Integrand size = 26, antiderivative size = 223 \[ \int \frac {x \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=-\frac {B \sqrt {b-\sqrt {b^2-4 a c}} \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b^2-4 a c}}+\frac {B \sqrt {b+\sqrt {b^2-4 a c}} \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b^2-4 a c}}-\frac {(2 A c-b C) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{2 c \sqrt {b^2-4 a c}}+\frac {C \log \left (a+b x^2+c x^4\right )}{4 c} \]
1/4*C*ln(c*x^4+b*x^2+a)/c-1/2*(2*A*c-C*b)*arctanh((2*c*x^2+b)/(-4*a*c+b^2) ^(1/2))/c/(-4*a*c+b^2)^(1/2)-1/2*B*arctan(x*2^(1/2)*c^(1/2)/(b-(-4*a*c+b^2 )^(1/2))^(1/2))*(b-(-4*a*c+b^2)^(1/2))^(1/2)*2^(1/2)/c^(1/2)/(-4*a*c+b^2)^ (1/2)+1/2*B*arctan(x*2^(1/2)*c^(1/2)/(b+(-4*a*c+b^2)^(1/2))^(1/2))*(b+(-4* a*c+b^2)^(1/2))^(1/2)*2^(1/2)/c^(1/2)/(-4*a*c+b^2)^(1/2)
Time = 0.22 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.08 \[ \int \frac {x \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\frac {-2 \sqrt {2} B \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}} \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )+2 \sqrt {2} B \sqrt {c} \sqrt {b+\sqrt {b^2-4 a c}} \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b+\sqrt {b^2-4 a c}}}\right )+\left (2 A c+\left (-b+\sqrt {b^2-4 a c}\right ) C\right ) \log \left (-b+\sqrt {b^2-4 a c}-2 c x^2\right )-\left (2 A c-\left (b+\sqrt {b^2-4 a c}\right ) C\right ) \log \left (b+\sqrt {b^2-4 a c}+2 c x^2\right )}{4 c \sqrt {b^2-4 a c}} \]
(-2*Sqrt[2]*B*Sqrt[c]*Sqrt[b - Sqrt[b^2 - 4*a*c]]*ArcTan[(Sqrt[2]*Sqrt[c]* x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]] + 2*Sqrt[2]*B*Sqrt[c]*Sqrt[b + Sqrt[b^2 - 4*a*c]]*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]] + (2*A*c + (-b + Sqrt[b^2 - 4*a*c])*C)*Log[-b + Sqrt[b^2 - 4*a*c] - 2*c*x^2] - (2*A* c - (b + Sqrt[b^2 - 4*a*c])*C)*Log[b + Sqrt[b^2 - 4*a*c] + 2*c*x^2])/(4*c* Sqrt[b^2 - 4*a*c])
Time = 0.50 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.05, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {2193, 27, 1450, 218, 1576, 1142, 1083, 219, 1103}
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.
| \(\displaystyle \int \frac {x \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx\) |
| \(\Big \downarrow \) 2193 |
| \(\displaystyle \int \frac {x \left (C x^2+A\right )}{c x^4+b x^2+a}dx+\int \frac {B x^2}{c x^4+b x^2+a}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {x \left (C x^2+A\right )}{c x^4+b x^2+a}dx+B \int \frac {x^2}{c x^4+b x^2+a}dx\) |
| \(\Big \downarrow \) 1450 |
| \(\displaystyle \int \frac {x \left (C x^2+A\right )}{c x^4+b x^2+a}dx+B \left (\frac {1}{2} \left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}dx+\frac {1}{2} \left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \int \frac {1}{c x^2+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}dx\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \int \frac {x \left (C x^2+A\right )}{c x^4+b x^2+a}dx+B \left (\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}\right )\) |
| \(\Big \downarrow \) 1576 |
| \(\displaystyle \frac {1}{2} \int \frac {C x^2+A}{c x^4+b x^2+a}dx^2+B \left (\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}\right )\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {1}{2} \left (\frac {(2 A c-b C) \int \frac {1}{c x^4+b x^2+a}dx^2}{2 c}+\frac {C \int \frac {2 c x^2+b}{c x^4+b x^2+a}dx^2}{2 c}\right )+B \left (\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}\right )\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {1}{2} \left (\frac {C \int \frac {2 c x^2+b}{c x^4+b x^2+a}dx^2}{2 c}-\frac {(2 A c-b C) \int \frac {1}{-x^4+b^2-4 a c}d\left (2 c x^2+b\right )}{c}\right )+B \left (\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{2} \left (\frac {C \int \frac {2 c x^2+b}{c x^4+b x^2+a}dx^2}{2 c}-\frac {(2 A c-b C) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}}\right )+B \left (\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{2} \left (\frac {C \log \left (a+b x^2+c x^4\right )}{2 c}-\frac {(2 A c-b C) \text {arctanh}\left (\frac {b+2 c x^2}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c}}\right )+B \left (\frac {\left (1-\frac {b}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {b-\sqrt {b^2-4 a c}}}\right )}{\sqrt {2} \sqrt {c} \sqrt {b-\sqrt {b^2-4 a c}}}+\frac {\left (\frac {b}{\sqrt {b^2-4 a c}}+1\right ) \arctan \left (\frac {\sqrt {2} \sqrt {c} x}{\sqrt {\sqrt {b^2-4 a c}+b}}\right )}{\sqrt {2} \sqrt {c} \sqrt {\sqrt {b^2-4 a c}+b}}\right )\) |
B*(((1 - b/Sqrt[b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b - Sqrt[b^2 - 4*a*c]]])/(Sqrt[2]*Sqrt[c]*Sqrt[b - Sqrt[b^2 - 4*a*c]]) + ((1 + b/Sqrt[ b^2 - 4*a*c])*ArcTan[(Sqrt[2]*Sqrt[c]*x)/Sqrt[b + Sqrt[b^2 - 4*a*c]]])/(Sq rt[2]*Sqrt[c]*Sqrt[b + Sqrt[b^2 - 4*a*c]])) + (-(((2*A*c - b*C)*ArcTanh[(b + 2*c*x^2)/Sqrt[b^2 - 4*a*c]])/(c*Sqrt[b^2 - 4*a*c])) + (C*Log[a + b*x^2 + c*x^4])/(2*c))/2
3.1.24.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> Wi th[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(d^2/2)*(b/q + 1) Int[(d*x)^(m - 2)/(b/ 2 + q/2 + c*x^2), x], x] - Simp[(d^2/2)*(b/q - 1) Int[(d*x)^(m - 2)/(b/2 - q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 2]
Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^( p_.), x_Symbol] :> Simp[1/2 Subst[Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x] , x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]
Int[(Pq_)*((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_S ymbol] :> Module[{q = Expon[Pq, x], k}, Int[Sum[Coeff[Pq, x, 2*k]*x^(2*k), {k, 0, q/2 + 1}]*(d*x)^m*(a + b*x^2 + c*x^4)^p, x] + Simp[1/d Int[Sum[Coe ff[Pq, x, 2*k + 1]*x^(2*k), {k, 0, (q + 1)/2}]*(d*x)^(m + 1)*(a + b*x^2 + c *x^4)^p, x], x]] /; FreeQ[{a, b, c, d, m, p}, x] && PolyQ[Pq, x] && !PolyQ [Pq, x^2]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.23
| method | result | size |
| risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (c \,\textit {\_Z}^{4}+\textit {\_Z}^{2} b +a \right )}{\sum }\frac {\left (C \,\textit {\_R}^{3}+B \,\textit {\_R}^{2}+A \textit {\_R} \right ) \ln \left (x -\textit {\_R} \right )}{2 c \,\textit {\_R}^{3}+\textit {\_R} b}\right )}{2}\) | \(52\) |
| default | \(4 c \left (\frac {\frac {\left (2 A c \sqrt {-4 a c +b^{2}}-C \sqrt {-4 a c +b^{2}}\, b +4 a c C -b^{2} C \right ) \ln \left (2 c \,x^{2}+\sqrt {-4 a c +b^{2}}+b \right )}{4 c}+\frac {\left (-B b \sqrt {-4 a c +b^{2}}+4 B a c -B \,b^{2}\right ) \sqrt {2}\, \arctan \left (\frac {c x \sqrt {2}}{\sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 \sqrt {\left (b +\sqrt {-4 a c +b^{2}}\right ) c}}}{4 c \left (4 a c -b^{2}\right )}+\frac {-\frac {\left (2 A c \sqrt {-4 a c +b^{2}}-C \sqrt {-4 a c +b^{2}}\, b -4 a c C +b^{2} C \right ) \ln \left (-2 c \,x^{2}+\sqrt {-4 a c +b^{2}}-b \right )}{4 c}+\frac {\left (-B b \sqrt {-4 a c +b^{2}}-4 B a c +B \,b^{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {c x \sqrt {2}}{\sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}\right )}{2 \sqrt {\left (-b +\sqrt {-4 a c +b^{2}}\right ) c}}}{4 c \left (4 a c -b^{2}\right )}\right )\) | \(313\) |
Result contains complex when optimal does not.
Time = 13.96 (sec) , antiderivative size = 845032, normalized size of antiderivative = 3789.38 \[ \int \frac {x \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\text {Too large to display} \]
Timed out. \[ \int \frac {x \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\text {Timed out} \]
\[ \int \frac {x \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\int { \frac {{\left (C x^{2} + B x + A\right )} x}{c x^{4} + b x^{2} + a} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 2368 vs. \(2 (179) = 358\).
Time = 1.46 (sec) , antiderivative size = 2368, normalized size of antiderivative = 10.62 \[ \int \frac {x \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\text {Too large to display} \]
1/4*C*log(abs(c*x^4 + b*x^2 + a))/c + 1/8*((2*b^4*c^2 - 16*a*b^2*c^3 + 32* a^2*c^4 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^4 + 8*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c + 2*sq rt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c - 16*sqrt(2) *sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a^2*c^2 - 8*sqrt(2)*sqr t(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b*c^2 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^2*c^2 + 4*sqrt(2)*sqrt(b^2 - 4* a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*c^3 - 2*(b^2 - 4*a*c)*b^2*c^2 + 8*( b^2 - 4*a*c)*a*c^3)*B*c^2 - (2*b^4*c^4 - 8*a*b^2*c^5 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^4*c^2 + 4*sqrt(2)*sqrt(b^2 - 4*a* c)*sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*a*b^2*c^3 + 2*sqrt(2)*sqrt(b^2 - 4*a*c) *sqrt(b*c + sqrt(b^2 - 4*a*c)*c)*b^3*c^3 - sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt( b*c + sqrt(b^2 - 4*a*c)*c)*b^2*c^4 - 2*(b^2 - 4*a*c)*b^2*c^4)*B)*arctan(2* sqrt(1/2)*x/sqrt((b*c + sqrt(b^2*c^2 - 4*a*c^3))/c^2))/((a*b^4*c^2 - 8*a^2 *b^2*c^3 - 2*a*b^3*c^3 + 16*a^3*c^4 + 8*a^2*b*c^4 + a*b^2*c^4 - 4*a^2*c^5) *c^2) - 1/8*((2*b^4*c^2 - 16*a*b^2*c^3 + 32*a^2*c^4 - sqrt(2)*sqrt(b^2 - 4 *a*c)*sqrt(b*c - sqrt(b^2 - 4*a*c)*c)*b^4 + 8*sqrt(2)*sqrt(b^2 - 4*a*c)*sq rt(b*c - sqrt(b^2 - 4*a*c)*c)*a*b^2*c + 2*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b *c - sqrt(b^2 - 4*a*c)*c)*b^3*c - 16*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c - sqrt(b^2 - 4*a*c)*c)*a^2*c^2 - 8*sqrt(2)*sqrt(b^2 - 4*a*c)*sqrt(b*c - s...
Time = 8.37 (sec) , antiderivative size = 5594, normalized size of antiderivative = 25.09 \[ \int \frac {x \left (A+B x+C x^2\right )}{a+b x^2+c x^4} \, dx=\text {Too large to display} \]
symsum(log(A^3*c^2*x - B^3*a*c - B*C^2*a*b - 8*root(128*a*b^2*c^3*z^4 - 16 *b^4*c^2*z^4 - 256*a^2*c^4*z^4 - 128*C*a*b^2*c^2*z^3 + 256*C*a^2*c^3*z^3 + 16*C*b^4*c*z^3 + 32*A*C*a*b*c^2*z^2 - 8*A*C*b^3*c*z^2 + 40*C^2*a*b^2*c*z^ 2 + 16*B^2*a*b*c^2*z^2 - 4*B^2*b^3*c*z^2 - 32*A^2*a*c^3*z^2 - 96*C^2*a^2*c ^2*z^2 + 8*A^2*b^2*c^2*z^2 - 4*C^2*b^4*z^2 - 16*A*C^2*a*b*c*z - 4*A^2*C*b^ 2*c*z + 16*A^2*C*a*c^2*z + 4*A*B^2*b^2*c*z - 16*A*B^2*a*c^2*z + 16*C^3*a^2 *c*z - 4*C^3*a*b^2*z + 4*A*C^2*b^3*z + 4*A*B^2*C*a*c - 2*A^2*C^2*a*c + 2*A ^3*C*b*c + 2*A*C^3*a*b - B^2*C^2*a*b - A^2*B^2*b*c - B^4*a*c - A^2*C^2*b^2 - C^4*a^2 - A^4*c^2, z, k)^3*b^3*c^2*x - C^3*a*b*x + A*C^2*b^2*x - 2*C^2* root(128*a*b^2*c^3*z^4 - 16*b^4*c^2*z^4 - 256*a^2*c^4*z^4 - 128*C*a*b^2*c^ 2*z^3 + 256*C*a^2*c^3*z^3 + 16*C*b^4*c*z^3 + 32*A*C*a*b*c^2*z^2 - 8*A*C*b^ 3*c*z^2 + 40*C^2*a*b^2*c*z^2 + 16*B^2*a*b*c^2*z^2 - 4*B^2*b^3*c*z^2 - 32*A ^2*a*c^3*z^2 - 96*C^2*a^2*c^2*z^2 + 8*A^2*b^2*c^2*z^2 - 4*C^2*b^4*z^2 - 16 *A*C^2*a*b*c*z - 4*A^2*C*b^2*c*z + 16*A^2*C*a*c^2*z + 4*A*B^2*b^2*c*z - 16 *A*B^2*a*c^2*z + 16*C^3*a^2*c*z - 4*C^3*a*b^2*z + 4*A*C^2*b^3*z + 4*A*B^2* C*a*c - 2*A^2*C^2*a*c + 2*A^3*C*b*c + 2*A*C^3*a*b - B^2*C^2*a*b - A^2*B^2* b*c - B^4*a*c - A^2*C^2*b^2 - C^4*a^2 - A^4*c^2, z, k)*b^3*x + 32*root(128 *a*b^2*c^3*z^4 - 16*b^4*c^2*z^4 - 256*a^2*c^4*z^4 - 128*C*a*b^2*c^2*z^3 + 256*C*a^2*c^3*z^3 + 16*C*b^4*c*z^3 + 32*A*C*a*b*c^2*z^2 - 8*A*C*b^3*c*z^2 + 40*C^2*a*b^2*c*z^2 + 16*B^2*a*b*c^2*z^2 - 4*B^2*b^3*c*z^2 - 32*A^2*a*...