Integrand size = 41, antiderivative size = 215 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) x+\frac {\left (10 a b B+a^2 (4 A+5 C)+b^2 (4 A+5 C)\right ) \sin (c+d x)}{5 d}+\frac {\left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (2 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {\left (2 A b^2+10 a b B+a^2 (4 A+5 C)\right ) \sin ^3(c+d x)}{15 d} \]
1/8*(6*A*a*b+3*B*a^2+4*B*b^2+8*C*a*b)*x+1/5*(10*B*a*b+a^2*(4*A+5*C)+b^2*(4 *A+5*C))*sin(d*x+c)/d+1/8*(6*A*a*b+3*B*a^2+4*B*b^2+8*C*a*b)*cos(d*x+c)*sin (d*x+c)/d+1/20*a*(2*A*b+5*B*a)*cos(d*x+c)^3*sin(d*x+c)/d+1/5*A*cos(d*x+c)^ 4*(a+b*sec(d*x+c))^2*sin(d*x+c)/d-1/15*(2*A*b^2+10*B*a*b+a^2*(4*A+5*C))*si n(d*x+c)^3/d
Time = 1.81 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.79 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {60 \left (6 a A b+3 a^2 B+4 b^2 B+8 a b C\right ) (c+d x)+60 \left (12 a b B+2 b^2 (3 A+4 C)+a^2 (5 A+6 C)\right ) \sin (c+d x)+120 \left (a^2 B+b^2 B+2 a b (A+C)\right ) \sin (2 (c+d x))+10 \left (4 A b^2+8 a b B+a^2 (5 A+4 C)\right ) \sin (3 (c+d x))+15 a (2 A b+a B) \sin (4 (c+d x))+6 a^2 A \sin (5 (c+d x))}{480 d} \]
(60*(6*a*A*b + 3*a^2*B + 4*b^2*B + 8*a*b*C)*(c + d*x) + 60*(12*a*b*B + 2*b ^2*(3*A + 4*C) + a^2*(5*A + 6*C))*Sin[c + d*x] + 120*(a^2*B + b^2*B + 2*a* b*(A + C))*Sin[2*(c + d*x)] + 10*(4*A*b^2 + 8*a*b*B + a^2*(5*A + 4*C))*Sin [3*(c + d*x)] + 15*a*(2*A*b + a*B)*Sin[4*(c + d*x)] + 6*a^2*A*Sin[5*(c + d *x)])/(480*d)
Time = 1.24 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.94, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.341, Rules used = {3042, 4582, 3042, 4562, 25, 3042, 4535, 3042, 3115, 24, 4532, 3042, 3492, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 4582 |
| \(\displaystyle \frac {1}{5} \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (b (2 A+5 C) \sec ^2(c+d x)+(4 a A+5 b B+5 a C) \sec (c+d x)+2 A b+5 a B\right )dx+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b (2 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+(4 a A+5 b B+5 a C) \csc \left (c+d x+\frac {\pi }{2}\right )+2 A b+5 a B\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4562 |
| \(\displaystyle \frac {1}{5} \left (\frac {a (5 a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac {1}{4} \int -\cos ^3(c+d x) \left (4 b^2 (2 A+5 C) \sec ^2(c+d x)+5 \left (3 B a^2+6 A b a+8 b C a+4 b^2 B\right ) \sec (c+d x)+4 \left ((4 A+5 C) a^2+10 b B a+2 A b^2\right )\right )dx\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \cos ^3(c+d x) \left (4 b^2 (2 A+5 C) \sec ^2(c+d x)+5 \left (3 B a^2+6 A b a+8 b C a+4 b^2 B\right ) \sec (c+d x)+4 \left ((4 A+5 C) a^2+10 b B a+2 A b^2\right )\right )dx+\frac {a (5 a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int \frac {4 b^2 (2 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 \left (3 B a^2+6 A b a+8 b C a+4 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+4 \left ((4 A+5 C) a^2+10 b B a+2 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {a (5 a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \int \cos ^2(c+d x)dx+\int \cos ^3(c+d x) \left (4 b^2 (2 A+5 C) \sec ^2(c+d x)+4 \left ((4 A+5 C) a^2+10 b B a+2 A b^2\right )\right )dx\right )+\frac {a (5 a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx+\int \frac {4 b^2 (2 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 \left ((4 A+5 C) a^2+10 b B a+2 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx\right )+\frac {a (5 a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\int \frac {4 b^2 (2 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 \left ((4 A+5 C) a^2+10 b B a+2 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+5 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )\right )+\frac {a (5 a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\int \frac {4 b^2 (2 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+4 \left ((4 A+5 C) a^2+10 b B a+2 A b^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+5 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a (5 a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 4532 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\int \cos (c+d x) \left (4 (2 A+5 C) b^2+4 \left ((4 A+5 C) a^2+10 b B a+2 A b^2\right ) \cos ^2(c+d x)\right )dx+5 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a (5 a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (\int \sin \left (c+d x+\frac {\pi }{2}\right ) \left (4 (2 A+5 C) b^2+4 \left ((4 A+5 C) a^2+10 b B a+2 A b^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+5 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a (5 a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 3492 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {\int \left (4 \left ((4 A+5 C) a^2+10 b B a+b^2 (4 A+5 C)\right )-4 \left ((4 A+5 C) a^2+10 b B a+2 A b^2\right ) \sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a (5 a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (\frac {1}{4} \left (5 \left (3 a^2 B+6 a A b+8 a b C+4 b^2 B\right ) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )-\frac {\frac {4}{3} \sin ^3(c+d x) \left (a^2 (4 A+5 C)+10 a b B+2 A b^2\right )-4 \sin (c+d x) \left (a^2 (4 A+5 C)+10 a b B+b^2 (4 A+5 C)\right )}{d}\right )+\frac {a (5 a B+2 A b) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d}\) |
(A*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) + ((a*(2*A*b + 5*a*B)*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) + (5*(6*a*A*b + 3*a^2*B + 4*b^ 2*B + 8*a*b*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)) - (-4*(10*a*b*B + a^2*(4*A + 5*C) + b^2*(4*A + 5*C))*Sin[c + d*x] + (4*(2*A*b^2 + 10*a*b*B + a^2*(4*A + 5*C))*Sin[c + d*x]^3)/3)/d)/4)/5
3.9.77.3.1 Defintions of rubi rules used
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] + Simp[b^2*((n - 1)/n) Int[(b*Sin [c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 2*n]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[-f^(-1) Subst[Int[(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]
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]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[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]
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] + Si mp[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]
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] - Simp[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)))*Cs c[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]
Time = 0.85 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.78
| method | result | size |
| parallelrisch | \(\frac {120 \left (B \,a^{2}+2 a b \left (A +C \right )+B \,b^{2}\right ) \sin \left (2 d x +2 c \right )+10 \left (\left (5 A +4 C \right ) a^{2}+8 B a b +4 A \,b^{2}\right ) \sin \left (3 d x +3 c \right )+15 \left (2 a A b +B \,a^{2}\right ) \sin \left (4 d x +4 c \right )+6 a^{2} A \sin \left (5 d x +5 c \right )+60 \left (a^{2} \left (5 A +6 C \right )+12 B a b +6 b^{2} \left (A +\frac {4 C}{3}\right )\right ) \sin \left (d x +c \right )+360 d x \left (\frac {B \,a^{2}}{2}+a \left (A +\frac {4 C}{3}\right ) b +\frac {2 B \,b^{2}}{3}\right )}{480 d}\) | \(167\) |
| derivativedivides | \(\frac {\frac {a^{2} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+2 a A b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {2 B a b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {C \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,b^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 C a b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \sin \left (d x +c \right ) b^{2}}{d}\) | \(244\) |
| default | \(\frac {\frac {a^{2} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+2 a A b \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+B \,a^{2} \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A \,b^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {2 B a b \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {C \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+B \,b^{2} \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 C a b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \sin \left (d x +c \right ) b^{2}}{d}\) | \(244\) |
| risch | \(\frac {3 a A b x}{4}+\frac {3 a^{2} B x}{8}+\frac {x B \,b^{2}}{2}+x C a b +\frac {5 \sin \left (d x +c \right ) a^{2} A}{8 d}+\frac {3 \sin \left (d x +c \right ) A \,b^{2}}{4 d}+\frac {3 \sin \left (d x +c \right ) B a b}{2 d}+\frac {3 \sin \left (d x +c \right ) C \,a^{2}}{4 d}+\frac {\sin \left (d x +c \right ) C \,b^{2}}{d}+\frac {a^{2} A \sin \left (5 d x +5 c \right )}{80 d}+\frac {\sin \left (4 d x +4 c \right ) a A b}{16 d}+\frac {\sin \left (4 d x +4 c \right ) B \,a^{2}}{32 d}+\frac {5 a^{2} A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) A \,b^{2}}{12 d}+\frac {\sin \left (3 d x +3 c \right ) B a b}{6 d}+\frac {\sin \left (3 d x +3 c \right ) C \,a^{2}}{12 d}+\frac {\sin \left (2 d x +2 c \right ) a A b}{2 d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{2}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) B \,b^{2}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C a b}{2 d}\) | \(294\) |
| norman | \(\frac {\left (-\frac {3}{4} a A b -\frac {3}{8} B \,a^{2}-\frac {1}{2} B \,b^{2}-C a b \right ) x +\left (-\frac {9}{2} a A b -\frac {9}{4} B \,a^{2}-3 B \,b^{2}-6 C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (-\frac {3}{2} a A b -\frac {3}{4} B \,a^{2}-B \,b^{2}-2 C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (-\frac {3}{2} a A b -\frac {3}{4} B \,a^{2}-B \,b^{2}-2 C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (\frac {3}{2} a A b +\frac {3}{4} B \,a^{2}+B \,b^{2}+2 C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (\frac {3}{2} a A b +\frac {3}{4} B \,a^{2}+B \,b^{2}+2 C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (\frac {3}{4} a A b +\frac {3}{8} B \,a^{2}+\frac {1}{2} B \,b^{2}+C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+\left (\frac {9}{2} a A b +\frac {9}{4} B \,a^{2}+3 B \,b^{2}+6 C a b \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {\left (8 a^{2} A -10 a A b +8 A \,b^{2}-5 B \,a^{2}+16 B a b -4 B \,b^{2}+8 C \,a^{2}-8 C a b +8 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{4 d}-\frac {\left (8 a^{2} A +10 a A b +8 A \,b^{2}+5 B \,a^{2}+16 B a b +4 B \,b^{2}+8 C \,a^{2}+8 C a b +8 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (40 a^{2} A -78 a A b +8 A \,b^{2}-39 B \,a^{2}+16 B a b -12 B \,b^{2}+8 C \,a^{2}-24 C a b -24 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{12 d}+\frac {\left (40 a^{2} A +78 a A b +8 A \,b^{2}+39 B \,a^{2}+16 B a b +12 B \,b^{2}+8 C \,a^{2}+24 C a b -24 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}+\frac {\left (344 a^{2} A -270 a A b -200 A \,b^{2}-135 B \,a^{2}-400 B a b +180 B \,b^{2}-200 C \,a^{2}+360 C a b -360 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{60 d}-\frac {\left (344 a^{2} A +270 a A b -200 A \,b^{2}+135 B \,a^{2}-400 B a b -180 B \,b^{2}-200 C \,a^{2}-360 C a b -360 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 d}-\frac {\left (872 a^{2} A -30 a A b +40 A \,b^{2}-15 B \,a^{2}+80 B a b +180 B \,b^{2}+40 C \,a^{2}+360 C a b +360 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 d}+\frac {\left (872 a^{2} A +30 a A b +40 A \,b^{2}+15 B \,a^{2}+80 B a b -180 B \,b^{2}+40 C \,a^{2}-360 C a b +360 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{60 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}\) | \(849\) |
1/480*(120*(B*a^2+2*a*b*(A+C)+B*b^2)*sin(2*d*x+2*c)+10*((5*A+4*C)*a^2+8*B* a*b+4*A*b^2)*sin(3*d*x+3*c)+15*(2*A*a*b+B*a^2)*sin(4*d*x+4*c)+6*a^2*A*sin( 5*d*x+5*c)+60*(a^2*(5*A+6*C)+12*B*a*b+6*b^2*(A+4/3*C))*sin(d*x+c)+360*d*x* (1/2*B*a^2+a*(A+4/3*C)*b+2/3*B*b^2))/d
Time = 0.28 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.80 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, B a^{2} + 2 \, {\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\right )} d x + {\left (24 \, A a^{2} \cos \left (d x + c\right )^{4} + 30 \, {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )^{3} + 16 \, {\left (4 \, A + 5 \, C\right )} a^{2} + 160 \, B a b + 40 \, {\left (2 \, A + 3 \, C\right )} b^{2} + 8 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{2} + 10 \, B a b + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, B a^{2} + 2 \, {\left (3 \, A + 4 \, C\right )} a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]
integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="fricas")
1/120*(15*(3*B*a^2 + 2*(3*A + 4*C)*a*b + 4*B*b^2)*d*x + (24*A*a^2*cos(d*x + c)^4 + 30*(B*a^2 + 2*A*a*b)*cos(d*x + c)^3 + 16*(4*A + 5*C)*a^2 + 160*B* a*b + 40*(2*A + 3*C)*b^2 + 8*((4*A + 5*C)*a^2 + 10*B*a*b + 5*A*b^2)*cos(d* x + c)^2 + 15*(3*B*a^2 + 2*(3*A + 4*C)*a*b + 4*B*b^2)*cos(d*x + c))*sin(d* x + c))/d
Timed out. \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Time = 0.25 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.08 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 30 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a b - 320 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a b + 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{2} + 480 \, C b^{2} \sin \left (d x + c\right )}{480 \, d} \]
integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="maxima")
1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^2 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*B*a^2 - 160*(s in(d*x + c)^3 - 3*sin(d*x + c))*C*a^2 + 30*(12*d*x + 12*c + sin(4*d*x + 4* c) + 8*sin(2*d*x + 2*c))*A*a*b - 320*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a *b + 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a*b - 160*(sin(d*x + c)^3 - 3* sin(d*x + c))*A*b^2 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*b^2 + 480*C*b ^2*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 720 vs. \(2 (203) = 406\).
Time = 0.33 (sec) , antiderivative size = 720, normalized size of antiderivative = 3.35 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]
integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2), x, algorithm="giac")
1/120*(15*(3*B*a^2 + 6*A*a*b + 8*C*a*b + 4*B*b^2)*(d*x + c) + 2*(120*A*a^2 *tan(1/2*d*x + 1/2*c)^9 - 75*B*a^2*tan(1/2*d*x + 1/2*c)^9 + 120*C*a^2*tan( 1/2*d*x + 1/2*c)^9 - 150*A*a*b*tan(1/2*d*x + 1/2*c)^9 + 240*B*a*b*tan(1/2* d*x + 1/2*c)^9 - 120*C*a*b*tan(1/2*d*x + 1/2*c)^9 + 120*A*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*B*b^2*tan(1/2*d*x + 1/2*c)^9 + 120*C*b^2*tan(1/2*d*x + 1/2 *c)^9 + 160*A*a^2*tan(1/2*d*x + 1/2*c)^7 - 30*B*a^2*tan(1/2*d*x + 1/2*c)^7 + 320*C*a^2*tan(1/2*d*x + 1/2*c)^7 - 60*A*a*b*tan(1/2*d*x + 1/2*c)^7 + 64 0*B*a*b*tan(1/2*d*x + 1/2*c)^7 - 240*C*a*b*tan(1/2*d*x + 1/2*c)^7 + 320*A* b^2*tan(1/2*d*x + 1/2*c)^7 - 120*B*b^2*tan(1/2*d*x + 1/2*c)^7 + 480*C*b^2* tan(1/2*d*x + 1/2*c)^7 + 464*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 400*C*a^2*tan( 1/2*d*x + 1/2*c)^5 + 800*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 400*A*b^2*tan(1/2* d*x + 1/2*c)^5 + 720*C*b^2*tan(1/2*d*x + 1/2*c)^5 + 160*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 30*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 320*C*a^2*tan(1/2*d*x + 1/2 *c)^3 + 60*A*a*b*tan(1/2*d*x + 1/2*c)^3 + 640*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 240*C*a*b*tan(1/2*d*x + 1/2*c)^3 + 320*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 1 20*B*b^2*tan(1/2*d*x + 1/2*c)^3 + 480*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 120*A *a^2*tan(1/2*d*x + 1/2*c) + 75*B*a^2*tan(1/2*d*x + 1/2*c) + 120*C*a^2*tan( 1/2*d*x + 1/2*c) + 150*A*a*b*tan(1/2*d*x + 1/2*c) + 240*B*a*b*tan(1/2*d*x + 1/2*c) + 120*C*a*b*tan(1/2*d*x + 1/2*c) + 120*A*b^2*tan(1/2*d*x + 1/2*c) + 60*B*b^2*tan(1/2*d*x + 1/2*c) + 120*C*b^2*tan(1/2*d*x + 1/2*c))/(tan...
Time = 17.08 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.19 \[ \int \cos ^5(c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\frac {25\,A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{2}+\frac {3\,A\,a^2\,\sin \left (5\,c+5\,d\,x\right )}{2}+30\,B\,a^2\,\sin \left (2\,c+2\,d\,x\right )+10\,A\,b^2\,\sin \left (3\,c+3\,d\,x\right )+\frac {15\,B\,a^2\,\sin \left (4\,c+4\,d\,x\right )}{4}+30\,B\,b^2\,\sin \left (2\,c+2\,d\,x\right )+10\,C\,a^2\,\sin \left (3\,c+3\,d\,x\right )+75\,A\,a^2\,\sin \left (c+d\,x\right )+90\,A\,b^2\,\sin \left (c+d\,x\right )+90\,C\,a^2\,\sin \left (c+d\,x\right )+120\,C\,b^2\,\sin \left (c+d\,x\right )+60\,A\,a\,b\,\sin \left (2\,c+2\,d\,x\right )+\frac {15\,A\,a\,b\,\sin \left (4\,c+4\,d\,x\right )}{2}+20\,B\,a\,b\,\sin \left (3\,c+3\,d\,x\right )+60\,C\,a\,b\,\sin \left (2\,c+2\,d\,x\right )+45\,B\,a^2\,d\,x+60\,B\,b^2\,d\,x+180\,B\,a\,b\,\sin \left (c+d\,x\right )+90\,A\,a\,b\,d\,x+120\,C\,a\,b\,d\,x}{120\,d} \]
((25*A*a^2*sin(3*c + 3*d*x))/2 + (3*A*a^2*sin(5*c + 5*d*x))/2 + 30*B*a^2*s in(2*c + 2*d*x) + 10*A*b^2*sin(3*c + 3*d*x) + (15*B*a^2*sin(4*c + 4*d*x))/ 4 + 30*B*b^2*sin(2*c + 2*d*x) + 10*C*a^2*sin(3*c + 3*d*x) + 75*A*a^2*sin(c + d*x) + 90*A*b^2*sin(c + d*x) + 90*C*a^2*sin(c + d*x) + 120*C*b^2*sin(c + d*x) + 60*A*a*b*sin(2*c + 2*d*x) + (15*A*a*b*sin(4*c + 4*d*x))/2 + 20*B* a*b*sin(3*c + 3*d*x) + 60*C*a*b*sin(2*c + 2*d*x) + 45*B*a^2*d*x + 60*B*b^2 *d*x + 180*B*a*b*sin(c + d*x) + 90*A*a*b*d*x + 120*C*a*b*d*x)/(120*d)