3.436 \(\int \cos ^6(c+d x) (a+a \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=235 \[ \frac {a^3 (34 A+38 B+45 C) \sin (c+d x)}{15 d}+\frac {a^3 (73 A+86 B+90 C) \sin (c+d x) \cos ^2(c+d x)}{120 d}+\frac {a^3 (23 A+26 B+30 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {(31 A+42 B+30 C) \sin (c+d x) \cos ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{120 d}+\frac {1}{16} a^3 x (23 A+26 B+30 C)+\frac {(A+2 B) \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{10 a d}+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^3}{6 d} \]

[Out]

1/16*a^3*(23*A+26*B+30*C)*x+1/15*a^3*(34*A+38*B+45*C)*sin(d*x+c)/d+1/16*a^3*(23*A+26*B+30*C)*cos(d*x+c)*sin(d*
x+c)/d+1/120*a^3*(73*A+86*B+90*C)*cos(d*x+c)^2*sin(d*x+c)/d+1/6*A*cos(d*x+c)^5*(a+a*sec(d*x+c))^3*sin(d*x+c)/d
+1/10*(A+2*B)*cos(d*x+c)^4*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/a/d+1/120*(31*A+42*B+30*C)*cos(d*x+c)^3*(a^3+a^3*
sec(d*x+c))*sin(d*x+c)/d

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Rubi [A]  time = 0.59, antiderivative size = 235, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4086, 4017, 3996, 3787, 2635, 8, 2637} \[ \frac {a^3 (34 A+38 B+45 C) \sin (c+d x)}{15 d}+\frac {a^3 (73 A+86 B+90 C) \sin (c+d x) \cos ^2(c+d x)}{120 d}+\frac {a^3 (23 A+26 B+30 C) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {(31 A+42 B+30 C) \sin (c+d x) \cos ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{120 d}+\frac {(A+2 B) \sin (c+d x) \cos ^4(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{10 a d}+\frac {1}{16} a^3 x (23 A+26 B+30 C)+\frac {A \sin (c+d x) \cos ^5(c+d x) (a \sec (c+d x)+a)^3}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(23*A + 26*B + 30*C)*x)/16 + (a^3*(34*A + 38*B + 45*C)*Sin[c + d*x])/(15*d) + (a^3*(23*A + 26*B + 30*C)*C
os[c + d*x]*Sin[c + d*x])/(16*d) + (a^3*(73*A + 86*B + 90*C)*Cos[c + d*x]^2*Sin[c + d*x])/(120*d) + (A*Cos[c +
d*x]^5*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(6*d) + ((A + 2*B)*Cos[c + d*x]^4*(a^2 + a^2*Sec[c + d*x])^2*Sin[
c + d*x])/(10*a*d) + ((31*A + 42*B + 30*C)*Cos[c + d*x]^3*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(120*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 2637

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

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 ^6(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 ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {\int \cos ^5(c+d x) (a+a \sec (c+d x))^3 (3 a (A+2 B)+2 a (A+3 C) \sec (c+d x)) \, dx}{6 a}\\ &=\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {(A+2 B) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{10 a d}+\frac {\int \cos ^4(c+d x) (a+a \sec (c+d x))^2 \left (a^2 (31 A+42 B+30 C)+2 a^2 (8 A+6 B+15 C) \sec (c+d x)\right ) \, dx}{30 a}\\ &=\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {(A+2 B) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{10 a d}+\frac {(31 A+42 B+30 C) \cos ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{120 d}+\frac {\int \cos ^3(c+d x) (a+a \sec (c+d x)) \left (3 a^3 (73 A+86 B+90 C)+6 a^3 (21 A+22 B+30 C) \sec (c+d x)\right ) \, dx}{120 a}\\ &=\frac {a^3 (73 A+86 B+90 C) \cos ^2(c+d x) \sin (c+d x)}{120 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {(A+2 B) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{10 a d}+\frac {(31 A+42 B+30 C) \cos ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{120 d}-\frac {\int \cos ^2(c+d x) \left (-45 a^4 (23 A+26 B+30 C)-24 a^4 (34 A+38 B+45 C) \sec (c+d x)\right ) \, dx}{360 a}\\ &=\frac {a^3 (73 A+86 B+90 C) \cos ^2(c+d x) \sin (c+d x)}{120 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {(A+2 B) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{10 a d}+\frac {(31 A+42 B+30 C) \cos ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{120 d}+\frac {1}{8} \left (a^3 (23 A+26 B+30 C)\right ) \int \cos ^2(c+d x) \, dx+\frac {1}{15} \left (a^3 (34 A+38 B+45 C)\right ) \int \cos (c+d x) \, dx\\ &=\frac {a^3 (34 A+38 B+45 C) \sin (c+d x)}{15 d}+\frac {a^3 (23 A+26 B+30 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^3 (73 A+86 B+90 C) \cos ^2(c+d x) \sin (c+d x)}{120 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {(A+2 B) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{10 a d}+\frac {(31 A+42 B+30 C) \cos ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{120 d}+\frac {1}{16} \left (a^3 (23 A+26 B+30 C)\right ) \int 1 \, dx\\ &=\frac {1}{16} a^3 (23 A+26 B+30 C) x+\frac {a^3 (34 A+38 B+45 C) \sin (c+d x)}{15 d}+\frac {a^3 (23 A+26 B+30 C) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a^3 (73 A+86 B+90 C) \cos ^2(c+d x) \sin (c+d x)}{120 d}+\frac {A \cos ^5(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {(A+2 B) \cos ^4(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{10 a d}+\frac {(31 A+42 B+30 C) \cos ^3(c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{120 d}\\ \end {align*}

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Mathematica [A]  time = 0.89, size = 170, normalized size = 0.72 \[ \frac {a^3 (120 (21 A+23 B+26 C) \sin (c+d x)+15 (63 A+64 (B+C)) \sin (2 (c+d x))+380 A \sin (3 (c+d x))+135 A \sin (4 (c+d x))+36 A \sin (5 (c+d x))+5 A \sin (6 (c+d x))+900 A c+1380 A d x+340 B \sin (3 (c+d x))+90 B \sin (4 (c+d x))+12 B \sin (5 (c+d x))+1560 B c+1560 B d x+240 C \sin (3 (c+d x))+30 C \sin (4 (c+d x))+1800 C d x)}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + a*Sec[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(900*A*c + 1560*B*c + 1380*A*d*x + 1560*B*d*x + 1800*C*d*x + 120*(21*A + 23*B + 26*C)*Sin[c + d*x] + 15*(
63*A + 64*(B + C))*Sin[2*(c + d*x)] + 380*A*Sin[3*(c + d*x)] + 340*B*Sin[3*(c + d*x)] + 240*C*Sin[3*(c + d*x)]
+ 135*A*Sin[4*(c + d*x)] + 90*B*Sin[4*(c + d*x)] + 30*C*Sin[4*(c + d*x)] + 36*A*Sin[5*(c + d*x)] + 12*B*Sin[5
*(c + d*x)] + 5*A*Sin[6*(c + d*x)]))/(960*d)

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fricas [A]  time = 0.45, size = 145, normalized size = 0.62 \[ \frac {15 \, {\left (23 \, A + 26 \, B + 30 \, C\right )} a^{3} d x + {\left (40 \, A a^{3} \cos \left (d x + c\right )^{5} + 48 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right )^{4} + 10 \, {\left (23 \, A + 18 \, B + 6 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 16 \, {\left (17 \, A + 19 \, B + 15 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 15 \, {\left (23 \, A + 26 \, B + 30 \, C\right )} a^{3} \cos \left (d x + c\right ) + 16 \, {\left (34 \, A + 38 \, B + 45 \, C\right )} a^{3}\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(15*(23*A + 26*B + 30*C)*a^3*d*x + (40*A*a^3*cos(d*x + c)^5 + 48*(3*A + B)*a^3*cos(d*x + c)^4 + 10*(23*A
+ 18*B + 6*C)*a^3*cos(d*x + c)^3 + 16*(17*A + 19*B + 15*C)*a^3*cos(d*x + c)^2 + 15*(23*A + 26*B + 30*C)*a^3*c
os(d*x + c) + 16*(34*A + 38*B + 45*C)*a^3)*sin(d*x + c))/d

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giac [A]  time = 0.33, size = 350, normalized size = 1.49 \[ \frac {15 \, {\left (23 \, A a^{3} + 26 \, B a^{3} + 30 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (345 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 390 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 450 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 1955 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 2210 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 2550 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 4554 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 5148 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 5940 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 5814 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 5988 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 7500 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3165 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4190 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5130 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 1575 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1530 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1470 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/240*(15*(23*A*a^3 + 26*B*a^3 + 30*C*a^3)*(d*x + c) + 2*(345*A*a^3*tan(1/2*d*x + 1/2*c)^11 + 390*B*a^3*tan(1/
2*d*x + 1/2*c)^11 + 450*C*a^3*tan(1/2*d*x + 1/2*c)^11 + 1955*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 2210*B*a^3*tan(1/2
*d*x + 1/2*c)^9 + 2550*C*a^3*tan(1/2*d*x + 1/2*c)^9 + 4554*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 5148*B*a^3*tan(1/2*d
*x + 1/2*c)^7 + 5940*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 5814*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 5988*B*a^3*tan(1/2*d*x
+ 1/2*c)^5 + 7500*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 3165*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 4190*B*a^3*tan(1/2*d*x +
1/2*c)^3 + 5130*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 1575*A*a^3*tan(1/2*d*x + 1/2*c) + 1530*B*a^3*tan(1/2*d*x + 1/2
*c) + 1470*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 = 3.29, size = 364, normalized size = 1.55 \[ \frac {A \,a^{3} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {a^{3} B \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+C \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {3 A \,a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+3 a^{3} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+C \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 A \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 C \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {A \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+a^{3} B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \,a^{3} \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

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

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maxima [A]  time = 0.35, size = 354, normalized size = 1.51 \[ \frac {192 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{3} - 5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} + 90 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} + 64 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} B a^{3} - 960 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{3} + 90 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 960 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 720 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} + 960 \, C a^{3} \sin \left (d x + c\right )}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+a*sec(d*x+c))^3*(A+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))*A*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))*A*a^3 - 320*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3 + 90*(12
*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^3 + 64*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin
(d*x + c))*B*a^3 - 960*(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 + 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^3 - 960*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3
+ 30*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^3 + 720*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a
^3 + 960*C*a^3*sin(d*x + c))/d

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mupad [B]  time = 5.78, size = 333, normalized size = 1.42 \[ \frac {\left (\frac {23\,A\,a^3}{8}+\frac {13\,B\,a^3}{4}+\frac {15\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {391\,A\,a^3}{24}+\frac {221\,B\,a^3}{12}+\frac {85\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {759\,A\,a^3}{20}+\frac {429\,B\,a^3}{10}+\frac {99\,C\,a^3}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {969\,A\,a^3}{20}+\frac {499\,B\,a^3}{10}+\frac {125\,C\,a^3}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {211\,A\,a^3}{8}+\frac {419\,B\,a^3}{12}+\frac {171\,C\,a^3}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {105\,A\,a^3}{8}+\frac {51\,B\,a^3}{4}+\frac {49\,C\,a^3}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{12}+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^3\,\mathrm {atan}\left (\frac {a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (23\,A+26\,B+30\,C\right )}{8\,\left (\frac {23\,A\,a^3}{8}+\frac {13\,B\,a^3}{4}+\frac {15\,C\,a^3}{4}\right )}\right )\,\left (23\,A+26\,B+30\,C\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6*(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)^11*((23*A*a^3)/8 + (13*B*a^3)/4 + (15*C*a^3)/4) + tan(c/2 + (d*x)/2)^9*((391*A*a^3)/24 + (
221*B*a^3)/12 + (85*C*a^3)/4) + tan(c/2 + (d*x)/2)^3*((211*A*a^3)/8 + (419*B*a^3)/12 + (171*C*a^3)/4) + tan(c/
2 + (d*x)/2)^7*((759*A*a^3)/20 + (429*B*a^3)/10 + (99*C*a^3)/2) + tan(c/2 + (d*x)/2)^5*((969*A*a^3)/20 + (499*
B*a^3)/10 + (125*C*a^3)/2) + tan(c/2 + (d*x)/2)*((105*A*a^3)/8 + (51*B*a^3)/4 + (49*C*a^3)/4))/(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*A + 26*B + 30*C))/(8*((23*A*a^3)/8
+ (13*B*a^3)/4 + (15*C*a^3)/4)))*(23*A + 26*B + 30*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)**6*(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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