Integrand size = 39, antiderivative size = 120 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {1}{2} a^2 (4 A+3 B+2 C) x+\frac {a^2 A \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 (2 A+3 B+2 C) \sin (c+d x)}{2 d}+\frac {C (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {(3 B+2 C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{6 d} \]
1/2*a^2*(4*A+3*B+2*C)*x+a^2*A*arctanh(sin(d*x+c))/d+1/2*a^2*(2*A+3*B+2*C)* sin(d*x+c)/d+1/3*C*(a+a*cos(d*x+c))^2*sin(d*x+c)/d+1/6*(3*B+2*C)*(a^2+a^2* cos(d*x+c))*sin(d*x+c)/d
Time = 1.95 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.01 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {a^2 \left (24 A d x+18 B d x+12 C d x-12 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 (4 A+8 B+7 C) \sin (c+d x)+3 (B+2 C) \sin (2 (c+d x))+C \sin (3 (c+d x))\right )}{12 d} \]
(a^2*(24*A*d*x + 18*B*d*x + 12*C*d*x - 12*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 12*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 3*(4*A + 8*B + 7*C)*Sin[c + d*x] + 3*(B + 2*C)*Sin[2*(c + d*x)] + C*Sin[3*(c + d*x)]))/ (12*d)
Time = 1.00 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.07, number of steps used = 13, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3524, 3042, 3455, 27, 3042, 3447, 3042, 3502, 3042, 3214, 3042, 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.
| \(\displaystyle \int \sec (c+d x) (a \cos (c+d x)+a)^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 3524 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^2 (3 a A+a (3 B+2 C) \cos (c+d x)) \sec (c+d x)dx}{3 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (3 a A+a (3 B+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{3 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{2} \int 3 (\cos (c+d x) a+a) \left (2 A a^2+(2 A+3 B+2 C) \cos (c+d x) a^2\right ) \sec (c+d x)dx+\frac {(3 B+2 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3}{2} \int (\cos (c+d x) a+a) \left (2 A a^2+(2 A+3 B+2 C) \cos (c+d x) a^2\right ) \sec (c+d x)dx+\frac {(3 B+2 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (2 A a^2+(2 A+3 B+2 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {(3 B+2 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {3}{2} \int \left ((2 A+3 B+2 C) \cos ^2(c+d x) a^3+2 A a^3+\left (2 A a^3+(2 A+3 B+2 C) a^3\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {(3 B+2 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{2} \int \frac {(2 A+3 B+2 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3+2 A a^3+\left (2 A a^3+(2 A+3 B+2 C) a^3\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {(3 B+2 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {3}{2} \left (\int \left (2 A a^3+(4 A+3 B+2 C) \cos (c+d x) a^3\right ) \sec (c+d x)dx+\frac {a^3 (2 A+3 B+2 C) \sin (c+d x)}{d}\right )+\frac {(3 B+2 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{2} \left (\int \frac {2 A a^3+(4 A+3 B+2 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {a^3 (2 A+3 B+2 C) \sin (c+d x)}{d}\right )+\frac {(3 B+2 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {3}{2} \left (2 a^3 A \int \sec (c+d x)dx+\frac {a^3 (2 A+3 B+2 C) \sin (c+d x)}{d}+a^3 x (4 A+3 B+2 C)\right )+\frac {(3 B+2 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{2} \left (2 a^3 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {a^3 (2 A+3 B+2 C) \sin (c+d x)}{d}+a^3 x (4 A+3 B+2 C)\right )+\frac {(3 B+2 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {3}{2} \left (\frac {2 a^3 A \text {arctanh}(\sin (c+d x))}{d}+\frac {a^3 (2 A+3 B+2 C) \sin (c+d x)}{d}+a^3 x (4 A+3 B+2 C)\right )+\frac {(3 B+2 C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{3 a}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d}\) |
(C*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(3*d) + (((3*B + 2*C)*(a^3 + a^3*C os[c + d*x])*Sin[c + d*x])/(2*d) + (3*(a^3*(4*A + 3*B + 2*C)*x + (2*a^3*A* ArcTanh[Sin[c + d*x]])/d + (a^3*(2*A + 3*B + 2*C)*Sin[c + d*x])/d))/2)/(3* a)
3.4.12.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, 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_)]), 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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1 ) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
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[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x] )^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Simp[1/(b*d*(m + n + 2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*Simp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n} , x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !Lt Q[m, -2^(-1)] && NeQ[m + n + 2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 5.48 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.81
| method | result | size |
| parallelrisch | \(-\frac {\left (A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-\frac {B}{4}-\frac {C}{2}\right ) \sin \left (2 d x +2 c \right )-\frac {\sin \left (3 d x +3 c \right ) C}{12}+\left (-A -2 B -\frac {7 C}{4}\right ) \sin \left (d x +c \right )-2 x \left (A +\frac {3 B}{4}+\frac {C}{2}\right ) d \right ) a^{2}}{d}\) | \(97\) |
| parts | \(\frac {A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (2 A \,a^{2}+B \,a^{2}\right ) \left (d x +c \right )}{d}+\frac {\left (B \,a^{2}+2 a^{2} C \right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (A \,a^{2}+2 B \,a^{2}+a^{2} C \right ) \sin \left (d x +c \right )}{d}+\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}\) | \(135\) |
| derivativedivides | \(\frac {A \,a^{2} \sin \left (d x +c \right )+B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 A \,a^{2} \left (d x +c \right )+2 B \sin \left (d x +c \right ) a^{2}+2 a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \left (d x +c \right )+a^{2} C \sin \left (d x +c \right )}{d}\) | \(157\) |
| default | \(\frac {A \,a^{2} \sin \left (d x +c \right )+B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+2 A \,a^{2} \left (d x +c \right )+2 B \sin \left (d x +c \right ) a^{2}+2 a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \left (d x +c \right )+a^{2} C \sin \left (d x +c \right )}{d}\) | \(157\) |
| risch | \(2 a^{2} x A +\frac {3 a^{2} B x}{2}+a^{2} C x -\frac {i {\mathrm e}^{i \left (d x +c \right )} A \,a^{2}}{2 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,a^{2}}{d}-\frac {7 i {\mathrm e}^{i \left (d x +c \right )} a^{2} C}{8 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{2}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{2}}{d}+\frac {7 i {\mathrm e}^{-i \left (d x +c \right )} a^{2} C}{8 d}+\frac {A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (3 d x +3 c \right ) a^{2} C}{12 d}+\frac {\sin \left (2 d x +2 c \right ) B \,a^{2}}{4 d}+\frac {\sin \left (2 d x +2 c \right ) a^{2} C}{2 d}\) | \(233\) |
| norman | \(\frac {\left (2 A \,a^{2}+\frac {3}{2} B \,a^{2}+a^{2} C \right ) x +\left (2 A \,a^{2}+\frac {3}{2} B \,a^{2}+a^{2} C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (8 A \,a^{2}+6 B \,a^{2}+4 a^{2} C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (8 A \,a^{2}+6 B \,a^{2}+4 a^{2} C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (12 A \,a^{2}+9 B \,a^{2}+6 a^{2} C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a^{2} \left (2 A +3 B +2 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{2} \left (2 A +5 B +6 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{2} \left (18 A +33 B +22 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {a^{2} \left (39 B +18 A +34 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {A \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {A \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(319\) |
-(A*ln(tan(1/2*d*x+1/2*c)-1)-A*ln(tan(1/2*d*x+1/2*c)+1)+(-1/4*B-1/2*C)*sin (2*d*x+2*c)-1/12*sin(3*d*x+3*c)*C+(-A-2*B-7/4*C)*sin(d*x+c)-2*x*(A+3/4*B+1 /2*C)*d)*a^2/d
Time = 0.28 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.90 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {3 \, {\left (4 \, A + 3 \, B + 2 \, C\right )} a^{2} d x + 3 \, A a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, A a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, C a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, {\left (3 \, A + 6 \, B + 5 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{6 \, d} \]
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="fricas")
1/6*(3*(4*A + 3*B + 2*C)*a^2*d*x + 3*A*a^2*log(sin(d*x + c) + 1) - 3*A*a^2 *log(-sin(d*x + c) + 1) + (2*C*a^2*cos(d*x + c)^2 + 3*(B + 2*C)*a^2*cos(d* x + c) + 2*(3*A + 6*B + 5*C)*a^2)*sin(d*x + c))/d
\[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=a^{2} \left (\int A \sec {\left (c + d x \right )}\, dx + \int 2 A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int A \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 2 B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 2 C \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(A*sec(c + d*x), x) + Integral(2*A*cos(c + d*x)*sec(c + d*x) , x) + Integral(A*cos(c + d*x)**2*sec(c + d*x), x) + Integral(B*cos(c + d* x)*sec(c + d*x), x) + Integral(2*B*cos(c + d*x)**2*sec(c + d*x), x) + Inte gral(B*cos(c + d*x)**3*sec(c + d*x), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x), x) + Integral(2*C*cos(c + d*x)**3*sec(c + d*x), x) + Integral(C*c os(c + d*x)**4*sec(c + d*x), x))
Time = 0.20 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.28 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {24 \, {\left (d x + c\right )} A a^{2} + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} + 12 \, {\left (d x + c\right )} B a^{2} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 6 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 12 \, A a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, A a^{2} \sin \left (d x + c\right ) + 24 \, B a^{2} \sin \left (d x + c\right ) + 12 \, C a^{2} \sin \left (d x + c\right )}{12 \, d} \]
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="maxima")
1/12*(24*(d*x + c)*A*a^2 + 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2 + 12*( d*x + c)*B*a^2 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^2 + 6*(2*d*x + 2* c + sin(2*d*x + 2*c))*C*a^2 + 12*A*a^2*log(sec(d*x + c) + tan(d*x + c)) + 12*A*a^2*sin(d*x + c) + 24*B*a^2*sin(d*x + c) + 12*C*a^2*sin(d*x + c))/d
Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (112) = 224\).
Time = 0.38 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.96 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {6 \, A a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, A a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (4 \, A a^{2} + 3 \, B a^{2} + 2 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 15 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]
1/6*(6*A*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*A*a^2*log(abs(tan(1/2* d*x + 1/2*c) - 1)) + 3*(4*A*a^2 + 3*B*a^2 + 2*C*a^2)*(d*x + c) + 2*(6*A*a^ 2*tan(1/2*d*x + 1/2*c)^5 + 9*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^2*tan(1/ 2*d*x + 1/2*c)^5 + 12*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 24*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 16*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*tan(1/2*d*x + 1/2*c ) + 15*B*a^2*tan(1/2*d*x + 1/2*c) + 18*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/ 2*d*x + 1/2*c)^2 + 1)^3)/d
Time = 1.84 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.88 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {A\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,B\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {7\,C\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {4\,A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {C\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}-\frac {A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d} \]
(A*a^2*sin(c + d*x))/d + (2*B*a^2*sin(c + d*x))/d + (7*C*a^2*sin(c + d*x)) /(4*d) + (4*A*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d - (A*a^2* atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i)/d + (3*B*a^2*atan(sin (c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^2*atan(sin(c/2 + (d*x)/2)/ cos(c/2 + (d*x)/2)))/d + (B*a^2*sin(2*c + 2*d*x))/(4*d) + (C*a^2*sin(2*c + 2*d*x))/(2*d) + (C*a^2*sin(3*c + 3*d*x))/(12*d)