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

Optimal. Leaf size=104 \[ -\frac {1}{4} (d-f) \log \left (x^2-x+1\right )+\frac {1}{4} (d-f) \log \left (x^2+x+1\right )-\frac {(d+f) \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {(d+f) \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {e \tan ^{-1}\left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}} \]

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Rubi [A]  time = 0.08, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {1673, 1169, 634, 618, 204, 628, 12, 1107} \begin {gather*} -\frac {1}{4} (d-f) \log \left (x^2-x+1\right )+\frac {1}{4} (d-f) \log \left (x^2+x+1\right )-\frac {(d+f) \tan ^{-1}\left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {(d+f) \tan ^{-1}\left (\frac {2 x+1}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {e \tan ^{-1}\left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((d + f)*ArcTan[(1 - 2*x)/Sqrt[3]])/(2*Sqrt[3]) + ((d + f)*ArcTan[(1 + 2*x)/Sqrt[3]])/(2*Sqrt[3]) + (e*ArcTan
[(1 + 2*x^2)/Sqrt[3]])/Sqrt[3] - ((d - f)*Log[1 - x + x^2])/4 + ((d - f)*Log[1 + x + x^2])/4

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 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 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 1107

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

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 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]

Rubi steps

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

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Mathematica [C]  time = 0.14, size = 121, normalized size = 1.16 \begin {gather*} \frac {\left (2 i d+\left (\sqrt {3}-i\right ) f\right ) \tan ^{-1}\left (\frac {1}{2} \left (\sqrt {3}-i\right ) x\right )}{\sqrt {6+6 i \sqrt {3}}}+\frac {\left (\left (\sqrt {3}+i\right ) f-2 i d\right ) \tan ^{-1}\left (\frac {1}{2} \left (\sqrt {3}+i\right ) x\right )}{\sqrt {6-6 i \sqrt {3}}}-\frac {e \tan ^{-1}\left (\frac {\sqrt {3}}{2 x^2+1}\right )}{\sqrt {3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {d+e x+f x^2}{1+x^2+x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.09, size = 75, normalized size = 0.72 \begin {gather*} \frac {1}{6} \, \sqrt {3} {\left (d - 2 \, e + f\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} {\left (d + 2 \, e + f\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{4} \, {\left (d - f\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, {\left (d - f\right )} \log \left (x^{2} - x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.23, size = 77, normalized size = 0.74 \begin {gather*} \frac {1}{6} \, \sqrt {3} {\left (d + f - 2 \, e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} {\left (d + f + 2 \, e\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{4} \, {\left (d - f\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, {\left (d - f\right )} \log \left (x^{2} - x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 148, normalized size = 1.42 \begin {gather*} \frac {\sqrt {3}\, d \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{6}+\frac {\sqrt {3}\, d \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}-\frac {d \ln \left (x^{2}-x +1\right )}{4}+\frac {d \ln \left (x^{2}+x +1\right )}{4}-\frac {\sqrt {3}\, e \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{3}+\frac {\sqrt {3}\, e \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}+\frac {\sqrt {3}\, f \arctan \left (\frac {\left (2 x +1\right ) \sqrt {3}}{3}\right )}{6}+\frac {\sqrt {3}\, f \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}+\frac {f \ln \left (x^{2}-x +1\right )}{4}-\frac {f \ln \left (x^{2}+x +1\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*d*ln(x^2+x+1)-1/4*ln(x^2+x+1)*f+1/6*3^(1/2)*d*arctan(1/3*(2*x+1)*3^(1/2))-1/3*3^(1/2)*e*arctan(1/3*(2*x+1)
*3^(1/2))+1/6*3^(1/2)*arctan(1/3*(2*x+1)*3^(1/2))*f+1/4*ln(x^2-x+1)*f-1/4*d*ln(x^2-x+1)+1/6*3^(1/2)*d*arctan(1
/3*(2*x-1)*3^(1/2))+1/3*3^(1/2)*e*arctan(1/3*(2*x-1)*3^(1/2))+1/6*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))*f

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maxima [A]  time = 2.58, size = 75, normalized size = 0.72 \begin {gather*} \frac {1}{6} \, \sqrt {3} {\left (d - 2 \, e + f\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} {\left (d + 2 \, e + f\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{4} \, {\left (d - f\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, {\left (d - f\right )} \log \left (x^{2} - x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.95, size = 159, normalized size = 1.53 \begin {gather*} -\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {d}{4}-\frac {f}{4}+\frac {\sqrt {3}\,d\,1{}\mathrm {i}}{12}+\frac {\sqrt {3}\,e\,1{}\mathrm {i}}{6}+\frac {\sqrt {3}\,f\,1{}\mathrm {i}}{12}\right )-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {f}{4}-\frac {d}{4}+\frac {\sqrt {3}\,d\,1{}\mathrm {i}}{12}-\frac {\sqrt {3}\,e\,1{}\mathrm {i}}{6}+\frac {\sqrt {3}\,f\,1{}\mathrm {i}}{12}\right )+\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {f}{4}-\frac {d}{4}+\frac {\sqrt {3}\,d\,1{}\mathrm {i}}{12}+\frac {\sqrt {3}\,e\,1{}\mathrm {i}}{6}+\frac {\sqrt {3}\,f\,1{}\mathrm {i}}{12}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {d}{4}-\frac {f}{4}+\frac {\sqrt {3}\,d\,1{}\mathrm {i}}{12}-\frac {\sqrt {3}\,e\,1{}\mathrm {i}}{6}+\frac {\sqrt {3}\,f\,1{}\mathrm {i}}{12}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + (3^(1/2)*1i)/2 - 1/2)*(f/4 - d/4 + (3^(1/2)*d*1i)/12 + (3^(1/2)*e*1i)/6 + (3^(1/2)*f*1i)/12) - log(x -
(3^(1/2)*1i)/2 + 1/2)*(f/4 - d/4 + (3^(1/2)*d*1i)/12 - (3^(1/2)*e*1i)/6 + (3^(1/2)*f*1i)/12) - log(x - (3^(1/
2)*1i)/2 - 1/2)*(d/4 - f/4 + (3^(1/2)*d*1i)/12 + (3^(1/2)*e*1i)/6 + (3^(1/2)*f*1i)/12) + log(x + (3^(1/2)*1i)/
2 + 1/2)*(d/4 - f/4 + (3^(1/2)*d*1i)/12 - (3^(1/2)*e*1i)/6 + (3^(1/2)*f*1i)/12)

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sympy [C]  time = 98.60, size = 3589, normalized size = 34.51

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12)*log(x + (-7*d**5*e + 6*d**5*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12
) + 25*d**4*e*f + 18*d**4*f*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12) - 15*d**3*e**3 - 18*d**3*e**2*(-d/4 + f/
4 - sqrt(3)*I*(d + 2*e + f)/12) - 25*d**3*e*f**2 + 60*d**3*e*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12)**2 - 42
*d**3*f**2*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12) + 72*d**3*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12)**3 +
108*d**2*e**2*f*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12) + 20*d**2*e*f**3 - 144*d**2*e*f*(-d/4 + f/4 - sqrt(3
)*I*(d + 2*e + f)/12)**2 - 12*d**2*f**3*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12) - 144*d**2*f*(-d/4 + f/4 - s
qrt(3)*I*(d + 2*e + f)/12)**3 + 4*d*e**5 + 24*d*e**4*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12) + 15*d*e**3*f**
2 + 48*d*e**3*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12)**2 - 54*d*e**2*f**2*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e +
f)/12) + 288*d*e**2*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12)**3 - 20*d*e*f**4 + 180*d*e*f**2*(-d/4 + f/4 - s
qrt(3)*I*(d + 2*e + f)/12)**2 + 36*d*f**4*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12) - 72*d*f**2*(-d/4 + f/4 -
sqrt(3)*I*(d + 2*e + f)/12)**3 - 8*e**5*f - 96*e**3*f*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12)**2 + 36*e**2*f
**3*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12) + 11*e*f**5 - 48*e*f**3*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12
)**2 - 6*f**5*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12) + 144*f**3*(-d/4 + f/4 - sqrt(3)*I*(d + 2*e + f)/12)**
3)/(3*d**6 - 3*d**5*f - 8*d**4*e**2 - 3*d**4*f**2 + 40*d**3*e**2*f + 6*d**3*f**3 - 16*d**2*e**4 - 48*d**2*e**2
*f**2 - 3*d**2*f**4 + 16*d*e**4*f + 40*d*e**2*f**3 - 3*d*f**5 - 16*e**4*f**2 - 8*e**2*f**4 + 3*f**6)) + (-d/4
+ f/4 + sqrt(3)*I*(d + 2*e + f)/12)*log(x + (-7*d**5*e + 6*d**5*(-d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12) + 25
*d**4*e*f + 18*d**4*f*(-d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12) - 15*d**3*e**3 - 18*d**3*e**2*(-d/4 + f/4 + sq
rt(3)*I*(d + 2*e + f)/12) - 25*d**3*e*f**2 + 60*d**3*e*(-d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12)**2 - 42*d**3*
f**2*(-d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12) + 72*d**3*(-d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12)**3 + 108*d*
*2*e**2*f*(-d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12) + 20*d**2*e*f**3 - 144*d**2*e*f*(-d/4 + f/4 + sqrt(3)*I*(d
+ 2*e + f)/12)**2 - 12*d**2*f**3*(-d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12) - 144*d**2*f*(-d/4 + f/4 + sqrt(3)
*I*(d + 2*e + f)/12)**3 + 4*d*e**5 + 24*d*e**4*(-d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12) + 15*d*e**3*f**2 + 48
*d*e**3*(-d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12)**2 - 54*d*e**2*f**2*(-d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12
) + 288*d*e**2*(-d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12)**3 - 20*d*e*f**4 + 180*d*e*f**2*(-d/4 + f/4 + sqrt(3)
*I*(d + 2*e + f)/12)**2 + 36*d*f**4*(-d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12) - 72*d*f**2*(-d/4 + f/4 + sqrt(3
)*I*(d + 2*e + f)/12)**3 - 8*e**5*f - 96*e**3*f*(-d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12)**2 + 36*e**2*f**3*(-
d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12) + 11*e*f**5 - 48*e*f**3*(-d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12)**2 -
6*f**5*(-d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12) + 144*f**3*(-d/4 + f/4 + sqrt(3)*I*(d + 2*e + f)/12)**3)/(3*
d**6 - 3*d**5*f - 8*d**4*e**2 - 3*d**4*f**2 + 40*d**3*e**2*f + 6*d**3*f**3 - 16*d**2*e**4 - 48*d**2*e**2*f**2
- 3*d**2*f**4 + 16*d*e**4*f + 40*d*e**2*f**3 - 3*d*f**5 - 16*e**4*f**2 - 8*e**2*f**4 + 3*f**6)) + (d/4 - f/4 -
sqrt(3)*I*(d - 2*e + f)/12)*log(x + (-7*d**5*e + 6*d**5*(d/4 - f/4 - sqrt(3)*I*(d - 2*e + f)/12) + 25*d**4*e*
f + 18*d**4*f*(d/4 - f/4 - sqrt(3)*I*(d - 2*e + f)/12) - 15*d**3*e**3 - 18*d**3*e**2*(d/4 - f/4 - sqrt(3)*I*(d
- 2*e + f)/12) - 25*d**3*e*f**2 + 60*d**3*e*(d/4 - f/4 - sqrt(3)*I*(d - 2*e + f)/12)**2 - 42*d**3*f**2*(d/4 -
f/4 - sqrt(3)*I*(d - 2*e + f)/12) + 72*d**3*(d/4 - f/4 - sqrt(3)*I*(d - 2*e + f)/12)**3 + 108*d**2*e**2*f*(d/
4 - f/4 - sqrt(3)*I*(d - 2*e + f)/12) + 20*d**2*e*f**3 - 144*d**2*e*f*(d/4 - f/4 - sqrt(3)*I*(d - 2*e + f)/12)
**2 - 12*d**2*f**3*(d/4 - f/4 - sqrt(3)*I*(d - 2*e + f)/12) - 144*d**2*f*(d/4 - f/4 - sqrt(3)*I*(d - 2*e + f)/
12)**3 + 4*d*e**5 + 24*d*e**4*(d/4 - f/4 - sqrt(3)*I*(d - 2*e + f)/12) + 15*d*e**3*f**2 + 48*d*e**3*(d/4 - f/4
- sqrt(3)*I*(d - 2*e + f)/12)**2 - 54*d*e**2*f**2*(d/4 - f/4 - sqrt(3)*I*(d - 2*e + f)/12) + 288*d*e**2*(d/4
- f/4 - sqrt(3)*I*(d - 2*e + f)/12)**3 - 20*d*e*f**4 + 180*d*e*f**2*(d/4 - f/4 - sqrt(3)*I*(d - 2*e + f)/12)**
2 + 36*d*f**4*(d/4 - f/4 - sqrt(3)*I*(d - 2*e + f)/12) - 72*d*f**2*(d/4 - f/4 - sqrt(3)*I*(d - 2*e + f)/12)**3
- 8*e**5*f - 96*e**3*f*(d/4 - f/4 - sqrt(3)*I*(d - 2*e + f)/12)**2 + 36*e**2*f**3*(d/4 - f/4 - sqrt(3)*I*(d -
2*e + f)/12) + 11*e*f**5 - 48*e*f**3*(d/4 - f/4 - sqrt(3)*I*(d - 2*e + f)/12)**2 - 6*f**5*(d/4 - f/4 - sqrt(3
)*I*(d - 2*e + f)/12) + 144*f**3*(d/4 - f/4 - sqrt(3)*I*(d - 2*e + f)/12)**3)/(3*d**6 - 3*d**5*f - 8*d**4*e**2
- 3*d**4*f**2 + 40*d**3*e**2*f + 6*d**3*f**3 - 16*d**2*e**4 - 48*d**2*e**2*f**2 - 3*d**2*f**4 + 16*d*e**4*f +
40*d*e**2*f**3 - 3*d*f**5 - 16*e**4*f**2 - 8*e**2*f**4 + 3*f**6)) + (d/4 - f/4 + sqrt(3)*I*(d - 2*e + f)/12)*
log(x + (-7*d**5*e + 6*d**5*(d/4 - f/4 + sqrt(3)*I*(d - 2*e + f)/12) + 25*d**4*e*f + 18*d**4*f*(d/4 - f/4 + sq
rt(3)*I*(d - 2*e + f)/12) - 15*d**3*e**3 - 18*d**3*e**2*(d/4 - f/4 + sqrt(3)*I*(d - 2*e + f)/12) - 25*d**3*e*f
**2 + 60*d**3*e*(d/4 - f/4 + sqrt(3)*I*(d - 2*e + f)/12)**2 - 42*d**3*f**2*(d/4 - f/4 + sqrt(3)*I*(d - 2*e + f
)/12) + 72*d**3*(d/4 - f/4 + sqrt(3)*I*(d - 2*e + f)/12)**3 + 108*d**2*e**2*f*(d/4 - f/4 + sqrt(3)*I*(d - 2*e
+ f)/12) + 20*d**2*e*f**3 - 144*d**2*e*f*(d/4 - f/4 + sqrt(3)*I*(d - 2*e + f)/12)**2 - 12*d**2*f**3*(d/4 - f/4
+ sqrt(3)*I*(d - 2*e + f)/12) - 144*d**2*f*(d/4 - f/4 + sqrt(3)*I*(d - 2*e + f)/12)**3 + 4*d*e**5 + 24*d*e**4
*(d/4 - f/4 + sqrt(3)*I*(d - 2*e + f)/12) + 15*d*e**3*f**2 + 48*d*e**3*(d/4 - f/4 + sqrt(3)*I*(d - 2*e + f)/12
)**2 - 54*d*e**2*f**2*(d/4 - f/4 + sqrt(3)*I*(d - 2*e + f)/12) + 288*d*e**2*(d/4 - f/4 + sqrt(3)*I*(d - 2*e +
f)/12)**3 - 20*d*e*f**4 + 180*d*e*f**2*(d/4 - f/4 + sqrt(3)*I*(d - 2*e + f)/12)**2 + 36*d*f**4*(d/4 - f/4 + sq
rt(3)*I*(d - 2*e + f)/12) - 72*d*f**2*(d/4 - f/4 + sqrt(3)*I*(d - 2*e + f)/12)**3 - 8*e**5*f - 96*e**3*f*(d/4
- f/4 + sqrt(3)*I*(d - 2*e + f)/12)**2 + 36*e**2*f**3*(d/4 - f/4 + sqrt(3)*I*(d - 2*e + f)/12) + 11*e*f**5 - 4
8*e*f**3*(d/4 - f/4 + sqrt(3)*I*(d - 2*e + f)/12)**2 - 6*f**5*(d/4 - f/4 + sqrt(3)*I*(d - 2*e + f)/12) + 144*f
**3*(d/4 - f/4 + sqrt(3)*I*(d - 2*e + f)/12)**3)/(3*d**6 - 3*d**5*f - 8*d**4*e**2 - 3*d**4*f**2 + 40*d**3*e**2
*f + 6*d**3*f**3 - 16*d**2*e**4 - 48*d**2*e**2*f**2 - 3*d**2*f**4 + 16*d*e**4*f + 40*d*e**2*f**3 - 3*d*f**5 -
16*e**4*f**2 - 8*e**2*f**4 + 3*f**6))

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