Integrand size = 26, antiderivative size = 179 \[ \int \frac {d+e x+f x^2+g x^3}{\left (1+x^2+x^4\right )^2} \, dx=\frac {x \left (d+f-(d-2 f) x^2\right )}{6 \left (1+x^2+x^4\right )}+\frac {e-2 g+(2 e-g) x^2}{6 \left (1+x^2+x^4\right )}-\frac {(4 d+f) \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{12 \sqrt {3}}+\frac {(4 d+f) \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{12 \sqrt {3}}+\frac {(2 e-g) \arctan \left (\frac {1+2 x^2}{\sqrt {3}}\right )}{3 \sqrt {3}}-\frac {1}{8} (2 d-f) \log \left (1-x+x^2\right )+\frac {1}{8} (2 d-f) \log \left (1+x+x^2\right ) \]
1/6*x*(d+f-(d-2*f)*x^2)/(x^4+x^2+1)+1/6*(e-2*g+(2*e-g)*x^2)/(x^4+x^2+1)-1/ 8*(2*d-f)*ln(x^2-x+1)+1/8*(2*d-f)*ln(x^2+x+1)-1/36*(4*d+f)*arctan(1/3*(1-2 *x)*3^(1/2))*3^(1/2)+1/36*(4*d+f)*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)+1/9* (2*e-g)*arctan(1/3*(2*x^2+1)*3^(1/2))*3^(1/2)
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.12 \[ \int \frac {d+e x+f x^2+g x^3}{\left (1+x^2+x^4\right )^2} \, dx=\frac {1}{36} \left (\frac {6 \left (e+2 e x^2-g \left (2+x^2\right )+x \left (d+f-d x^2+2 f x^2\right )\right )}{1+x^2+x^4}-\frac {\left (\left (-11 i+\sqrt {3}\right ) d-2 \left (-2 i+\sqrt {3}\right ) f\right ) \arctan \left (\frac {1}{2} \left (-i+\sqrt {3}\right ) x\right )}{\sqrt {\frac {1}{6} \left (1+i \sqrt {3}\right )}}-\frac {\left (\left (11 i+\sqrt {3}\right ) d-2 \left (2 i+\sqrt {3}\right ) f\right ) \arctan \left (\frac {1}{2} \left (i+\sqrt {3}\right ) x\right )}{\sqrt {\frac {1}{6} \left (1-i \sqrt {3}\right )}}-4 \sqrt {3} (2 e-g) \arctan \left (\frac {\sqrt {3}}{1+2 x^2}\right )\right ) \]
((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
Time = 0.49 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.08, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {2202, 1492, 1483, 1142, 25, 1083, 217, 1103, 1576, 1159, 1083, 217}
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.
| \(\displaystyle \int \frac {d+e x+f x^2+g x^3}{\left (x^4+x^2+1\right )^2} \, dx\) |
| \(\Big \downarrow \) 2202 |
| \(\displaystyle \int \frac {f x^2+d}{\left (x^4+x^2+1\right )^2}dx+\int \frac {x \left (g x^2+e\right )}{\left (x^4+x^2+1\right )^2}dx\) |
| \(\Big \downarrow \) 1492 |
| \(\displaystyle \frac {1}{6} \int \frac {-\left ((d-2 f) x^2\right )+5 d-f}{x^4+x^2+1}dx+\int \frac {x \left (g x^2+e\right )}{\left (x^4+x^2+1\right )^2}dx+\frac {x \left (-\left (x^2 (d-2 f)\right )+d+f\right )}{6 \left (x^4+x^2+1\right )}\) |
| \(\Big \downarrow \) 1483 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{2} \int \frac {5 d-f-3 (2 d-f) x}{x^2-x+1}dx+\frac {1}{2} \int \frac {5 d-f+3 (2 d-f) x}{x^2+x+1}dx\right )+\int \frac {x \left (g x^2+e\right )}{\left (x^4+x^2+1\right )^2}dx+\frac {x \left (-\left (x^2 (d-2 f)\right )+d+f\right )}{6 \left (x^4+x^2+1\right )}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{2} \left (\frac {1}{2} (4 d+f) \int \frac {1}{x^2-x+1}dx-\frac {3}{2} (2 d-f) \int -\frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{2} \left (\frac {1}{2} (4 d+f) \int \frac {1}{x^2+x+1}dx+\frac {3}{2} (2 d-f) \int \frac {2 x+1}{x^2+x+1}dx\right )\right )+\int \frac {x \left (g x^2+e\right )}{\left (x^4+x^2+1\right )^2}dx+\frac {x \left (-\left (x^2 (d-2 f)\right )+d+f\right )}{6 \left (x^4+x^2+1\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{2} \left (\frac {1}{2} (4 d+f) \int \frac {1}{x^2-x+1}dx+\frac {3}{2} (2 d-f) \int \frac {1-2 x}{x^2-x+1}dx\right )+\frac {1}{2} \left (\frac {1}{2} (4 d+f) \int \frac {1}{x^2+x+1}dx+\frac {3}{2} (2 d-f) \int \frac {2 x+1}{x^2+x+1}dx\right )\right )+\int \frac {x \left (g x^2+e\right )}{\left (x^4+x^2+1\right )^2}dx+\frac {x \left (-\left (x^2 (d-2 f)\right )+d+f\right )}{6 \left (x^4+x^2+1\right )}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{2} \left (\frac {3}{2} (2 d-f) \int \frac {1-2 x}{x^2-x+1}dx-(4 d+f) \int \frac {1}{-(2 x-1)^2-3}d(2 x-1)\right )+\frac {1}{2} \left (\frac {3}{2} (2 d-f) \int \frac {2 x+1}{x^2+x+1}dx-(4 d+f) \int \frac {1}{-(2 x+1)^2-3}d(2 x+1)\right )\right )+\int \frac {x \left (g x^2+e\right )}{\left (x^4+x^2+1\right )^2}dx+\frac {x \left (-\left (x^2 (d-2 f)\right )+d+f\right )}{6 \left (x^4+x^2+1\right )}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{2} \left (\frac {3}{2} (2 d-f) \int \frac {1-2 x}{x^2-x+1}dx+\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right ) (4 d+f)}{\sqrt {3}}\right )+\frac {1}{2} \left (\frac {3}{2} (2 d-f) \int \frac {2 x+1}{x^2+x+1}dx+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right ) (4 d+f)}{\sqrt {3}}\right )\right )+\int \frac {x \left (g x^2+e\right )}{\left (x^4+x^2+1\right )^2}dx+\frac {x \left (-\left (x^2 (d-2 f)\right )+d+f\right )}{6 \left (x^4+x^2+1\right )}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \int \frac {x \left (g x^2+e\right )}{\left (x^4+x^2+1\right )^2}dx+\frac {1}{6} \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right ) (4 d+f)}{\sqrt {3}}-\frac {3}{2} (2 d-f) \log \left (x^2-x+1\right )\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right ) (4 d+f)}{\sqrt {3}}+\frac {3}{2} (2 d-f) \log \left (x^2+x+1\right )\right )\right )+\frac {x \left (-\left (x^2 (d-2 f)\right )+d+f\right )}{6 \left (x^4+x^2+1\right )}\) |
| \(\Big \downarrow \) 1576 |
| \(\displaystyle \frac {1}{2} \int \frac {g x^2+e}{\left (x^4+x^2+1\right )^2}dx^2+\frac {1}{6} \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right ) (4 d+f)}{\sqrt {3}}-\frac {3}{2} (2 d-f) \log \left (x^2-x+1\right )\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right ) (4 d+f)}{\sqrt {3}}+\frac {3}{2} (2 d-f) \log \left (x^2+x+1\right )\right )\right )+\frac {x \left (-\left (x^2 (d-2 f)\right )+d+f\right )}{6 \left (x^4+x^2+1\right )}\) |
| \(\Big \downarrow \) 1159 |
| \(\displaystyle \frac {1}{2} \left (\frac {1}{3} (2 e-g) \int \frac {1}{x^4+x^2+1}dx^2+\frac {x^2 (2 e-g)+e-2 g}{3 \left (x^4+x^2+1\right )}\right )+\frac {1}{6} \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right ) (4 d+f)}{\sqrt {3}}-\frac {3}{2} (2 d-f) \log \left (x^2-x+1\right )\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right ) (4 d+f)}{\sqrt {3}}+\frac {3}{2} (2 d-f) \log \left (x^2+x+1\right )\right )\right )+\frac {x \left (-\left (x^2 (d-2 f)\right )+d+f\right )}{6 \left (x^4+x^2+1\right )}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {1}{2} \left (\frac {x^2 (2 e-g)+e-2 g}{3 \left (x^4+x^2+1\right )}-\frac {2}{3} (2 e-g) \int \frac {1}{-x^4-3}d\left (2 x^2+1\right )\right )+\frac {1}{6} \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right ) (4 d+f)}{\sqrt {3}}-\frac {3}{2} (2 d-f) \log \left (x^2-x+1\right )\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right ) (4 d+f)}{\sqrt {3}}+\frac {3}{2} (2 d-f) \log \left (x^2+x+1\right )\right )\right )+\frac {x \left (-\left (x^2 (d-2 f)\right )+d+f\right )}{6 \left (x^4+x^2+1\right )}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right ) (4 d+f)}{\sqrt {3}}-\frac {3}{2} (2 d-f) \log \left (x^2-x+1\right )\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right ) (4 d+f)}{\sqrt {3}}+\frac {3}{2} (2 d-f) \log \left (x^2+x+1\right )\right )\right )+\frac {1}{2} \left (\frac {2 \arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right ) (2 e-g)}{3 \sqrt {3}}+\frac {x^2 (2 e-g)+e-2 g}{3 \left (x^4+x^2+1\right )}\right )+\frac {x \left (-\left (x^2 (d-2 f)\right )+d+f\right )}{6 \left (x^4+x^2+1\right )}\) |
(x*(d + f - (d - 2*f)*x^2))/(6*(1 + x^2 + x^4)) + ((e - 2*g + (2*e - g)*x^ 2)/(3*(1 + x^2 + x^4)) + (2*(2*e - g)*ArcTan[(1 + 2*x^2)/Sqrt[3]])/(3*Sqrt [3]))/2 + ((((4*d + f)*ArcTan[(-1 + 2*x)/Sqrt[3]])/Sqrt[3] - (3*(2*d - f)* Log[1 - x + x^2])/2)/2 + (((4*d + f)*ArcTan[(1 + 2*x)/Sqrt[3]])/Sqrt[3] + (3*(2*d - f)*Log[1 + x + x^2])/2)/2)/6
3.1.33.3.1 Defintions of rubi rules used
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])
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_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[((b*d - 2*a*e + (2*c*d - b*e)*x)/((p + 1)*(b^2 - 4*a*c)))*(a + b* x + c*x^2)^(p + 1), x] - Simp[(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] & & LtQ[p, -1] && NeQ[p, -3/2]
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]
Int[((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symb ol] :> 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] + Simp[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]
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[(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]
Time = 0.28 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.96
| method | result | size |
| default | \(-\frac {\left (\frac {d}{3}-\frac {e}{3}-\frac {g}{3}-\frac {2 f}{3}\right ) x -\frac {2 d}{3}-\frac {e}{3}+\frac {2 g}{3}+\frac {f}{3}}{4 \left (x^{2}-x +1\right )}-\frac {\left (6 d -3 f \right ) \ln \left (x^{2}-x +1\right )}{24}-\frac {\left (-2 d -4 e -\frac {f}{2}+2 g \right ) \sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{18}+\frac {\left (-\frac {d}{3}-\frac {e}{3}-\frac {g}{3}+\frac {2 f}{3}\right ) x -\frac {2 d}{3}+\frac {e}{3}-\frac {2 g}{3}+\frac {f}{3}}{4 x^{2}+4 x +4}+\frac {\left (6 d -3 f \right ) \ln \left (x^{2}+x +1\right )}{24}+\frac {\left (2 d -4 e +\frac {f}{2}+2 g \right ) \arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{18}\) | \(172\) |
| risch | \(\text {Expression too large to display}\) | \(28327\) |
-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)-1/24* (6*d-3*f)*ln(x^2-x+1)-1/18*(-2*d-4*e-1/2*f+2*g)*3^(1/2)*arctan(1/3*(2*x-1) *3^(1/2))+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)+1/24*(6*d-3*f)*ln(x^2+x+1)+1/18*(2*d-4*e+1/2*f+2*g)*arctan(1/3*(1+2*x )*3^(1/2))*3^(1/2)
Time = 0.46 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.34 \[ \int \frac {d+e x+f x^2+g x^3}{\left (1+x^2+x^4\right )^2} \, dx=-\frac {12 \, {\left (d - 2 \, f\right )} x^{3} - 12 \, {\left (2 \, e - g\right )} x^{2} - 2 \, \sqrt {3} {\left ({\left (4 \, d - 8 \, e + f + 4 \, g\right )} x^{4} + {\left (4 \, d - 8 \, e + f + 4 \, g\right )} x^{2} + 4 \, d - 8 \, e + f + 4 \, g\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) - 2 \, \sqrt {3} {\left ({\left (4 \, d + 8 \, e + f - 4 \, g\right )} x^{4} + {\left (4 \, d + 8 \, e + f - 4 \, g\right )} x^{2} + 4 \, d + 8 \, e + f - 4 \, g\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - 12 \, {\left (d + f\right )} x - 9 \, {\left ({\left (2 \, d - f\right )} x^{4} + {\left (2 \, d - f\right )} x^{2} + 2 \, d - f\right )} \log \left (x^{2} + x + 1\right ) + 9 \, {\left ({\left (2 \, d - f\right )} x^{4} + {\left (2 \, d - f\right )} x^{2} + 2 \, d - f\right )} \log \left (x^{2} - x + 1\right ) - 12 \, e + 24 \, g}{72 \, {\left (x^{4} + x^{2} + 1\right )}} \]
-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)
Timed out. \[ \int \frac {d+e x+f x^2+g x^3}{\left (1+x^2+x^4\right )^2} \, dx=\text {Timed out} \]
Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.75 \[ \int \frac {d+e x+f x^2+g x^3}{\left (1+x^2+x^4\right )^2} \, dx=\frac {1}{36} \, \sqrt {3} {\left (4 \, d - 8 \, e + f + 4 \, g\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{36} \, \sqrt {3} {\left (4 \, d + 8 \, e + f - 4 \, g\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{8} \, {\left (2 \, d - f\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{8} \, {\left (2 \, d - f\right )} \log \left (x^{2} - x + 1\right ) - \frac {{\left (d - 2 \, f\right )} x^{3} - {\left (2 \, e - g\right )} x^{2} - {\left (d + f\right )} x - e + 2 \, g}{6 \, {\left (x^{4} + x^{2} + 1\right )}} \]
1/36*sqrt(3)*(4*d - 8*e + f + 4*g)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/36*sq rt(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)
Time = 0.33 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.77 \[ \int \frac {d+e x+f x^2+g x^3}{\left (1+x^2+x^4\right )^2} \, dx=\frac {1}{36} \, \sqrt {3} {\left (4 \, d - 8 \, e + f + 4 \, g\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{36} \, \sqrt {3} {\left (4 \, d + 8 \, e + f - 4 \, g\right )} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) + \frac {1}{8} \, {\left (2 \, d - f\right )} \log \left (x^{2} + x + 1\right ) - \frac {1}{8} \, {\left (2 \, d - f\right )} \log \left (x^{2} - x + 1\right ) - \frac {d x^{3} - 2 \, f x^{3} - 2 \, e x^{2} + g x^{2} - d x - f x - e + 2 \, g}{6 \, {\left (x^{4} + x^{2} + 1\right )}} \]
1/36*sqrt(3)*(4*d - 8*e + f + 4*g)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/36*sq rt(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*x^3 - 2*f*x^3 - 2*e*x^2 + g*x^2 - d*x - f*x - e + 2*g)/(x^4 + x^2 + 1)
Time = 8.14 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.32 \[ \int \frac {d+e x+f x^2+g x^3}{\left (1+x^2+x^4\right )^2} \, dx=\frac {\left (\frac {f}{3}-\frac {d}{6}\right )\,x^3+\left (\frac {e}{3}-\frac {g}{6}\right )\,x^2+\left (\frac {d}{6}+\frac {f}{6}\right )\,x+\frac {e}{6}-\frac {g}{3}}{x^4+x^2+1}-\ln \left (x-\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {d}{4}-\frac {f}{8}+\frac {\sqrt {3}\,d\,1{}\mathrm {i}}{18}+\frac {\sqrt {3}\,e\,1{}\mathrm {i}}{9}+\frac {\sqrt {3}\,f\,1{}\mathrm {i}}{72}-\frac {\sqrt {3}\,g\,1{}\mathrm {i}}{18}\right )-\ln \left (x+\frac {1}{2}-\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {f}{8}-\frac {d}{4}+\frac {\sqrt {3}\,d\,1{}\mathrm {i}}{18}-\frac {\sqrt {3}\,e\,1{}\mathrm {i}}{9}+\frac {\sqrt {3}\,f\,1{}\mathrm {i}}{72}+\frac {\sqrt {3}\,g\,1{}\mathrm {i}}{18}\right )+\ln \left (x-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {f}{8}-\frac {d}{4}+\frac {\sqrt {3}\,d\,1{}\mathrm {i}}{18}+\frac {\sqrt {3}\,e\,1{}\mathrm {i}}{9}+\frac {\sqrt {3}\,f\,1{}\mathrm {i}}{72}-\frac {\sqrt {3}\,g\,1{}\mathrm {i}}{18}\right )+\ln \left (x+\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {d}{4}-\frac {f}{8}+\frac {\sqrt {3}\,d\,1{}\mathrm {i}}{18}-\frac {\sqrt {3}\,e\,1{}\mathrm {i}}{9}+\frac {\sqrt {3}\,f\,1{}\mathrm {i}}{72}+\frac {\sqrt {3}\,g\,1{}\mathrm {i}}{18}\right ) \]
(e/6 - g/3 - x^3*(d/6 - f/3) + x^2*(e/3 - g/6) + x*(d/6 + f/6))/(x^2 + x^4 + 1) - log(x - (3^(1/2)*1i)/2 - 1/2)*(d/4 - f/8 + (3^(1/2)*d*1i)/18 + (3^ (1/2)*e*1i)/9 + (3^(1/2)*f*1i)/72 - (3^(1/2)*g*1i)/18) - log(x - (3^(1/2)* 1i)/2 + 1/2)*(f/8 - d/4 + (3^(1/2)*d*1i)/18 - (3^(1/2)*e*1i)/9 + (3^(1/2)* f*1i)/72 + (3^(1/2)*g*1i)/18) + log(x + (3^(1/2)*1i)/2 - 1/2)*(f/8 - d/4 + (3^(1/2)*d*1i)/18 + (3^(1/2)*e*1i)/9 + (3^(1/2)*f*1i)/72 - (3^(1/2)*g*1i) /18) + log(x + (3^(1/2)*1i)/2 + 1/2)*(d/4 - f/8 + (3^(1/2)*d*1i)/18 - (3^( 1/2)*e*1i)/9 + (3^(1/2)*f*1i)/72 + (3^(1/2)*g*1i)/18)