Integrand size = 19, antiderivative size = 334 \[ \int \frac {\sec ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=-\frac {2 c \left (b^3-3 a b c+\sqrt {b^2-4 a c} \left (b^2-a c\right )\right ) \arctan \left (\frac {\sqrt {b-2 c-\sqrt {b^2-4 a c}} \tan \left (\frac {x}{2}\right )}{\sqrt {b+2 c-\sqrt {b^2-4 a c}}}\right )}{a^3 \sqrt {b^2-4 a c} \sqrt {b-2 c-\sqrt {b^2-4 a c}} \sqrt {b+2 c-\sqrt {b^2-4 a c}}}+\frac {2 c \left (b^3-3 a b c-\sqrt {b^2-4 a c} \left (b^2-a c\right )\right ) \arctan \left (\frac {\sqrt {b-2 c+\sqrt {b^2-4 a c}} \tan \left (\frac {x}{2}\right )}{\sqrt {b+2 c+\sqrt {b^2-4 a c}}}\right )}{a^3 \sqrt {b^2-4 a c} \sqrt {b-2 c+\sqrt {b^2-4 a c}} \sqrt {b+2 c+\sqrt {b^2-4 a c}}}+\frac {\text {arctanh}(\sin (x))}{2 a}+\frac {\left (b^2-a c\right ) \text {arctanh}(\sin (x))}{a^3}-\frac {b \tan (x)}{a^2}+\frac {\sec (x) \tan (x)}{2 a} \]
1/2*arctanh(sin(x))/a+(-a*c+b^2)*arctanh(sin(x))/a^3-2*c*arctan((b-2*c-(-4 *a*c+b^2)^(1/2))^(1/2)*tan(1/2*x)/(b+2*c-(-4*a*c+b^2)^(1/2))^(1/2))*(b^3-3 *a*b*c+(-a*c+b^2)*(-4*a*c+b^2)^(1/2))/a^3/(-4*a*c+b^2)^(1/2)/(b-2*c-(-4*a* c+b^2)^(1/2))^(1/2)/(b+2*c-(-4*a*c+b^2)^(1/2))^(1/2)+2*c*arctan((b-2*c+(-4 *a*c+b^2)^(1/2))^(1/2)*tan(1/2*x)/(b+2*c+(-4*a*c+b^2)^(1/2))^(1/2))*(b^3-3 *a*b*c-(-a*c+b^2)*(-4*a*c+b^2)^(1/2))/a^3/(-4*a*c+b^2)^(1/2)/(b-2*c+(-4*a* c+b^2)^(1/2))^(1/2)/(b+2*c+(-4*a*c+b^2)^(1/2))^(1/2)-b*tan(x)/a^2+1/2*sec( x)*tan(x)/a
Time = 3.61 (sec) , antiderivative size = 446, normalized size of antiderivative = 1.34 \[ \int \frac {\sec ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=-\frac {\frac {4 \sqrt {2} c \left (b^3-3 a b c-b^2 \sqrt {b^2-4 a c}+a c \sqrt {b^2-4 a c}\right ) \text {arctanh}\left (\frac {\left (b-2 c+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 c (a+c)-2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {-b^2+2 c (a+c)-b \sqrt {b^2-4 a c}}}+\frac {4 \sqrt {2} c \left (b^3-3 a b c+b^2 \sqrt {b^2-4 a c}-a c \sqrt {b^2-4 a c}\right ) \text {arctanh}\left (\frac {\left (-b+2 c+\sqrt {b^2-4 a c}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {-2 b^2+4 c (a+c)+2 b \sqrt {b^2-4 a c}}}\right )}{\sqrt {b^2-4 a c} \sqrt {-b^2+2 c (a+c)+b \sqrt {b^2-4 a c}}}+2 \left (a^2+2 b^2-2 a c\right ) \log \left (\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )\right )-2 \left (a^2+2 b^2-2 a c\right ) \log \left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )+\frac {4 a b \sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )-\sin \left (\frac {x}{2}\right )}+\frac {a^2}{\left (\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )\right )^2}+\frac {4 a b \sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )+\sin \left (\frac {x}{2}\right )}+\frac {a^2}{-1+\sin (x)}}{4 a^3} \]
-1/4*((4*Sqrt[2]*c*(b^3 - 3*a*b*c - b^2*Sqrt[b^2 - 4*a*c] + a*c*Sqrt[b^2 - 4*a*c])*ArcTanh[((b - 2*c + Sqrt[b^2 - 4*a*c])*Tan[x/2])/Sqrt[-2*b^2 + 4* c*(a + c) - 2*b*Sqrt[b^2 - 4*a*c]]])/(Sqrt[b^2 - 4*a*c]*Sqrt[-b^2 + 2*c*(a + c) - b*Sqrt[b^2 - 4*a*c]]) + (4*Sqrt[2]*c*(b^3 - 3*a*b*c + b^2*Sqrt[b^2 - 4*a*c] - a*c*Sqrt[b^2 - 4*a*c])*ArcTanh[((-b + 2*c + Sqrt[b^2 - 4*a*c]) *Tan[x/2])/Sqrt[-2*b^2 + 4*c*(a + c) + 2*b*Sqrt[b^2 - 4*a*c]]])/(Sqrt[b^2 - 4*a*c]*Sqrt[-b^2 + 2*c*(a + c) + b*Sqrt[b^2 - 4*a*c]]) + 2*(a^2 + 2*b^2 - 2*a*c)*Log[Cos[x/2] - Sin[x/2]] - 2*(a^2 + 2*b^2 - 2*a*c)*Log[Cos[x/2] + Sin[x/2]] + (4*a*b*Sin[x/2])/(Cos[x/2] - Sin[x/2]) + a^2/(Cos[x/2] + Sin[ x/2])^2 + (4*a*b*Sin[x/2])/(Cos[x/2] + Sin[x/2]) + a^2/(-1 + Sin[x]))/a^3
Time = 3.07 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3042, 3738, 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.
| \(\displaystyle \int \frac {\sec ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (x)^3 \left (a+b \cos (x)+c \cos (x)^2\right )}dx\) |
| \(\Big \downarrow \) 3738 |
| \(\displaystyle \int \left (\frac {\sec (x) \left (b^2-a c\right )}{a^3}+\frac {-b^2 c \cos (x) \left (1-\frac {a c}{b^2}\right )-\left (b^3 \left (1-\frac {2 a c}{b^2}\right )\right )}{a^3 \left (a+b \cos (x)+c \cos ^2(x)\right )}-\frac {b \sec ^2(x)}{a^2}+\frac {\sec ^3(x)}{a}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 c \left (\sqrt {b^2-4 a c} \left (b^2-a c\right )-3 a b c+b^3\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \sqrt {-\sqrt {b^2-4 a c}+b-2 c}}{\sqrt {-\sqrt {b^2-4 a c}+b+2 c}}\right )}{a^3 \sqrt {b^2-4 a c} \sqrt {-\sqrt {b^2-4 a c}+b-2 c} \sqrt {-\sqrt {b^2-4 a c}+b+2 c}}+\frac {2 c \left (-\sqrt {b^2-4 a c} \left (b^2-a c\right )-3 a b c+b^3\right ) \arctan \left (\frac {\tan \left (\frac {x}{2}\right ) \sqrt {\sqrt {b^2-4 a c}+b-2 c}}{\sqrt {\sqrt {b^2-4 a c}+b+2 c}}\right )}{a^3 \sqrt {b^2-4 a c} \sqrt {\sqrt {b^2-4 a c}+b-2 c} \sqrt {\sqrt {b^2-4 a c}+b+2 c}}+\frac {\left (b^2-a c\right ) \text {arctanh}(\sin (x))}{a^3}-\frac {b \tan (x)}{a^2}+\frac {\text {arctanh}(\sin (x))}{2 a}+\frac {\tan (x) \sec (x)}{2 a}\) |
(-2*c*(b^3 - 3*a*b*c + Sqrt[b^2 - 4*a*c]*(b^2 - a*c))*ArcTan[(Sqrt[b - 2*c - Sqrt[b^2 - 4*a*c]]*Tan[x/2])/Sqrt[b + 2*c - Sqrt[b^2 - 4*a*c]]])/(a^3*S qrt[b^2 - 4*a*c]*Sqrt[b - 2*c - Sqrt[b^2 - 4*a*c]]*Sqrt[b + 2*c - Sqrt[b^2 - 4*a*c]]) + (2*c*(b^3 - 3*a*b*c - Sqrt[b^2 - 4*a*c]*(b^2 - a*c))*ArcTan[ (Sqrt[b - 2*c + Sqrt[b^2 - 4*a*c]]*Tan[x/2])/Sqrt[b + 2*c + Sqrt[b^2 - 4*a *c]]])/(a^3*Sqrt[b^2 - 4*a*c]*Sqrt[b - 2*c + Sqrt[b^2 - 4*a*c]]*Sqrt[b + 2 *c + Sqrt[b^2 - 4*a*c]]) + ArcTanh[Sin[x]]/(2*a) + ((b^2 - a*c)*ArcTanh[Si n[x]])/a^3 - (b*Tan[x])/a^2 + (Sec[x]*Tan[x])/(2*a)
3.1.20.3.1 Defintions of rubi rules used
Int[cos[(d_.) + (e_.)*(x_)]^(m_.)*((a_.) + cos[(d_.) + (e_.)*(x_)]^(n_.)*(b _.) + cos[(d_.) + (e_.)*(x_)]^(n2_.)*(c_.))^(p_), x_Symbol] :> Int[ExpandTr ig[cos[d + e*x]^m*(a + b*cos[d + e*x]^n + c*cos[d + e*x]^(2*n))^p, x], x] / ; FreeQ[{a, b, c, d, e}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && Integ ersQ[m, n, p]
Time = 7.69 (sec) , antiderivative size = 481, normalized size of antiderivative = 1.44
| method | result | size |
| default | \(\frac {2 \left (a -b +c \right ) \left (\frac {\left (2 c a b \sqrt {-4 a c +b^{2}}-a \,c^{2} \sqrt {-4 a c +b^{2}}-b^{3} \sqrt {-4 a c +b^{2}}+b^{2} c \sqrt {-4 a c +b^{2}}+2 a^{2} c^{2}-4 a \,b^{2} c +3 a b \,c^{2}+b^{4}-b^{3} c \right ) \operatorname {arctanh}\left (\frac {\left (-a +b -c \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (\sqrt {-4 a c +b^{2}}-a +c \right ) \left (a -b +c \right )}}\right )}{2 \sqrt {-4 a c +b^{2}}\, \left (a -b +c \right ) \sqrt {\left (\sqrt {-4 a c +b^{2}}-a +c \right ) \left (a -b +c \right )}}+\frac {\left (2 c a b \sqrt {-4 a c +b^{2}}-a \,c^{2} \sqrt {-4 a c +b^{2}}-b^{3} \sqrt {-4 a c +b^{2}}+b^{2} c \sqrt {-4 a c +b^{2}}-2 a^{2} c^{2}+4 a \,b^{2} c -3 a b \,c^{2}-b^{4}+b^{3} c \right ) \arctan \left (\frac {\left (a -b +c \right ) \tan \left (\frac {x}{2}\right )}{\sqrt {\left (\sqrt {-4 a c +b^{2}}+a -c \right ) \left (a -b +c \right )}}\right )}{2 \sqrt {-4 a c +b^{2}}\, \left (a -b +c \right ) \sqrt {\left (\sqrt {-4 a c +b^{2}}+a -c \right ) \left (a -b +c \right )}}\right )}{a^{3}}-\frac {1}{2 a \left (\tan \left (\frac {x}{2}\right )+1\right )^{2}}-\frac {-a -2 b}{2 a^{2} \left (\tan \left (\frac {x}{2}\right )+1\right )}+\frac {\left (a^{2}-2 a c +2 b^{2}\right ) \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{2 a^{3}}+\frac {1}{2 a \left (\tan \left (\frac {x}{2}\right )-1\right )^{2}}-\frac {-a -2 b}{2 a^{2} \left (\tan \left (\frac {x}{2}\right )-1\right )}+\frac {\left (-a^{2}+2 a c -2 b^{2}\right ) \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{2 a^{3}}\) | \(481\) |
| risch | \(\text {Expression too large to display}\) | \(4131\) |
2/a^3*(a-b+c)*(1/2*(2*c*a*b*(-4*a*c+b^2)^(1/2)-a*c^2*(-4*a*c+b^2)^(1/2)-b^ 3*(-4*a*c+b^2)^(1/2)+b^2*c*(-4*a*c+b^2)^(1/2)+2*a^2*c^2-4*a*b^2*c+3*a*b*c^ 2+b^4-b^3*c)/(-4*a*c+b^2)^(1/2)/(a-b+c)/(((-4*a*c+b^2)^(1/2)-a+c)*(a-b+c)) ^(1/2)*arctanh((-a+b-c)*tan(1/2*x)/(((-4*a*c+b^2)^(1/2)-a+c)*(a-b+c))^(1/2 ))+1/2*(2*c*a*b*(-4*a*c+b^2)^(1/2)-a*c^2*(-4*a*c+b^2)^(1/2)-b^3*(-4*a*c+b^ 2)^(1/2)+b^2*c*(-4*a*c+b^2)^(1/2)-2*a^2*c^2+4*a*b^2*c-3*a*b*c^2-b^4+b^3*c) /(-4*a*c+b^2)^(1/2)/(a-b+c)/(((-4*a*c+b^2)^(1/2)+a-c)*(a-b+c))^(1/2)*arcta n((a-b+c)*tan(1/2*x)/(((-4*a*c+b^2)^(1/2)+a-c)*(a-b+c))^(1/2)))-1/2/a/(tan (1/2*x)+1)^2-1/2*(-a-2*b)/a^2/(tan(1/2*x)+1)+1/2*(a^2-2*a*c+2*b^2)/a^3*ln( tan(1/2*x)+1)+1/2/a/(tan(1/2*x)-1)^2-1/2*(-a-2*b)/a^2/(tan(1/2*x)-1)+1/2/a ^3*(-a^2+2*a*c-2*b^2)*ln(tan(1/2*x)-1)
Timed out. \[ \int \frac {\sec ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Timed out} \]
\[ \int \frac {\sec ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\int \frac {\sec ^{3}{\left (x \right )}}{a + b \cos {\left (x \right )} + c \cos ^{2}{\left (x \right )}}\, dx \]
\[ \int \frac {\sec ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\int { \frac {\sec \left (x\right )^{3}}{c \cos \left (x\right )^{2} + b \cos \left (x\right ) + a} \,d x } \]
-1/4*(8*a^2*cos(3*x)*sin(2*x) + 8*a^2*cos(2*x)*sin(x) + 4*a^2*sin(x) - 4*( a^2*sin(3*x) + 2*a*b*sin(2*x) - a^2*sin(x))*cos(4*x) - 4*(a^3*cos(4*x)^2 + 4*a^3*cos(2*x)^2 + a^3*sin(4*x)^2 + 4*a^3*sin(4*x)*sin(2*x) + 4*a^3*sin(2 *x)^2 + 4*a^3*cos(2*x) + a^3 + 2*(2*a^3*cos(2*x) + a^3)*cos(4*x))*integrat e(-2*(2*(b^3*c - a*b*c^2)*cos(3*x)^2 + 4*(2*a*b^3 - 2*a*b*c^2 - (4*a^2*b - b^3)*c)*cos(2*x)^2 + 2*(b^3*c - a*b*c^2)*cos(x)^2 + 2*(b^3*c - a*b*c^2)*s in(3*x)^2 + 4*(2*a*b^3 - 2*a*b*c^2 - (4*a^2*b - b^3)*c)*sin(2*x)^2 + 2*(2* b^4 - 2*a*b^2*c - a*c^3 - (2*a^2 - b^2)*c^2)*sin(2*x)*sin(x) + 2*(b^3*c - a*b*c^2)*sin(x)^2 + ((b^2*c^2 - a*c^3)*cos(3*x) + 2*(b^3*c - 2*a*b*c^2)*co s(2*x) + (b^2*c^2 - a*c^3)*cos(x))*cos(4*x) + (b^2*c^2 - a*c^3 + 2*(2*b^4 - 2*a*b^2*c - a*c^3 - (2*a^2 - b^2)*c^2)*cos(2*x) + 4*(b^3*c - a*b*c^2)*co s(x))*cos(3*x) + 2*(b^3*c - 2*a*b*c^2 + (2*b^4 - 2*a*b^2*c - a*c^3 - (2*a^ 2 - b^2)*c^2)*cos(x))*cos(2*x) + (b^2*c^2 - a*c^3)*cos(x) + ((b^2*c^2 - a* c^3)*sin(3*x) + 2*(b^3*c - 2*a*b*c^2)*sin(2*x) + (b^2*c^2 - a*c^3)*sin(x)) *sin(4*x) + 2*((2*b^4 - 2*a*b^2*c - a*c^3 - (2*a^2 - b^2)*c^2)*sin(2*x) + 2*(b^3*c - a*b*c^2)*sin(x))*sin(3*x))/(a^3*c^2*cos(4*x)^2 + 4*a^3*b^2*cos( 3*x)^2 + 4*a^3*b^2*cos(x)^2 + a^3*c^2*sin(4*x)^2 + 4*a^3*b^2*sin(3*x)^2 + 4*a^3*b^2*sin(x)^2 + 4*a^3*b*c*cos(x) + a^3*c^2 + 4*(4*a^5 + 4*a^4*c + a^3 *c^2)*cos(2*x)^2 + 4*(4*a^5 + 4*a^4*c + a^3*c^2)*sin(2*x)^2 + 8*(2*a^4*b + a^3*b*c)*sin(2*x)*sin(x) + 2*(2*a^3*b*c*cos(3*x) + 2*a^3*b*c*cos(x) + ...
Leaf count of result is larger than twice the leaf count of optimal. 12615 vs. \(2 (286) = 572\).
Time = 2.84 (sec) , antiderivative size = 12615, normalized size of antiderivative = 37.77 \[ \int \frac {\sec ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Too large to display} \]
((2*a^2*b^7 - 4*a*b^8 + 2*b^9 - 20*a^3*b^5*c + 38*a^2*b^6*c - 12*a*b^7*c - 6*b^8*c + 64*a^4*b^3*c^2 - 110*a^3*b^4*c^2 - 12*a^2*b^5*c^2 + 54*a*b^6*c^ 2 + 6*b^7*c^2 - 64*a^5*b*c^3 + 80*a^4*b^2*c^3 + 160*a^3*b^3*c^3 - 140*a^2* b^4*c^3 - 56*a*b^5*c^3 - 2*b^6*c^3 + 32*a^5*c^4 - 192*a^4*b*c^4 + 64*a^3*b ^2*c^4 + 160*a^2*b^3*c^4 + 18*a*b^4*c^4 + 64*a^4*c^5 - 128*a^3*b*c^5 - 48* a^2*b^2*c^5 + 32*a^3*c^6 + 3*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c )*(a - b + c))*sqrt(b^2 - 4*a*c)*a^2*b^5 - 2*(b^2 - 4*a*c)*a^2*b^5 - 2*sqr t(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c) *a*b^6 + 4*(b^2 - 4*a*c)*a*b^6 - 5*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c)*b^7 - 2*(b^2 - 4*a*c)*b^7 - 18*sqrt (a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c)* a^3*b^3*c + 12*(b^2 - 4*a*c)*a^3*b^3*c + 9*sqrt(a^2 - a*b + b*c - c^2 + sq rt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c)*a^2*b^4*c - 22*(b^2 - 4*a*c )*a^2*b^4*c + 46*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c ))*sqrt(b^2 - 4*a*c)*a*b^5*c + 4*(b^2 - 4*a*c)*a*b^5*c + 11*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c)*b^6*c + 6* (b^2 - 4*a*c)*b^6*c + 24*sqrt(a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c)*a^4*b*c^2 - 16*(b^2 - 4*a*c)*a^4*b*c^2 - sqrt (a^2 - a*b + b*c - c^2 + sqrt(b^2 - 4*a*c)*(a - b + c))*sqrt(b^2 - 4*a*c)* a^3*b^2*c^2 + 22*(b^2 - 4*a*c)*a^3*b^2*c^2 - 134*sqrt(a^2 - a*b + b*c -...
Time = 16.99 (sec) , antiderivative size = 45255, normalized size of antiderivative = 135.49 \[ \int \frac {\sec ^3(x)}{a+b \cos (x)+c \cos ^2(x)} \, dx=\text {Too large to display} \]
((tan(x/2)^3*(a + 2*b))/a^2 + (tan(x/2)*(a - 2*b))/a^2)/(tan(x/2)^4 - 2*ta n(x/2)^2 + 1) - atan(((((((2048*(26*a^9*b^7 - 12*a^8*b^8 - 18*a^10*b^6 + 6 *a^11*b^5 - 2*a^12*b^4 + 48*a^10*c^6 + 176*a^11*c^5 + 176*a^12*c^4 + 16*a^ 13*c^3 - 32*a^14*c^2 + 20*a^8*b^7*c + 74*a^9*b^6*c - 144*a^10*b*c^5 - 192* a^10*b^5*c - 352*a^11*b*c^4 + 122*a^11*b^4*c - 144*a^12*b*c^3 - 40*a^12*b^ 3*c + 64*a^13*b*c^2 + 16*a^13*b^2*c + 8*a^8*b^4*c^4 - 20*a^8*b^5*c^3 + 4*a ^8*b^6*c^2 - 44*a^9*b^2*c^5 + 116*a^9*b^3*c^4 + 10*a^9*b^4*c^3 - 182*a^9*b ^5*c^2 - 148*a^10*b^2*c^4 + 496*a^10*b^3*c^3 - 50*a^10*b^4*c^2 - 260*a^11* b^2*c^3 + 388*a^11*b^3*c^2 - 204*a^12*b^2*c^2))/a^8 - (2048*tan(x/2)*((8*a ^4*c^6 - b^10 + 8*a^5*c^5 - b^7*(-(4*a*c - b^2)^3)^(1/2) + b^8*c^2 - 10*a* b^6*c^3 + 33*a^2*b^4*c^4 - 52*a^2*b^6*c^2 - 38*a^3*b^2*c^5 + 96*a^3*b^4*c^ 3 - 66*a^4*b^2*c^4 + b^5*c^2*(-(4*a*c - b^2)^3)^(1/2) + 12*a*b^8*c - 4*a*b ^3*c^3*(-(4*a*c - b^2)^3)^(1/2) + 3*a^2*b*c^4*(-(4*a*c - b^2)^3)^(1/2) + 4 *a^3*b*c^3*(-(4*a*c - b^2)^3)^(1/2) - 10*a^2*b^3*c^2*(-(4*a*c - b^2)^3)^(1 /2) + 6*a*b^5*c*(-(4*a*c - b^2)^3)^(1/2))/(2*(a^8*b^4 - a^6*b^6 + 16*a^8*c ^4 + 32*a^9*c^3 + 16*a^10*c^2 + 10*a^7*b^4*c - 8*a^9*b^2*c + a^6*b^4*c^2 - 8*a^7*b^2*c^3 - 32*a^8*b^2*c^2)))^(1/2)*(32*a^16*c + 8*a^10*b^7 - 24*a^11 *b^6 + 32*a^12*b^5 - 32*a^13*b^4 + 24*a^14*b^3 - 8*a^15*b^2 + 96*a^12*c^5 + 64*a^13*c^4 - 128*a^14*c^3 - 64*a^15*c^2 - 8*a^10*b^6*c - 56*a^11*b^5*c - 32*a^12*b*c^4 + 184*a^12*b^4*c + 352*a^13*b*c^3 - 200*a^13*b^3*c + 28...