Integrand size = 33, antiderivative size = 112 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {1}{2} a^2 (2 A+3 C) x+\frac {2 a^2 A \text {arctanh}(\sin (c+d x))}{d}-\frac {a^2 (2 A-3 C) \sin (c+d x)}{2 d}-\frac {(2 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {A (a+a \cos (c+d x))^2 \tan (c+d x)}{d} \]
1/2*a^2*(2*A+3*C)*x+2*a^2*A*arctanh(sin(d*x+c))/d-1/2*a^2*(2*A-3*C)*sin(d* x+c)/d-1/2*(2*A-C)*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d+A*(a+a*cos(d*x+c))^2* tan(d*x+c)/d
Time = 1.63 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.97 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {a^2 \left (4 A c+6 c C+4 A d x+6 C d x-8 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 C \sin (c+d x)+C \sin (2 (c+d x))+4 A \tan (c+d x)\right )}{4 d} \]
(a^2*(4*A*c + 6*c*C + 4*A*d*x + 6*C*d*x - 8*A*Log[Cos[(c + d*x)/2] - Sin[( c + d*x)/2]] + 8*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 8*C*Sin[c + d*x] + C*Sin[2*(c + d*x)] + 4*A*Tan[c + d*x]))/(4*d)
Time = 0.97 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 3523, 3042, 3455, 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.
| \(\displaystyle \int \sec ^2(c+d x) (a \cos (c+d x)+a)^2 \left (A+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^2}dx\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \frac {\int (\cos (c+d x) a+a)^2 (2 a A-a (2 A-C) \cos (c+d x)) \sec (c+d x)dx}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (2 a A-a (2 A-C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3455 |
| \(\displaystyle \frac {\frac {1}{2} \int (\cos (c+d x) a+a) \left (4 a^2 A-a^2 (2 A-3 C) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(2 A-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (4 a^2 A-a^2 (2 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(2 A-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\frac {1}{2} \int \left (-\left ((2 A-3 C) \cos ^2(c+d x) a^3\right )+4 A a^3+\left (4 a^3 A-a^3 (2 A-3 C)\right ) \cos (c+d x)\right ) \sec (c+d x)dx-\frac {(2 A-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {-\left ((2 A-3 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2 a^3\right )+4 A a^3+\left (4 a^3 A-a^3 (2 A-3 C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {(2 A-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {1}{2} \left (\int \left (4 A a^3+(2 A+3 C) \cos (c+d x) a^3\right ) \sec (c+d x)dx-\frac {a^3 (2 A-3 C) \sin (c+d x)}{d}\right )-\frac {(2 A-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (\int \frac {4 A a^3+(2 A+3 C) \sin \left (c+d x+\frac {\pi }{2}\right ) a^3}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {a^3 (2 A-3 C) \sin (c+d x)}{d}\right )-\frac {(2 A-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {1}{2} \left (4 a^3 A \int \sec (c+d x)dx-\frac {a^3 (2 A-3 C) \sin (c+d x)}{d}+a^3 x (2 A+3 C)\right )-\frac {(2 A-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {1}{2} \left (4 a^3 A \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {a^3 (2 A-3 C) \sin (c+d x)}{d}+a^3 x (2 A+3 C)\right )-\frac {(2 A-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {1}{2} \left (\frac {4 a^3 A \text {arctanh}(\sin (c+d x))}{d}-\frac {a^3 (2 A-3 C) \sin (c+d x)}{d}+a^3 x (2 A+3 C)\right )-\frac {(2 A-C) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{2 d}}{a}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^2}{d}\) |
(-1/2*((2*A - C)*(a^3 + a^3*Cos[c + d*x])*Sin[c + d*x])/d + (a^3*(2*A + 3* C)*x + (4*a^3*A*ArcTanh[Sin[c + d*x]])/d - (a^3*(2*A - 3*C)*Sin[c + d*x])/ d)/2)/a + (A*(a + a*Cos[c + d*x])^2*Tan[c + d*x])/d
3.1.13.3.1 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)*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_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + 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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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 = 4.18 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.86
| method | result | size |
| derivativedivides | \(\frac {A \,a^{2} \left (d x +c \right )+a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a^{2} C \sin \left (d x +c \right )+A \,a^{2} \tan \left (d x +c \right )+a^{2} C \left (d x +c \right )}{d}\) | \(96\) |
| default | \(\frac {A \,a^{2} \left (d x +c \right )+a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a^{2} C \sin \left (d x +c \right )+A \,a^{2} \tan \left (d x +c \right )+a^{2} C \left (d x +c \right )}{d}\) | \(96\) |
| parts | \(\frac {a^{2} A \tan \left (d x +c \right )}{d}+\frac {\left (A \,a^{2}+a^{2} C \right ) \left (d x +c \right )}{d}+\frac {a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 \sin \left (d x +c \right ) a^{2} C}{d}\) | \(104\) |
| parallelrisch | \(-\frac {2 \left (A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (d x +c \right )-A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (d x +c \right )-\frac {\sin \left (2 d x +2 c \right ) C}{2}-\frac {\sin \left (3 d x +3 c \right ) C}{16}-\frac {\left (A +\frac {3 C}{2}\right ) x d \cos \left (d x +c \right )}{2}-\frac {\sin \left (d x +c \right ) \left (A +\frac {C}{8}\right )}{2}\right ) a^{2}}{d \cos \left (d x +c \right )}\) | \(111\) |
| risch | \(a^{2} x A +\frac {3 a^{2} C x}{2}-\frac {i a^{2} C \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2} C}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2} C}{d}+\frac {i a^{2} C \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i A \,a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {2 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {2 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(158\) |
| norman | \(\frac {\left (-A \,a^{2}-\frac {3}{2} a^{2} C \right ) x +\left (-3 A \,a^{2}-\frac {9}{2} a^{2} C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (A \,a^{2}+\frac {3}{2} a^{2} C \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 A \,a^{2}+\frac {9}{2} a^{2} C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 A \,a^{2}-3 a^{2} C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 A \,a^{2}+3 a^{2} C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {8 a^{2} \left (A -C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {8 a^{2} \left (A +C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (2 A -3 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (2 A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {2 a^{2} \left (6 A -C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2 A \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}+\frac {2 A \,a^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(342\) |
1/d*(A*a^2*(d*x+c)+a^2*C*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+2*A*a^2 *ln(sec(d*x+c)+tan(d*x+c))+2*a^2*C*sin(d*x+c)+A*a^2*tan(d*x+c)+a^2*C*(d*x+ c))
Time = 0.28 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.04 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {{\left (2 \, A + 3 \, C\right )} a^{2} d x \cos \left (d x + c\right ) + 2 \, A a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \, A a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (C a^{2} \cos \left (d x + c\right )^{2} + 4 \, C a^{2} \cos \left (d x + c\right ) + 2 \, A a^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]
1/2*((2*A + 3*C)*a^2*d*x*cos(d*x + c) + 2*A*a^2*cos(d*x + c)*log(sin(d*x + c) + 1) - 2*A*a^2*cos(d*x + c)*log(-sin(d*x + c) + 1) + (C*a^2*cos(d*x + c)^2 + 4*C*a^2*cos(d*x + c) + 2*A*a^2)*sin(d*x + c))/(d*cos(d*x + c))
\[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=a^{2} \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 A \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int A \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 C \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]
a**2*(Integral(A*sec(c + d*x)**2, x) + Integral(2*A*cos(c + d*x)*sec(c + d *x)**2, x) + Integral(A*cos(c + d*x)**2*sec(c + d*x)**2, x) + Integral(C*c os(c + d*x)**2*sec(c + d*x)**2, x) + Integral(2*C*cos(c + d*x)**3*sec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**4*sec(c + d*x)**2, x))
Time = 0.28 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.90 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} A a^{2} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 4 \, {\left (d x + c\right )} C a^{2} + 4 \, A a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 8 \, C a^{2} \sin \left (d x + c\right ) + 4 \, A a^{2} \tan \left (d x + c\right )}{4 \, d} \]
1/4*(4*(d*x + c)*A*a^2 + (2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2 + 4*(d*x + c)*C*a^2 + 4*A*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 8*C* a^2*sin(d*x + c) + 4*A*a^2*tan(d*x + c))/d
Time = 0.29 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.28 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {4 \, A a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 4 \, A a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + {\left (2 \, A a^{2} + 3 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]
1/2*(4*A*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 4*A*a^2*log(abs(tan(1/2* d*x + 1/2*c) - 1)) - 4*A*a^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) + (2*A*a^2 + 3*C*a^2)*(d*x + c) + 2*(3*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 5*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d
Time = 1.10 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.36 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2\,C\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {2\,A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {C\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d} \]