Integrand size = 41, antiderivative size = 123 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=a^2 (B+2 C) x+\frac {a^2 (3 A+4 B+2 C) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {a^2 (3 A+2 B-2 C) \sin (c+d x)}{2 d}+\frac {(A+B) \left (a^2+a^2 \cos (c+d x)\right ) \tan (c+d x)}{d}+\frac {A (a+a \cos (c+d x))^2 \sec (c+d x) \tan (c+d x)}{2 d} \]
a^2*(B+2*C)*x+1/2*a^2*(3*A+4*B+2*C)*arctanh(sin(d*x+c))/d-1/2*a^2*(3*A+2*B -2*C)*sin(d*x+c)/d+(A+B)*(a^2+a^2*cos(d*x+c))*tan(d*x+c)/d+1/2*A*(a+a*cos( d*x+c))^2*sec(d*x+c)*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(259\) vs. \(2(123)=246\).
Time = 4.12 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.11 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {a^2 (1+\cos (c+d x))^2 \sec ^4\left (\frac {1}{2} (c+d x)\right ) \left (4 (B+2 C) (c+d x)-2 (3 A+4 B+2 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 (3 A+4 B+2 C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (2 A+B) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {4 (2 A+B) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+4 C \sin (c+d x)\right )}{16 d} \]
(a^2*(1 + Cos[c + d*x])^2*Sec[(c + d*x)/2]^4*(4*(B + 2*C)*(c + d*x) - 2*(3 *A + 4*B + 2*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*(3*A + 4*B + 2*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + A/(Cos[(c + d*x)/2] - Sin[ (c + d*x)/2])^2 + (4*(2*A + B)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[( c + d*x)/2]) - A/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (4*(2*A + B)*Si n[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 4*C*Sin[c + d*x])) /(16*d)
Time = 1.06 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.293, Rules used = {3042, 3522, 3042, 3454, 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.
| \(\displaystyle \int \sec ^3(c+d x) (a \cos (c+d x)+a)^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 3522 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^2 (2 a (A+B)-a (A-2 C) \cos (c+d x)) \sec ^2(c+d x)dx}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 a (A+B)-a (A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3454 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a) \left (a^2 (3 A+4 B+2 C)-a^2 (3 A+2 B-2 C) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 (A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (a^2 (3 A+4 B+2 C)-a^2 (3 A+2 B-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 (A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\int \left (-\left ((3 A+2 B-2 C) \cos ^2(c+d x) a^3\right )+(3 A+4 B+2 C) a^3+\left (a^3 (3 A+4 B+2 C)-a^3 (3 A+2 B-2 C)\right ) \cos (c+d x)\right ) \sec (c+d x)dx+\frac {2 (A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {-\left ((3 A+2 B-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3\right )+(3 A+4 B+2 C) a^3+\left (a^3 (3 A+4 B+2 C)-a^3 (3 A+2 B-2 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 (A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\int \left ((3 A+4 B+2 C) a^3+2 (B+2 C) \cos (c+d x) a^3\right ) \sec (c+d x)dx-\frac {a^3 (3 A+2 B-2 C) \sin (c+d x)}{d}+\frac {2 (A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(3 A+4 B+2 C) a^3+2 (B+2 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {a^3 (3 A+2 B-2 C) \sin (c+d x)}{d}+\frac {2 (A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {a^3 (3 A+4 B+2 C) \int \sec (c+d x)dx-\frac {a^3 (3 A+2 B-2 C) \sin (c+d x)}{d}+\frac {2 (A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}+2 a^3 x (B+2 C)}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a^3 (3 A+4 B+2 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a^3 (3 A+2 B-2 C) \sin (c+d x)}{d}+\frac {2 (A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}+2 a^3 x (B+2 C)}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {a^3 (3 A+4 B+2 C) \text {arctanh}(\sin (c+d x))}{d}-\frac {a^3 (3 A+2 B-2 C) \sin (c+d x)}{d}+\frac {2 (A+B) \tan (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{d}+2 a^3 x (B+2 C)}{2 a}+\frac {A \tan (c+d x) \sec (c+d x) (a \cos (c+d x)+a)^2}{2 d}\) |
(A*(a + a*Cos[c + d*x])^2*Sec[c + d*x]*Tan[c + d*x])/(2*d) + (2*a^3*(B + 2 *C)*x + (a^3*(3*A + 4*B + 2*C)*ArcTanh[Sin[c + d*x]])/d - (a^3*(3*A + 2*B - 2*C)*Sin[c + d*x])/d + (2*(A + B)*(a^3 + a^3*Cos[c + d*x])*Tan[c + d*x]) /d)/(2*a)
3.4.14.3.1 Defintions of rubi rules used
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^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[(-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^2*C - B*c*d + A*d^2))*Cos[e + f*x ]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m* (c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C - B*d)*( a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2* (n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ [m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 7.01 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.08
| method | result | size |
| parts | \(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (2 A \,a^{2}+B \,a^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (B \,a^{2}+2 a^{2} C \right ) \left (d x +c \right )}{d}+\frac {\left (A \,a^{2}+2 B \,a^{2}+a^{2} C \right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\sin \left (d x +c \right ) a^{2} C}{d}\) | \(133\) |
| derivativedivides | \(\frac {A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \left (d x +c \right )+a^{2} C \sin \left (d x +c \right )+2 A \,a^{2} \tan \left (d x +c \right )+2 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a^{2} C \left (d x +c \right )+A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \tan \left (d x +c \right )+a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(155\) |
| default | \(\frac {A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{2} \left (d x +c \right )+a^{2} C \sin \left (d x +c \right )+2 A \,a^{2} \tan \left (d x +c \right )+2 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a^{2} C \left (d x +c \right )+A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \tan \left (d x +c \right )+a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\) | \(155\) |
| parallelrisch | \(\frac {2 \left (-\frac {3 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {4 B}{3}+\frac {2 C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4}+\frac {3 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {4 B}{3}+\frac {2 C}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4}+\frac {d x \left (B +2 C \right ) \cos \left (2 d x +2 c \right )}{2}+\left (A +\frac {B}{2}\right ) \sin \left (2 d x +2 c \right )+\frac {\sin \left (3 d x +3 c \right ) C}{4}+\left (\frac {A}{2}+\frac {C}{4}\right ) \sin \left (d x +c \right )+\frac {d x \left (B +2 C \right )}{2}\right ) a^{2}}{d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(157\) |
| risch | \(a^{2} B x +2 a^{2} C x -\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2} C}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2} C}{2 d}-\frac {i a^{2} \left (A \,{\mathrm e}^{3 i \left (d x +c \right )}-4 A \,{\mathrm e}^{2 i \left (d x +c \right )}-2 B \,{\mathrm e}^{2 i \left (d x +c \right )}-A \,{\mathrm e}^{i \left (d x +c \right )}-4 A -2 B \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {3 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}\) | \(260\) |
| norman | \(\frac {\left (B \,a^{2}+2 a^{2} C \right ) x +\left (-4 B \,a^{2}-8 a^{2} C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-B \,a^{2}-2 a^{2} C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-B \,a^{2}-2 a^{2} C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (B \,a^{2}+2 a^{2} C \right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 B \,a^{2}+4 a^{2} C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 B \,a^{2}+4 a^{2} C \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a^{2} \left (5 A +2 B +2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{2} \left (17 A +6 B +2 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} \left (A -2 B -2 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (3 A +2 B -2 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (7 A +6 B -2 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {2 a^{2} \left (9 A +2 B -2 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {a^{2} \left (3 A +4 B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{2} \left (3 A +4 B +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(431\) |
A*a^2/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+(2*A*a^2 +B*a^2)/d*tan(d*x+c)+(B*a^2+2*C*a^2)/d*(d*x+c)+(A*a^2+2*B*a^2+C*a^2)/d*ln( sec(d*x+c)+tan(d*x+c))+1/d*sin(d*x+c)*a^2*C
Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.16 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {4 \, {\left (B + 2 \, C\right )} a^{2} d x \cos \left (d x + c\right )^{2} + {\left (3 \, A + 4 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (3 \, A + 4 \, B + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, C a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3, x, algorithm="fricas")
1/4*(4*(B + 2*C)*a^2*d*x*cos(d*x + c)^2 + (3*A + 4*B + 2*C)*a^2*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (3*A + 4*B + 2*C)*a^2*cos(d*x + c)^2*log(-sin (d*x + c) + 1) + 2*(2*C*a^2*cos(d*x + c)^2 + 2*(2*A + B)*a^2*cos(d*x + c) + A*a^2)*sin(d*x + c))/(d*cos(d*x + c)^2)
\[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=a^{2} \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 A \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int A \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 B \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \cos ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 C \cos ^{3}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int C \cos ^{4}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(A*sec(c + d*x)**3, x) + Integral(2*A*cos(c + d*x)*sec(c + d *x)**3, x) + Integral(A*cos(c + d*x)**2*sec(c + d*x)**3, x) + Integral(B*c os(c + d*x)*sec(c + d*x)**3, x) + Integral(2*B*cos(c + d*x)**2*sec(c + d*x )**3, x) + Integral(B*cos(c + d*x)**3*sec(c + d*x)**3, x) + Integral(C*cos (c + d*x)**2*sec(c + d*x)**3, x) + Integral(2*C*cos(c + d*x)**3*sec(c + d* x)**3, x) + Integral(C*cos(c + d*x)**4*sec(c + d*x)**3, x))
Time = 0.20 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.56 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} B a^{2} + 8 \, {\left (d x + c\right )} C a^{2} - A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, C a^{2} \sin \left (d x + c\right ) + 8 \, A a^{2} \tan \left (d x + c\right ) + 4 \, B a^{2} \tan \left (d x + c\right )}{4 \, d} \]
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3, x, algorithm="maxima")
1/4*(4*(d*x + c)*B*a^2 + 8*(d*x + c)*C*a^2 - A*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)) + 2*A*a^2*( log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*B*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 2*C*a^2*(log(sin(d*x + c) + 1) - log(sin( d*x + c) - 1)) + 4*C*a^2*sin(d*x + c) + 8*A*a^2*tan(d*x + c) + 4*B*a^2*tan (d*x + c))/d
Time = 0.33 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.66 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {\frac {4 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 2 \, {\left (B a^{2} + 2 \, C a^{2}\right )} {\left (d x + c\right )} + {\left (3 \, A a^{2} + 4 \, B a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (3 \, A a^{2} + 4 \, B a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 5 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3, x, algorithm="giac")
1/2*(4*C*a^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 2*(B*a^2 + 2*C*a^2)*(d*x + c) + (3*A*a^2 + 4*B*a^2 + 2*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - (3*A*a^2 + 4*B*a^2 + 2*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^2*tan(1/2*d*x + 1/2*c)^ 3 - 5*A*a^2*tan(1/2*d*x + 1/2*c) - 2*B*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2* d*x + 1/2*c)^2 - 1)^2)/d
Time = 2.29 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.98 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {A\,a^2\,\sin \left (2\,c+2\,d\,x\right )+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {C\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{2}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{4}}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )}-\frac {2\,\left (\frac {A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,3{}\mathrm {i}}{2}-B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}-2\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,1{}\mathrm {i}\right )}{d} \]
(A*a^2*sin(2*c + 2*d*x) + (B*a^2*sin(2*c + 2*d*x))/2 + (C*a^2*sin(3*c + 3* d*x))/4 + (A*a^2*sin(c + d*x))/2 + (C*a^2*sin(c + d*x))/4)/(d*(cos(2*c + 2 *d*x)/2 + 1/2)) - (2*((A*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/ 2))*3i)/2 - B*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + B*a^2*atan ((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*2i - 2*C*a^2*atan(sin(c/2 + ( d*x)/2)/cos(c/2 + (d*x)/2)) + C*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*1i))/d