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

Optimal. Leaf size=269 \[ \frac{\sin (c+d x) \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right )}{15 d}+\frac{a \sin (c+d x) \cos ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{30 d}+\frac{\sin (c+d x) \cos (c+d x) \left (15 a^2 b (3 A+4 C)+15 a^3 B+50 a b^2 B+6 A b^3\right )}{40 d}+\frac{1}{8} x \left (3 a^2 b (3 A+4 C)+3 a^3 B+12 a b^2 B+4 b^3 (A+2 C)\right )+\frac{(5 a B+3 A b) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{20 d}+\frac{A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d} \]

[Out]

((3*a^3*B + 12*a*b^2*B + 4*b^3*(A + 2*C) + 3*a^2*b*(3*A + 4*C))*x)/8 + ((30*a^2*b*B + 15*b^3*B + 15*a*b^2*(2*A
+ 3*C) + 2*a^3*(4*A + 5*C))*Sin[c + d*x])/(15*d) + ((6*A*b^3 + 15*a^3*B + 50*a*b^2*B + 15*a^2*b*(3*A + 4*C))*
Cos[c + d*x]*Sin[c + d*x])/(40*d) + (a*(3*A*b^2 + 15*a*b*B + 2*a^2*(4*A + 5*C))*Cos[c + d*x]^2*Sin[c + d*x])/(
30*d) + ((3*A*b + 5*a*B)*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(20*d) + (A*Cos[c + d*x]^4*(a + b
*Sec[c + d*x])^3*Sin[c + d*x])/(5*d)

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Rubi [A]  time = 0.751157, antiderivative size = 269, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.146, Rules used = {4094, 4074, 4047, 2637, 4045, 8} \[ \frac{\sin (c+d x) \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right )}{15 d}+\frac{a \sin (c+d x) \cos ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{30 d}+\frac{\sin (c+d x) \cos (c+d x) \left (15 a^2 b (3 A+4 C)+15 a^3 B+50 a b^2 B+6 A b^3\right )}{40 d}+\frac{1}{8} x \left (3 a^2 b (3 A+4 C)+3 a^3 B+12 a b^2 B+4 b^3 (A+2 C)\right )+\frac{(5 a B+3 A b) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{20 d}+\frac{A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^3*B + 12*a*b^2*B + 4*b^3*(A + 2*C) + 3*a^2*b*(3*A + 4*C))*x)/8 + ((30*a^2*b*B + 15*b^3*B + 15*a*b^2*(2*A
+ 3*C) + 2*a^3*(4*A + 5*C))*Sin[c + d*x])/(15*d) + ((6*A*b^3 + 15*a^3*B + 50*a*b^2*B + 15*a^2*b*(3*A + 4*C))*
Cos[c + d*x]*Sin[c + d*x])/(40*d) + (a*(3*A*b^2 + 15*a*b*B + 2*a^2*(4*A + 5*C))*Cos[c + d*x]^2*Sin[c + d*x])/(
30*d) + ((3*A*b + 5*a*B)*Cos[c + d*x]^3*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(20*d) + (A*Cos[c + d*x]^4*(a + b
*Sec[c + d*x])^3*Sin[c + d*x])/(5*d)

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 ^5(c+d x) (a+b \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+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{1}{5} \int \cos ^4(c+d x) (a+b \sec (c+d x))^2 \left (3 A b+5 a B+(4 a A+5 b B+5 a C) \sec (c+d x)+b (A+5 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{(3 A b+5 a B) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}+\frac{1}{20} \int \cos ^3(c+d x) (a+b \sec (c+d x)) \left (2 \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right )+\left (15 a^2 B+20 b^2 B+a b (29 A+40 C)\right ) \sec (c+d x)+b (7 A b+5 a B+20 b C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac{(3 A b+5 a B) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}-\frac{1}{60} \int \cos ^2(c+d x) \left (-3 \left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right )-4 \left (30 a^2 b B+15 b^3 B+15 a b^2 (2 A+3 C)+2 a^3 (4 A+5 C)\right ) \sec (c+d x)-3 b^2 (7 A b+5 a B+20 b C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac{(3 A b+5 a B) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}-\frac{1}{60} \int \cos ^2(c+d x) \left (-3 \left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right )-3 b^2 (7 A b+5 a B+20 b C) \sec ^2(c+d x)\right ) \, dx-\frac{1}{15} \left (-30 a^2 b B-15 b^3 B-15 a b^2 (2 A+3 C)-2 a^3 (4 A+5 C)\right ) \int \cos (c+d x) \, dx\\ &=\frac{\left (30 a^2 b B+15 b^3 B+15 a b^2 (2 A+3 C)+2 a^3 (4 A+5 C)\right ) \sin (c+d x)}{15 d}+\frac{\left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac{a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac{(3 A b+5 a B) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}-\frac{1}{8} \left (-3 a^3 B-12 a b^2 B-4 b^3 (A+2 C)-3 a^2 b (3 A+4 C)\right ) \int 1 \, dx\\ &=\frac{1}{8} \left (3 a^3 B+12 a b^2 B+4 b^3 (A+2 C)+3 a^2 b (3 A+4 C)\right ) x+\frac{\left (30 a^2 b B+15 b^3 B+15 a b^2 (2 A+3 C)+2 a^3 (4 A+5 C)\right ) \sin (c+d x)}{15 d}+\frac{\left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac{a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac{(3 A b+5 a B) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac{A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 1.0612, size = 288, normalized size = 1.07 \[ \frac{60 \sin (c+d x) \left (a^3 (5 A+6 C)+18 a^2 b B+6 a b^2 (3 A+4 C)+8 b^3 B\right )+120 \sin (2 (c+d x)) \left (3 a^2 b (A+C)+a^3 B+3 a b^2 B+A b^3\right )+45 a^2 A b \sin (4 (c+d x))+540 a^2 A b c+540 a^2 A b d x+50 a^3 A \sin (3 (c+d x))+6 a^3 A \sin (5 (c+d x))+120 a^2 b B \sin (3 (c+d x))+720 a^2 b c C+720 a^2 b C d x+15 a^3 B \sin (4 (c+d x))+180 a^3 B c+180 a^3 B d x+40 a^3 C \sin (3 (c+d x))+120 a A b^2 \sin (3 (c+d x))+720 a b^2 B c+720 a b^2 B d x+240 A b^3 c+240 A b^3 d x+480 b^3 c C+480 b^3 C d x}{480 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(540*a^2*A*b*c + 240*A*b^3*c + 180*a^3*B*c + 720*a*b^2*B*c + 720*a^2*b*c*C + 480*b^3*c*C + 540*a^2*A*b*d*x + 2
40*A*b^3*d*x + 180*a^3*B*d*x + 720*a*b^2*B*d*x + 720*a^2*b*C*d*x + 480*b^3*C*d*x + 60*(18*a^2*b*B + 8*b^3*B +
6*a*b^2*(3*A + 4*C) + a^3*(5*A + 6*C))*Sin[c + d*x] + 120*(A*b^3 + a^3*B + 3*a*b^2*B + 3*a^2*b*(A + C))*Sin[2*
(c + d*x)] + 50*a^3*A*Sin[3*(c + d*x)] + 120*a*A*b^2*Sin[3*(c + d*x)] + 120*a^2*b*B*Sin[3*(c + d*x)] + 40*a^3*
C*Sin[3*(c + d*x)] + 45*a^2*A*b*Sin[4*(c + d*x)] + 15*a^3*B*Sin[4*(c + d*x)] + 6*a^3*A*Sin[5*(c + d*x)])/(480*
d)

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Maple [A]  time = 0.08, size = 301, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{A{a}^{3}\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 ) }+3\,A{a}^{2}b \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 ) +B{a}^{3} \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 ) +Aa{b}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +B{a}^{2}b \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) +{\frac{{a}^{3}C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+A{b}^{3} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +3\,Ba{b}^{2} \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +3\,{a}^{2}bC \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +B{b}^{3}\sin \left ( dx+c \right ) +3\,Ca{b}^{2}\sin \left ( dx+c \right ) +C{b}^{3} \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]  time = 1.06811, size = 389, normalized size = 1.45 \begin{align*} \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} + 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} - 160 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C 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^{2} b - 480 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} b + 360 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b - 480 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b^{2} + 360 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{2} + 120 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{3} + 480 \,{\left (d x + c\right )} C b^{3} + 1440 \, C a b^{2} \sin \left (d x + c\right ) + 480 \, B b^{3} \sin \left (d x + c\right )}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 0.554683, size = 498, normalized size = 1.85 \begin{align*} \frac{15 \,{\left (3 \, B a^{3} + 3 \,{\left (3 \, A + 4 \, C\right )} a^{2} b + 12 \, B a b^{2} + 4 \,{\left (A + 2 \, C\right )} b^{3}\right )} d x +{\left (24 \, A a^{3} \cos \left (d x + c\right )^{4} + 16 \,{\left (4 \, A + 5 \, C\right )} a^{3} + 240 \, B a^{2} b + 120 \,{\left (2 \, A + 3 \, C\right )} a b^{2} + 120 \, B b^{3} + 30 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )^{3} + 8 \,{\left ({\left (4 \, A + 5 \, C\right )} a^{3} + 15 \, B a^{2} b + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (3 \, B a^{3} + 3 \,{\left (3 \, A + 4 \, C\right )} a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(3*B*a^3 + 3*(3*A + 4*C)*a^2*b + 12*B*a*b^2 + 4*(A + 2*C)*b^3)*d*x + (24*A*a^3*cos(d*x + c)^4 + 16*(
4*A + 5*C)*a^3 + 240*B*a^2*b + 120*(2*A + 3*C)*a*b^2 + 120*B*b^3 + 30*(B*a^3 + 3*A*a^2*b)*cos(d*x + c)^3 + 8*(
(4*A + 5*C)*a^3 + 15*B*a^2*b + 15*A*a*b^2)*cos(d*x + c)^2 + 15*(3*B*a^3 + 3*(3*A + 4*C)*a^2*b + 12*B*a*b^2 + 4
*A*b^3)*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)**5*(a+b*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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Giac [B]  time = 1.39036, size = 1250, normalized size = 4.65 \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)^5*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(15*(3*B*a^3 + 9*A*a^2*b + 12*C*a^2*b + 12*B*a*b^2 + 4*A*b^3 + 8*C*b^3)*(d*x + c) + 2*(120*A*a^3*tan(1/2
*d*x + 1/2*c)^9 - 75*B*a^3*tan(1/2*d*x + 1/2*c)^9 + 120*C*a^3*tan(1/2*d*x + 1/2*c)^9 - 225*A*a^2*b*tan(1/2*d*x
+ 1/2*c)^9 + 360*B*a^2*b*tan(1/2*d*x + 1/2*c)^9 - 180*C*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*A*a*b^2*tan(1/2*d*
x + 1/2*c)^9 - 180*B*a*b^2*tan(1/2*d*x + 1/2*c)^9 + 360*C*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*A*b^3*tan(1/2*d*x
+ 1/2*c)^9 + 120*B*b^3*tan(1/2*d*x + 1/2*c)^9 + 160*A*a^3*tan(1/2*d*x + 1/2*c)^7 - 30*B*a^3*tan(1/2*d*x + 1/2*
c)^7 + 320*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 90*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 960*B*a^2*b*tan(1/2*d*x + 1/2*c)
^7 - 360*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 960*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 360*B*a*b^2*tan(1/2*d*x + 1/2*c
)^7 + 1440*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 120*A*b^3*tan(1/2*d*x + 1/2*c)^7 + 480*B*b^3*tan(1/2*d*x + 1/2*c)^
7 + 464*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 400*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 1200*B*a^2*b*tan(1/2*d*x + 1/2*c)^5
+ 1200*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 2160*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 720*B*b^3*tan(1/2*d*x + 1/2*c)^5
+ 160*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 30*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 320*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 90*
A*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 960*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 360*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 960
*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 360*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 1440*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 1
20*A*b^3*tan(1/2*d*x + 1/2*c)^3 + 480*B*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*A*a^3*tan(1/2*d*x + 1/2*c) + 75*B*a^3
*tan(1/2*d*x + 1/2*c) + 120*C*a^3*tan(1/2*d*x + 1/2*c) + 225*A*a^2*b*tan(1/2*d*x + 1/2*c) + 360*B*a^2*b*tan(1/
2*d*x + 1/2*c) + 180*C*a^2*b*tan(1/2*d*x + 1/2*c) + 360*A*a*b^2*tan(1/2*d*x + 1/2*c) + 180*B*a*b^2*tan(1/2*d*x
+ 1/2*c) + 360*C*a*b^2*tan(1/2*d*x + 1/2*c) + 60*A*b^3*tan(1/2*d*x + 1/2*c) + 120*B*b^3*tan(1/2*d*x + 1/2*c))
/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d