3.435 \(\int \cos ^5(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=179 \[ -\frac {a^3 (13 A+15 B+20 C) \sin ^3(c+d x)}{60 d}+\frac {a^3 (13 A+15 B+20 C) \sin (c+d x)}{5 d}+\frac {3 a^3 (13 A+15 B+20 C) \sin (c+d x) \cos (c+d x)}{40 d}+\frac {1}{8} a^3 x (13 A+15 B+20 C)+\frac {(3 A+5 B) \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{20 d}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^3}{5 d} \]

[Out]

1/8*a^3*(13*A+15*B+20*C)*x+1/5*a^3*(13*A+15*B+20*C)*sin(d*x+c)/d+3/40*a^3*(13*A+15*B+20*C)*cos(d*x+c)*sin(d*x+
c)/d+1/20*(3*A+5*B)*cos(d*x+c)^3*(a+a*sec(d*x+c))^3*sin(d*x+c)/d+1/5*A*cos(d*x+c)^4*(a+a*sec(d*x+c))^3*sin(d*x
+c)/d-1/60*a^3*(13*A+15*B+20*C)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.37, antiderivative size = 179, 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, 4013, 3791, 2637, 2635, 8, 2633} \[ -\frac {a^3 (13 A+15 B+20 C) \sin ^3(c+d x)}{60 d}+\frac {a^3 (13 A+15 B+20 C) \sin (c+d x)}{5 d}+\frac {3 a^3 (13 A+15 B+20 C) \sin (c+d x) \cos (c+d x)}{40 d}+\frac {1}{8} a^3 x (13 A+15 B+20 C)+\frac {(3 A+5 B) \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{20 d}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^3}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(13*A + 15*B + 20*C)*x)/8 + (a^3*(13*A + 15*B + 20*C)*Sin[c + d*x])/(5*d) + (3*a^3*(13*A + 15*B + 20*C)*C
os[c + d*x]*Sin[c + d*x])/(40*d) + ((3*A + 5*B)*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(20*d) + (
A*Cos[c + d*x]^4*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(5*d) - (a^3*(13*A + 15*B + 20*C)*Sin[c + d*x]^3)/(60*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 2637

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

Rule 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rule 4013

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), 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[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -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 ^5(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 ^4(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {\int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (a (3 A+5 B)+a (A+5 C) \sec (c+d x)) \, dx}{5 a}\\ &=\frac {(3 A+5 B) \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {1}{20} (13 A+15 B+20 C) \int \cos ^3(c+d x) (a+a \sec (c+d x))^3 \, dx\\ &=\frac {(3 A+5 B) \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {1}{20} (13 A+15 B+20 C) \int \left (a^3+3 a^3 \cos (c+d x)+3 a^3 \cos ^2(c+d x)+a^3 \cos ^3(c+d x)\right ) \, dx\\ &=\frac {1}{20} a^3 (13 A+15 B+20 C) x+\frac {(3 A+5 B) \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {1}{20} \left (a^3 (13 A+15 B+20 C)\right ) \int \cos ^3(c+d x) \, dx+\frac {1}{20} \left (3 a^3 (13 A+15 B+20 C)\right ) \int \cos (c+d x) \, dx+\frac {1}{20} \left (3 a^3 (13 A+15 B+20 C)\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {1}{20} a^3 (13 A+15 B+20 C) x+\frac {3 a^3 (13 A+15 B+20 C) \sin (c+d x)}{20 d}+\frac {3 a^3 (13 A+15 B+20 C) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {(3 A+5 B) \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac {1}{40} \left (3 a^3 (13 A+15 B+20 C)\right ) \int 1 \, dx-\frac {\left (a^3 (13 A+15 B+20 C)\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{20 d}\\ &=\frac {1}{8} a^3 (13 A+15 B+20 C) x+\frac {a^3 (13 A+15 B+20 C) \sin (c+d x)}{5 d}+\frac {3 a^3 (13 A+15 B+20 C) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {(3 A+5 B) \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{5 d}-\frac {a^3 (13 A+15 B+20 C) \sin ^3(c+d x)}{60 d}\\ \end {align*}

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Mathematica [A]  time = 0.50, size = 130, normalized size = 0.73 \[ \frac {a^3 (60 (23 A+26 B+30 C) \sin (c+d x)+120 (4 A+4 B+3 C) \sin (2 (c+d x))+170 A \sin (3 (c+d x))+45 A \sin (4 (c+d x))+6 A \sin (5 (c+d x))+780 A d x+120 B \sin (3 (c+d x))+15 B \sin (4 (c+d x))+900 B d x+40 C \sin (3 (c+d x))+1200 C d x)}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(780*A*d*x + 900*B*d*x + 1200*C*d*x + 60*(23*A + 26*B + 30*C)*Sin[c + d*x] + 120*(4*A + 4*B + 3*C)*Sin[2*
(c + d*x)] + 170*A*Sin[3*(c + d*x)] + 120*B*Sin[3*(c + d*x)] + 40*C*Sin[3*(c + d*x)] + 45*A*Sin[4*(c + d*x)] +
15*B*Sin[4*(c + d*x)] + 6*A*Sin[5*(c + d*x)]))/(480*d)

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fricas [A]  time = 0.45, size = 122, normalized size = 0.68 \[ \frac {15 \, {\left (13 \, A + 15 \, B + 20 \, C\right )} a^{3} d x + {\left (24 \, A a^{3} \cos \left (d x + c\right )^{4} + 30 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \, {\left (19 \, A + 15 \, B + 5 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 15 \, {\left (13 \, A + 15 \, B + 12 \, C\right )} a^{3} \cos \left (d x + c\right ) + 8 \, {\left (38 \, A + 45 \, B + 55 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/120*(15*(13*A + 15*B + 20*C)*a^3*d*x + (24*A*a^3*cos(d*x + c)^4 + 30*(3*A + B)*a^3*cos(d*x + c)^3 + 8*(19*A
+ 15*B + 5*C)*a^3*cos(d*x + c)^2 + 15*(13*A + 15*B + 12*C)*a^3*cos(d*x + c) + 8*(38*A + 45*B + 55*C)*a^3)*sin(
d*x + c))/d

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giac [A]  time = 0.30, size = 299, normalized size = 1.67 \[ \frac {15 \, {\left (13 \, A a^{3} + 15 \, B a^{3} + 20 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (195 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 225 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 300 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 910 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1050 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1400 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1664 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1920 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2560 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1330 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1830 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2120 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 765 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 735 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 660 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(15*(13*A*a^3 + 15*B*a^3 + 20*C*a^3)*(d*x + c) + 2*(195*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 225*B*a^3*tan(1/2
*d*x + 1/2*c)^9 + 300*C*a^3*tan(1/2*d*x + 1/2*c)^9 + 910*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 1050*B*a^3*tan(1/2*d*x
+ 1/2*c)^7 + 1400*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 1664*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 1920*B*a^3*tan(1/2*d*x +
1/2*c)^5 + 2560*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 1330*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 1830*B*a^3*tan(1/2*d*x + 1
/2*c)^3 + 2120*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 765*A*a^3*tan(1/2*d*x + 1/2*c) + 735*B*a^3*tan(1/2*d*x + 1/2*c)
+ 660*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d

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maple [A]  time = 2.24, size = 295, normalized size = 1.65 \[ \frac {\frac {A \,a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 A \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+A \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+a^{3} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\frac {C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{3} B \sin \left (d x +c \right )+3 C \,a^{3} \sin \left (d x +c \right )+C \,a^{3} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

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

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maxima [A]  time = 0.35, size = 282, normalized size = 1.58 \[ \frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{3} - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 480 \, {\left (d x + c\right )} C a^{3} + 480 \, B a^{3} \sin \left (d x + c\right ) + 1440 \, C a^{3} \sin \left (d x + c\right )}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^3 - 480*(sin(d*x + c)^3 - 3*sin(d*x + c
))*A*a^3 + 45*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^3 + 120*(2*d*x + 2*c + sin(2*d*x + 2
*c))*A*a^3 - 480*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a^3 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x
+ 2*c))*B*a^3 + 360*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3 + 360
*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 + 480*(d*x + c)*C*a^3 + 480*B*a^3*sin(d*x + c) + 1440*C*a^3*sin(d*x +
c))/d

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mupad [B]  time = 5.88, size = 289, normalized size = 1.61 \[ \frac {\left (\frac {13\,A\,a^3}{4}+\frac {15\,B\,a^3}{4}+5\,C\,a^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {91\,A\,a^3}{6}+\frac {35\,B\,a^3}{2}+\frac {70\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {416\,A\,a^3}{15}+32\,B\,a^3+\frac {128\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {133\,A\,a^3}{6}+\frac {61\,B\,a^3}{2}+\frac {106\,C\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {51\,A\,a^3}{4}+\frac {49\,B\,a^3}{4}+11\,C\,a^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^3\,\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (13\,A+15\,B+20\,C\right )}{4\,\left (\frac {13\,A\,a^3}{4}+\frac {15\,B\,a^3}{4}+5\,C\,a^3\right )}\right )\,\left (13\,A+15\,B+20\,C\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(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)^9*((13*A*a^3)/4 + (15*B*a^3)/4 + 5*C*a^3) + tan(c/2 + (d*x)/2)^7*((91*A*a^3)/6 + (35*B*a^3
)/2 + (70*C*a^3)/3) + tan(c/2 + (d*x)/2)^3*((133*A*a^3)/6 + (61*B*a^3)/2 + (106*C*a^3)/3) + tan(c/2 + (d*x)/2)
^5*((416*A*a^3)/15 + 32*B*a^3 + (128*C*a^3)/3) + tan(c/2 + (d*x)/2)*((51*A*a^3)/4 + (49*B*a^3)/4 + 11*C*a^3))/
(d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(
c/2 + (d*x)/2)^10 + 1)) + (a^3*atan((a^3*tan(c/2 + (d*x)/2)*(13*A + 15*B + 20*C))/(4*((13*A*a^3)/4 + (15*B*a^3
)/4 + 5*C*a^3)))*(13*A + 15*B + 20*C))/(4*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)**5*(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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