3.671 \(\int \cos ^6(c+d x) (a+b \sec (c+d x))^4 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=298 \[ \frac{4 a b \left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \sin (c+d x)}{15 d}+\frac{a b \left (a^2 (39 A+50 C)+4 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{60 d}+\frac{\left (10 a^2 b^2 (49 A+66 C)+15 a^4 (5 A+6 C)+24 A b^4\right ) \sin (c+d x) \cos (c+d x)}{240 d}+\frac{\left (5 a^2 (5 A+6 C)+12 A b^2\right ) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{120 d}+\frac{1}{16} x \left (12 a^2 b^2 (3 A+4 C)+a^4 (5 A+6 C)+8 b^4 (A+2 C)\right )+\frac{A \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^4}{6 d}+\frac{2 A b \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{15 d} \]

[Out]

((8*b^4*(A + 2*C) + 12*a^2*b^2*(3*A + 4*C) + a^4*(5*A + 6*C))*x)/16 + (4*a*b*(5*b^2*(2*A + 3*C) + 2*a^2*(4*A +
 5*C))*Sin[c + d*x])/(15*d) + ((24*A*b^4 + 15*a^4*(5*A + 6*C) + 10*a^2*b^2*(49*A + 66*C))*Cos[c + d*x]*Sin[c +
 d*x])/(240*d) + (a*b*(4*A*b^2 + a^2*(39*A + 50*C))*Cos[c + d*x]^2*Sin[c + d*x])/(60*d) + ((12*A*b^2 + 5*a^2*(
5*A + 6*C))*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(120*d) + (2*A*b*Cos[c + d*x]^4*(a + b*Sec[c +
 d*x])^3*Sin[c + d*x])/(15*d) + (A*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(6*d)

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Rubi [A]  time = 1.03726, antiderivative size = 298, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4095, 4094, 4074, 4047, 2637, 4045, 8} \[ \frac{4 a b \left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \sin (c+d x)}{15 d}+\frac{a b \left (a^2 (39 A+50 C)+4 A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{60 d}+\frac{\left (10 a^2 b^2 (49 A+66 C)+15 a^4 (5 A+6 C)+24 A b^4\right ) \sin (c+d x) \cos (c+d x)}{240 d}+\frac{\left (5 a^2 (5 A+6 C)+12 A b^2\right ) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{120 d}+\frac{1}{16} x \left (12 a^2 b^2 (3 A+4 C)+a^4 (5 A+6 C)+8 b^4 (A+2 C)\right )+\frac{A \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^4}{6 d}+\frac{2 A b \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{15 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

((8*b^4*(A + 2*C) + 12*a^2*b^2*(3*A + 4*C) + a^4*(5*A + 6*C))*x)/16 + (4*a*b*(5*b^2*(2*A + 3*C) + 2*a^2*(4*A +
 5*C))*Sin[c + d*x])/(15*d) + ((24*A*b^4 + 15*a^4*(5*A + 6*C) + 10*a^2*b^2*(49*A + 66*C))*Cos[c + d*x]*Sin[c +
 d*x])/(240*d) + (a*b*(4*A*b^2 + a^2*(39*A + 50*C))*Cos[c + d*x]^2*Sin[c + d*x])/(60*d) + ((12*A*b^2 + 5*a^2*(
5*A + 6*C))*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(120*d) + (2*A*b*Cos[c + d*x]^4*(a + b*Sec[c +
 d*x])^3*Sin[c + d*x])/(15*d) + (A*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(6*d)

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]

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 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 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 2637

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

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
 + f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 8

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

Rubi steps

\begin{align*} \int \cos ^6(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos ^5(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{6 d}+\frac{1}{6} \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (4 A b+a (5 A+6 C) \sec (c+d x)+b (A+6 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{2 A b \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{15 d}+\frac{A \cos ^5(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{6 d}+\frac{1}{30} \int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (12 A b^2+5 a^2 (5 A+6 C)+2 a b (23 A+30 C) \sec (c+d x)+3 b^2 (3 A+10 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{\left (12 A b^2+5 a^2 (5 A+6 C)\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{120 d}+\frac{2 A b \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{15 d}+\frac{A \cos ^5(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{6 d}+\frac{1}{120} \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (6 b \left (4 A b^2+a^2 (39 A+50 C)\right )+a \left (15 a^2 (5 A+6 C)+8 b^2 (32 A+45 C)\right ) \sec (c+d x)+b \left (24 b^2 (2 A+5 C)+5 a^2 (5 A+6 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a b \left (4 A b^2+a^2 (39 A+50 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{60 d}+\frac{\left (12 A b^2+5 a^2 (5 A+6 C)\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{120 d}+\frac{2 A b \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{15 d}+\frac{A \cos ^5(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{6 d}-\frac{1}{360} \int \cos ^2(c+d x) \left (-3 \left (24 A b^4+15 a^4 (5 A+6 C)+10 a^2 b^2 (49 A+66 C)\right )-96 a b \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \sec (c+d x)-3 b^2 \left (24 b^2 (2 A+5 C)+5 a^2 (5 A+6 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a b \left (4 A b^2+a^2 (39 A+50 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{60 d}+\frac{\left (12 A b^2+5 a^2 (5 A+6 C)\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{120 d}+\frac{2 A b \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{15 d}+\frac{A \cos ^5(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{6 d}-\frac{1}{360} \int \cos ^2(c+d x) \left (-3 \left (24 A b^4+15 a^4 (5 A+6 C)+10 a^2 b^2 (49 A+66 C)\right )-3 b^2 \left (24 b^2 (2 A+5 C)+5 a^2 (5 A+6 C)\right ) \sec ^2(c+d x)\right ) \, dx+\frac{1}{15} \left (4 a b \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right )\right ) \int \cos (c+d x) \, dx\\ &=\frac{4 a b \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \sin (c+d x)}{15 d}+\frac{\left (24 A b^4+15 a^4 (5 A+6 C)+10 a^2 b^2 (49 A+66 C)\right ) \cos (c+d x) \sin (c+d x)}{240 d}+\frac{a b \left (4 A b^2+a^2 (39 A+50 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{60 d}+\frac{\left (12 A b^2+5 a^2 (5 A+6 C)\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{120 d}+\frac{2 A b \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{15 d}+\frac{A \cos ^5(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{6 d}-\frac{1}{16} \left (-8 b^4 (A+2 C)-12 a^2 b^2 (3 A+4 C)-a^4 (5 A+6 C)\right ) \int 1 \, dx\\ &=\frac{1}{16} \left (8 b^4 (A+2 C)+12 a^2 b^2 (3 A+4 C)+a^4 (5 A+6 C)\right ) x+\frac{4 a b \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \sin (c+d x)}{15 d}+\frac{\left (24 A b^4+15 a^4 (5 A+6 C)+10 a^2 b^2 (49 A+66 C)\right ) \cos (c+d x) \sin (c+d x)}{240 d}+\frac{a b \left (4 A b^2+a^2 (39 A+50 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{60 d}+\frac{\left (12 A b^2+5 a^2 (5 A+6 C)\right ) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{120 d}+\frac{2 A b \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{15 d}+\frac{A \cos ^5(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.906359, size = 302, normalized size = 1.01 \[ \frac{480 a b \left (a^2 (5 A+6 C)+2 b^2 (3 A+4 C)\right ) \sin (c+d x)+15 \left (96 a^2 b^2 (A+C)+a^4 (15 A+16 C)+16 A b^4\right ) \sin (2 (c+d x))+180 a^2 A b^2 \sin (4 (c+d x))+2160 a^2 A b^2 c+2160 a^2 A b^2 d x+400 a^3 A b \sin (3 (c+d x))+48 a^3 A b \sin (5 (c+d x))+45 a^4 A \sin (4 (c+d x))+5 a^4 A \sin (6 (c+d x))+300 a^4 A c+300 a^4 A d x+2880 a^2 b^2 c C+2880 a^2 b^2 C d x+320 a^3 b C \sin (3 (c+d x))+30 a^4 C \sin (4 (c+d x))+360 a^4 c C+360 a^4 C d x+320 a A b^3 \sin (3 (c+d x))+480 A b^4 c+480 A b^4 d x+960 b^4 c C+960 b^4 C d x}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

(300*a^4*A*c + 2160*a^2*A*b^2*c + 480*A*b^4*c + 360*a^4*c*C + 2880*a^2*b^2*c*C + 960*b^4*c*C + 300*a^4*A*d*x +
 2160*a^2*A*b^2*d*x + 480*A*b^4*d*x + 360*a^4*C*d*x + 2880*a^2*b^2*C*d*x + 960*b^4*C*d*x + 480*a*b*(2*b^2*(3*A
 + 4*C) + a^2*(5*A + 6*C))*Sin[c + d*x] + 15*(16*A*b^4 + 96*a^2*b^2*(A + C) + a^4*(15*A + 16*C))*Sin[2*(c + d*
x)] + 400*a^3*A*b*Sin[3*(c + d*x)] + 320*a*A*b^3*Sin[3*(c + d*x)] + 320*a^3*b*C*Sin[3*(c + d*x)] + 45*a^4*A*Si
n[4*(c + d*x)] + 180*a^2*A*b^2*Sin[4*(c + d*x)] + 30*a^4*C*Sin[4*(c + d*x)] + 48*a^3*A*b*Sin[5*(c + d*x)] + 5*
a^4*A*Sin[6*(c + d*x)])/(960*d)

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Maple [A]  time = 0.084, size = 294, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( A{a}^{4} \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) +{\frac{4\,A{a}^{3}b\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+6\,A{a}^{2}{b}^{2} \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +{a}^{4}C \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +{\frac{4\,Aa{b}^{3} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{4\,{a}^{3}bC \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+A{b}^{4} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +6\,C{a}^{2}{b}^{2} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +4\,Ca{b}^{3}\sin \left ( dx+c \right ) +C{b}^{4} \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x)

[Out]

1/d*(A*a^4*(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)+4/5*A*a^3*b*(8/3+c
os(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+6*A*a^2*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)+a^4*C*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+4/3*A*a*b^3*(2+cos(d*x+c)^2)*sin(d*x+c)
+4/3*a^3*b*C*(2+cos(d*x+c)^2)*sin(d*x+c)+A*b^4*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+6*C*a^2*b^2*(1/2*cos(
d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+4*C*a*b^3*sin(d*x+c)+C*b^4*(d*x+c))

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Maxima [A]  time = 1.01971, size = 382, normalized size = 1.28 \begin{align*} -\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^{4} - 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^{4} - 256 \,{\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} b + 1280 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} b - 180 \,{\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^{2} b^{2} - 1440 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b^{2} + 1280 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b^{3} - 240 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{4} - 960 \,{\left (d x + c\right )} C b^{4} - 3840 \, C a b^{3} \sin \left (d x + c\right )}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^4*(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^4 - 30*(12*d*x
 + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^4 - 256*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*
x + c))*A*a^3*b + 1280*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^3*b - 180*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*s
in(2*d*x + 2*c))*A*a^2*b^2 - 1440*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2*b^2 + 1280*(sin(d*x + c)^3 - 3*sin(d*
x + c))*A*a*b^3 - 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*b^4 - 960*(d*x + c)*C*b^4 - 3840*C*a*b^3*sin(d*x + c)
)/d

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Fricas [A]  time = 0.567601, size = 504, normalized size = 1.69 \begin{align*} \frac{15 \,{\left ({\left (5 \, A + 6 \, C\right )} a^{4} + 12 \,{\left (3 \, A + 4 \, C\right )} a^{2} b^{2} + 8 \,{\left (A + 2 \, C\right )} b^{4}\right )} d x +{\left (40 \, A a^{4} \cos \left (d x + c\right )^{5} + 192 \, A a^{3} b \cos \left (d x + c\right )^{4} + 128 \,{\left (4 \, A + 5 \, C\right )} a^{3} b + 320 \,{\left (2 \, A + 3 \, C\right )} a b^{3} + 10 \,{\left ({\left (5 \, A + 6 \, C\right )} a^{4} + 36 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} + 64 \,{\left ({\left (4 \, A + 5 \, C\right )} a^{3} b + 5 \, A a b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left ({\left (5 \, A + 6 \, C\right )} a^{4} + 12 \,{\left (3 \, A + 4 \, C\right )} a^{2} b^{2} + 8 \, A b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(15*((5*A + 6*C)*a^4 + 12*(3*A + 4*C)*a^2*b^2 + 8*(A + 2*C)*b^4)*d*x + (40*A*a^4*cos(d*x + c)^5 + 192*A*
a^3*b*cos(d*x + c)^4 + 128*(4*A + 5*C)*a^3*b + 320*(2*A + 3*C)*a*b^3 + 10*((5*A + 6*C)*a^4 + 36*A*a^2*b^2)*cos
(d*x + c)^3 + 64*((4*A + 5*C)*a^3*b + 5*A*a*b^3)*cos(d*x + c)^2 + 15*((5*A + 6*C)*a^4 + 12*(3*A + 4*C)*a^2*b^2
 + 8*A*b^4)*cos(d*x + c))*sin(d*x + c))/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*(a+b*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.27712, size = 1396, normalized size = 4.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/240*(15*(5*A*a^4 + 6*C*a^4 + 36*A*a^2*b^2 + 48*C*a^2*b^2 + 8*A*b^4 + 16*C*b^4)*(d*x + c) - 2*(165*A*a^4*tan(
1/2*d*x + 1/2*c)^11 + 150*C*a^4*tan(1/2*d*x + 1/2*c)^11 - 960*A*a^3*b*tan(1/2*d*x + 1/2*c)^11 - 960*C*a^3*b*ta
n(1/2*d*x + 1/2*c)^11 + 900*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 + 720*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 - 960*A*
a*b^3*tan(1/2*d*x + 1/2*c)^11 - 960*C*a*b^3*tan(1/2*d*x + 1/2*c)^11 + 120*A*b^4*tan(1/2*d*x + 1/2*c)^11 - 25*A
*a^4*tan(1/2*d*x + 1/2*c)^9 + 210*C*a^4*tan(1/2*d*x + 1/2*c)^9 - 2240*A*a^3*b*tan(1/2*d*x + 1/2*c)^9 - 3520*C*
a^3*b*tan(1/2*d*x + 1/2*c)^9 + 1260*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 + 2160*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 -
 3520*A*a*b^3*tan(1/2*d*x + 1/2*c)^9 - 4800*C*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 360*A*b^4*tan(1/2*d*x + 1/2*c)^9
+ 450*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 60*C*a^4*tan(1/2*d*x + 1/2*c)^7 - 4992*A*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 5
760*C*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 360*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 1440*C*a^2*b^2*tan(1/2*d*x + 1/2*c
)^7 - 5760*A*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 9600*C*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 240*A*b^4*tan(1/2*d*x + 1/2*
c)^7 - 450*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 60*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 4992*A*a^3*b*tan(1/2*d*x + 1/2*c)^
5 - 5760*C*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 360*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 1440*C*a^2*b^2*tan(1/2*d*x +
1/2*c)^5 - 5760*A*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 9600*C*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 240*A*b^4*tan(1/2*d*x +
 1/2*c)^5 + 25*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 210*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 2240*A*a^3*b*tan(1/2*d*x + 1/
2*c)^3 - 3520*C*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 1260*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 2160*C*a^2*b^2*tan(1/2*
d*x + 1/2*c)^3 - 3520*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 4800*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 360*A*b^4*tan(1/2
*d*x + 1/2*c)^3 - 165*A*a^4*tan(1/2*d*x + 1/2*c) - 150*C*a^4*tan(1/2*d*x + 1/2*c) - 960*A*a^3*b*tan(1/2*d*x +
1/2*c) - 960*C*a^3*b*tan(1/2*d*x + 1/2*c) - 900*A*a^2*b^2*tan(1/2*d*x + 1/2*c) - 720*C*a^2*b^2*tan(1/2*d*x + 1
/2*c) - 960*A*a*b^3*tan(1/2*d*x + 1/2*c) - 960*C*a*b^3*tan(1/2*d*x + 1/2*c) - 120*A*b^4*tan(1/2*d*x + 1/2*c))/
(tan(1/2*d*x + 1/2*c)^2 + 1)^6)/d