Integrand size = 39, antiderivative size = 106 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {(b B-a C) \text {arctanh}(\sin (c+d x))}{b^2 d}+\frac {2 \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b^2 \sqrt {a+b} d}+\frac {C \tan (c+d x)}{b d} \]
(B*b-C*a)*arctanh(sin(d*x+c))/b^2/d+2*(A*b^2-a*(B*b-C*a))*arctanh((a-b)^(1 /2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/b^2/d/(a-b)^(1/2)/(a+b)^(1/2)+C*tan(d* x+c)/b/d
Result contains complex when optimal does not.
Time = 5.93 (sec) , antiderivative size = 365, normalized size of antiderivative = 3.44 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\frac {2 \cos (c+d x) (b+a \cos (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-\left ((b B-a C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+(b B-a C) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-\frac {2 i \left (A b^2+a (-b B+a C)\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (a \sin (c)+(-b+a \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right ) (\cos (c)-i \sin (c))}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {b C \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {b C \sin \left (\frac {d x}{2}\right )}{\left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )}{b^2 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a+b \sec (c+d x))} \]
(2*Cos[c + d*x]*(b + a*Cos[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^ 2)*(-((b*B - a*C)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]) + (b*B - a*C)* Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - ((2*I)*(A*b^2 + a*(-(b*B) + a*C ))*ArcTan[((I*Cos[c] + Sin[c])*(a*Sin[c] + (-b + a*Cos[c])*Tan[(d*x)/2]))/ (Sqrt[a^2 - b^2]*Sqrt[(Cos[c] - I*Sin[c])^2])]*(Cos[c] - I*Sin[c]))/(Sqrt[ a^2 - b^2]*Sqrt[(Cos[c] - I*Sin[c])^2]) + (b*C*Sin[(d*x)/2])/((Cos[c/2] - Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])) + (b*C*Sin[(d*x)/2])/((Co s[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))))/(b^2*d*(A + 2* C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x]))
Time = 0.69 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {3042, 4570, 3042, 4486, 3042, 4257, 4318, 3042, 3138, 221}
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.
| \(\displaystyle \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4570 |
| \(\displaystyle \frac {\int \frac {\sec (c+d x) (A b+(b B-a C) \sec (c+d x))}{a+b \sec (c+d x)}dx}{b}+\frac {C \tan (c+d x)}{b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (A b+(b B-a C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {C \tan (c+d x)}{b d}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle \frac {\frac {\left (A b^2-a (b B-a C)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b}+\frac {(b B-a C) \int \sec (c+d x)dx}{b}}{b}+\frac {C \tan (c+d x)}{b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (A b^2-a (b B-a C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {(b B-a C) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}}{b}+\frac {C \tan (c+d x)}{b d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {\frac {\left (A b^2-a (b B-a C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}+\frac {(b B-a C) \text {arctanh}(\sin (c+d x))}{b d}}{b}+\frac {C \tan (c+d x)}{b d}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {\frac {\left (A b^2-a (b B-a C)\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b^2}+\frac {(b B-a C) \text {arctanh}(\sin (c+d x))}{b d}}{b}+\frac {C \tan (c+d x)}{b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (A b^2-a (b B-a C)\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b^2}+\frac {(b B-a C) \text {arctanh}(\sin (c+d x))}{b d}}{b}+\frac {C \tan (c+d x)}{b d}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {2 \left (A b^2-a (b B-a C)\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b^2 d}+\frac {(b B-a C) \text {arctanh}(\sin (c+d x))}{b d}}{b}+\frac {C \tan (c+d x)}{b d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {2 \left (A b^2-a (b B-a C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}+\frac {(b B-a C) \text {arctanh}(\sin (c+d x))}{b d}}{b}+\frac {C \tan (c+d x)}{b d}\) |
(((b*B - a*C)*ArcTanh[Sin[c + d*x]])/(b*d) + (2*(A*b^2 - a*(b*B - a*C))*Ar cTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d))/b + (C*Tan[c + d*x])/(b*d)
3.10.2.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[( e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[B/b Int[Csc[e + f*x], x], x] + Simp[(A*b - a*B)/b Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x ] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e _.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) )), x] + Simp[1/(b*(m + 2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Time = 0.40 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.43
| method | result | size |
| derivativedivides | \(\frac {-\frac {C}{b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (B b -C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}-\frac {C}{b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}-\frac {2 \left (-A \,b^{2}+B a b -C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(152\) |
| default | \(\frac {-\frac {C}{b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (B b -C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}-\frac {C}{b \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (-B b +C a \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}-\frac {2 \left (-A \,b^{2}+B a b -C \,a^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{2} \sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(152\) |
| risch | \(\frac {2 i C}{d b \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B a}{\sqrt {a^{2}-b^{2}}\, d b}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C \,a^{2}}{\sqrt {a^{2}-b^{2}}\, d \,b^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B a}{\sqrt {a^{2}-b^{2}}\, d b}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C \,a^{2}}{\sqrt {a^{2}-b^{2}}\, d \,b^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{b d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{b^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{b d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{b^{2} d}\) | \(550\) |
1/d*(-C/b/(tan(1/2*d*x+1/2*c)+1)+(B*b-C*a)/b^2*ln(tan(1/2*d*x+1/2*c)+1)-C/ b/(tan(1/2*d*x+1/2*c)-1)+1/b^2*(-B*b+C*a)*ln(tan(1/2*d*x+1/2*c)-1)-2/b^2*( -A*b^2+B*a*b-C*a^2)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/( (a+b)*(a-b))^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (97) = 194\).
Time = 2.75 (sec) , antiderivative size = 482, normalized size of antiderivative = 4.55 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\left [\frac {{\left (C a^{2} - B a b + A b^{2}\right )} \sqrt {a^{2} - b^{2}} \cos \left (d x + c\right ) \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a^{2} b - C b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right )}, \frac {2 \, {\left (C a^{2} - B a b + A b^{2}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (C a^{3} - B a^{2} b - C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C a^{2} b - C b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d \cos \left (d x + c\right )}\right ] \]
[1/2*((C*a^2 - B*a*b + A*b^2)*sqrt(a^2 - b^2)*cos(d*x + c)*log((2*a*b*cos( d*x + c) - (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c ) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c ) + b^2)) - (C*a^3 - B*a^2*b - C*a*b^2 + B*b^3)*cos(d*x + c)*log(sin(d*x + c) + 1) + (C*a^3 - B*a^2*b - C*a*b^2 + B*b^3)*cos(d*x + c)*log(-sin(d*x + c) + 1) + 2*(C*a^2*b - C*b^3)*sin(d*x + c))/((a^2*b^2 - b^4)*d*cos(d*x + c)), 1/2*(2*(C*a^2 - B*a*b + A*b^2)*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b ^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c)))*cos(d*x + c) - (C*a^3 - B*a^2*b - C*a*b^2 + B*b^3)*cos(d*x + c)*log(sin(d*x + c) + 1) + (C*a^3 - B*a^2*b - C*a*b^2 + B*b^3)*cos(d*x + c)*log(-sin(d*x + c) + 1) + 2*(C*a^ 2*b - C*b^3)*sin(d*x + c))/((a^2*b^2 - b^4)*d*cos(d*x + c))]
\[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Time = 0.34 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.70 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=-\frac {\frac {{\left (C a - B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} - \frac {{\left (C a - B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}} + \frac {2 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} b} + \frac {2 \, {\left (C a^{2} - B a b + A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} b^{2}}}{d} \]
-((C*a - B*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^2 - (C*a - B*b)*log(abs (tan(1/2*d*x + 1/2*c) - 1))/b^2 + 2*C*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 - 1)*b) + 2*(C*a^2 - B*a*b + A*b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*d*x + 1/2*c) - b*tan(1/2*d*x + 1/ 2*c))/sqrt(-a^2 + b^2)))/(sqrt(-a^2 + b^2)*b^2))/d
Time = 23.78 (sec) , antiderivative size = 3452, normalized size of antiderivative = 32.57 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx=\text {Too large to display} \]
- (atan(-(((B*b - C*a)*(((B*b - C*a)*((32*(A*b^7 + B*b^7 + A*a^2*b^5 + B*a ^2*b^5 + 2*C*a^2*b^5 - C*a^3*b^4 - 2*A*a*b^6 - 2*B*a*b^6 - C*a*b^6))/b^3 - (32*tan(c/2 + (d*x)/2)*(B*b - C*a)*(2*a*b^6 - 4*a^2*b^5 + 2*a^3*b^4))/b^4 ))/b^2 - (32*tan(c/2 + (d*x)/2)*(A^2*b^5 + B^2*b^5 - 2*C^2*a^5 - A^2*a*b^4 - 3*B^2*a*b^4 + 4*C^2*a^4*b + 4*B^2*a^2*b^3 - 2*B^2*a^3*b^2 + C^2*a^2*b^3 - 3*C^2*a^3*b^2 - 2*A*B*a*b^4 - 2*B*C*a*b^4 + 4*B*C*a^4*b + 2*A*B*a^2*b^3 + 2*A*C*a^2*b^3 - 2*A*C*a^3*b^2 + 6*B*C*a^2*b^3 - 8*B*C*a^3*b^2))/b^2)*1i )/b^2 - ((B*b - C*a)*(((B*b - C*a)*((32*(A*b^7 + B*b^7 + A*a^2*b^5 + B*a^2 *b^5 + 2*C*a^2*b^5 - C*a^3*b^4 - 2*A*a*b^6 - 2*B*a*b^6 - C*a*b^6))/b^3 + ( 32*tan(c/2 + (d*x)/2)*(B*b - C*a)*(2*a*b^6 - 4*a^2*b^5 + 2*a^3*b^4))/b^4)) /b^2 + (32*tan(c/2 + (d*x)/2)*(A^2*b^5 + B^2*b^5 - 2*C^2*a^5 - A^2*a*b^4 - 3*B^2*a*b^4 + 4*C^2*a^4*b + 4*B^2*a^2*b^3 - 2*B^2*a^3*b^2 + C^2*a^2*b^3 - 3*C^2*a^3*b^2 - 2*A*B*a*b^4 - 2*B*C*a*b^4 + 4*B*C*a^4*b + 2*A*B*a^2*b^3 + 2*A*C*a^2*b^3 - 2*A*C*a^3*b^2 + 6*B*C*a^2*b^3 - 8*B*C*a^3*b^2))/b^2)*1i)/ b^2)/((64*(C^3*a^5 - A*B^2*b^5 + A^2*B*b^5 + B^3*a*b^4 - C^3*a^4*b - B^3*a ^2*b^3 - A^2*B*a*b^4 + A*C^2*a^4*b - A^2*C*a*b^4 - 3*B*C^2*a^4*b + A*B^2*a ^2*b^3 - A*C^2*a^2*b^3 + A^2*C*a^2*b^3 + 3*B*C^2*a^3*b^2 - 3*B^2*C*a^2*b^3 + 3*B^2*C*a^3*b^2 + 2*A*B*C*a*b^4 - 2*A*B*C*a^3*b^2))/b^3 + ((B*b - C*a)* (((B*b - C*a)*((32*(A*b^7 + B*b^7 + A*a^2*b^5 + B*a^2*b^5 + 2*C*a^2*b^5 - C*a^3*b^4 - 2*A*a*b^6 - 2*B*a*b^6 - C*a*b^6))/b^3 - (32*tan(c/2 + (d*x)...