\(\int \cos ^5(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [885]

Optimal result

Integrand size = 41, antiderivative size = 269 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} \left (3 a^3 B+12 a b^2 B+4 b^3 (A+2 C)+3 a^2 b (3 A+4 C)\right ) x+\frac {\left (30 a^2 b B+15 b^3 B+15 a b^2 (2 A+3 C)+2 a^3 (4 A+5 C)\right ) \sin (c+d x)}{15 d}+\frac {\left (6 A b^3+15 a^3 B+50 a b^2 B+15 a^2 b (3 A+4 C)\right ) \cos (c+d x) \sin (c+d x)}{40 d}+\frac {a \left (3 A b^2+15 a b B+2 a^2 (4 A+5 C)\right ) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {(3 A b+5 a B) \cos ^3(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{5 d} \]

[Out]

1/8*(3*B*a^3+12*B*a*b^2+4*b^3*(A+2*C)+3*a^2*b*(3*A+4*C))*x+1/15*(30*B*a^2*b+15*B*b^3+15*a*b^2*(2*A+3*C)+2*a^3*
(4*A+5*C))*sin(d*x+c)/d+1/40*(6*A*b^3+15*B*a^3+50*B*a*b^2+15*a^2*b*(3*A+4*C))*cos(d*x+c)*sin(d*x+c)/d+1/30*a*(
3*A*b^2+15*B*a*b+2*a^2*(4*A+5*C))*cos(d*x+c)^2*sin(d*x+c)/d+1/20*(3*A*b+5*B*a)*cos(d*x+c)^3*(a+b*sec(d*x+c))^2
*sin(d*x+c)/d+1/5*A*cos(d*x+c)^4*(a+b*sec(d*x+c))^3*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {4179, 4159, 4132, 2717, 4130, 8} \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a \sin (c+d x) \cos ^2(c+d x) \left (2 a^2 (4 A+5 C)+15 a b B+3 A b^2\right )}{30 d}+\frac {\sin (c+d x) \left (2 a^3 (4 A+5 C)+30 a^2 b B+15 a b^2 (2 A+3 C)+15 b^3 B\right )}{15 d}+\frac {\sin (c+d x) \cos (c+d x) \left (15 a^3 B+15 a^2 b (3 A+4 C)+50 a b^2 B+6 A b^3\right )}{40 d}+\frac {1}{8} x \left (3 a^3 B+3 a^2 b (3 A+4 C)+12 a b^2 B+4 b^3 (A+2 C)\right )+\frac {(5 a B+3 A b) \sin (c+d x) \cos ^3(c+d x) (a+b \sec (c+d x))^2}{20 d}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^3}{5 d} \]

[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 8

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

Rule 2717

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

Rule 4130

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 4132

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 4159

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 4179

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 2.78 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.07 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {540 a^2 A b c+240 A b^3 c+180 a^3 B c+720 a b^2 B c+720 a^2 b c C+480 b^3 c C+540 a^2 A b d x+240 A b^3 d x+180 a^3 B d x+720 a b^2 B d x+720 a^2 b C d x+480 b^3 C d x+60 \left (18 a^2 b B+8 b^3 B+6 a b^2 (3 A+4 C)+a^3 (5 A+6 C)\right ) \sin (c+d x)+120 \left (A b^3+a^3 B+3 a b^2 B+3 a^2 b (A+C)\right ) \sin (2 (c+d x))+50 a^3 A \sin (3 (c+d x))+120 a A b^2 \sin (3 (c+d x))+120 a^2 b B \sin (3 (c+d x))+40 a^3 C \sin (3 (c+d x))+45 a^2 A b \sin (4 (c+d x))+15 a^3 B \sin (4 (c+d x))+6 a^3 A \sin (5 (c+d x))}{480 d} \]

[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)

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.75

method	result	size
parallelrisch	\(\frac {120 \left (B \,a^{3}+3 b \left (A +C \right ) a^{2}+3 B a \,b^{2}+A \,b^{3}\right ) \sin \left (2 d x +2 c \right )+10 \left (\left (5 A +4 C \right ) a^{3}+12 B \,a^{2} b +12 a A \,b^{2}\right ) \sin \left (3 d x +3 c \right )+15 \left (3 A \,a^{2} b +B \,a^{3}\right ) \sin \left (4 d x +4 c \right )+6 a^{3} A \sin \left (5 d x +5 c \right )+60 \left (a^{3} \left (5 A +6 C \right )+18 B \,a^{2} b +18 \left (A +\frac {4 C}{3}\right ) a \,b^{2}+8 B \,b^{3}\right ) \sin \left (d x +c \right )+540 \left (\frac {B \,a^{3}}{3}+b \left (A +\frac {4 C}{3}\right ) a^{2}+\frac {4 B a \,b^{2}}{3}+\frac {4 b^{3} \left (A +2 C \right )}{9}\right ) d x}{480 d}\)	\(203\)
derivativedivides	\(\frac {\frac {a^{3} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+3 A \,a^{2} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B \,a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a A \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+B \,a^{2} b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+\frac {a^{3} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+A \,b^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 B a \,b^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b C \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,b^{3} \sin \left (d x +c \right )+3 C a \,b^{2} \sin \left (d x +c \right )+C \,b^{3} \left (d x +c \right )}{d}\)	\(301\)
default	\(\frac {\frac {a^{3} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+3 A \,a^{2} b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B \,a^{3} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a A \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+B \,a^{2} b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )+\frac {a^{3} C \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+A \,b^{3} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 B a \,b^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{2} b C \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B \,b^{3} \sin \left (d x +c \right )+3 C a \,b^{2} \sin \left (d x +c \right )+C \,b^{3} \left (d x +c \right )}{d}\)	\(301\)
risch	\(\frac {9 a^{2} A b x}{8}+\frac {x A \,b^{3}}{2}+\frac {3 a^{3} B x}{8}+x C \,b^{3}+\frac {3 x B a \,b^{2}}{2}+\frac {3 x \,a^{2} b C}{2}+\frac {5 a^{3} A \sin \left (d x +c \right )}{8 d}+\frac {9 \sin \left (d x +c \right ) a A \,b^{2}}{4 d}+\frac {9 \sin \left (d x +c \right ) B \,a^{2} b}{4 d}+\frac {\sin \left (d x +c \right ) B \,b^{3}}{d}+\frac {3 a^{3} C \sin \left (d x +c \right )}{4 d}+\frac {3 \sin \left (d x +c \right ) C a \,b^{2}}{d}+\frac {a^{3} A \sin \left (5 d x +5 c \right )}{80 d}+\frac {3 \sin \left (4 d x +4 c \right ) A \,a^{2} b}{32 d}+\frac {\sin \left (4 d x +4 c \right ) B \,a^{3}}{32 d}+\frac {5 a^{3} A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) a A \,b^{2}}{4 d}+\frac {\sin \left (3 d x +3 c \right ) B \,a^{2} b}{4 d}+\frac {\sin \left (3 d x +3 c \right ) a^{3} C}{12 d}+\frac {3 \sin \left (2 d x +2 c \right ) A \,a^{2} b}{4 d}+\frac {\sin \left (2 d x +2 c \right ) A \,b^{3}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{3}}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) B a \,b^{2}}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) a^{2} b C}{4 d}\)	\(360\)

[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,method=_RETURNVERBOSE)

[Out]

1/480*(120*(B*a^3+3*b*(A+C)*a^2+3*B*a*b^2+A*b^3)*sin(2*d*x+2*c)+10*((5*A+4*C)*a^3+12*B*a^2*b+12*a*A*b^2)*sin(3
*d*x+3*c)+15*(3*A*a^2*b+B*a^3)*sin(4*d*x+4*c)+6*a^3*A*sin(5*d*x+5*c)+60*(a^3*(5*A+6*C)+18*B*a^2*b+18*(A+4/3*C)
*a*b^2+8*B*b^3)*sin(d*x+c)+540*(1/3*B*a^3+b*(A+4/3*C)*a^2+4/3*B*a*b^2+4/9*b^3*(A+2*C))*d*x)/d

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.77 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, B a^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 12 \, B a b^{2} + 4 \, {\left (A + 2 \, C\right )} b^{3}\right )} d x + {\left (24 \, A a^{3} \cos \left (d x + c\right )^{4} + 16 \, {\left (4 \, A + 5 \, C\right )} a^{3} + 240 \, B a^{2} b + 120 \, {\left (2 \, A + 3 \, C\right )} a b^{2} + 120 \, B b^{3} + 30 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{3} + 15 \, B a^{2} b + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, B a^{3} + 3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

[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

Sympy [F(-1)]

Timed out. \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.07 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{3} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{3} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} b + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b^{2} + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{3} + 480 \, {\left (d x + c\right )} C b^{3} + 1440 \, C a b^{2} \sin \left (d x + c\right ) + 480 \, B b^{3} \sin \left (d x + c\right )}{480 \, d} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 926 vs. \(2 (257) = 514\).

Time = 0.35 (sec) , antiderivative size = 926, normalized size of antiderivative = 3.44 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[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

Mupad [B] (verification not implemented)

Time = 17.73 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.33 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {A\,b^3\,x}{2}+\frac {3\,B\,a^3\,x}{8}+C\,b^3\,x+\frac {9\,A\,a^2\,b\,x}{8}+\frac {3\,B\,a\,b^2\,x}{2}+\frac {3\,C\,a^2\,b\,x}{2}+\frac {5\,A\,a^3\,\sin \left (c+d\,x\right )}{8\,d}+\frac {B\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {5\,A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {A\,a^3\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {A\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {C\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {3\,A\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,A\,a^2\,b\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,B\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,C\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {9\,A\,a\,b^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {9\,B\,a^2\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,C\,a\,b^2\,\sin \left (c+d\,x\right )}{d} \]

[In]

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

[Out]

(A*b^3*x)/2 + (3*B*a^3*x)/8 + C*b^3*x + (9*A*a^2*b*x)/8 + (3*B*a*b^2*x)/2 + (3*C*a^2*b*x)/2 + (5*A*a^3*sin(c +
d*x))/(8*d) + (B*b^3*sin(c + d*x))/d + (3*C*a^3*sin(c + d*x))/(4*d) + (5*A*a^3*sin(3*c + 3*d*x))/(48*d) + (A*
a^3*sin(5*c + 5*d*x))/(80*d) + (A*b^3*sin(2*c + 2*d*x))/(4*d) + (B*a^3*sin(2*c + 2*d*x))/(4*d) + (B*a^3*sin(4*
c + 4*d*x))/(32*d) + (C*a^3*sin(3*c + 3*d*x))/(12*d) + (3*A*a^2*b*sin(2*c + 2*d*x))/(4*d) + (A*a*b^2*sin(3*c +
3*d*x))/(4*d) + (3*A*a^2*b*sin(4*c + 4*d*x))/(32*d) + (3*B*a*b^2*sin(2*c + 2*d*x))/(4*d) + (B*a^2*b*sin(3*c +
3*d*x))/(4*d) + (3*C*a^2*b*sin(2*c + 2*d*x))/(4*d) + (9*A*a*b^2*sin(c + d*x))/(4*d) + (9*B*a^2*b*sin(c + d*x)
)/(4*d) + (3*C*a*b^2*sin(c + d*x))/d