Integrand size = 25, antiderivative size = 125 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {A x}{a^2}-\frac {2 b \left (2 a^2 A-A b^2+a^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {\left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \]
A*x/a^2-2*b*(2*A*a^2-A*b^2+C*a^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/( a+b)^(1/2))/a^2/(a-b)^(3/2)/(a+b)^(3/2)/d+(A*b^2+C*a^2)*tan(d*x+c)/a/(a^2- b^2)/d/(a+b*sec(d*x+c))
Result contains complex when optimal does not.
Time = 2.81 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.16 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {2 (b+a \cos (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \left (A x (b+a \cos (c+d x))+\frac {2 b \left (-A b^2+a^2 (2 A+C)\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (a \sin (c)+(-b+a \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right ) (b+a \cos (c+d x)) (i \cos (c)+\sin (c))}{\left (a^2-b^2\right )^{3/2} d \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {\left (A b^2+a^2 C\right ) (-b \sin (c)+a \sin (d x))}{(a-b) (a+b) d \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )}\right )}{a^2 (A+2 C+A \cos (2 (c+d x))) (a+b \sec (c+d x))^2} \]
(2*(b + a*Cos[c + d*x])*(A + C*Sec[c + d*x]^2)*(A*x*(b + a*Cos[c + d*x]) + (2*b*(-(A*b^2) + a^2*(2*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] )]*(b + a*Cos[c + d*x])*(I*Cos[c] + Sin[c]))/((a^2 - b^2)^(3/2)*d*Sqrt[(Co s[c] - I*Sin[c])^2]) + ((A*b^2 + a^2*C)*(-(b*Sin[c]) + a*Sin[d*x]))/((a - b)*(a + b)*d*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2]))))/(a^2*(A + 2*C + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^2)
Time = 0.68 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.18, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4549, 25, 3042, 4407, 3042, 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.
| \(\displaystyle \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 4549 |
| \(\displaystyle \frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {\int -\frac {A \left (a^2-b^2\right )-a b (A+C) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a \left (a^2-b^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {A \left (a^2-b^2\right )-a b (A+C) \sec (c+d x)}{a+b \sec (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {A \left (a^2-b^2\right )-a b (A+C) \csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4407 |
| \(\displaystyle \frac {\frac {b \left (A b^2-a^2 (2 A+C)\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a}+\frac {A x \left (a^2-b^2\right )}{a}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {b \left (A b^2-a^2 (2 A+C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a}+\frac {A x \left (a^2-b^2\right )}{a}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 4318 |
| \(\displaystyle \frac {\frac {\left (A b^2-a^2 (2 A+C)\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a}+\frac {A x \left (a^2-b^2\right )}{a}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\left (A b^2-a^2 (2 A+C)\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a}+\frac {A x \left (a^2-b^2\right )}{a}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {2 \left (A b^2-a^2 (2 A+C)\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}+\frac {A x \left (a^2-b^2\right )}{a}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {2 b \left (A b^2-a^2 (2 A+C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}+\frac {A x \left (a^2-b^2\right )}{a}}{a \left (a^2-b^2\right )}+\frac {\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}\) |
((A*(a^2 - b^2)*x)/a + (2*b*(A*b^2 - a^2*(2*A + C))*ArcTanh[(Sqrt[a - b]*T an[(c + d*x)/2])/Sqrt[a + b]])/(a*Sqrt[a - b]*Sqrt[a + b]*d))/(a*(a^2 - b^ 2)) + ((A*b^2 + a^2*C)*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x]) )
3.7.87.3.1 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[(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_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a Int[Csc[e + f* x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_. ) + (a_))^(m_), x_Symbol] :> Simp[(A*b^2 + a^2*C)*Cot[e + f*x]*((a + b*Csc[ e + f*x])^(m + 1)/(a*f*(m + 1)*(a^2 - b^2))), x] + Simp[1/(a*(m + 1)*(a^2 - b^2)) Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*(a^2 - b^2)*(m + 1) - a*b* (A + C)*(m + 1)*Csc[e + f*x] + (A*b^2 + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m ] && LtQ[m, -1]
Time = 0.25 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.35
| method | result | size |
| derivativedivides | \(\frac {\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {-\frac {2 \left (A \,b^{2}+C \,a^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {2 \left (2 a^{2} A -A \,b^{2}+C \,a^{2}\right ) b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{a^{2}}}{d}\) | \(169\) |
| default | \(\frac {\frac {2 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {-\frac {2 \left (A \,b^{2}+C \,a^{2}\right ) a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {2 \left (2 a^{2} A -A \,b^{2}+C \,a^{2}\right ) b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{a^{2}}}{d}\) | \(169\) |
| risch | \(\frac {A x}{a^{2}}+\frac {2 i \left (A \,b^{2}+C \,a^{2}\right ) \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{a^{2} \left (a^{2}-b^{2}\right ) d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}\) | \(579\) |
1/d*(2*A/a^2*arctan(tan(1/2*d*x+1/2*c))+2/a^2*(-(A*b^2+C*a^2)*a/(a^2-b^2)* tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)-(2* A*a^2-A*b^2+C*a^2)*b/(a+b)/(a-b)/((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. 241 vs. \(2 (116) = 232\).
Time = 0.28 (sec) , antiderivative size = 543, normalized size of antiderivative = 4.34 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\left [\frac {2 \, {\left (A a^{5} - 2 \, A a^{3} b^{2} + A a b^{4}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (A a^{4} b - 2 \, A a^{2} b^{3} + A b^{5}\right )} d x - {\left ({\left (2 \, A + C\right )} a^{2} b^{2} - A b^{4} + {\left ({\left (2 \, A + C\right )} a^{3} b - A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + 2 \, {\left (C a^{5} + {\left (A - C\right )} a^{3} b^{2} - A a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}}, \frac {{\left (A a^{5} - 2 \, A a^{3} b^{2} + A a b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (A a^{4} b - 2 \, A a^{2} b^{3} + A b^{5}\right )} d x - {\left ({\left (2 \, A + C\right )} a^{2} b^{2} - A b^{4} + {\left ({\left (2 \, A + C\right )} a^{3} b - A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) + {\left (C a^{5} + {\left (A - C\right )} a^{3} b^{2} - A a b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}\right ] \]
[1/2*(2*(A*a^5 - 2*A*a^3*b^2 + A*a*b^4)*d*x*cos(d*x + c) + 2*(A*a^4*b - 2* A*a^2*b^3 + A*b^5)*d*x - ((2*A + C)*a^2*b^2 - A*b^4 + ((2*A + C)*a^3*b - A *a*b^3)*cos(d*x + c))*sqrt(a^2 - b^2)*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)) + 2*(C*a^5 + (A - C)*a^3*b^2 - A*a*b^4)*sin(d*x + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d* cos(d*x + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d), ((A*a^5 - 2*A*a^3*b^2 + A *a*b^4)*d*x*cos(d*x + c) + (A*a^4*b - 2*A*a^2*b^3 + A*b^5)*d*x - ((2*A + C )*a^2*b^2 - A*b^4 + ((2*A + C)*a^3*b - A*a*b^3)*cos(d*x + c))*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))) + (C*a^5 + (A - C)*a^3*b^2 - A*a*b^4)*sin(d*x + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d)]
\[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\int \frac {A + C \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
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 = 0.31 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.64 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (2 \, A a^{2} b + C a^{2} b - A b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {{\left (d x + c\right )} A}{a^{2}} + \frac {2 \, {\left (C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{3} - a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}}}{d} \]
-(2*(2*A*a^2*b + C*a^2*b - A*b^3)*(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)))/((a^4 - a^2*b^2)*sqrt(-a^2 + b^2)) - (d*x + c)*A/a^2 + 2*(C *a^2*tan(1/2*d*x + 1/2*c) + A*b^2*tan(1/2*d*x + 1/2*c))/((a^3 - a*b^2)*(a* tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)))/d
Time = 24.13 (sec) , antiderivative size = 3850, normalized size of antiderivative = 30.80 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \]
(2*A*atan(((A*((A*((32*(A*a^4*b^5 - A*a^9 - 3*A*a^6*b^3 + A*a^7*b^2 + C*a^ 5*b^4 - C*a^6*b^3 - C*a^7*b^2 + 2*A*a^8*b + C*a^8*b))/(a^5*b + a^6 - a^3*b ^3 - a^4*b^2) - (A*tan(c/2 + (d*x)/2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4 *a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2)*32i)/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b ^2)))*1i)/a^2 + (32*tan(c/2 + (d*x)/2)*(A^2*a^6 + 2*A^2*b^6 - 2*A^2*a*b^5 - 2*A^2*a^5*b - 5*A^2*a^2*b^4 + 4*A^2*a^3*b^3 + 3*A^2*a^4*b^2 + C^2*a^4*b^ 2 - 2*A*C*a^2*b^4 + 4*A*C*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))/a^ 2 - (A*((A*((32*(A*a^4*b^5 - A*a^9 - 3*A*a^6*b^3 + A*a^7*b^2 + C*a^5*b^4 - C*a^6*b^3 - C*a^7*b^2 + 2*A*a^8*b + C*a^8*b))/(a^5*b + a^6 - a^3*b^3 - a^ 4*b^2) + (A*tan(c/2 + (d*x)/2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^ 4 - 4*a^7*b^3 - 2*a^8*b^2)*32i)/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))*1 i)/a^2 - (32*tan(c/2 + (d*x)/2)*(A^2*a^6 + 2*A^2*b^6 - 2*A^2*a*b^5 - 2*A^2 *a^5*b - 5*A^2*a^2*b^4 + 4*A^2*a^3*b^3 + 3*A^2*a^4*b^2 + C^2*a^4*b^2 - 2*A *C*a^2*b^4 + 4*A*C*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))/a^2)/((64 *(A^3*b^5 - A^3*a*b^4 + 2*A^3*a^4*b - 3*A^3*a^2*b^3 + 2*A^3*a^3*b^2 - A^2* C*a*b^4 + A^2*C*a^4*b + A*C^2*a^3*b^2 - A^2*C*a^2*b^3 + 3*A^2*C*a^3*b^2))/ (a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (A*((A*((32*(A*a^4*b^5 - A*a^9 - 3*A*a ^6*b^3 + A*a^7*b^2 + C*a^5*b^4 - C*a^6*b^3 - C*a^7*b^2 + 2*A*a^8*b + C*a^8 *b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (A*tan(c/2 + (d*x)/2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2)*32i)/(a^2*(a...