\(\int \frac {d+e x+f x^2}{1+x^2+x^4} \, dx\) [28]

Optimal result

Integrand size = 21, antiderivative size = 74 \[ \int \frac {d+e x+f x^2}{1+x^2+x^4} \, dx=-\frac {(d+2 e+f) \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {(d-2 e+f) \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {1}{2} (d-f) \text {arctanh}\left (\frac {x}{1+x^2}\right ) \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.17 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.64 \[ \int \frac {d+e x+f x^2}{1+x^2+x^4} \, dx=\frac {\left (2 i d+\left (-i+\sqrt {3}\right ) f\right ) \arctan \left (\frac {1}{2} \left (-i+\sqrt {3}\right ) x\right )}{\sqrt {6+6 i \sqrt {3}}}+\frac {\left (-2 i d+\left (i+\sqrt {3}\right ) f\right ) \arctan \left (\frac {1}{2} \left (i+\sqrt {3}\right ) x\right )}{\sqrt {6-6 i \sqrt {3}}}-\frac {e \arctan \left (\frac {\sqrt {3}}{1+2 x^2}\right )}{\sqrt {3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.46, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {2202, 27, 1432, 1083, 217, 1483, 1142, 25, 1083, 217, 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.

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

\(\Big \downarrow \) 2202

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1432

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

\(\Big \downarrow \) 1083

\(\displaystyle \int \frac {f x^2+d}{x^4+x^2+1}dx-e \int \frac {1}{-x^4-3}d\left (2 x^2+1\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \int \frac {f x^2+d}{x^4+x^2+1}dx+\frac {e \arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}\)

\(\Big \downarrow \) 1483

\(\displaystyle \frac {1}{2} \int \frac {d-(d-f) x}{x^2-x+1}dx+\frac {1}{2} \int \frac {d+(d-f) x}{x^2+x+1}dx+\frac {e \arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} (d+f) \int \frac {1}{x^2-x+1}dx-\frac {1}{2} (d-f) \int -\frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{2} \left (\frac {1}{2} (d+f) \int \frac {1}{x^2+x+1}dx+\frac {1}{2} (d-f) \int \frac {2 x+1}{x^2+x+1}dx\right )+\frac {e \arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} (d+f) \int \frac {1}{x^2-x+1}dx+\frac {1}{2} (d-f) \int \frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{2} \left (\frac {1}{2} (d+f) \int \frac {1}{x^2+x+1}dx+\frac {1}{2} (d-f) \int \frac {2 x+1}{x^2+x+1}dx\right )+\frac {e \arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} (d-f) \int \frac {1-2 x}{x^2-x+1}dx-(d+f) \int \frac {1}{-(2 x-1)^2-3}d(2 x-1)\right )+\frac {1}{2} \left (\frac {1}{2} (d-f) \int \frac {2 x+1}{x^2+x+1}dx-(d+f) \int \frac {1}{-(2 x+1)^2-3}d(2 x+1)\right )+\frac {e \arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{2} \left (\frac {1}{2} (d-f) \int \frac {1-2 x}{x^2-x+1}dx+\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right ) (d+f)}{\sqrt {3}}\right )+\frac {1}{2} \left (\frac {1}{2} (d-f) \int \frac {2 x+1}{x^2+x+1}dx+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right ) (d+f)}{\sqrt {3}}\right )+\frac {e \arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right ) (d+f)}{\sqrt {3}}-\frac {1}{2} (d-f) \log \left (x^2-x+1\right )\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right ) (d+f)}{\sqrt {3}}+\frac {1}{2} (d-f) \log \left (x^2+x+1\right )\right )+\frac {e \arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{\sqrt {3}}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 217

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

rule 1083

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]
 

rule 1103

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]
 

rule 1142

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]
 

rule 1432

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

rule 1483

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]}, Simp[1/(2*c*q*r)   In 
t[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Simp[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] && N 
eQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]
 

rule 2202

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

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.11

method	result	size
default	\(\frac {\left (d -f \right ) \ln \left (x^{2}+x +1\right )}{4}+\frac {\left (\frac {d}{2}-e +\frac {f}{2}\right ) \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{3}+\frac {\left (f -d \right ) \ln \left (x^{2}-x +1\right )}{4}+\frac {\left (\frac {d}{2}+e +\frac {f}{2}\right ) \sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{3}\)	\(82\)
risch	\(\text {Expression too large to display}\)	\(7878\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01 \[ \int \frac {d+e x+f x^2}{1+x^2+x^4} \, dx=\frac {1}{6} \, \sqrt {3} {\left (d - 2 \, e + f\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} {\left (d + 2 \, e + f\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{4} \, {\left (d - f\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, {\left (d - f\right )} \log \left (x^{2} - x + 1\right ) \] Input:

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

Output:

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)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 56.50 (sec) , antiderivative size = 3589, normalized size of antiderivative = 48.50 \[ \int \frac {d+e x+f x^2}{1+x^2+x^4} \, dx=\text {Too large to display} \] Input:

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

Output:

(-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 - s 
qrt(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 - 1 
44*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*...
                                                                                    
                                                                                    
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01 \[ \int \frac {d+e x+f x^2}{1+x^2+x^4} \, dx=\frac {1}{6} \, \sqrt {3} {\left (d - 2 \, e + f\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} {\left (d + 2 \, e + f\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{4} \, {\left (d - f\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, {\left (d - f\right )} \log \left (x^{2} - x + 1\right ) \] Input:

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

Output:

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)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01 \[ \int \frac {d+e x+f x^2}{1+x^2+x^4} \, dx=\frac {1}{6} \, \sqrt {3} {\left (d - 2 \, e + f\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} {\left (d + 2 \, e + f\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{4} \, {\left (d - f\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{4} \, {\left (d - f\right )} \log \left (x^{2} - x + 1\right ) \] Input:

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

Output:

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)
 

Mupad [B] (verification not implemented)

Time = 18.27 (sec) , antiderivative size = 159, normalized size of antiderivative = 2.15 \[ \int \frac {d+e x+f x^2}{1+x^2+x^4} \, dx=-\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {d}{4}-\frac {f}{4}+\frac {\sqrt {3}\,d\,1{}\mathrm {i}}{12}+\frac {\sqrt {3}\,e\,1{}\mathrm {i}}{6}+\frac {\sqrt {3}\,f\,1{}\mathrm {i}}{12}\right )-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {f}{4}-\frac {d}{4}+\frac {\sqrt {3}\,d\,1{}\mathrm {i}}{12}-\frac {\sqrt {3}\,e\,1{}\mathrm {i}}{6}+\frac {\sqrt {3}\,f\,1{}\mathrm {i}}{12}\right )+\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {f}{4}-\frac {d}{4}+\frac {\sqrt {3}\,d\,1{}\mathrm {i}}{12}+\frac {\sqrt {3}\,e\,1{}\mathrm {i}}{6}+\frac {\sqrt {3}\,f\,1{}\mathrm {i}}{12}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {d}{4}-\frac {f}{4}+\frac {\sqrt {3}\,d\,1{}\mathrm {i}}{12}-\frac {\sqrt {3}\,e\,1{}\mathrm {i}}{6}+\frac {\sqrt {3}\,f\,1{}\mathrm {i}}{12}\right ) \] Input:

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

Output:

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)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.91 \[ \int \frac {d+e x+f x^2}{1+x^2+x^4} \, dx=\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right ) d}{6}+\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right ) e}{3}+\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right ) f}{6}+\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right ) d}{6}-\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right ) e}{3}+\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right ) f}{6}-\frac {\mathrm {log}\left (x^{2}-x +1\right ) d}{4}+\frac {\mathrm {log}\left (x^{2}-x +1\right ) f}{4}+\frac {\mathrm {log}\left (x^{2}+x +1\right ) d}{4}-\frac {\mathrm {log}\left (x^{2}+x +1\right ) f}{4} \] Input:

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

Output:

(2*sqrt(3)*atan((2*x - 1)/sqrt(3))*d + 4*sqrt(3)*atan((2*x - 1)/sqrt(3))*e 
 + 2*sqrt(3)*atan((2*x - 1)/sqrt(3))*f + 2*sqrt(3)*atan((2*x + 1)/sqrt(3)) 
*d - 4*sqrt(3)*atan((2*x + 1)/sqrt(3))*e + 2*sqrt(3)*atan((2*x + 1)/sqrt(3 
))*f - 3*log(x**2 - x + 1)*d + 3*log(x**2 - x + 1)*f + 3*log(x**2 + x + 1) 
*d - 3*log(x**2 + x + 1)*f)/12