Integrand size = 40, antiderivative size = 113 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {a^2 (2 B+3 C) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {a^2 (5 B+6 C) \tan (c+d x)}{3 d}+\frac {a^2 (4 B+3 C) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {B \left (a^2+a^2 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{3 d} \]
1/2*a^2*(2*B+3*C)*arctanh(sin(d*x+c))/d+1/3*a^2*(5*B+6*C)*tan(d*x+c)/d+1/6 *a^2*(4*B+3*C)*sec(d*x+c)*tan(d*x+c)/d+1/3*B*(a^2+a^2*cos(d*x+c))*sec(d*x+ c)^2*tan(d*x+c)/d
Time = 0.95 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.56 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {a^2 \left ((6 B+9 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (12 (B+C)+3 (2 B+C) \sec (c+d x)+2 B \tan ^2(c+d x)\right )\right )}{6 d} \]
(a^2*((6*B + 9*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(12*(B + C) + 3*(2* B + C)*Sec[c + d*x] + 2*B*Tan[c + d*x]^2)))/(6*d)
Time = 1.00 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {3042, 3508, 3042, 3454, 3042, 3447, 3042, 3500, 3042, 3227, 3042, 4254, 24, 4257}
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.
| \(\displaystyle \int \sec ^5(c+d x) (a \cos (c+d x)+a)^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 3508 |
| \(\displaystyle \int \sec ^4(c+d x) (a \cos (c+d x)+a)^2 (B+C \cos (c+d x))dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (B+C \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {1}{3} \int (\cos (c+d x) a+a) (a (4 B+3 C)+a (B+3 C) \cos (c+d x)) \sec ^3(c+d x)dx+\frac {B \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a (4 B+3 C)+a (B+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {B \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {1}{3} \int \left ((B+3 C) \cos ^2(c+d x) a^2+(4 B+3 C) a^2+\left ((B+3 C) a^2+(4 B+3 C) a^2\right ) \cos (c+d x)\right ) \sec ^3(c+d x)dx+\frac {B \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \int \frac {(B+3 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^2+(4 B+3 C) a^2+\left ((B+3 C) a^2+(4 B+3 C) a^2\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx+\frac {B \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \left (2 (5 B+6 C) a^2+3 (2 B+3 C) \cos (c+d x) a^2\right ) \sec ^2(c+d x)dx+\frac {a^2 (4 B+3 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {B \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int \frac {2 (5 B+6 C) a^2+3 (2 B+3 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx+\frac {a^2 (4 B+3 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {B \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (2 a^2 (5 B+6 C) \int \sec ^2(c+d x)dx+3 a^2 (2 B+3 C) \int \sec (c+d x)dx\right )+\frac {a^2 (4 B+3 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {B \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 a^2 (2 B+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+2 a^2 (5 B+6 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {a^2 (4 B+3 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {B \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 a^2 (2 B+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {2 a^2 (5 B+6 C) \int 1d(-\tan (c+d x))}{d}\right )+\frac {a^2 (4 B+3 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {B \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (3 a^2 (2 B+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 a^2 (5 B+6 C) \tan (c+d x)}{d}\right )+\frac {a^2 (4 B+3 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {B \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{2} \left (\frac {3 a^2 (2 B+3 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {2 a^2 (5 B+6 C) \tan (c+d x)}{d}\right )+\frac {a^2 (4 B+3 C) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {B \tan (c+d x) \sec ^2(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}\) |
(B*(a^2 + a^2*Cos[c + d*x])*Sec[c + d*x]^2*Tan[c + d*x])/(3*d) + ((a^2*(4* B + 3*C)*Sec[c + d*x]*Tan[c + d*x])/(2*d) + ((3*a^2*(2*B + 3*C)*ArcTanh[Si n[c + d*x]])/d + (2*a^2*(5*B + 6*C)*Tan[c + d*x])/d)/2)/3
3.3.41.3.1 Defintions of rubi rules used
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-b^2)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[ e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a*d))), x] - Simp[b/(d*(n + 1)*(b*c + a*d)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp [a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n + 1) - B *(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0 ])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_ .) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b^2 Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 7.98 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.06
| method | result | size |
| parts | \(\frac {\left (B \,a^{2}+2 a^{2} C \right ) \tan \left (d x +c \right )}{d}+\frac {\left (2 B \,a^{2}+a^{2} C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {B \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right ) a^{2}}{d}\) | \(120\) |
| derivativedivides | \(\frac {a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \tan \left (d x +c \right )+2 a^{2} C \tan \left (d x +c \right )+2 B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(145\) |
| default | \(\frac {a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \tan \left (d x +c \right )+2 a^{2} C \tan \left (d x +c \right )+2 B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{2} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\) | \(145\) |
| parallelrisch | \(\frac {2 \left (-\frac {3 \left (B +\frac {3 C}{2}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {3 \left (B +\frac {3 C}{2}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\left (B +\frac {C}{2}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {5 B}{6}+C \right ) \sin \left (3 d x +3 c \right )+\frac {3 \left (B +\frac {2 C}{3}\right ) \sin \left (d x +c \right )}{2}\right ) a^{2}}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\) | \(147\) |
| risch | \(-\frac {i a^{2} \left (6 B \,{\mathrm e}^{5 i \left (d x +c \right )}+3 C \,{\mathrm e}^{5 i \left (d x +c \right )}-6 B \,{\mathrm e}^{4 i \left (d x +c \right )}-12 C \,{\mathrm e}^{4 i \left (d x +c \right )}-24 B \,{\mathrm e}^{2 i \left (d x +c \right )}-24 C \,{\mathrm e}^{2 i \left (d x +c \right )}-6 B \,{\mathrm e}^{i \left (d x +c \right )}-3 C \,{\mathrm e}^{i \left (d x +c \right )}-10 B -12 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}\) | \(214\) |
| norman | \(\frac {\frac {a^{2} \left (6 B +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{2} \left (6 B +13 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (2 B +3 C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a^{2} \left (2 B +3 C \right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (2 B +11 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (10 B -21 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {a^{2} \left (38 B +21 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a^{2} \left (50 B +51 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {a^{2} \left (2 B +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{2} \left (2 B +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(292\) |
(B*a^2+2*C*a^2)/d*tan(d*x+c)+(2*B*a^2+C*a^2)/d*(1/2*sec(d*x+c)*tan(d*x+c)+ 1/2*ln(sec(d*x+c)+tan(d*x+c)))-B*a^2/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+ 1/d*C*ln(sec(d*x+c)+tan(d*x+c))*a^2
Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.11 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (2 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (5 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (2 \, B + C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, B a^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]
integrate((a+a*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="fricas")
1/12*(3*(2*B + 3*C)*a^2*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(2*B + 3* C)*a^2*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(5*B + 6*C)*a^2*cos(d* x + c)^2 + 3*(2*B + C)*a^2*cos(d*x + c) + 2*B*a^2)*sin(d*x + c))/(d*cos(d* x + c)^3)
Timed out. \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.54 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} - 6 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, B a^{2} \tan \left (d x + c\right ) + 24 \, C a^{2} \tan \left (d x + c\right )}{12 \, d} \]
integrate((a+a*cos(d*x+c))^2*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^5,x, algorithm="maxima")
1/12*(4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 - 6*B*a^2*(2*sin(d*x + c)/ (sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 3* C*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(s in(d*x + c) - 1)) + 6*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1) ) + 12*B*a^2*tan(d*x + c) + 24*C*a^2*tan(d*x + c))/d
Time = 0.36 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.58 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (2 \, B a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (2 \, B a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (6 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 16 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]
1/6*(3*(2*B*a^2 + 3*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(2*B*a^2 + 3*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(6*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 9*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 16*B*a^2*tan(1/2*d*x + 1/2*c)^ 3 - 24*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 18*B*a^2*tan(1/2*d*x + 1/2*c) + 15*C *a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)/d
Time = 3.47 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.28 \[ \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx=\frac {2\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (B+\frac {3\,C}{2}\right )}{d}-\frac {\left (2\,B\,a^2+3\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {16\,B\,a^2}{3}-8\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (6\,B\,a^2+5\,C\,a^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]