Integrand size = 19, antiderivative size = 118 \[ \int \frac {\sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=-\frac {2 b \left (2 a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {\text {arctanh}(\sin (c+d x))}{a^2 d}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \cos (c+d x))} \]
-2*b*(2*a^2-b^2)*arctan((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+arctanh(sin(d*x+c))/a^2/d+b^2*sin(d*x+c)/a/(a^2-b^ 2)/d/(a+b*cos(d*x+c))
Time = 0.40 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.24 \[ \int \frac {\sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {2 b \left (-2 a^2+b^2\right ) \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2+b^2}}\right )}{\left (-a^2+b^2\right )^{3/2}}-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {a b^2 \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))}}{a^2 d} \]
((2*b*(-2*a^2 + b^2)*ArcTanh[((a - b)*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]]) /(-a^2 + b^2)^(3/2) - Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + Log[Cos[( c + d*x)/2] + Sin[(c + d*x)/2]] + (a*b^2*Sin[c + d*x])/((a - b)*(a + b)*(a + b*Cos[c + d*x])))/(a^2*d)
Time = 0.65 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.21, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {3042, 3281, 3042, 3480, 3042, 3138, 218, 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.
\(\displaystyle \int \frac {\sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 3281 |
\(\displaystyle \frac {\int \frac {\left (a^2-b \cos (c+d x) a-b^2\right ) \sec (c+d x)}{a+b \cos (c+d x)}dx}{a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a^2-b \sin \left (c+d x+\frac {\pi }{2}\right ) a-b^2}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3480 |
\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \int \sec (c+d x)dx}{a}-\frac {b \left (2 a^2-b^2\right ) \int \frac {1}{a+b \cos (c+d x)}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {b \left (2 a^2-b^2\right ) \int \frac {1}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}}{a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \left (2 a^2-b^2\right ) \int \frac {1}{(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}d\tan \left (\frac {1}{2} (c+d x)\right )}{a d}}{a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{a}-\frac {2 b \left (2 a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\frac {\left (a^2-b^2\right ) \text {arctanh}(\sin (c+d x))}{a d}-\frac {2 b \left (2 a^2-b^2\right ) \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a d \sqrt {a-b} \sqrt {a+b}}}{a \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \cos (c+d x))}\) |
((-2*b*(2*a^2 - b^2)*ArcTan[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/( a*Sqrt[a - b]*Sqrt[a + b]*d) + ((a^2 - b^2)*ArcTanh[Sin[c + d*x]])/(a*d))/ (a*(a^2 - b^2)) + (b^2*Sin[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Cos[c + d*x]) )
3.5.65.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2 ))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] && !IntegerQ[n]) || !(IntegerQ[2* n] && LtQ[n, -1] && ((IntegerQ[n] && !IntegerQ[m]) || EqQ[a, 0])))
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_ .)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(A*b - a*B)/(b*c - a*d) Int[1/(a + b*Sin[e + f*x]), x], x] + Simp[(B*c - A*d)/ (b*c - a*d) Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f , A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 1.49 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.39
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}-\frac {2 b \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}}{d}\) | \(164\) |
default | \(\frac {-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2}}-\frac {2 b \left (-\frac {a b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b \right )}+\frac {\left (2 a^{2}-b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{\left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2}}}{d}\) | \(164\) |
risch | \(-\frac {2 i b \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{\left (-a^{2}+b^{2}\right ) d a \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}+\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}\) | \(441\) |
1/d*(-1/a^2*ln(tan(1/2*d*x+1/2*c)-1)-2*b/a^2*(-a*b/(a^2-b^2)*tan(1/2*d*x+1 /2*c)/(tan(1/2*d*x+1/2*c)^2*a-b*tan(1/2*d*x+1/2*c)^2+a+b)+(2*a^2-b^2)/(a-b )/(a+b)/((a-b)*(a+b))^(1/2)*arctan((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^ (1/2)))+1/a^2*ln(tan(1/2*d*x+1/2*c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 262 vs. \(2 (109) = 218\).
Time = 0.47 (sec) , antiderivative size = 592, normalized size of antiderivative = 5.02 \[ \int \frac {\sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\left [-\frac {{\left (2 \, a^{3} b - a b^{3} + {\left (2 \, a^{2} b^{2} - b^{4}\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 (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a^{2} + b^{2}} {\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d\right )}}, -\frac {2 \, {\left (2 \, a^{3} b - a b^{3} + {\left (2 \, a^{2} b^{2} - b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \cos \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) - {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d \cos \left (d x + c\right ) + {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d\right )}}\right ] \]
[-1/2*((2*a^3*b - a*b^3 + (2*a^2*b^2 - b^4)*cos(d*x + c))*sqrt(-a^2 + b^2) *log((2*a*b*cos(d*x + c) + (2*a^2 - b^2)*cos(d*x + c)^2 - 2*sqrt(-a^2 + b^ 2)*(a*cos(d*x + c) + b)*sin(d*x + c) - a^2 + 2*b^2)/(b^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + a^2)) - (a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c))*log(sin(d*x + c) + 1) + (a^5 - 2*a^3*b^2 + a*b^4 + ( a^4*b - 2*a^2*b^3 + b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(a^3*b^2 - a*b^4)*sin(d*x + c))/((a^6*b - 2*a^4*b^3 + a^2*b^5)*d*cos(d*x + c) + (a ^7 - 2*a^5*b^2 + a^3*b^4)*d), -1/2*(2*(2*a^3*b - a*b^3 + (2*a^2*b^2 - b^4) *cos(d*x + c))*sqrt(a^2 - b^2)*arctan(-(a*cos(d*x + c) + b)/(sqrt(a^2 - b^ 2)*sin(d*x + c))) - (a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)*c os(d*x + c))*log(sin(d*x + c) + 1) + (a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2 *a^2*b^3 + b^5)*cos(d*x + c))*log(-sin(d*x + c) + 1) - 2*(a^3*b^2 - a*b^4) *sin(d*x + c))/((a^6*b - 2*a^4*b^3 + a^2*b^5)*d*cos(d*x + c) + (a^7 - 2*a^ 5*b^2 + a^3*b^4)*d)]
\[ \int \frac {\sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\int \frac {\sec {\left (c + d x \right )}}{\left (a + b \cos {\left (c + d x \right )}\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
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*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.31 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.68 \[ \int \frac {\sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\frac {\frac {2 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{3} - a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}} - \frac {2 \, {\left (2 \, a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}}}{d} \]
(2*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*t an(1/2*d*x + 1/2*c)^2 + a + b)) - 2*(2*a^2*b - 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)) + log(abs( tan(1/2*d*x + 1/2*c) + 1))/a^2 - log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^2)/d
Time = 20.08 (sec) , antiderivative size = 2886, normalized size of antiderivative = 24.46 \[ \int \frac {\sec (c+d x)}{(a+b \cos (c+d x))^2} \, dx=\text {Too large to display} \]
- (atan((((((32*(2*a^8*b - a^9 + a^4*b^5 - 3*a^6*b^3 + a^7*b^2))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (32*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))/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))/a^2 - (32*tan(c/2 + (d*x)/2)*(a^6 - 2*a^5*b - 2*a*b^5 + 2*b^ 6 - 5*a^2*b^4 + 4*a^3*b^3 + 3*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2)) *1i)/a^2 - ((((32*(2*a^8*b - a^9 + a^4*b^5 - 3*a^6*b^3 + a^7*b^2))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (32*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))/(a^2*(a^4*b + a^5 - a^2*b ^3 - a^3*b^2)))/a^2 + (32*tan(c/2 + (d*x)/2)*(a^6 - 2*a^5*b - 2*a*b^5 + 2* b^6 - 5*a^2*b^4 + 4*a^3*b^3 + 3*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2 ))*1i)/a^2)/((((32*(2*a^8*b - a^9 + a^4*b^5 - 3*a^6*b^3 + a^7*b^2))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (32*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))/(a^2*(a^4*b + a^5 - a^2* b^3 - a^3*b^2)))/a^2 - (32*tan(c/2 + (d*x)/2)*(a^6 - 2*a^5*b - 2*a*b^5 + 2 *b^6 - 5*a^2*b^4 + 4*a^3*b^3 + 3*a^4*b^2))/(a^4*b + a^5 - a^2*b^3 - a^3*b^ 2))/a^2 - (64*(2*a^4*b - a*b^4 + b^5 - 3*a^2*b^3 + 2*a^3*b^2))/(a^5*b + a^ 6 - a^3*b^3 - a^4*b^2) + (((32*(2*a^8*b - a^9 + a^4*b^5 - 3*a^6*b^3 + a^7* b^2))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (32*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))/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))/a^2 + (32*tan(c/2 + (d*x)/2)*(a^6 - 2*a^5*...