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

Optimal. Leaf size=257 \[ -\frac {b \left (3 a^2 (4 A+5 C)+A b^2\right ) \sin ^3(c+d x)}{15 d}+\frac {b \left (9 a^2 (4 A+5 C)+b^2 (11 A+15 C)\right ) \sin (c+d x)}{15 d}+\frac {a \left (5 a^2 (5 A+6 C)+6 A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{120 d}+\frac {a \left (a^2 (5 A+6 C)+6 b^2 (3 A+4 C)\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} a x \left (a^2 (5 A+6 C)+6 b^2 (3 A+4 C)\right )+\frac {A \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^3}{6 d}+\frac {A b \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{10 d} \]

[Out]

1/16*a*(6*b^2*(3*A+4*C)+a^2*(5*A+6*C))*x+1/15*b*(9*a^2*(4*A+5*C)+b^2*(11*A+15*C))*sin(d*x+c)/d+1/16*a*(6*b^2*(
3*A+4*C)+a^2*(5*A+6*C))*cos(d*x+c)*sin(d*x+c)/d+1/120*a*(6*A*b^2+5*a^2*(5*A+6*C))*cos(d*x+c)^3*sin(d*x+c)/d+1/
10*A*b*cos(d*x+c)^4*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+1/6*A*cos(d*x+c)^5*(a+b*sec(d*x+c))^3*sin(d*x+c)/d-1/15*b*
(A*b^2+3*a^2*(4*A+5*C))*sin(d*x+c)^3/d

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Rubi [A]  time = 0.75, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ -\frac {b \left (3 a^2 (4 A+5 C)+A b^2\right ) \sin ^3(c+d x)}{15 d}+\frac {b \left (9 a^2 (4 A+5 C)+b^2 (11 A+15 C)\right ) \sin (c+d x)}{15 d}+\frac {a \left (5 a^2 (5 A+6 C)+6 A b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{120 d}+\frac {a \left (a^2 (5 A+6 C)+6 b^2 (3 A+4 C)\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac {1}{16} a x \left (a^2 (5 A+6 C)+6 b^2 (3 A+4 C)\right )+\frac {A \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^3}{6 d}+\frac {A b \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{10 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(6*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*x)/16 + (b*(9*a^2*(4*A + 5*C) + b^2*(11*A + 15*C))*Sin[c + d*x])/(15*
d) + (a*(6*b^2*(3*A + 4*C) + a^2*(5*A + 6*C))*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (a*(6*A*b^2 + 5*a^2*(5*A + 6
*C))*Cos[c + d*x]^3*Sin[c + d*x])/(120*d) + (A*b*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(10*d) +
(A*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*Sin[c + d*x])/(6*d) - (b*(A*b^2 + 3*a^2*(4*A + 5*C))*Sin[c + d*x]^3)/
(15*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 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]

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 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 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 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 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]

Rubi steps

\begin {align*} \int \cos ^6(c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {1}{6} \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (3 A b+a (5 A+6 C) \sec (c+d x)+2 b (A+3 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {A b \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac {A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {1}{30} \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (6 A b^2+5 a^2 (5 A+6 C)+a b (47 A+60 C) \sec (c+d x)+2 b^2 (8 A+15 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a \left (6 A b^2+5 a^2 (5 A+6 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac {A b \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac {A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}-\frac {1}{120} \int \cos ^3(c+d x) \left (-24 b \left (A b^2+3 a^2 (4 A+5 C)\right )-15 a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \sec (c+d x)-8 b^3 (8 A+15 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a \left (6 A b^2+5 a^2 (5 A+6 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac {A b \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac {A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}-\frac {1}{120} \int \cos ^3(c+d x) \left (-24 b \left (A b^2+3 a^2 (4 A+5 C)\right )-8 b^3 (8 A+15 C) \sec ^2(c+d x)\right ) \, dx+\frac {1}{8} \left (a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right )\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a \left (6 A b^2+5 a^2 (5 A+6 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac {A b \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac {A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}-\frac {1}{120} \int \cos (c+d x) \left (-8 b^3 (8 A+15 C)-24 b \left (A b^2+3 a^2 (4 A+5 C)\right ) \cos ^2(c+d x)\right ) \, dx+\frac {1}{16} \left (a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right )\right ) \int 1 \, dx\\ &=\frac {1}{16} a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) x+\frac {a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a \left (6 A b^2+5 a^2 (5 A+6 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac {A b \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac {A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac {\operatorname {Subst}\left (\int \left (-8 b^3 (8 A+15 C)-24 b \left (A b^2+3 a^2 (4 A+5 C)\right )+24 b \left (A b^2+3 a^2 (4 A+5 C)\right ) x^2\right ) \, dx,x,-\sin (c+d x)\right )}{120 d}\\ &=\frac {1}{16} a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) x+\frac {b \left (9 a^2 (4 A+5 C)+b^2 (11 A+15 C)\right ) \sin (c+d x)}{15 d}+\frac {a \left (6 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac {a \left (6 A b^2+5 a^2 (5 A+6 C)\right ) \cos ^3(c+d x) \sin (c+d x)}{120 d}+\frac {A b \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{10 d}+\frac {A \cos ^5(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{6 d}-\frac {b \left (A b^2+3 a^2 (4 A+5 C)\right ) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 1.09, size = 253, normalized size = 0.98 \[ \frac {45 a^3 A \sin (4 (c+d x))+5 a^3 A \sin (6 (c+d x))+300 a^3 A c+300 a^3 A d x+30 a^3 C \sin (4 (c+d x))+360 a^3 c C+360 a^3 C d x+15 a \left (a^2 (15 A+16 C)+48 b^2 (A+C)\right ) \sin (2 (c+d x))+120 b \left (3 a^2 (5 A+6 C)+2 b^2 (3 A+4 C)\right ) \sin (c+d x)+300 a^2 A b \sin (3 (c+d x))+36 a^2 A b \sin (5 (c+d x))+240 a^2 b C \sin (3 (c+d x))+90 a A b^2 \sin (4 (c+d x))+1080 a A b^2 c+1080 a A b^2 d x+1440 a b^2 c C+1440 a b^2 C d x+80 A b^3 \sin (3 (c+d x))}{960 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(300*a^3*A*c + 1080*a*A*b^2*c + 360*a^3*c*C + 1440*a*b^2*c*C + 300*a^3*A*d*x + 1080*a*A*b^2*d*x + 360*a^3*C*d*
x + 1440*a*b^2*C*d*x + 120*b*(2*b^2*(3*A + 4*C) + 3*a^2*(5*A + 6*C))*Sin[c + d*x] + 15*a*(48*b^2*(A + C) + a^2
*(15*A + 16*C))*Sin[2*(c + d*x)] + 300*a^2*A*b*Sin[3*(c + d*x)] + 80*A*b^3*Sin[3*(c + d*x)] + 240*a^2*b*C*Sin[
3*(c + d*x)] + 45*a^3*A*Sin[4*(c + d*x)] + 90*a*A*b^2*Sin[4*(c + d*x)] + 30*a^3*C*Sin[4*(c + d*x)] + 36*a^2*A*
b*Sin[5*(c + d*x)] + 5*a^3*A*Sin[6*(c + d*x)])/(960*d)

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fricas [A]  time = 0.46, size = 189, normalized size = 0.74 \[ \frac {15 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 6 \, {\left (3 \, A + 4 \, C\right )} a b^{2}\right )} d x + {\left (40 \, A a^{3} \cos \left (d x + c\right )^{5} + 144 \, A a^{2} b \cos \left (d x + c\right )^{4} + 96 \, {\left (4 \, A + 5 \, C\right )} a^{2} b + 80 \, {\left (2 \, A + 3 \, C\right )} b^{3} + 10 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 18 \, A a b^{2}\right )} \cos \left (d x + c\right )^{3} + 16 \, {\left (3 \, {\left (4 \, A + 5 \, C\right )} a^{2} b + 5 \, A b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left ({\left (5 \, A + 6 \, C\right )} a^{3} + 6 \, {\left (3 \, A + 4 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*((5*A + 6*C)*a^3 + 6*(3*A + 4*C)*a*b^2)*d*x + (40*A*a^3*cos(d*x + c)^5 + 144*A*a^2*b*cos(d*x + c)^4
+ 96*(4*A + 5*C)*a^2*b + 80*(2*A + 3*C)*b^3 + 10*((5*A + 6*C)*a^3 + 18*A*a*b^2)*cos(d*x + c)^3 + 16*(3*(4*A +
5*C)*a^2*b + 5*A*b^3)*cos(d*x + c)^2 + 15*((5*A + 6*C)*a^3 + 6*(3*A + 4*C)*a*b^2)*cos(d*x + c))*sin(d*x + c))/
d

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giac [B]  time = 0.30, size = 882, normalized size = 3.43 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(5*A*a^3 + 6*C*a^3 + 18*A*a*b^2 + 24*C*a*b^2)*(d*x + c) - 2*(165*A*a^3*tan(1/2*d*x + 1/2*c)^11 + 150
*C*a^3*tan(1/2*d*x + 1/2*c)^11 - 720*A*a^2*b*tan(1/2*d*x + 1/2*c)^11 - 720*C*a^2*b*tan(1/2*d*x + 1/2*c)^11 + 4
50*A*a*b^2*tan(1/2*d*x + 1/2*c)^11 + 360*C*a*b^2*tan(1/2*d*x + 1/2*c)^11 - 240*A*b^3*tan(1/2*d*x + 1/2*c)^11 -
240*C*b^3*tan(1/2*d*x + 1/2*c)^11 - 25*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 210*C*a^3*tan(1/2*d*x + 1/2*c)^9 - 1680
*A*a^2*b*tan(1/2*d*x + 1/2*c)^9 - 2640*C*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 630*A*a*b^2*tan(1/2*d*x + 1/2*c)^9 + 1
080*C*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 880*A*b^3*tan(1/2*d*x + 1/2*c)^9 - 1200*C*b^3*tan(1/2*d*x + 1/2*c)^9 + 45
0*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 60*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 3744*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 4320*
C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 180*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 720*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 144
0*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 2400*C*b^3*tan(1/2*d*x + 1/2*c)^7 - 450*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 60*C*a
^3*tan(1/2*d*x + 1/2*c)^5 - 3744*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 4320*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 180*A*
a*b^2*tan(1/2*d*x + 1/2*c)^5 - 720*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 1440*A*b^3*tan(1/2*d*x + 1/2*c)^5 - 2400*C
*b^3*tan(1/2*d*x + 1/2*c)^5 + 25*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 210*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 1680*A*a^2*
b*tan(1/2*d*x + 1/2*c)^3 - 2640*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 630*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 1080*C*a
*b^2*tan(1/2*d*x + 1/2*c)^3 - 880*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 1200*C*b^3*tan(1/2*d*x + 1/2*c)^3 - 165*A*a^3
*tan(1/2*d*x + 1/2*c) - 150*C*a^3*tan(1/2*d*x + 1/2*c) - 720*A*a^2*b*tan(1/2*d*x + 1/2*c) - 720*C*a^2*b*tan(1/
2*d*x + 1/2*c) - 450*A*a*b^2*tan(1/2*d*x + 1/2*c) - 360*C*a*b^2*tan(1/2*d*x + 1/2*c) - 240*A*b^3*tan(1/2*d*x +
1/2*c) - 240*C*b^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 = 1.79, size = 249, normalized size = 0.97 \[ \frac {A \,a^{3} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+C \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {3 A \,a^{2} b \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+C \,a^{2} b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 A a \,b^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 C a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {A \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+b^{3} C \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+b*sec(d*x+c))^3*(A+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)+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^2*b*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+C*
a^2*b*(2+cos(d*x+c)^2)*sin(d*x+c)+3*A*a*b^2*(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*C*a
*b^2*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+1/3*A*b^3*(2+cos(d*x+c)^2)*sin(d*x+c)+b^3*C*sin(d*x+c))

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maxima [A]  time = 0.36, size = 243, normalized size = 0.95 \[ -\frac {5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 192 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{2} b + 960 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} b - 90 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a b^{2} - 720 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{2} + 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{3} - 960 \, C b^{3} \sin \left (d x + c\right )}{960 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/960*(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 - 30*(12*d*x
+ 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^3 - 192*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*
x + c))*A*a^2*b + 960*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^2*b - 90*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin
(2*d*x + 2*c))*A*a*b^2 - 720*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a*b^2 + 320*(sin(d*x + c)^3 - 3*sin(d*x + c))*
A*b^3 - 960*C*b^3*sin(d*x + c))/d

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mupad [B]  time = 6.78, size = 573, normalized size = 2.23 \[ \frac {\left (2\,A\,b^3-\frac {11\,A\,a^3}{8}-\frac {5\,C\,a^3}{4}+2\,C\,b^3-\frac {15\,A\,a\,b^2}{4}+6\,A\,a^2\,b-3\,C\,a\,b^2+6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {5\,A\,a^3}{24}+\frac {22\,A\,b^3}{3}-\frac {7\,C\,a^3}{4}+10\,C\,b^3-\frac {21\,A\,a\,b^2}{4}+14\,A\,a^2\,b-9\,C\,a\,b^2+22\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (12\,A\,b^3-\frac {15\,A\,a^3}{4}-\frac {C\,a^3}{2}+20\,C\,b^3-\frac {3\,A\,a\,b^2}{2}+\frac {156\,A\,a^2\,b}{5}-6\,C\,a\,b^2+36\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {15\,A\,a^3}{4}+12\,A\,b^3+\frac {C\,a^3}{2}+20\,C\,b^3+\frac {3\,A\,a\,b^2}{2}+\frac {156\,A\,a^2\,b}{5}+6\,C\,a\,b^2+36\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {22\,A\,b^3}{3}-\frac {5\,A\,a^3}{24}+\frac {7\,C\,a^3}{4}+10\,C\,b^3+\frac {21\,A\,a\,b^2}{4}+14\,A\,a^2\,b+9\,C\,a\,b^2+22\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {11\,A\,a^3}{8}+2\,A\,b^3+\frac {5\,C\,a^3}{4}+2\,C\,b^3+\frac {15\,A\,a\,b^2}{4}+6\,A\,a^2\,b+3\,C\,a\,b^2+6\,C\,a^2\,b\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\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (5\,A\,a^2+18\,A\,b^2+6\,C\,a^2+24\,C\,b^2\right )}{8\,\left (\frac {5\,A\,a^3}{8}+\frac {3\,C\,a^3}{4}+\frac {9\,A\,a\,b^2}{4}+3\,C\,a\,b^2\right )}\right )\,\left (5\,A\,a^2+18\,A\,b^2+6\,C\,a^2+24\,C\,b^2\right )}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((11*A*a^3)/8 + 2*A*b^3 + (5*C*a^3)/4 + 2*C*b^3 + (15*A*a*b^2)/4 + 6*A*a^2*b + 3*C*a*b^2 +
6*C*a^2*b) - tan(c/2 + (d*x)/2)^11*((11*A*a^3)/8 - 2*A*b^3 + (5*C*a^3)/4 - 2*C*b^3 + (15*A*a*b^2)/4 - 6*A*a^2
*b + 3*C*a*b^2 - 6*C*a^2*b) + tan(c/2 + (d*x)/2)^3*((22*A*b^3)/3 - (5*A*a^3)/24 + (7*C*a^3)/4 + 10*C*b^3 + (21
*A*a*b^2)/4 + 14*A*a^2*b + 9*C*a*b^2 + 22*C*a^2*b) + tan(c/2 + (d*x)/2)^9*((5*A*a^3)/24 + (22*A*b^3)/3 - (7*C*
a^3)/4 + 10*C*b^3 - (21*A*a*b^2)/4 + 14*A*a^2*b - 9*C*a*b^2 + 22*C*a^2*b) + tan(c/2 + (d*x)/2)^5*((15*A*a^3)/4
+ 12*A*b^3 + (C*a^3)/2 + 20*C*b^3 + (3*A*a*b^2)/2 + (156*A*a^2*b)/5 + 6*C*a*b^2 + 36*C*a^2*b) - tan(c/2 + (d*
x)/2)^7*((15*A*a^3)/4 - 12*A*b^3 + (C*a^3)/2 - 20*C*b^3 + (3*A*a*b^2)/2 - (156*A*a^2*b)/5 + 6*C*a*b^2 - 36*C*a
^2*b))/(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*atan((a*tan(c/2 + (d*x)/2)*(5*A*a^2 + 18*A*b^2
+ 6*C*a^2 + 24*C*b^2))/(8*((5*A*a^3)/8 + (3*C*a^3)/4 + (9*A*a*b^2)/4 + 3*C*a*b^2)))*(5*A*a^2 + 18*A*b^2 + 6*C*
a^2 + 24*C*b^2))/(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+b*sec(d*x+c))**3*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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