (d*x)/2)^5*((3011*A*a^3)/40 + (2993*B*
a^3)/40 + (5089*C*a^3)/60) + tan(c/2 + (d*x)/2)^9*((1981*A*a^3)/40 + (6509*B*a^3)/120 + (3679*C*a^3)/60) + tan
(c/2 + (d*x)/2)*((107*A*a^3)/8 + (105*B*a^3)/8 + (51*C*a^3)/4))/(d*(7*tan(c/2 + (d*x)/2)^2 + 21*tan(c/2 + (d*x
)/2)^4 + 35*tan(c/2 + (d*x)/2)^6 + 35*tan(c/2 + (d*x)/2)^8 + 21*tan(c/2 + (d*x)/2)^10 + 7*tan(c/2 + (d*x)/2)^1
2 + tan(c/2 + (d*x)/2)^14 + 1)) + (a^3*atan((a^3*tan(c/2 + (d*x)/2)*(21*A + 23*B + 26*C))/(8*((21*A*a^3)/8 + (
23*B*a^3)/8 + (13*C*a^3)/4)))*(21*A + 23*B + 26*C))/(8*d)

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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)**7*(a+a*sec(d*x+c))**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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