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

Optimal. Leaf size=201 \[ -\frac {a^3 (17 B+19 C) \sin ^3(c+d x)}{15 d}+\frac {a^3 (17 B+19 C) \sin (c+d x)}{5 d}+\frac {a^3 (21 B+22 C) \sin (c+d x) \cos ^3(c+d x)}{40 d}+\frac {a^3 (23 B+26 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {(4 B+3 C) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac {1}{16} a^3 x (23 B+26 C)+\frac {a B \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d} \]

[Out]

1/16*a^3*(23*B+26*C)*x+1/5*a^3*(17*B+19*C)*sin(d*x+c)/d+1/16*a^3*(23*B+26*C)*cos(d*x+c)*sin(d*x+c)/d+1/40*a^3*
(21*B+22*C)*cos(d*x+c)^3*sin(d*x+c)/d+1/6*a*B*cos(d*x+c)^5*(a+a*sec(d*x+c))^2*sin(d*x+c)/d+1/15*(4*B+3*C)*cos(
d*x+c)^4*(a^3+a^3*sec(d*x+c))*sin(d*x+c)/d-1/15*a^3*(17*B+19*C)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.48, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {4072, 4017, 3996, 3787, 2633, 2635, 8} \[ -\frac {a^3 (17 B+19 C) \sin ^3(c+d x)}{15 d}+\frac {a^3 (17 B+19 C) \sin (c+d x)}{5 d}+\frac {a^3 (21 B+22 C) \sin (c+d x) \cos ^3(c+d x)}{40 d}+\frac {a^3 (23 B+26 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {(4 B+3 C) \sin (c+d x) \cos ^4(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{15 d}+\frac {1}{16} a^3 x (23 B+26 C)+\frac {a B \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^2}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(23*B + 26*C)*x)/16 + (a^3*(17*B + 19*C)*Sin[c + d*x])/(5*d) + (a^3*(23*B + 26*C)*Cos[c + d*x]*Sin[c + d*
x])/(16*d) + (a^3*(21*B + 22*C)*Cos[c + d*x]^3*Sin[c + d*x])/(40*d) + (a*B*Cos[c + d*x]^5*(a + a*Sec[c + d*x])
^2*Sin[c + d*x])/(6*d) + ((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*(17*
B + 19*C)*Sin[c + d*x]^3)/(15*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 4072

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

\begin {align*} \int \cos ^7(c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \cos ^6(c+d x) (a+a \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx\\ &=\frac {a B \cos ^5(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {1}{6} \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 (2 a (4 B+3 C)+3 a (B+2 C) \sec (c+d x)) \, dx\\ &=\frac {a B \cos ^5(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(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}{30} \int \cos ^4(c+d x) (a+a \sec (c+d x)) \left (3 a^2 (21 B+22 C)+3 a^2 (13 B+16 C) \sec (c+d x)\right ) \, dx\\ &=\frac {a^3 (21 B+22 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {a B \cos ^5(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(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}{120} \int \cos ^3(c+d x) \left (-24 a^3 (17 B+19 C)-15 a^3 (23 B+26 C) \sec (c+d x)\right ) \, dx\\ &=\frac {a^3 (21 B+22 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {a B \cos ^5(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(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}{5} \left (a^3 (17 B+19 C)\right ) \int \cos ^3(c+d x) \, dx+\frac {1}{8} \left (a^3 (23 B+26 C)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {a^3 (23 B+26 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^3 (21 B+22 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {a B \cos ^5(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(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 (23 B+26 C)\right ) \int 1 \, dx-\frac {\left (a^3 (17 B+19 C)\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac {1}{16} a^3 (23 B+26 C) x+\frac {a^3 (17 B+19 C) \sin (c+d x)}{5 d}+\frac {a^3 (23 B+26 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^3 (21 B+22 C) \cos ^3(c+d x) \sin (c+d x)}{40 d}+\frac {a B \cos ^5(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac {(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 (17 B+19 C) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.53, size = 134, normalized size = 0.67 \[ \frac {a^3 (120 (21 B+23 C) \sin (c+d x)+15 (63 B+64 C) \sin (2 (c+d x))+380 B \sin (3 (c+d x))+135 B \sin (4 (c+d x))+36 B \sin (5 (c+d x))+5 B \sin (6 (c+d x))+1380 B c+1380 B d x+340 C \sin (3 (c+d x))+90 C \sin (4 (c+d x))+12 C \sin (5 (c+d x))+1560 C d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1380*B*c + 1380*B*d*x + 1560*C*d*x + 120*(21*B + 23*C)*Sin[c + d*x] + 15*(63*B + 64*C)*Sin[2*(c + d*x)]
+ 380*B*Sin[3*(c + d*x)] + 340*C*Sin[3*(c + d*x)] + 135*B*Sin[4*(c + d*x)] + 90*C*Sin[4*(c + d*x)] + 36*B*Sin[
5*(c + d*x)] + 12*C*Sin[5*(c + d*x)] + 5*B*Sin[6*(c + d*x)]))/(960*d)

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fricas [A]  time = 0.51, size = 130, normalized size = 0.65 \[ \frac {15 \, {\left (23 \, B + 26 \, C\right )} a^{3} d x + {\left (40 \, B a^{3} \cos \left (d x + c\right )^{5} + 48 \, {\left (3 \, B + C\right )} a^{3} \cos \left (d x + c\right )^{4} + 10 \, {\left (23 \, B + 18 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 16 \, {\left (17 \, B + 19 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 15 \, {\left (23 \, B + 26 \, C\right )} a^{3} \cos \left (d x + c\right ) + 32 \, {\left (17 \, B + 19 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(23*B + 26*C)*a^3*d*x + (40*B*a^3*cos(d*x + c)^5 + 48*(3*B + C)*a^3*cos(d*x + c)^4 + 10*(23*B + 18*C
)*a^3*cos(d*x + c)^3 + 16*(17*B + 19*C)*a^3*cos(d*x + c)^2 + 15*(23*B + 26*C)*a^3*cos(d*x + c) + 32*(17*B + 19
*C)*a^3)*sin(d*x + c))/d

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giac [A]  time = 0.40, size = 244, normalized size = 1.21 \[ \frac {15 \, {\left (23 \, B a^{3} + 26 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (345 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 390 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 1955 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 2210 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 4554 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 5148 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 5814 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5988 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3165 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4190 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1575 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1530 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(23*B*a^3 + 26*C*a^3)*(d*x + c) + 2*(345*B*a^3*tan(1/2*d*x + 1/2*c)^11 + 390*C*a^3*tan(1/2*d*x + 1/2
*c)^11 + 1955*B*a^3*tan(1/2*d*x + 1/2*c)^9 + 2210*C*a^3*tan(1/2*d*x + 1/2*c)^9 + 4554*B*a^3*tan(1/2*d*x + 1/2*
c)^7 + 5148*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 5814*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 5988*C*a^3*tan(1/2*d*x + 1/2*c)
^5 + 3165*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 4190*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 1575*B*a^3*tan(1/2*d*x + 1/2*c) +
1530*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d

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maple [A]  time = 2.30, size = 266, normalized size = 1.32 \[ \frac {a^{3} B \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {C \,a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {3 a^{3} B \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 C \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 a^{3} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\frac {a^{3} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(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)+1/5*C*a^3*(8/3+cos
(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+3/5*a^3*B*(8/3+cos(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*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)+1/3*a^3*B*(2+cos(d*x+c)^2)*sin(d*x+c)+C*a^3*(1/2*cos(d*x+c)*si
n(d*x+c)+1/2*d*x+1/2*c))

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maxima [A]  time = 0.36, size = 262, normalized size = 1.30 \[ \frac {192 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{3} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} + 90 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a^{3} - 960 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 90 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3}}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(192*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*B*a^3 - 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))*B*a^3 - 320*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 + 90*(12
*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^3 + 64*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin
(d*x + c))*C*a^3 - 960*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3 + 90*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(
2*d*x + 2*c))*C*a^3 + 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3)/d

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mupad [B]  time = 5.71, size = 285, normalized size = 1.42 \[ \frac {\left (\frac {23\,B\,a^3}{8}+\frac {13\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {391\,B\,a^3}{24}+\frac {221\,C\,a^3}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {759\,B\,a^3}{20}+\frac {429\,C\,a^3}{10}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {969\,B\,a^3}{20}+\frac {499\,C\,a^3}{10}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {211\,B\,a^3}{8}+\frac {419\,C\,a^3}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {105\,B\,a^3}{8}+\frac {51\,C\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^3\,\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (23\,B+26\,C\right )}{8\,\left (\frac {23\,B\,a^3}{8}+\frac {13\,C\,a^3}{4}\right )}\right )\,\left (23\,B+26\,C\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((105*B*a^3)/8 + (51*C*a^3)/4) + tan(c/2 + (d*x)/2)^11*((23*B*a^3)/8 + (13*C*a^3)/4) + tan
(c/2 + (d*x)/2)^3*((211*B*a^3)/8 + (419*C*a^3)/12) + tan(c/2 + (d*x)/2)^9*((391*B*a^3)/24 + (221*C*a^3)/12) +
tan(c/2 + (d*x)/2)^7*((759*B*a^3)/20 + (429*C*a^3)/10) + tan(c/2 + (d*x)/2)^5*((969*B*a^3)/20 + (499*C*a^3)/10
))/(d*(6*tan(c/2 + (d*x)/2)^2 + 15*tan(c/2 + (d*x)/2)^4 + 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x)/2)^8 +
6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1)) + (a^3*atan((a^3*tan(c/2 + (d*x)/2)*(23*B + 26*C))/(8*((
23*B*a^3)/8 + (13*C*a^3)/4)))*(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*(B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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