Integrand size = 20, antiderivative size = 520 \[ \int \frac {x^{11/2}}{\left (a+b x^2+c x^4\right )^2} \, dx=-\frac {b \sqrt {x}}{2 c \left (b^2-4 a c\right )}+\frac {x^{5/2} \left (2 a+b x^2\right )}{2 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\left (b^2-10 a c+\frac {b \left (b^2-12 a c\right )}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{4 \sqrt [4]{2} c^{5/4} \left (b^2-4 a c\right ) \left (-b-\sqrt {b^2-4 a c}\right )^{3/4}}-\frac {\left (b^2-10 a c-\frac {b \left (b^2-12 a c\right )}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{4 \sqrt [4]{2} c^{5/4} \left (b^2-4 a c\right ) \left (-b+\sqrt {b^2-4 a c}\right )^{3/4}}-\frac {\left (b^2-10 a c+\frac {b \left (b^2-12 a c\right )}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{4 \sqrt [4]{2} c^{5/4} \left (b^2-4 a c\right ) \left (-b-\sqrt {b^2-4 a c}\right )^{3/4}}-\frac {\left (b^2-10 a c-\frac {b \left (b^2-12 a c\right )}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{4 \sqrt [4]{2} c^{5/4} \left (b^2-4 a c\right ) \left (-b+\sqrt {b^2-4 a c}\right )^{3/4}} \]
1/2*x^(5/2)*(b*x^2+2*a)/(-4*a*c+b^2)/(c*x^4+b*x^2+a)-1/8*arctan(2^(1/4)*c^ (1/4)*x^(1/2)/(-b-(-4*a*c+b^2)^(1/2))^(1/4))*(b^2-10*a*c+b*(-12*a*c+b^2)/( -4*a*c+b^2)^(1/2))*2^(3/4)/c^(5/4)/(-4*a*c+b^2)/(-b-(-4*a*c+b^2)^(1/2))^(3 /4)-1/8*arctanh(2^(1/4)*c^(1/4)*x^(1/2)/(-b-(-4*a*c+b^2)^(1/2))^(1/4))*(b^ 2-10*a*c+b*(-12*a*c+b^2)/(-4*a*c+b^2)^(1/2))*2^(3/4)/c^(5/4)/(-4*a*c+b^2)/ (-b-(-4*a*c+b^2)^(1/2))^(3/4)-1/8*arctan(2^(1/4)*c^(1/4)*x^(1/2)/(-b+(-4*a *c+b^2)^(1/2))^(1/4))*(b^2-10*a*c-b*(-12*a*c+b^2)/(-4*a*c+b^2)^(1/2))*2^(3 /4)/c^(5/4)/(-4*a*c+b^2)/(-b+(-4*a*c+b^2)^(1/2))^(3/4)-1/8*arctanh(2^(1/4) *c^(1/4)*x^(1/2)/(-b+(-4*a*c+b^2)^(1/2))^(1/4))*(b^2-10*a*c-b*(-12*a*c+b^2 )/(-4*a*c+b^2)^(1/2))*2^(3/4)/c^(5/4)/(-4*a*c+b^2)/(-b+(-4*a*c+b^2)^(1/2)) ^(3/4)-1/2*b*x^(1/2)/c/(-4*a*c+b^2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.40 (sec) , antiderivative size = 237, normalized size of antiderivative = 0.46 \[ \int \frac {x^{11/2}}{\left (a+b x^2+c x^4\right )^2} \, dx=-\frac {4 \text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {b \log \left (\sqrt {x}-\text {$\#$1}\right )-c \log \left (\sqrt {x}-\text {$\#$1}\right ) \text {$\#$1}^4}{b \text {$\#$1}^3+2 c \text {$\#$1}^7}\&\right ]+\frac {\frac {4 c \sqrt {x} \left (a b+b^2 x^2-2 a c x^2\right )}{a+b x^2+c x^4}+\text {RootSum}\left [a+b \text {$\#$1}^4+c \text {$\#$1}^8\&,\frac {-4 b^3 \log \left (\sqrt {x}-\text {$\#$1}\right )+15 a b c \log \left (\sqrt {x}-\text {$\#$1}\right )+3 b^2 c \log \left (\sqrt {x}-\text {$\#$1}\right ) \text {$\#$1}^4-6 a c^2 \log \left (\sqrt {x}-\text {$\#$1}\right ) \text {$\#$1}^4}{b \text {$\#$1}^3+2 c \text {$\#$1}^7}\&\right ]}{b^2-4 a c}}{8 c^2} \]
-1/8*(4*RootSum[a + b*#1^4 + c*#1^8 & , (b*Log[Sqrt[x] - #1] - c*Log[Sqrt[ x] - #1]*#1^4)/(b*#1^3 + 2*c*#1^7) & ] + ((4*c*Sqrt[x]*(a*b + b^2*x^2 - 2* a*c*x^2))/(a + b*x^2 + c*x^4) + RootSum[a + b*#1^4 + c*#1^8 & , (-4*b^3*Lo g[Sqrt[x] - #1] + 15*a*b*c*Log[Sqrt[x] - #1] + 3*b^2*c*Log[Sqrt[x] - #1]*# 1^4 - 6*a*c^2*Log[Sqrt[x] - #1]*#1^4)/(b*#1^3 + 2*c*#1^7) & ])/(b^2 - 4*a* c))/c^2
Time = 0.64 (sec) , antiderivative size = 431, normalized size of antiderivative = 0.83, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {1435, 1701, 1826, 1752, 756, 218, 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.
| \(\displaystyle \int \frac {x^{11/2}}{\left (a+b x^2+c x^4\right )^2} \, dx\) |
| \(\Big \downarrow \) 1435 |
| \(\displaystyle 2 \int \frac {x^6}{\left (c x^4+b x^2+a\right )^2}d\sqrt {x}\) |
| \(\Big \downarrow \) 1701 |
| \(\displaystyle 2 \left (\frac {x^{5/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\int \frac {x^2 \left (b x^2+10 a\right )}{c x^4+b x^2+a}d\sqrt {x}}{4 \left (b^2-4 a c\right )}\right )\) |
| \(\Big \downarrow \) 1826 |
| \(\displaystyle 2 \left (\frac {x^{5/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\frac {b \sqrt {x}}{c}-\frac {\int \frac {\left (b^2-10 a c\right ) x^2+a b}{c x^4+b x^2+a}d\sqrt {x}}{c}}{4 \left (b^2-4 a c\right )}\right )\) |
| \(\Big \downarrow \) 1752 |
| \(\displaystyle 2 \left (\frac {x^{5/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\frac {b \sqrt {x}}{c}-\frac {\frac {1}{2} \left (-\frac {b \left (b^2-12 a c\right )}{\sqrt {b^2-4 a c}}-10 a c+b^2\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}d\sqrt {x}+\frac {1}{2} \left (\frac {b \left (b^2-12 a c\right )}{\sqrt {b^2-4 a c}}-10 a c+b^2\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}d\sqrt {x}}{c}}{4 \left (b^2-4 a c\right )}\right )\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle 2 \left (\frac {x^{5/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\frac {b \sqrt {x}}{c}-\frac {\frac {1}{2} \left (\frac {b \left (b^2-12 a c\right )}{\sqrt {b^2-4 a c}}-10 a c+b^2\right ) \left (-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{\sqrt {-\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x+\sqrt {-b-\sqrt {b^2-4 a c}}}d\sqrt {x}}{\sqrt {-\sqrt {b^2-4 a c}-b}}\right )+\frac {1}{2} \left (-\frac {b \left (b^2-12 a c\right )}{\sqrt {b^2-4 a c}}-10 a c+b^2\right ) \left (-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{\sqrt {\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {2} \sqrt {c} x+\sqrt {\sqrt {b^2-4 a c}-b}}d\sqrt {x}}{\sqrt {\sqrt {b^2-4 a c}-b}}\right )}{c}}{4 \left (b^2-4 a c\right )}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle 2 \left (\frac {x^{5/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\frac {b \sqrt {x}}{c}-\frac {\frac {1}{2} \left (\frac {b \left (b^2-12 a c\right )}{\sqrt {b^2-4 a c}}-10 a c+b^2\right ) \left (-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{\sqrt {-\sqrt {b^2-4 a c}-b}}-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (-\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )+\frac {1}{2} \left (-\frac {b \left (b^2-12 a c\right )}{\sqrt {b^2-4 a c}}-10 a c+b^2\right ) \left (-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {c} x}d\sqrt {x}}{\sqrt {\sqrt {b^2-4 a c}-b}}-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )}{c}}{4 \left (b^2-4 a c\right )}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle 2 \left (\frac {x^{5/2} \left (2 a+b x^2\right )}{4 \left (b^2-4 a c\right ) \left (a+b x^2+c x^4\right )}-\frac {\frac {b \sqrt {x}}{c}-\frac {\frac {1}{2} \left (\frac {b \left (b^2-12 a c\right )}{\sqrt {b^2-4 a c}}-10 a c+b^2\right ) \left (-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (-\sqrt {b^2-4 a c}-b\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (-\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )+\frac {1}{2} \left (-\frac {b \left (b^2-12 a c\right )}{\sqrt {b^2-4 a c}}-10 a c+b^2\right ) \left (-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (\sqrt {b^2-4 a c}-b\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{c} \left (\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )}{c}}{4 \left (b^2-4 a c\right )}\right )\) |
2*((x^(5/2)*(2*a + b*x^2))/(4*(b^2 - 4*a*c)*(a + b*x^2 + c*x^4)) - ((b*Sqr t[x])/c - (((b^2 - 10*a*c + (b*(b^2 - 12*a*c))/Sqrt[b^2 - 4*a*c])*(-(ArcTa n[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b - Sqrt[b^2 - 4*a*c])^(1/4)]/(2^(1/4)*c^(1/ 4)*(-b - Sqrt[b^2 - 4*a*c])^(3/4))) - ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(- b - Sqrt[b^2 - 4*a*c])^(1/4)]/(2^(1/4)*c^(1/4)*(-b - Sqrt[b^2 - 4*a*c])^(3 /4))))/2 + ((b^2 - 10*a*c - (b*(b^2 - 12*a*c))/Sqrt[b^2 - 4*a*c])*(-(ArcTa n[(2^(1/4)*c^(1/4)*Sqrt[x])/(-b + Sqrt[b^2 - 4*a*c])^(1/4)]/(2^(1/4)*c^(1/ 4)*(-b + Sqrt[b^2 - 4*a*c])^(3/4))) - ArcTanh[(2^(1/4)*c^(1/4)*Sqrt[x])/(- b + Sqrt[b^2 - 4*a*c])^(1/4)]/(2^(1/4)*c^(1/4)*(-b + Sqrt[b^2 - 4*a*c])^(3 /4))))/2)/c)/(4*(b^2 - 4*a*c)))
3.11.72.3.1 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_.)*(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_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/d Subst[Int[x^(k*(m + 1) - 1)*(a + b *(x^(2*k)/d^2) + c*(x^(4*k)/d^4))^p, x], x, (d*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b^2 - 4*a*c, 0] && FractionQ[m] && IntegerQ[p]
Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x _Symbol] :> Simp[(-d^(2*n - 1))*(d*x)^(m - 2*n + 1)*(2*a + b*x^n)*((a + b*x ^n + c*x^(2*n))^(p + 1)/(n*(p + 1)*(b^2 - 4*a*c))), x] + Simp[d^(2*n)/(n*(p + 1)*(b^2 - 4*a*c)) Int[(d*x)^(m - 2*n)*(2*a*(m - 2*n + 1) + b*(m + n*(2 *p + 1) + 1)*x^n)*(a + b*x^n + c*x^(2*n))^(p + 1), x], x] /; FreeQ[{a, b, c , d}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && ILtQ[p, -1 ] && GtQ[m, 2*n - 1]
Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x _Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/(b/2 - q/2 + c*x^n), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) I nt[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2 , 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 - 4*a*c] || !IGtQ[n/2, 0])
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + ( c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Simp[e*f^(n - 1)*(f*x)^(m - n + 1)*((a + b*x^n + c*x^(2*n))^(p + 1)/(c*(m + n*(2*p + 1) + 1))), x] - Simp[f^n/(c*( m + n*(2*p + 1) + 1)) Int[(f*x)^(m - n)*(a + b*x^n + c*x^(2*n))^p*Simp[a* e*(m - n + 1) + (b*e*(m + n*p + 1) - c*d*(m + n*(2*p + 1) + 1))*x^n, x], x] , x] /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*(2*p + 1) + 1, 0] && Intege rQ[p]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.44 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.28
| method | result | size |
| derivativedivides | \(\frac {-\frac {\left (2 a c -b^{2}\right ) x^{\frac {5}{2}}}{2 c \left (4 a c -b^{2}\right )}+\frac {a b \sqrt {x}}{2 c \left (4 a c -b^{2}\right )}}{c \,x^{4}+b \,x^{2}+a}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (\left (10 a c -b^{2}\right ) \textit {\_R}^{4}-a b \right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{8 c \left (4 a c -b^{2}\right )}\) | \(146\) |
| default | \(\frac {-\frac {\left (2 a c -b^{2}\right ) x^{\frac {5}{2}}}{2 c \left (4 a c -b^{2}\right )}+\frac {a b \sqrt {x}}{2 c \left (4 a c -b^{2}\right )}}{c \,x^{4}+b \,x^{2}+a}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{8}+\textit {\_Z}^{4} b +a \right )}{\sum }\frac {\left (\left (10 a c -b^{2}\right ) \textit {\_R}^{4}-a b \right ) \ln \left (\sqrt {x}-\textit {\_R} \right )}{2 \textit {\_R}^{7} c +\textit {\_R}^{3} b}}{8 c \left (4 a c -b^{2}\right )}\) | \(146\) |
2*(-1/4*(2*a*c-b^2)/c/(4*a*c-b^2)*x^(5/2)+1/4*a*b/c/(4*a*c-b^2)*x^(1/2))/( c*x^4+b*x^2+a)+1/8/c/(4*a*c-b^2)*sum(((10*a*c-b^2)*_R^4-a*b)/(2*_R^7*c+_R^ 3*b)*ln(x^(1/2)-_R),_R=RootOf(_Z^8*c+_Z^4*b+a))
Leaf count of result is larger than twice the leaf count of optimal. 9180 vs. \(2 (424) = 848\).
Time = 2.21 (sec) , antiderivative size = 9180, normalized size of antiderivative = 17.65 \[ \int \frac {x^{11/2}}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {x^{11/2}}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {x^{11/2}}{\left (a+b x^2+c x^4\right )^2} \, dx=\int { \frac {x^{\frac {11}{2}}}{{\left (c x^{4} + b x^{2} + a\right )}^{2}} \,d x } \]
1/2*(b*x^(9/2) + 2*a*x^(5/2))/((b^2*c - 4*a*c^2)*x^4 + a*b^2 - 4*a^2*c + ( b^3 - 4*a*b*c)*x^2) + integrate(-1/4*(b*x^(7/2) + 10*a*x^(3/2))/((b^2*c - 4*a*c^2)*x^4 + a*b^2 - 4*a^2*c + (b^3 - 4*a*b*c)*x^2), x)
\[ \int \frac {x^{11/2}}{\left (a+b x^2+c x^4\right )^2} \, dx=\int { \frac {x^{\frac {11}{2}}}{{\left (c x^{4} + b x^{2} + a\right )}^{2}} \,d x } \]
Time = 20.59 (sec) , antiderivative size = 31964, normalized size of antiderivative = 61.47 \[ \int \frac {x^{11/2}}{\left (a+b x^2+c x^4\right )^2} \, dx=\text {Too large to display} \]
2*atan(((((9*a^3*b^9 - 397*a^4*b^7*c + 130000*a^7*b*c^4 + 6549*a^5*b^5*c^2 - 47800*a^6*b^3*c^3)/(2*(b^8*c + 256*a^4*c^5 - 16*a*b^6*c^2 + 96*a^2*b^4* c^3 - 256*a^3*b^2*c^4)) + ((x^(1/2)*(1006632960*a^10*b*c^11 + 4096*a^3*b^1 5*c^4 + 147456*a^4*b^13*c^5 - 4915200*a^5*b^11*c^6 + 53739520*a^6*b^9*c^7 - 298844160*a^7*b^7*c^8 + 918552576*a^8*b^5*c^9 - 1493172224*a^9*b^3*c^10) )/(16*(b^12*c + 4096*a^6*c^7 - 24*a*b^10*c^2 + 240*a^2*b^8*c^3 - 1280*a^3* b^6*c^4 + 3840*a^4*b^4*c^5 - 6144*a^5*b^2*c^6)) - ((-(b^21 + b^6*(-(4*a*c - b^2)^15)^(1/2) + 73728000*a^10*b*c^10 + 2085*a^2*b^17*c^2 - 36320*a^3*b^ 15*c^3 + 404160*a^4*b^13*c^4 - 3001344*a^5*b^11*c^5 + 15064576*a^6*b^9*c^6 - 50503680*a^7*b^7*c^7 + 108380160*a^8*b^5*c^8 - 134676480*a^9*b^3*c^9 - 2500*a^3*c^3*(-(4*a*c - b^2)^15)^(1/2) - 69*a*b^19*c + 525*a^2*b^2*c^2*(-( 4*a*c - b^2)^15)^(1/2) - 39*a*b^4*c*(-(4*a*c - b^2)^15)^(1/2))/(8192*(1677 7216*a^12*c^17 + b^24*c^5 - 48*a*b^22*c^6 + 1056*a^2*b^20*c^7 - 14080*a^3* b^18*c^8 + 126720*a^4*b^16*c^9 - 811008*a^5*b^14*c^10 + 3784704*a^6*b^12*c ^11 - 12976128*a^7*b^10*c^12 + 32440320*a^8*b^8*c^13 - 57671680*a^9*b^6*c^ 14 + 69206016*a^10*b^4*c^15 - 50331648*a^11*b^2*c^16)))^(1/4)*(167772160*a ^9*c^11 + 40960*a^3*b^12*c^5 - 983040*a^4*b^10*c^6 + 9830400*a^5*b^8*c^7 - 52428800*a^6*b^6*c^8 + 157286400*a^7*b^4*c^9 - 251658240*a^8*b^2*c^10)*1i )/(2*(b^8*c + 256*a^4*c^5 - 16*a*b^6*c^2 + 96*a^2*b^4*c^3 - 256*a^3*b^2*c^ 4)))*(-(b^21 + b^6*(-(4*a*c - b^2)^15)^(1/2) + 73728000*a^10*b*c^10 + 2...