3.33 \(\int \frac{d+e x+f x^2+g x^3}{(1+x^2+x^4)^2} \, dx\)

Optimal. Leaf size=179 \[ \frac{x \left (x^2 (-(d-2 f))+d+f\right )}{6 \left (x^4+x^2+1\right )}-\frac{1}{8} (2 d-f) \log \left (x^2-x+1\right )+\frac{1}{8} (2 d-f) \log \left (x^2+x+1\right )-\frac{(4 d+f) \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{12 \sqrt{3}}+\frac{(4 d+f) \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{12 \sqrt{3}}+\frac{x^2 (2 e-g)+e-2 g}{6 \left (x^4+x^2+1\right )}+\frac{(2 e-g) \tan ^{-1}\left (\frac{2 x^2+1}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

[Out]

(x*(d + f - (d - 2*f)*x^2))/(6*(1 + x^2 + x^4)) + (e - 2*g + (2*e - g)*x^2)/(6*(1 + x^2 + x^4)) - ((4*d + f)*A
rcTan[(1 - 2*x)/Sqrt[3]])/(12*Sqrt[3]) + ((4*d + f)*ArcTan[(1 + 2*x)/Sqrt[3]])/(12*Sqrt[3]) + ((2*e - g)*ArcTa
n[(1 + 2*x^2)/Sqrt[3]])/(3*Sqrt[3]) - ((2*d - f)*Log[1 - x + x^2])/8 + ((2*d - f)*Log[1 + x + x^2])/8

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Rubi [A]  time = 0.141012, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {1673, 1178, 1169, 634, 618, 204, 628, 1247, 638} \[ \frac{x \left (x^2 (-(d-2 f))+d+f\right )}{6 \left (x^4+x^2+1\right )}-\frac{1}{8} (2 d-f) \log \left (x^2-x+1\right )+\frac{1}{8} (2 d-f) \log \left (x^2+x+1\right )-\frac{(4 d+f) \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{12 \sqrt{3}}+\frac{(4 d+f) \tan ^{-1}\left (\frac{2 x+1}{\sqrt{3}}\right )}{12 \sqrt{3}}+\frac{x^2 (2 e-g)+e-2 g}{6 \left (x^4+x^2+1\right )}+\frac{(2 e-g) \tan ^{-1}\left (\frac{2 x^2+1}{\sqrt{3}}\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*x^2 + g*x^3)/(1 + x^2 + x^4)^2,x]

[Out]

(x*(d + f - (d - 2*f)*x^2))/(6*(1 + x^2 + x^4)) + (e - 2*g + (2*e - g)*x^2)/(6*(1 + x^2 + x^4)) - ((4*d + f)*A
rcTan[(1 - 2*x)/Sqrt[3]])/(12*Sqrt[3]) + ((4*d + f)*ArcTan[(1 + 2*x)/Sqrt[3]])/(12*Sqrt[3]) + ((2*e - g)*ArcTa
n[(1 + 2*x^2)/Sqrt[3]])/(3*Sqrt[3]) - ((2*d - f)*Log[1 - x + x^2])/8 + ((2*d - f)*Log[1 + x + x^2])/8

Rule 1673

Int[(Pq_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], k}, Int[Sum[Coeff[
Pq, x, 2*k]*x^(2*k), {k, 0, q/2}]*(a + b*x^2 + c*x^4)^p, x] + Int[x*Sum[Coeff[Pq, x, 2*k + 1]*x^(2*k), {k, 0,
(q - 1)/2}]*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] &&  !PolyQ[Pq, x^2]

Rule 1178

Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[(x*(a*b*e - d*(b^2 - 2*
a*c) - c*(b*d - 2*a*e)*x^2)*(a + b*x^2 + c*x^4)^(p + 1))/(2*a*(p + 1)*(b^2 - 4*a*c)), x] + Dist[1/(2*a*(p + 1)
*(b^2 - 4*a*c)), Int[Simp[(2*p + 3)*d*b^2 - a*b*e - 2*a*c*d*(4*p + 5) + (4*p + 7)*(d*b - 2*a*e)*c*x^2, x]*(a +
b*x^2 + c*x^4)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e
^2, 0] && LtQ[p, -1] && IntegerQ[2*p]

Rule 1169

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

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(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]

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[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]

Rule 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rubi steps

\begin{align*} \int \frac{d+e x+f x^2+g x^3}{\left (1+x^2+x^4\right )^2} \, dx &=\int \frac{d+f x^2}{\left (1+x^2+x^4\right )^2} \, dx+\int \frac{x \left (e+g x^2\right )}{\left (1+x^2+x^4\right )^2} \, dx\\ &=\frac{x \left (d+f-(d-2 f) x^2\right )}{6 \left (1+x^2+x^4\right )}+\frac{1}{6} \int \frac{5 d-f+(-d+2 f) x^2}{1+x^2+x^4} \, dx+\frac{1}{2} \operatorname{Subst}\left (\int \frac{e+g x}{\left (1+x+x^2\right )^2} \, dx,x,x^2\right )\\ &=\frac{x \left (d+f-(d-2 f) x^2\right )}{6 \left (1+x^2+x^4\right )}+\frac{e-2 g+(2 e-g) x^2}{6 \left (1+x^2+x^4\right )}+\frac{1}{12} \int \frac{5 d-f-(6 d-3 f) x}{1-x+x^2} \, dx+\frac{1}{12} \int \frac{5 d-f+(6 d-3 f) x}{1+x+x^2} \, dx+\frac{1}{6} (2 e-g) \operatorname{Subst}\left (\int \frac{1}{1+x+x^2} \, dx,x,x^2\right )\\ &=\frac{x \left (d+f-(d-2 f) x^2\right )}{6 \left (1+x^2+x^4\right )}+\frac{e-2 g+(2 e-g) x^2}{6 \left (1+x^2+x^4\right )}+\frac{1}{8} (2 d-f) \int \frac{1+2 x}{1+x+x^2} \, dx+\frac{1}{8} (-2 d+f) \int \frac{-1+2 x}{1-x+x^2} \, dx+\frac{1}{24} (4 d+f) \int \frac{1}{1-x+x^2} \, dx+\frac{1}{24} (4 d+f) \int \frac{1}{1+x+x^2} \, dx+\frac{1}{3} (-2 e+g) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 x^2\right )\\ &=\frac{x \left (d+f-(d-2 f) x^2\right )}{6 \left (1+x^2+x^4\right )}+\frac{e-2 g+(2 e-g) x^2}{6 \left (1+x^2+x^4\right )}+\frac{(2 e-g) \tan ^{-1}\left (\frac{1+2 x^2}{\sqrt{3}}\right )}{3 \sqrt{3}}-\frac{1}{8} (2 d-f) \log \left (1-x+x^2\right )+\frac{1}{8} (2 d-f) \log \left (1+x+x^2\right )+\frac{1}{12} (-4 d-f) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 x\right )+\frac{1}{12} (-4 d-f) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 x\right )\\ &=\frac{x \left (d+f-(d-2 f) x^2\right )}{6 \left (1+x^2+x^4\right )}+\frac{e-2 g+(2 e-g) x^2}{6 \left (1+x^2+x^4\right )}-\frac{(4 d+f) \tan ^{-1}\left (\frac{1-2 x}{\sqrt{3}}\right )}{12 \sqrt{3}}+\frac{(4 d+f) \tan ^{-1}\left (\frac{1+2 x}{\sqrt{3}}\right )}{12 \sqrt{3}}+\frac{(2 e-g) \tan ^{-1}\left (\frac{1+2 x^2}{\sqrt{3}}\right )}{3 \sqrt{3}}-\frac{1}{8} (2 d-f) \log \left (1-x+x^2\right )+\frac{1}{8} (2 d-f) \log \left (1+x+x^2\right )\\ \end{align*}

Mathematica [C]  time = 0.434399, size = 200, normalized size = 1.12 \[ \frac{1}{36} \left (\frac{6 \left (x \left (-d x^2+d+2 f x^2+f\right )+2 e x^2+e-g \left (x^2+2\right )\right )}{x^4+x^2+1}-\frac{\left (\left (\sqrt{3}-11 i\right ) d-2 \left (\sqrt{3}-2 i\right ) f\right ) \tan ^{-1}\left (\frac{1}{2} \left (\sqrt{3}-i\right ) x\right )}{\sqrt{\frac{1}{6} \left (1+i \sqrt{3}\right )}}-\frac{\left (\left (\sqrt{3}+11 i\right ) d-2 \left (\sqrt{3}+2 i\right ) f\right ) \tan ^{-1}\left (\frac{1}{2} \left (\sqrt{3}+i\right ) x\right )}{\sqrt{\frac{1}{6} \left (1-i \sqrt{3}\right )}}-4 \sqrt{3} (2 e-g) \tan ^{-1}\left (\frac{\sqrt{3}}{2 x^2+1}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d + e*x + f*x^2 + g*x^3)/(1 + x^2 + x^4)^2,x]

[Out]

((6*(e + 2*e*x^2 - g*(2 + x^2) + x*(d + f - d*x^2 + 2*f*x^2)))/(1 + x^2 + x^4) - (((-11*I + Sqrt[3])*d - 2*(-2
*I + Sqrt[3])*f)*ArcTan[((-I + Sqrt[3])*x)/2])/Sqrt[(1 + I*Sqrt[3])/6] - (((11*I + Sqrt[3])*d - 2*(2*I + Sqrt[
3])*f)*ArcTan[((I + Sqrt[3])*x)/2])/Sqrt[(1 - I*Sqrt[3])/6] - 4*Sqrt[3]*(2*e - g)*ArcTan[Sqrt[3]/(1 + 2*x^2)])
/36

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Maple [A]  time = 0.013, size = 260, normalized size = 1.5 \begin{align*}{\frac{1}{4\,{x}^{2}+4\,x+4} \left ( \left ( -{\frac{d}{3}}-{\frac{e}{3}}-{\frac{g}{3}}+{\frac{2\,f}{3}} \right ) x-{\frac{2\,d}{3}}+{\frac{e}{3}}-{\frac{2\,g}{3}}+{\frac{f}{3}} \right ) }+{\frac{d\ln \left ({x}^{2}+x+1 \right ) }{4}}-{\frac{\ln \left ({x}^{2}+x+1 \right ) f}{8}}+{\frac{d\sqrt{3}}{9}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) }-{\frac{2\,\sqrt{3}e}{9}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) }+{\frac{\sqrt{3}f}{36}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) }+{\frac{\sqrt{3}g}{9}\arctan \left ({\frac{ \left ( 1+2\,x \right ) \sqrt{3}}{3}} \right ) }-{\frac{1}{4\,{x}^{2}-4\,x+4} \left ( \left ({\frac{d}{3}}-{\frac{e}{3}}-{\frac{g}{3}}-{\frac{2\,f}{3}} \right ) x-{\frac{2\,d}{3}}-{\frac{e}{3}}+{\frac{2\,g}{3}}+{\frac{f}{3}} \right ) }-{\frac{d\ln \left ({x}^{2}-x+1 \right ) }{4}}+{\frac{\ln \left ({x}^{2}-x+1 \right ) f}{8}}+{\frac{d\sqrt{3}}{9}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) }+{\frac{2\,\sqrt{3}e}{9}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) }+{\frac{\sqrt{3}f}{36}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) }-{\frac{\sqrt{3}g}{9}\arctan \left ({\frac{ \left ( 2\,x-1 \right ) \sqrt{3}}{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*x^3+f*x^2+e*x+d)/(x^4+x^2+1)^2,x)

[Out]

1/4*((-1/3*d-1/3*e-1/3*g+2/3*f)*x-2/3*d+1/3*e-2/3*g+1/3*f)/(x^2+x+1)+1/4*d*ln(x^2+x+1)-1/8*ln(x^2+x+1)*f+1/9*d
*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)-2/9*3^(1/2)*arctan(1/3*(1+2*x)*3^(1/2))*e+1/36*3^(1/2)*arctan(1/3*(1+2*x)
*3^(1/2))*f+1/9*3^(1/2)*arctan(1/3*(1+2*x)*3^(1/2))*g-1/4*((1/3*d-1/3*e-1/3*g-2/3*f)*x-2/3*d-1/3*e+2/3*g+1/3*f
)/(x^2-x+1)-1/4*d*ln(x^2-x+1)+1/8*ln(x^2-x+1)*f+1/9*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))*d+2/9*3^(1/2)*arctan(1
/3*(2*x-1)*3^(1/2))*e+1/36*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))*f-1/9*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))*g

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Maxima [A]  time = 1.45662, size = 182, normalized size = 1.02 \begin{align*} \frac{1}{36} \, \sqrt{3}{\left (4 \, d - 8 \, e + f + 4 \, g\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) + \frac{1}{36} \, \sqrt{3}{\left (4 \, d + 8 \, e + f - 4 \, g\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + \frac{1}{8} \,{\left (2 \, d - f\right )} \log \left (x^{2} + x + 1\right ) - \frac{1}{8} \,{\left (2 \, d - f\right )} \log \left (x^{2} - x + 1\right ) - \frac{{\left (d - 2 \, f\right )} x^{3} -{\left (2 \, e - g\right )} x^{2} -{\left (d + f\right )} x - e + 2 \, g}{6 \,{\left (x^{4} + x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^3+f*x^2+e*x+d)/(x^4+x^2+1)^2,x, algorithm="maxima")

[Out]

1/36*sqrt(3)*(4*d - 8*e + f + 4*g)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/36*sqrt(3)*(4*d + 8*e + f - 4*g)*arctan(1
/3*sqrt(3)*(2*x - 1)) + 1/8*(2*d - f)*log(x^2 + x + 1) - 1/8*(2*d - f)*log(x^2 - x + 1) - 1/6*((d - 2*f)*x^3 -
(2*e - g)*x^2 - (d + f)*x - e + 2*g)/(x^4 + x^2 + 1)

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Fricas [A]  time = 3.15898, size = 616, normalized size = 3.44 \begin{align*} -\frac{12 \,{\left (d - 2 \, f\right )} x^{3} - 12 \,{\left (2 \, e - g\right )} x^{2} - 2 \, \sqrt{3}{\left ({\left (4 \, d - 8 \, e + f + 4 \, g\right )} x^{4} +{\left (4 \, d - 8 \, e + f + 4 \, g\right )} x^{2} + 4 \, d - 8 \, e + f + 4 \, g\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) - 2 \, \sqrt{3}{\left ({\left (4 \, d + 8 \, e + f - 4 \, g\right )} x^{4} +{\left (4 \, d + 8 \, e + f - 4 \, g\right )} x^{2} + 4 \, d + 8 \, e + f - 4 \, g\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) - 12 \,{\left (d + f\right )} x - 9 \,{\left ({\left (2 \, d - f\right )} x^{4} +{\left (2 \, d - f\right )} x^{2} + 2 \, d - f\right )} \log \left (x^{2} + x + 1\right ) + 9 \,{\left ({\left (2 \, d - f\right )} x^{4} +{\left (2 \, d - f\right )} x^{2} + 2 \, d - f\right )} \log \left (x^{2} - x + 1\right ) - 12 \, e + 24 \, g}{72 \,{\left (x^{4} + x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^3+f*x^2+e*x+d)/(x^4+x^2+1)^2,x, algorithm="fricas")

[Out]

-1/72*(12*(d - 2*f)*x^3 - 12*(2*e - g)*x^2 - 2*sqrt(3)*((4*d - 8*e + f + 4*g)*x^4 + (4*d - 8*e + f + 4*g)*x^2
+ 4*d - 8*e + f + 4*g)*arctan(1/3*sqrt(3)*(2*x + 1)) - 2*sqrt(3)*((4*d + 8*e + f - 4*g)*x^4 + (4*d + 8*e + f -
4*g)*x^2 + 4*d + 8*e + f - 4*g)*arctan(1/3*sqrt(3)*(2*x - 1)) - 12*(d + f)*x - 9*((2*d - f)*x^4 + (2*d - f)*x
^2 + 2*d - f)*log(x^2 + x + 1) + 9*((2*d - f)*x^4 + (2*d - f)*x^2 + 2*d - f)*log(x^2 - x + 1) - 12*e + 24*g)/(
x^4 + x^2 + 1)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x**3+f*x**2+e*x+d)/(x**4+x**2+1)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.08282, size = 192, normalized size = 1.07 \begin{align*} \frac{1}{36} \, \sqrt{3}{\left (4 \, d + f + 4 \, g - 8 \, e\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x + 1\right )}\right ) + \frac{1}{36} \, \sqrt{3}{\left (4 \, d + f - 4 \, g + 8 \, e\right )} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \, x - 1\right )}\right ) + \frac{1}{8} \,{\left (2 \, d - f\right )} \log \left (x^{2} + x + 1\right ) - \frac{1}{8} \,{\left (2 \, d - f\right )} \log \left (x^{2} - x + 1\right ) - \frac{d x^{3} - 2 \, f x^{3} + g x^{2} - 2 \, x^{2} e - d x - f x + 2 \, g - e}{6 \,{\left (x^{4} + x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*x^3+f*x^2+e*x+d)/(x^4+x^2+1)^2,x, algorithm="giac")

[Out]

1/36*sqrt(3)*(4*d + f + 4*g - 8*e)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/36*sqrt(3)*(4*d + f - 4*g + 8*e)*arctan(1
/3*sqrt(3)*(2*x - 1)) + 1/8*(2*d - f)*log(x^2 + x + 1) - 1/8*(2*d - f)*log(x^2 - x + 1) - 1/6*(d*x^3 - 2*f*x^3
+ g*x^2 - 2*x^2*e - d*x - f*x + 2*g - e)/(x^4 + x^2 + 1)