Integrand size = 33, antiderivative size = 169 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {1}{4} a^2 (3 A+4 C) x+\frac {a^2 (18 A+25 C) \sin (c+d x)}{15 d}+\frac {a^2 (3 A+4 C) \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a^2 (9 A+10 C) \cos ^2(c+d x) \sin (c+d x)}{30 d}+\frac {A \cos ^4(c+d x) (a+a \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {A \cos ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \sin (c+d x)}{10 d} \] Output:
1/4*a^2*(3*A+4*C)*x+1/15*a^2*(18*A+25*C)*sin(d*x+c)/d+1/4*a^2*(3*A+4*C)*co s(d*x+c)*sin(d*x+c)/d+1/30*a^2*(9*A+10*C)*cos(d*x+c)^2*sin(d*x+c)/d+1/5*A* cos(d*x+c)^4*(a+a*sec(d*x+c))^2*sin(d*x+c)/d+1/10*A*cos(d*x+c)^3*(a^2+a^2* sec(d*x+c))*sin(d*x+c)/d
Time = 0.29 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.57 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (120 A c+180 A d x+240 C d x+30 (11 A+14 C) \sin (c+d x)+120 (A+C) \sin (2 (c+d x))+45 A \sin (3 (c+d x))+20 C \sin (3 (c+d x))+15 A \sin (4 (c+d x))+3 A \sin (5 (c+d x)))}{240 d} \] Input:
Integrate[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]
Output:
(a^2*(120*A*c + 180*A*d*x + 240*C*d*x + 30*(11*A + 14*C)*Sin[c + d*x] + 12 0*(A + C)*Sin[2*(c + d*x)] + 45*A*Sin[3*(c + d*x)] + 20*C*Sin[3*(c + d*x)] + 15*A*Sin[4*(c + d*x)] + 3*A*Sin[5*(c + d*x)]))/(240*d)
Time = 1.07 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 4575, 3042, 4505, 27, 3042, 4484, 25, 3042, 4274, 3042, 3115, 24, 3117}
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.
| \(\displaystyle \int \cos ^5(c+d x) (a \sec (c+d x)+a)^2 \left (A+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 4575 |
| \(\displaystyle \frac {\int \cos ^4(c+d x) (\sec (c+d x) a+a)^2 (2 a A+a (2 A+5 C) \sec (c+d x))dx}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 a A+a (2 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^4}dx}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4505 |
| \(\displaystyle \frac {\frac {1}{4} \int 2 \cos ^3(c+d x) (\sec (c+d x) a+a) \left ((9 A+10 C) a^2+2 (3 A+5 C) \sec (c+d x) a^2\right )dx+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {1}{2} \int \cos ^3(c+d x) (\sec (c+d x) a+a) \left ((9 A+10 C) a^2+2 (3 A+5 C) \sec (c+d x) a^2\right )dx+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left ((9 A+10 C) a^2+2 (3 A+5 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4484 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {a^3 (9 A+10 C) \sin (c+d x) \cos ^2(c+d x)}{3 d}-\frac {1}{3} \int -\cos ^2(c+d x) \left (15 (3 A+4 C) a^3+2 (18 A+25 C) \sec (c+d x) a^3\right )dx\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \int \cos ^2(c+d x) \left (15 (3 A+4 C) a^3+2 (18 A+25 C) \sec (c+d x) a^3\right )dx+\frac {a^3 (9 A+10 C) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \int \frac {15 (3 A+4 C) a^3+2 (18 A+25 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a^3}{\csc \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^3 (9 A+10 C) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \left (15 a^3 (3 A+4 C) \int \cos ^2(c+d x)dx+2 a^3 (18 A+25 C) \int \cos (c+d x)dx\right )+\frac {a^3 (9 A+10 C) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \left (2 a^3 (18 A+25 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+15 a^3 (3 A+4 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {a^3 (9 A+10 C) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3115 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \left (2 a^3 (18 A+25 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+15 a^3 (3 A+4 C) \left (\frac {\int 1dx}{2}+\frac {\sin (c+d x) \cos (c+d x)}{2 d}\right )\right )+\frac {a^3 (9 A+10 C) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {1}{3} \left (2 a^3 (18 A+25 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )dx+15 a^3 (3 A+4 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )+\frac {a^3 (9 A+10 C) \sin (c+d x) \cos ^2(c+d x)}{3 d}\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d}\) |
| \(\Big \downarrow \) 3117 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {a^3 (9 A+10 C) \sin (c+d x) \cos ^2(c+d x)}{3 d}+\frac {1}{3} \left (\frac {2 a^3 (18 A+25 C) \sin (c+d x)}{d}+15 a^3 (3 A+4 C) \left (\frac {\sin (c+d x) \cos (c+d x)}{2 d}+\frac {x}{2}\right )\right )\right )+\frac {A \sin (c+d x) \cos ^3(c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}}{5 a}+\frac {A \sin (c+d x) \cos ^4(c+d x) (a \sec (c+d x)+a)^2}{5 d}\) |
Input:
Int[Cos[c + d*x]^5*(a + a*Sec[c + d*x])^2*(A + C*Sec[c + d*x]^2),x]
Output:
(A*Cos[c + d*x]^4*(a + a*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) + ((A*Cos[c + d*x]^3*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(2*d) + ((a^3*(9*A + 10*C)* Cos[c + d*x]^2*Sin[c + d*x])/(3*d) + ((2*a^3*(18*A + 25*C)*Sin[c + d*x])/d + 15*a^3*(3*A + 4*C)*(x/2 + (Cos[c + d*x]*Sin[c + d*x])/(2*d)))/3)/2)/(5* a)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[A*a*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*n)), x] + Simp[1/(d*n) Int[(d*Csc[e + f*x])^( n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot [e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x] - Sim p[b/(a*d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Sim p[a*A*(m - n - 1) - b*B*n - (a*B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0 ] && GtQ[m, 1/2] && LtQ[n, -1]
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*Co t[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/( b*d*n) Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b *(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 0])
Time = 0.51 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.53
| method | result | size |
| parallelrisch | \(\frac {\left (\left (8 A +8 C \right ) \sin \left (2 d x +2 c \right )+\left (3 A +\frac {4 C}{3}\right ) \sin \left (3 d x +3 c \right )+A \sin \left (4 d x +4 c \right )+\frac {A \sin \left (5 d x +5 c \right )}{5}+\left (22 A +28 C \right ) \sin \left (d x +c \right )+12 x d \left (A +\frac {4 C}{3}\right )\right ) a^{2}}{16 d}\) | \(90\) |
| risch | \(\frac {3 a^{2} A x}{4}+a^{2} x C +\frac {11 \sin \left (d x +c \right ) a^{2} A}{8 d}+\frac {7 \sin \left (d x +c \right ) C \,a^{2}}{4 d}+\frac {a^{2} A \sin \left (5 d x +5 c \right )}{80 d}+\frac {a^{2} A \sin \left (4 d x +4 c \right )}{16 d}+\frac {3 a^{2} A \sin \left (3 d x +3 c \right )}{16 d}+\frac {\sin \left (3 d x +3 c \right ) C \,a^{2}}{12 d}+\frac {a^{2} A \sin \left (2 d x +2 c \right )}{2 d}+\frac {\sin \left (2 d x +2 c \right ) C \,a^{2}}{2 d}\) | \(153\) |
| derivativedivides | \(\frac {\frac {a^{2} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {C \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+2 a^{2} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 C \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {a^{2} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \,a^{2} \sin \left (d x +c \right )}{d}\) | \(160\) |
| default | \(\frac {\frac {a^{2} A \left (\frac {8}{3}+\cos \left (d x +c \right )^{4}+\frac {4 \cos \left (d x +c \right )^{2}}{3}\right ) \sin \left (d x +c \right )}{5}+\frac {C \,a^{2} \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+2 a^{2} A \left (\frac {\left (\cos \left (d x +c \right )^{3}+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+2 C \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {a^{2} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+C \,a^{2} \sin \left (d x +c \right )}{d}\) | \(160\) |
| norman | \(\frac {-\frac {a^{2} \left (3 A +4 C \right ) x}{4}+\frac {5 a^{2} \left (3 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{6 d}+\frac {a^{2} \left (3 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{2 d}-\frac {a^{2} \left (3 A +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2}+\frac {a^{2} \left (3 A +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{2}+\frac {3 a^{2} \left (3 A +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{2}-\frac {3 a^{2} \left (3 A +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2}-\frac {a^{2} \left (3 A +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}}{2}+\frac {a^{2} \left (3 A +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}}{2}+\frac {a^{2} \left (3 A +4 C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}}{4}-\frac {23 a^{2} \left (9 A -20 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{30 d}-\frac {a^{2} \left (13 A +12 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}+\frac {a^{2} \left (63 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}-\frac {a^{2} \left (63 A +100 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{30 d}-\frac {a^{2} \left (441 A +380 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{30 d}+\frac {a^{2} \left (471 A +20 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{30 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}\) | \(420\) |
Input:
int(cos(d*x+c)^5*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x,method=_RETURNVER BOSE)
Output:
1/16*((8*A+8*C)*sin(2*d*x+2*c)+(3*A+4/3*C)*sin(3*d*x+3*c)+A*sin(4*d*x+4*c) +1/5*A*sin(5*d*x+5*c)+(22*A+28*C)*sin(d*x+c)+12*x*d*(A+4/3*C))*a^2/d
Time = 0.08 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.63 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, A + 4 \, C\right )} a^{2} d x + {\left (12 \, A a^{2} \cos \left (d x + c\right )^{4} + 30 \, A a^{2} \cos \left (d x + c\right )^{3} + 4 \, {\left (9 \, A + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, A + 4 \, C\right )} a^{2} \cos \left (d x + c\right ) + 4 \, {\left (18 \, A + 25 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{60 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm= "fricas")
Output:
1/60*(15*(3*A + 4*C)*a^2*d*x + (12*A*a^2*cos(d*x + c)^4 + 30*A*a^2*cos(d*x + c)^3 + 4*(9*A + 5*C)*a^2*cos(d*x + c)^2 + 15*(3*A + 4*C)*a^2*cos(d*x + c) + 4*(18*A + 25*C)*a^2)*sin(d*x + c))/d
Timed out. \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**5*(a+a*sec(d*x+c))**2*(A+C*sec(d*x+c)**2),x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.92 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {16 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{2} - 80 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{2} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} - 80 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 240 \, C a^{2} \sin \left (d x + c\right )}{240 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm= "maxima")
Output:
1/240*(16*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^2 - 80*(sin(d*x + c)^3 - 3*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))*A*a^2 - 80*(sin(d*x + c)^3 - 3*sin(d*x + c) )*C*a^2 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2 + 240*C*a^2*sin(d*x + c))/d
Time = 0.35 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.24 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (3 \, A a^{2} + 4 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (45 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 210 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 280 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 432 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 560 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 270 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 520 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 195 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 180 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{60 \, d} \] Input:
integrate(cos(d*x+c)^5*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x, algorithm= "giac")
Output:
1/60*(15*(3*A*a^2 + 4*C*a^2)*(d*x + c) + 2*(45*A*a^2*tan(1/2*d*x + 1/2*c)^ 9 + 60*C*a^2*tan(1/2*d*x + 1/2*c)^9 + 210*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 2 80*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 432*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 560*C *a^2*tan(1/2*d*x + 1/2*c)^5 + 270*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 520*C*a^2 *tan(1/2*d*x + 1/2*c)^3 + 195*A*a^2*tan(1/2*d*x + 1/2*c) + 180*C*a^2*tan(1 /2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d
Time = 14.15 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.46 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (\frac {3\,A\,a^2}{2}+2\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (7\,A\,a^2+\frac {28\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {72\,A\,a^2}{5}+\frac {56\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (9\,A\,a^2+\frac {52\,C\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a^2}{2}+6\,C\,a^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^2\,\mathrm {atan}\left (\frac {a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,A+4\,C\right )}{2\,\left (\frac {3\,A\,a^2}{2}+2\,C\,a^2\right )}\right )\,\left (3\,A+4\,C\right )}{2\,d} \] Input:
int(cos(c + d*x)^5*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^2,x)
Output:
(tan(c/2 + (d*x)/2)*((13*A*a^2)/2 + 6*C*a^2) + tan(c/2 + (d*x)/2)^9*((3*A* a^2)/2 + 2*C*a^2) + tan(c/2 + (d*x)/2)^7*(7*A*a^2 + (28*C*a^2)/3) + tan(c/ 2 + (d*x)/2)^3*(9*A*a^2 + (52*C*a^2)/3) + tan(c/2 + (d*x)/2)^5*((72*A*a^2) /5 + (56*C*a^2)/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^2*atan((a^2*tan(c/2 + (d*x)/2)*(3*A + 4*C))/(2*((3*A*a^2)/2 + 2 *C*a^2)))*(3*A + 4*C))/(2*d)
Time = 0.16 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.69 \[ \int \cos ^5(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{2} \left (-30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a +75 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +60 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c +12 \sin \left (d x +c \right )^{5} a -60 \sin \left (d x +c \right )^{3} a -20 \sin \left (d x +c \right )^{3} c +120 \sin \left (d x +c \right ) a +120 \sin \left (d x +c \right ) c +45 a d x +60 c d x \right )}{60 d} \] Input:
int(cos(d*x+c)^5*(a+a*sec(d*x+c))^2*(A+C*sec(d*x+c)^2),x)
Output:
(a**2*( - 30*cos(c + d*x)*sin(c + d*x)**3*a + 75*cos(c + d*x)*sin(c + d*x) *a + 60*cos(c + d*x)*sin(c + d*x)*c + 12*sin(c + d*x)**5*a - 60*sin(c + d* x)**3*a - 20*sin(c + d*x)**3*c + 120*sin(c + d*x)*a + 120*sin(c + d*x)*c + 45*a*d*x + 60*c*d*x))/(60*d)