Integrand size = 39, antiderivative size = 126 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a (2 A b+a B) x+\frac {\left (2 A b^2+4 a b B+2 a^2 C+b^2 C\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {A (a+b \sec (c+d x))^2 \sin (c+d x)}{d}-\frac {b (2 a A-b B-2 a C) \tan (c+d x)}{d}-\frac {b^2 (2 A-C) \sec (c+d x) \tan (c+d x)}{2 d} \]
a*(2*A*b+B*a)*x+1/2*(2*A*b^2+4*B*a*b+2*C*a^2+C*b^2)*arctanh(sin(d*x+c))/d+ A*(a+b*sec(d*x+c))^2*sin(d*x+c)/d-b*(2*A*a-B*b-2*C*a)*tan(d*x+c)/d-1/2*b^2 *(2*A-C)*sec(d*x+c)*tan(d*x+c)/d
Leaf count is larger than twice the leaf count of optimal. \(453\) vs. \(2(126)=252\).
Time = 4.67 (sec) , antiderivative size = 453, normalized size of antiderivative = 3.60 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\sec ^2(c+d x) \left (4 a A b c+2 a^2 B c+4 a A b d x+2 a^2 B d x-2 A b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 a b B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 a^2 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-b^2 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 A b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 a b B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a^2 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+b^2 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\cos (2 (c+d x)) \left (2 a (2 A b+a B) (c+d x)-\left (2 A b^2+4 a b B+2 a^2 C+b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\left (2 A b^2+4 a b B+2 a^2 C+b^2 C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\left (a^2 A+2 b^2 C\right ) \sin (c+d x)+2 b^2 B \sin (2 (c+d x))+4 a b C \sin (2 (c+d x))+a^2 A \sin (3 (c+d x))\right )}{4 d} \]
(Sec[c + d*x]^2*(4*a*A*b*c + 2*a^2*B*c + 4*a*A*b*d*x + 2*a^2*B*d*x - 2*A*b ^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 4*a*b*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 2*a^2*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - b^2*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 2*A*b^2*Log[Cos[(c + d*x) /2] + Sin[(c + d*x)/2]] + 4*a*b*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 2*a^2*C*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + b^2*C*Log[Cos[(c + d *x)/2] + Sin[(c + d*x)/2]] + Cos[2*(c + d*x)]*(2*a*(2*A*b + a*B)*(c + d*x) - (2*A*b^2 + 4*a*b*B + 2*a^2*C + b^2*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d *x)/2]] + (2*A*b^2 + 4*a*b*B + 2*a^2*C + b^2*C)*Log[Cos[(c + d*x)/2] + Sin [(c + d*x)/2]]) + (a^2*A + 2*b^2*C)*Sin[c + d*x] + 2*b^2*B*Sin[2*(c + d*x) ] + 4*a*b*C*Sin[2*(c + d*x)] + a^2*A*Sin[3*(c + d*x)]))/(4*d)
Time = 0.51 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {3042, 4582, 3042, 4536, 2009}
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.
| \(\displaystyle \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4582 |
| \(\displaystyle \int (a+b \sec (c+d x)) \left (-b (2 A-C) \sec ^2(c+d x)+(b B+a C) \sec (c+d x)+2 A b+a B\right )dx+\frac {A \sin (c+d x) (a+b \sec (c+d x))^2}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-b (2 A-C) \csc \left (c+d x+\frac {\pi }{2}\right )^2+(b B+a C) \csc \left (c+d x+\frac {\pi }{2}\right )+2 A b+a B\right )dx+\frac {A \sin (c+d x) (a+b \sec (c+d x))^2}{d}\) |
| \(\Big \downarrow \) 4536 |
| \(\displaystyle \frac {1}{2} \int \left (-2 b (2 a A-b B-2 a C) \sec ^2(c+d x)+\left (2 C a^2+4 b B a+2 A b^2+b^2 C\right ) \sec (c+d x)+2 a (2 A b+a B)\right )dx+\frac {A \sin (c+d x) (a+b \sec (c+d x))^2}{d}-\frac {b^2 (2 A-C) \tan (c+d x) \sec (c+d x)}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {\left (2 a^2 C+4 a b B+2 A b^2+b^2 C\right ) \text {arctanh}(\sin (c+d x))}{d}-\frac {2 b \tan (c+d x) (2 a A-2 a C-b B)}{d}+2 a x (a B+2 A b)\right )+\frac {A \sin (c+d x) (a+b \sec (c+d x))^2}{d}-\frac {b^2 (2 A-C) \tan (c+d x) \sec (c+d x)}{2 d}\) |
(A*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/d - (b^2*(2*A - C)*Sec[c + d*x]*Ta n[c + d*x])/(2*d) + (2*a*(2*A*b + a*B)*x + ((2*A*b^2 + 4*a*b*B + 2*a^2*C + b^2*C)*ArcTanh[Sin[c + d*x]])/d - (2*b*(2*a*A - b*B - 2*a*C)*Tan[c + d*x] )/d)/2
3.9.73.3.1 Defintions of rubi rules used
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-b)*C*Csc[e + f*x]*(Cot[e + f*x]/(2*f)), x] + Simp[1/2 Int[Simp[2*A*a + (2*B*a + b*(2* A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, e, f, A, B, C}, x]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. ))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a _))^(m_), x_Symbol] :> Simp[A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Simp[1/(d*n) Int[(a + b*Csc[e + f*x])^(m - 1)*(d* Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Cs c[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a , b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]
Time = 0.76 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.21
| method | result | size |
| derivativedivides | \(\frac {a^{2} A \sin \left (d x +c \right )+B \,a^{2} \left (d x +c \right )+C \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a A b \left (d x +c \right )+2 B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 C \tan \left (d x +c \right ) a b +A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) b^{2}+C \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(152\) |
| default | \(\frac {a^{2} A \sin \left (d x +c \right )+B \,a^{2} \left (d x +c \right )+C \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 a A b \left (d x +c \right )+2 B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 C \tan \left (d x +c \right ) a b +A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \tan \left (d x +c \right ) b^{2}+C \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\) | \(152\) |
| parallelrisch | \(\frac {-2 \left (\left (A +\frac {C}{2}\right ) b^{2}+2 B a b +C \,a^{2}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \left (\left (A +\frac {C}{2}\right ) b^{2}+2 B a b +C \,a^{2}\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+4 d a x \left (A b +\frac {a B}{2}\right ) \cos \left (2 d x +2 c \right )+\left (2 B \,b^{2}+4 C a b \right ) \sin \left (2 d x +2 c \right )+a^{2} A \sin \left (3 d x +3 c \right )+\left (a^{2} A +2 C \,b^{2}\right ) \sin \left (d x +c \right )+4 d a x \left (A b +\frac {a B}{2}\right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(200\) |
| risch | \(2 a A b x +a^{2} B x -\frac {i a^{2} A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{2} A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i b \left (C b \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B b \,{\mathrm e}^{2 i \left (d x +c \right )}-4 C a \,{\mathrm e}^{2 i \left (d x +c \right )}-C b \,{\mathrm e}^{i \left (d x +c \right )}-2 B b -4 C a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{d}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a b}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,b^{2}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a b}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,b^{2}}{2 d}\) | \(304\) |
| norman | \(\frac {\left (-2 a A b -B \,a^{2}\right ) x +\left (-4 a A b -2 B \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (2 a A b +B \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (4 a A b +2 B \,a^{2}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\frac {\left (2 a^{2} A -2 B \,b^{2}-4 C a b +C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{d}+\frac {\left (6 a^{2} A +2 B \,b^{2}+4 C a b -C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-\frac {\left (2 a^{2} A +2 B \,b^{2}+4 C a b +C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (6 a^{2} A -2 B \,b^{2}-4 C a b -C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {\left (2 A \,b^{2}+4 B a b +2 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {\left (2 A \,b^{2}+4 B a b +2 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(356\) |
1/d*(a^2*A*sin(d*x+c)+B*a^2*(d*x+c)+C*a^2*ln(sec(d*x+c)+tan(d*x+c))+2*a*A* b*(d*x+c)+2*B*a*b*ln(sec(d*x+c)+tan(d*x+c))+2*C*tan(d*x+c)*a*b+A*b^2*ln(se c(d*x+c)+tan(d*x+c))+B*tan(d*x+c)*b^2+C*b^2*(1/2*sec(d*x+c)*tan(d*x+c)+1/2 *ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.29 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.31 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left (B a^{2} + 2 \, A a b\right )} d x \cos \left (d x + c\right )^{2} + {\left (2 \, C a^{2} + 4 \, B a b + {\left (2 \, A + C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, C a^{2} + 4 \, B a b + {\left (2 \, A + C\right )} b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{2} \cos \left (d x + c\right )^{2} + C b^{2} + 2 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
integrate(cos(d*x+c)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")
1/4*(4*(B*a^2 + 2*A*a*b)*d*x*cos(d*x + c)^2 + (2*C*a^2 + 4*B*a*b + (2*A + C)*b^2)*cos(d*x + c)^2*log(sin(d*x + c) + 1) - (2*C*a^2 + 4*B*a*b + (2*A + C)*b^2)*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(2*A*a^2*cos(d*x + c)^2 + C*b^2 + 2*(2*C*a*b + B*b^2)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) ^2)
\[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right )^{2} \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos {\left (c + d x \right )}\, dx \]
Time = 0.22 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.50 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {4 \, {\left (d x + c\right )} B a^{2} + 8 \, {\left (d x + c\right )} A a b - C b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, A b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, A a^{2} \sin \left (d x + c\right ) + 8 \, C a b \tan \left (d x + c\right ) + 4 \, B b^{2} \tan \left (d x + c\right )}{4 \, d} \]
integrate(cos(d*x+c)*(a+b*sec(d*x+c))^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")
1/4*(4*(d*x + c)*B*a^2 + 8*(d*x + c)*A*a*b - C*b^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*C*a^2*( log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 4*B*a*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 2*A*b^2*(log(sin(d*x + c) + 1) - log(sin( d*x + c) - 1)) + 4*A*a^2*sin(d*x + c) + 8*C*a*b*tan(d*x + c) + 4*B*b^2*tan (d*x + c))/d
Time = 0.32 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.91 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\frac {4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} {\left (d x + c\right )} + {\left (2 \, C a^{2} + 4 \, B a b + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (2 \, C a^{2} + 4 \, B a b + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
1/2*(4*A*a^2*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 2*(B*a^2 + 2*A*a*b)*(d*x + c) + (2*C*a^2 + 4*B*a*b + 2*A*b^2 + C*b^2)*log(abs(tan(1 /2*d*x + 1/2*c) + 1)) - (2*C*a^2 + 4*B*a*b + 2*A*b^2 + C*b^2)*log(abs(tan( 1/2*d*x + 1/2*c) - 1)) - 2*(4*C*a*b*tan(1/2*d*x + 1/2*c)^3 + 2*B*b^2*tan(1 /2*d*x + 1/2*c)^3 - C*b^2*tan(1/2*d*x + 1/2*c)^3 - 4*C*a*b*tan(1/2*d*x + 1 /2*c) - 2*B*b^2*tan(1/2*d*x + 1/2*c) - C*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/ 2*d*x + 1/2*c)^2 - 1)^2)/d
Time = 18.03 (sec) , antiderivative size = 257, normalized size of antiderivative = 2.04 \[ \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,\left (B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+A\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+\frac {C\,b^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{2}+2\,A\,a\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+2\,B\,a\,b\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\right )}{d}+\frac {\frac {A\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{4}+\frac {B\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{4}+\frac {C\,b^2\,\sin \left (c+d\,x\right )}{2}+C\,a\,b\,\sin \left (2\,c+2\,d\,x\right )}{d\,\left (\frac {\cos \left (2\,c+2\,d\,x\right )}{2}+\frac {1}{2}\right )} \]
(2*(B*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + A*b^2*atanh(sin(c/ 2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + C*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + (C*b^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + 2*A *a*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 2*B*a*b*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))))/d + ((A*a^2*sin(3*c + 3*d*x))/4 + (B*b^2*s in(2*c + 2*d*x))/2 + (A*a^2*sin(c + d*x))/4 + (C*b^2*sin(c + d*x))/2 + C*a *b*sin(2*c + 2*d*x))/(d*(cos(2*c + 2*d*x)/2 + 1/2))