\(\int \frac {1}{(-3+2 x^2-2 x^4)^{3/2}} \, dx\) [268]

Optimal result

Integrand size = 16, antiderivative size = 250 \[ \int \frac {1}{\left (-3+2 x^2-2 x^4\right )^{3/2}} \, dx=-\frac {x \left (2+x^2\right )}{15 \sqrt {-3+2 x^2-2 x^4}}-\frac {x \sqrt {-3+2 x^2-2 x^4}}{15 \left (\sqrt {6}+2 x^2\right )}-\frac {\left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3-2 x^2+2 x^4}{\left (3+\sqrt {6} x^2\right )^2}} E\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right )|\frac {1}{12} \left (6+\sqrt {6}\right )\right )}{5\ 6^{3/4} \sqrt {-3+2 x^2-2 x^4}}+\frac {\left (1-\sqrt {6}\right ) \left (3+\sqrt {6} x^2\right ) \sqrt {\frac {3-2 x^2+2 x^4}{\left (3+\sqrt {6} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{12} \left (6+\sqrt {6}\right )\right )}{10\ 6^{3/4} \sqrt {-3+2 x^2-2 x^4}} \] Output:

-1/15*x*(x^2+2)/(-2*x^4+2*x^2-3)^(1/2)-x*(-2*x^4+2*x^2-3)^(1/2)/(15*6^(1/2 
)+30*x^2)-1/30*(3+6^(1/2)*x^2)*((2*x^4-2*x^2+3)/(3+6^(1/2)*x^2)^2)^(1/2)*E 
llipticE(sin(2*arctan(1/3*2^(1/4)*3^(3/4)*x)),1/6*(18+3*6^(1/2))^(1/2))*6^ 
(1/4)/(-2*x^4+2*x^2-3)^(1/2)+1/60*(1-6^(1/2))*(3+6^(1/2)*x^2)*((2*x^4-2*x^ 
2+3)/(3+6^(1/2)*x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(1/3*2^(1/4)*3^(3/4) 
*x),1/6*(18+3*6^(1/2))^(1/2))*6^(1/4)/(-2*x^4+2*x^2-3)^(1/2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 6.46 (sec) , antiderivative size = 326, normalized size of antiderivative = 1.30 \[ \int \frac {1}{\left (-3+2 x^2-2 x^4\right )^{3/2}} \, dx=\frac {-4 \sqrt {-\frac {i}{i+\sqrt {5}}} x \left (2+x^2\right )-\sqrt {2} \left (-i+\sqrt {5}\right ) \sqrt {\frac {i+\sqrt {5}-2 i x^2}{i+\sqrt {5}}} \sqrt {\frac {-i+\sqrt {5}+2 i x^2}{-i+\sqrt {5}}} E\left (i \text {arcsinh}\left (\sqrt {-\frac {2 i}{i+\sqrt {5}}} x\right )|\frac {i+\sqrt {5}}{i-\sqrt {5}}\right )+\sqrt {2} \left (5 i+\sqrt {5}\right ) \sqrt {\frac {i+\sqrt {5}-2 i x^2}{i+\sqrt {5}}} \sqrt {\frac {-i+\sqrt {5}+2 i x^2}{-i+\sqrt {5}}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {-\frac {2 i}{i+\sqrt {5}}} x\right ),\frac {i+\sqrt {5}}{i-\sqrt {5}}\right )}{60 \sqrt {-\frac {i}{i+\sqrt {5}}} \sqrt {-3+2 x^2-2 x^4}} \] Input:

Integrate[(-3 + 2*x^2 - 2*x^4)^(-3/2),x]
 

Output:

(-4*Sqrt[(-I)/(I + Sqrt[5])]*x*(2 + x^2) - Sqrt[2]*(-I + Sqrt[5])*Sqrt[(I 
+ Sqrt[5] - (2*I)*x^2)/(I + Sqrt[5])]*Sqrt[(-I + Sqrt[5] + (2*I)*x^2)/(-I 
+ Sqrt[5])]*EllipticE[I*ArcSinh[Sqrt[(-2*I)/(I + Sqrt[5])]*x], (I + Sqrt[5 
])/(I - Sqrt[5])] + Sqrt[2]*(5*I + Sqrt[5])*Sqrt[(I + Sqrt[5] - (2*I)*x^2) 
/(I + Sqrt[5])]*Sqrt[(-I + Sqrt[5] + (2*I)*x^2)/(-I + Sqrt[5])]*EllipticF[ 
I*ArcSinh[Sqrt[(-2*I)/(I + Sqrt[5])]*x], (I + Sqrt[5])/(I - Sqrt[5])])/(60 
*Sqrt[(-I)/(I + Sqrt[5])]*Sqrt[-3 + 2*x^2 - 2*x^4])
 

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1405, 27, 1511, 27, 1416, 1509}

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.

\(\displaystyle \int \frac {1}{\left (-2 x^4+2 x^2-3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1405

\(\displaystyle -\frac {1}{60} \int \frac {4 \left (3-x^2\right )}{\sqrt {-2 x^4+2 x^2-3}}dx-\frac {x \left (x^2+2\right )}{15 \sqrt {-2 x^4+2 x^2-3}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{15} \int \frac {3-x^2}{\sqrt {-2 x^4+2 x^2-3}}dx-\frac {x \left (x^2+2\right )}{15 \sqrt {-2 x^4+2 x^2-3}}\)

\(\Big \downarrow \) 1511

\(\displaystyle \frac {1}{15} \left (-\frac {1}{2} \left (6-\sqrt {6}\right ) \int \frac {1}{\sqrt {-2 x^4+2 x^2-3}}dx-\sqrt {\frac {3}{2}} \int \frac {3-\sqrt {6} x^2}{3 \sqrt {-2 x^4+2 x^2-3}}dx\right )-\frac {x \left (x^2+2\right )}{15 \sqrt {-2 x^4+2 x^2-3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{15} \left (-\frac {1}{2} \left (6-\sqrt {6}\right ) \int \frac {1}{\sqrt {-2 x^4+2 x^2-3}}dx-\frac {\int \frac {3-\sqrt {6} x^2}{\sqrt {-2 x^4+2 x^2-3}}dx}{\sqrt {6}}\right )-\frac {x \left (x^2+2\right )}{15 \sqrt {-2 x^4+2 x^2-3}}\)

\(\Big \downarrow \) 1416

\(\displaystyle \frac {1}{15} \left (-\frac {\int \frac {3-\sqrt {6} x^2}{\sqrt {-2 x^4+2 x^2-3}}dx}{\sqrt {6}}-\frac {\left (6-\sqrt {6}\right ) \left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4-2 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{12} \left (6+\sqrt {6}\right )\right )}{4 \sqrt [4]{6} \sqrt {-2 x^4+2 x^2-3}}\right )-\frac {x \left (x^2+2\right )}{15 \sqrt {-2 x^4+2 x^2-3}}\)

\(\Big \downarrow \) 1509

\(\displaystyle \frac {1}{15} \left (-\frac {\left (6-\sqrt {6}\right ) \left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4-2 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right ),\frac {1}{12} \left (6+\sqrt {6}\right )\right )}{4 \sqrt [4]{6} \sqrt {-2 x^4+2 x^2-3}}-\frac {\frac {3^{3/4} \left (\sqrt {6} x^2+3\right ) \sqrt {\frac {2 x^4-2 x^2+3}{\left (\sqrt {6} x^2+3\right )^2}} E\left (2 \arctan \left (\sqrt [4]{\frac {2}{3}} x\right )|\frac {1}{12} \left (6+\sqrt {6}\right )\right )}{\sqrt [4]{2} \sqrt {-2 x^4+2 x^2-3}}+\frac {3 \sqrt {-2 x^4+2 x^2-3} x}{\sqrt {6} x^2+3}}{\sqrt {6}}\right )-\frac {x \left (x^2+2\right )}{15 \sqrt {-2 x^4+2 x^2-3}}\)

Input:

Int[(-3 + 2*x^2 - 2*x^4)^(-3/2),x]
 

Output:

-1/15*(x*(2 + x^2))/Sqrt[-3 + 2*x^2 - 2*x^4] + (-(((3*x*Sqrt[-3 + 2*x^2 - 
2*x^4])/(3 + Sqrt[6]*x^2) + (3^(3/4)*(3 + Sqrt[6]*x^2)*Sqrt[(3 - 2*x^2 + 2 
*x^4)/(3 + Sqrt[6]*x^2)^2]*EllipticE[2*ArcTan[(2/3)^(1/4)*x], (6 + Sqrt[6] 
)/12])/(2^(1/4)*Sqrt[-3 + 2*x^2 - 2*x^4]))/Sqrt[6]) - ((6 - Sqrt[6])*(3 + 
Sqrt[6]*x^2)*Sqrt[(3 - 2*x^2 + 2*x^4)/(3 + Sqrt[6]*x^2)^2]*EllipticF[2*Arc 
Tan[(2/3)^(1/4)*x], (6 + Sqrt[6])/12])/(4*6^(1/4)*Sqrt[-3 + 2*x^2 - 2*x^4] 
))/15
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 1405

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(-x)*(b^2 
- 2*a*c + b*c*x^2)*((a + b*x^2 + c*x^4)^(p + 1)/(2*a*(p + 1)*(b^2 - 4*a*c)) 
), x] + Simp[1/(2*a*(p + 1)*(b^2 - 4*a*c))   Int[(b^2 - 2*a*c + 2*(p + 1)*( 
b^2 - 4*a*c) + b*c*(4*p + 7)*x^2)*(a + b*x^2 + c*x^4)^(p + 1), x], x] /; Fr 
eeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && IntegerQ[2*p]
 

rule 1416

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c 
/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/ 
(2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) 
], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
 

rule 1509

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + b*x^2 + c*x^4]/(a*(1 + q 
^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2* 
x^2)^2)]/(q*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[2*ArcTan[q*x], 1/2 - b*(q^2 
/(4*c))], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 
- 4*a*c, 0] && PosQ[c/a]
 

rule 1511

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo 
l] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q   Int[1/Sqrt[a + b*x^2 + c*x^ 
4], x], x] - Simp[e/q   Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x] /; 
NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && Pos 
Q[c/a]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.75 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.94

method	result	size
risch	\(-\frac {x \left (x^{2}+2\right )}{15 \sqrt {-2 x^{4}+2 x^{2}-3}}-\frac {3 \sqrt {1-\left (\frac {1}{3}-\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{3}+\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3-3 i \sqrt {5}}\, x}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )}{5 \sqrt {3-3 i \sqrt {5}}\, \sqrt {-2 x^{4}+2 x^{2}-3}}+\frac {6 \sqrt {1-\left (\frac {1}{3}-\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{3}+\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {3-3 i \sqrt {5}}\, x}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )-\operatorname {EllipticE}\left (\frac {\sqrt {3-3 i \sqrt {5}}\, x}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )\right )}{5 \sqrt {3-3 i \sqrt {5}}\, \sqrt {-2 x^{4}+2 x^{2}-3}\, \left (2+2 i \sqrt {5}\right )}\)	\(235\)
default	\(\frac {-\frac {2}{15} x -\frac {1}{15} x^{3}}{\sqrt {-2 x^{4}+2 x^{2}-3}}-\frac {3 \sqrt {1-\left (\frac {1}{3}-\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{3}+\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3-3 i \sqrt {5}}\, x}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )}{5 \sqrt {3-3 i \sqrt {5}}\, \sqrt {-2 x^{4}+2 x^{2}-3}}+\frac {6 \sqrt {1-\left (\frac {1}{3}-\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{3}+\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {3-3 i \sqrt {5}}\, x}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )-\operatorname {EllipticE}\left (\frac {\sqrt {3-3 i \sqrt {5}}\, x}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )\right )}{5 \sqrt {3-3 i \sqrt {5}}\, \sqrt {-2 x^{4}+2 x^{2}-3}\, \left (2+2 i \sqrt {5}\right )}\)	\(238\)
elliptic	\(\frac {-\frac {2}{15} x -\frac {1}{15} x^{3}}{\sqrt {-2 x^{4}+2 x^{2}-3}}-\frac {3 \sqrt {1-\left (\frac {1}{3}-\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{3}+\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \operatorname {EllipticF}\left (\frac {\sqrt {3-3 i \sqrt {5}}\, x}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )}{5 \sqrt {3-3 i \sqrt {5}}\, \sqrt {-2 x^{4}+2 x^{2}-3}}+\frac {6 \sqrt {1-\left (\frac {1}{3}-\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \sqrt {1-\left (\frac {1}{3}+\frac {i \sqrt {5}}{3}\right ) x^{2}}\, \left (\operatorname {EllipticF}\left (\frac {\sqrt {3-3 i \sqrt {5}}\, x}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )-\operatorname {EllipticE}\left (\frac {\sqrt {3-3 i \sqrt {5}}\, x}{3}, \frac {\sqrt {-6+3 i \sqrt {5}}}{3}\right )\right )}{5 \sqrt {3-3 i \sqrt {5}}\, \sqrt {-2 x^{4}+2 x^{2}-3}\, \left (2+2 i \sqrt {5}\right )}\)	\(238\)

Input:

int(1/(-2*x^4+2*x^2-3)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/15*x*(x^2+2)/(-2*x^4+2*x^2-3)^(1/2)-3/5/(3-3*I*5^(1/2))^(1/2)*(1-(1/3-1 
/3*I*5^(1/2))*x^2)^(1/2)*(1-(1/3+1/3*I*5^(1/2))*x^2)^(1/2)/(-2*x^4+2*x^2-3 
)^(1/2)*EllipticF(1/3*(3-3*I*5^(1/2))^(1/2)*x,1/3*(-6+3*I*5^(1/2))^(1/2))+ 
6/5/(3-3*I*5^(1/2))^(1/2)*(1-(1/3-1/3*I*5^(1/2))*x^2)^(1/2)*(1-(1/3+1/3*I* 
5^(1/2))*x^2)^(1/2)/(-2*x^4+2*x^2-3)^(1/2)/(2+2*I*5^(1/2))*(EllipticF(1/3* 
(3-3*I*5^(1/2))^(1/2)*x,1/3*(-6+3*I*5^(1/2))^(1/2))-EllipticE(1/3*(3-3*I*5 
^(1/2))^(1/2)*x,1/3*(-6+3*I*5^(1/2))^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\left (-3+2 x^2-2 x^4\right )^{3/2}} \, dx=\frac {\sqrt {-3} {\left (2 \, x^{4} - 2 \, x^{2} + \sqrt {-5} {\left (2 \, x^{4} - 2 \, x^{2} + 3\right )} + 3\right )} \sqrt {\frac {1}{3} \, \sqrt {-5} + \frac {1}{3}} E(\arcsin \left (x \sqrt {\frac {1}{3} \, \sqrt {-5} + \frac {1}{3}}\right )\,|\,-\frac {1}{3} \, \sqrt {-5} - \frac {2}{3}) + 2 \, \sqrt {-3} {\left (2 \, x^{4} - 2 \, x^{2} - 2 \, \sqrt {-5} {\left (2 \, x^{4} - 2 \, x^{2} + 3\right )} + 3\right )} \sqrt {\frac {1}{3} \, \sqrt {-5} + \frac {1}{3}} F(\arcsin \left (x \sqrt {\frac {1}{3} \, \sqrt {-5} + \frac {1}{3}}\right )\,|\,-\frac {1}{3} \, \sqrt {-5} - \frac {2}{3}) + 6 \, \sqrt {-2 \, x^{4} + 2 \, x^{2} - 3} {\left (x^{3} + 2 \, x\right )}}{90 \, {\left (2 \, x^{4} - 2 \, x^{2} + 3\right )}} \] Input:

integrate(1/(-2*x^4+2*x^2-3)^(3/2),x, algorithm="fricas")
 

Output:

1/90*(sqrt(-3)*(2*x^4 - 2*x^2 + sqrt(-5)*(2*x^4 - 2*x^2 + 3) + 3)*sqrt(1/3 
*sqrt(-5) + 1/3)*elliptic_e(arcsin(x*sqrt(1/3*sqrt(-5) + 1/3)), -1/3*sqrt( 
-5) - 2/3) + 2*sqrt(-3)*(2*x^4 - 2*x^2 - 2*sqrt(-5)*(2*x^4 - 2*x^2 + 3) + 
3)*sqrt(1/3*sqrt(-5) + 1/3)*elliptic_f(arcsin(x*sqrt(1/3*sqrt(-5) + 1/3)), 
 -1/3*sqrt(-5) - 2/3) + 6*sqrt(-2*x^4 + 2*x^2 - 3)*(x^3 + 2*x))/(2*x^4 - 2 
*x^2 + 3)
 

Sympy [F]

\[ \int \frac {1}{\left (-3+2 x^2-2 x^4\right )^{3/2}} \, dx=\int \frac {1}{\left (- 2 x^{4} + 2 x^{2} - 3\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/(-2*x**4+2*x**2-3)**(3/2),x)
 

Output:

Integral((-2*x**4 + 2*x**2 - 3)**(-3/2), x)
 

Maxima [F]

\[ \int \frac {1}{\left (-3+2 x^2-2 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-2 \, x^{4} + 2 \, x^{2} - 3\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(-2*x^4+2*x^2-3)^(3/2),x, algorithm="maxima")
 

Output:

integrate((-2*x^4 + 2*x^2 - 3)^(-3/2), x)
 

Giac [F]

\[ \int \frac {1}{\left (-3+2 x^2-2 x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-2 \, x^{4} + 2 \, x^{2} - 3\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(-2*x^4+2*x^2-3)^(3/2),x, algorithm="giac")
 

Output:

integrate((-2*x^4 + 2*x^2 - 3)^(-3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (-3+2 x^2-2 x^4\right )^{3/2}} \, dx=\int \frac {1}{{\left (-2\,x^4+2\,x^2-3\right )}^{3/2}} \,d x \] Input:

int(1/(2*x^2 - 2*x^4 - 3)^(3/2),x)
 

Output:

int(1/(2*x^2 - 2*x^4 - 3)^(3/2), x)
 

Reduce [F]

\[ \int \frac {1}{\left (-3+2 x^2-2 x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {-2 x^{4}+2 x^{2}-3}}{4 x^{8}-8 x^{6}+16 x^{4}-12 x^{2}+9}d x \] Input:

int(1/(-2*x^4+2*x^2-3)^(3/2),x)
 

Output:

int(sqrt( - 2*x**4 + 2*x**2 - 3)/(4*x**8 - 8*x**6 + 16*x**4 - 12*x**2 + 9) 
,x)