Integrand size = 39, antiderivative size = 141 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{8} a (3 (A+B)+4 C) x+\frac {a (4 A+5 (B+C)) \sin (c+d x)}{5 d}+\frac {a (3 (A+B)+4 C) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a (A+B) \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac {a A \cos ^4(c+d x) \sin (c+d x)}{5 d}-\frac {a (4 A+5 (B+C)) \sin ^3(c+d x)}{15 d} \]
1/8*a*(3*A+3*B+4*C)*x+1/5*a*(4*A+5*B+5*C)*sin(d*x+c)/d+1/8*a*(3*A+3*B+4*C) *cos(d*x+c)*sin(d*x+c)/d+1/4*a*(A+B)*cos(d*x+c)^3*sin(d*x+c)/d+1/5*a*A*cos (d*x+c)^4*sin(d*x+c)/d-1/15*a*(4*A+5*B+5*C)*sin(d*x+c)^3/d
Time = 0.31 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.67 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a \left (480 (A+B+C) \sin (c+d x)-160 (2 A+B+C) \sin ^3(c+d x)+96 A \sin ^5(c+d x)+15 (4 (3 A+3 B+4 C) (c+d x)+8 (A+B+C) \sin (2 (c+d x))+(A+B) \sin (4 (c+d x)))\right )}{480 d} \]
(a*(480*(A + B + C)*Sin[c + d*x] - 160*(2*A + B + C)*Sin[c + d*x]^3 + 96*A *Sin[c + d*x]^5 + 15*(4*(3*A + 3*B + 4*C)*(c + d*x) + 8*(A + B + C)*Sin[2* (c + d*x)] + (A + B)*Sin[4*(c + d*x)])))/(480*d)
Time = 0.72 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.91, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 4562, 25, 3042, 4535, 3042, 3113, 2009, 4533, 3042, 3115, 24}
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.
| \(\displaystyle \int \cos ^5(c+d x) (a \sec (c+d x)+a) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 4562 |
| \(\displaystyle \frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}-\frac {1}{5} \int -\cos ^4(c+d x) \left (5 a C \sec ^2(c+d x)+a (4 A+5 (B+C)) \sec (c+d x)+5 a (A+B)\right )dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{5} \int \cos ^4(c+d x) \left (5 a C \sec ^2(c+d x)+a (4 A+5 (B+C)) \sec (c+d x)+5 a (A+B)\right )dx+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {5 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+a (4 A+5 (B+C)) \csc \left (c+d x+\frac {\pi }{2}\right )+5 a (A+B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4535 |
| \(\displaystyle \frac {1}{5} \left (a (4 A+5 (B+C)) \int \cos ^3(c+d x)dx+\int \cos ^4(c+d x) \left (5 a C \sec ^2(c+d x)+5 a (A+B)\right )dx\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (a (4 A+5 (B+C)) \int \sin \left (c+d x+\frac {\pi }{2}\right )^3dx+\int \frac {5 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a (A+B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3113 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a (A+B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx-\frac {a (4 A+5 (B+C)) \int \left (1-\sin ^2(c+d x)\right )d(-\sin (c+d x))}{d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} \left (\int \frac {5 a C \csc \left (c+d x+\frac {\pi }{2}\right )^2+5 a (A+B)}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx-\frac {a (4 A+5 (B+C)) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 4533 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} a (3 (A+B)+4 C) \int \cos ^2(c+d x)dx-\frac {a (4 A+5 (B+C)) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5 a (A+B) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} a (3 (A+B)+4 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx-\frac {a (4 A+5 (B+C)) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5 a (A+B) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {1}{5} \left (\frac {5}{4} a (3 (A+B)+4 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )-\frac {a (4 A+5 (B+C)) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5 a (A+B) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{5} \left (-\frac {a (4 A+5 (B+C)) \left (\frac {1}{3} \sin ^3(c+d x)-\sin (c+d x)\right )}{d}+\frac {5}{4} a (3 (A+B)+4 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )+\frac {5 a (A+B) \sin (c+d x) \cos ^3(c+d x)}{4 d}\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x)}{5 d}\) |
(a*A*Cos[c + d*x]^4*Sin[c + d*x])/(5*d) + ((5*a*(A + B)*Cos[c + d*x]^3*Sin [c + d*x])/(4*d) + (5*a*(3*(A + B) + 4*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x ])/(2*d)))/4 - (a*(4*A + 5*(B + C))*(-Sin[c + d*x] + Sin[c + d*x]^3/3))/d) /5
3.5.16.3.1 Defintions of rubi rules used
Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp and[(1 - x^2)^((n - 1)/2), x], x], x, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]
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[(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] + Simp[(C*m + A*(m + 1))/(b^2*m) Int[(b*Csc[e + f*x])^(m + 2), x], x] /; Fr eeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]
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]
Time = 0.48 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.66
| method | result | size |
| parallelrisch | \(\frac {5 \left (\frac {12 \left (A +B +C \right ) \sin \left (2 d x +2 c \right )}{5}+\left (A +\frac {4 B}{5}+\frac {4 C}{5}\right ) \sin \left (3 d x +3 c \right )+\frac {3 \left (A +B \right ) \sin \left (4 d x +4 c \right )}{10}+\frac {3 A \sin \left (5 d x +5 c \right )}{25}+6 \left (A +\frac {6 B}{5}+\frac {6 C}{5}\right ) \sin \left (d x +c \right )+\frac {18 x \left (A +B +\frac {4 C}{3}\right ) d}{5}\right ) a}{48 d}\) | \(93\) |
| derivativedivides | \(\frac {\frac {a A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+a A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a B \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a B \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {C a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(173\) |
| default | \(\frac {\frac {a A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+a A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a B \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {a B \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+\frac {C a \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C a \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\) | \(173\) |
| risch | \(\frac {3 a A x}{8}+\frac {3 a B x}{8}+\frac {a x C}{2}+\frac {5 a A \sin \left (d x +c \right )}{8 d}+\frac {3 a B \sin \left (d x +c \right )}{4 d}+\frac {3 \sin \left (d x +c \right ) C a}{4 d}+\frac {a A \sin \left (5 d x +5 c \right )}{80 d}+\frac {a A \sin \left (4 d x +4 c \right )}{32 d}+\frac {\sin \left (4 d x +4 c \right ) a B}{32 d}+\frac {5 a A \sin \left (3 d x +3 c \right )}{48 d}+\frac {\sin \left (3 d x +3 c \right ) a B}{12 d}+\frac {\sin \left (3 d x +3 c \right ) C a}{12 d}+\frac {a A \sin \left (2 d x +2 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a B}{4 d}+\frac {\sin \left (2 d x +2 c \right ) C a}{4 d}\) | \(200\) |
| norman | \(\frac {\frac {a \left (3 A +3 B +4 C \right ) x}{8}+\frac {2 a \left (A +5 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}+\frac {a \left (3 A +3 B +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{4 d}+\frac {3 a \left (3 A +3 B +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8}+\frac {a \left (3 A +3 B +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8}-\frac {5 a \left (3 A +3 B +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{8}-\frac {5 a \left (3 A +3 B +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{8}+\frac {a \left (3 A +3 B +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{8}+\frac {3 a \left (3 A +3 B +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{8}+\frac {a \left (3 A +3 B +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{8}-\frac {2 a \left (5 A +B -2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {a \left (13 A +13 B +12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {8 a \left (19 A +5 B +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{15 d}+\frac {a \left (83 A -45 B +20 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{20 d}+\frac {a \left (93 A -35 B -100 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2}}\) | \(405\) |
5/48*(12/5*(A+B+C)*sin(2*d*x+2*c)+(A+4/5*B+4/5*C)*sin(3*d*x+3*c)+3/10*(A+B )*sin(4*d*x+4*c)+3/25*A*sin(5*d*x+5*c)+6*(A+6/5*B+6/5*C)*sin(d*x+c)+18/5*x *(A+B+4/3*C)*d)*a/d
Time = 0.25 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.77 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, A + 3 \, B + 4 \, C\right )} a d x + {\left (24 \, A a \cos \left (d x + c\right )^{4} + 30 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, A + 5 \, B + 5 \, C\right )} a \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, A + 3 \, B + 4 \, C\right )} a \cos \left (d x + c\right ) + 16 \, {\left (4 \, A + 5 \, B + 5 \, C\right )} a\right )} \sin \left (d x + c\right )}{120 \, d} \]
integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
1/120*(15*(3*A + 3*B + 4*C)*a*d*x + (24*A*a*cos(d*x + c)^4 + 30*(A + B)*a* cos(d*x + c)^3 + 8*(4*A + 5*B + 5*C)*a*cos(d*x + c)^2 + 15*(3*A + 3*B + 4* C)*a*cos(d*x + c) + 16*(4*A + 5*B + 5*C)*a)*sin(d*x + c))/d
Timed out. \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.18 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 160 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a}{480 \, d} \]
integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))*(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 + 1 5*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a - 160*(sin(d *x + c)^3 - 3*sin(d*x + c))*B*a + 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8 *sin(2*d*x + 2*c))*B*a - 160*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a + 120*( 2*d*x + 2*c + sin(2*d*x + 2*c))*C*a)/d
Time = 0.34 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.87 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, A a + 3 \, B a + 4 \, C a\right )} {\left (d x + c\right )} + \frac {2 \, {\left (45 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 130 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 290 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 200 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 190 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 350 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 440 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 195 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 195 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 180 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \]
1/120*(15*(3*A*a + 3*B*a + 4*C*a)*(d*x + c) + 2*(45*A*a*tan(1/2*d*x + 1/2* c)^9 + 45*B*a*tan(1/2*d*x + 1/2*c)^9 + 60*C*a*tan(1/2*d*x + 1/2*c)^9 + 130 *A*a*tan(1/2*d*x + 1/2*c)^7 + 290*B*a*tan(1/2*d*x + 1/2*c)^7 + 200*C*a*tan (1/2*d*x + 1/2*c)^7 + 464*A*a*tan(1/2*d*x + 1/2*c)^5 + 400*B*a*tan(1/2*d*x + 1/2*c)^5 + 400*C*a*tan(1/2*d*x + 1/2*c)^5 + 190*A*a*tan(1/2*d*x + 1/2*c )^3 + 350*B*a*tan(1/2*d*x + 1/2*c)^3 + 440*C*a*tan(1/2*d*x + 1/2*c)^3 + 19 5*A*a*tan(1/2*d*x + 1/2*c) + 195*B*a*tan(1/2*d*x + 1/2*c) + 180*C*a*tan(1/ 2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d
Time = 19.39 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.76 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (\frac {3\,A\,a}{4}+\frac {3\,B\,a}{4}+C\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {13\,A\,a}{6}+\frac {29\,B\,a}{6}+\frac {10\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a}{15}+\frac {20\,B\,a}{3}+\frac {20\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {19\,A\,a}{6}+\frac {35\,B\,a}{6}+\frac {22\,C\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a}{4}+\frac {13\,B\,a}{4}+3\,C\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atan}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,A+3\,B+4\,C\right )}{4\,\left (\frac {3\,A\,a}{4}+\frac {3\,B\,a}{4}+C\,a\right )}\right )\,\left (3\,A+3\,B+4\,C\right )}{4\,d} \]
(tan(c/2 + (d*x)/2)*((13*A*a)/4 + (13*B*a)/4 + 3*C*a) + tan(c/2 + (d*x)/2) ^9*((3*A*a)/4 + (3*B*a)/4 + C*a) + tan(c/2 + (d*x)/2)^7*((13*A*a)/6 + (29* B*a)/6 + (10*C*a)/3) + tan(c/2 + (d*x)/2)^3*((19*A*a)/6 + (35*B*a)/6 + (22 *C*a)/3) + tan(c/2 + (d*x)/2)^5*((116*A*a)/15 + (20*B*a)/3 + (20*C*a)/3))/ (d*(5*tan(c/2 + (d*x)/2)^2 + 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/ 2)^6 + 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) + (a*atan((a*t an(c/2 + (d*x)/2)*(3*A + 3*B + 4*C))/(4*((3*A*a)/4 + (3*B*a)/4 + C*a)))*(3 *A + 3*B + 4*C))/(4*d)