Integrand size = 14, antiderivative size = 57 \[ \int \frac {x^2}{1+x^2+x^4} \, dx=-\frac {\arctan \left (\frac {1-2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}+\frac {\arctan \left (\frac {1+2 x}{\sqrt {3}}\right )}{2 \sqrt {3}}-\frac {1}{2} \text {arctanh}\left (\frac {x}{1+x^2}\right ) \] Output:
-1/6*arctan(1/3*(1-2*x)*3^(1/2))*3^(1/2)+1/6*arctan(1/3*(1+2*x)*3^(1/2))*3 ^(1/2)-1/2*arctanh(x/(x^2+1))
Result contains complex when optimal does not.
Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.58 \[ \int \frac {x^2}{1+x^2+x^4} \, dx=\frac {\sqrt {1-i \sqrt {3}} \left (-i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (-i+\sqrt {3}\right ) x\right )+\sqrt {1+i \sqrt {3}} \left (i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (i+\sqrt {3}\right ) x\right )}{2 \sqrt {6}} \] Input:
Integrate[x^2/(1 + x^2 + x^4),x]
Output:
(Sqrt[1 - I*Sqrt[3]]*(-I + Sqrt[3])*ArcTan[((-I + Sqrt[3])*x)/2] + Sqrt[1 + I*Sqrt[3]]*(I + Sqrt[3])*ArcTan[((I + Sqrt[3])*x)/2])/(2*Sqrt[6])
Time = 0.42 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.25, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {1447, 1475, 1083, 217, 1478, 25, 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.
\(\displaystyle \int \frac {x^2}{x^4+x^2+1} \, dx\) |
\(\Big \downarrow \) 1447 |
\(\displaystyle \frac {1}{2} \int \frac {x^2+1}{x^4+x^2+1}dx-\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx\) |
\(\Big \downarrow \) 1475 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2-x+1}dx+\frac {1}{2} \int \frac {1}{x^2+x+1}dx\right )-\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (-\int \frac {1}{-(2 x-1)^2-3}d(2 x-1)-\int \frac {1}{-(2 x+1)^2-3}d(2 x+1)\right )-\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )-\frac {1}{2} \int \frac {1-x^2}{x^4+x^2+1}dx\) |
\(\Big \downarrow \) 1478 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int -\frac {1-2 x}{x^2-x+1}dx+\frac {1}{2} \int -\frac {2 x+1}{x^2+x+1}dx\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (-\frac {1}{2} \int \frac {1-2 x}{x^2-x+1}dx-\frac {1}{2} \int \frac {2 x+1}{x^2+x+1}dx\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\frac {\arctan \left (\frac {2 x-1}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {\arctan \left (\frac {2 x+1}{\sqrt {3}}\right )}{\sqrt {3}}\right )+\frac {1}{2} \left (\frac {1}{2} \log \left (x^2-x+1\right )-\frac {1}{2} \log \left (x^2+x+1\right )\right )\) |
Input:
Int[x^2/(1 + x^2 + x^4),x]
Output:
(ArcTan[(-1 + 2*x)/Sqrt[3]]/Sqrt[3] + ArcTan[(1 + 2*x)/Sqrt[3]]/Sqrt[3])/2 + (Log[1 - x + x^2]/2 - Log[1 + x + x^2]/2)/2
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[(x_)^2/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ a/c, 2]}, Simp[1/2 Int[(q + x^2)/(a + b*x^2 + c*x^4), x], x] - Simp[1/2 Int[(q - x^2)/(a + b*x^2 + c*x^4), x], x]] /; FreeQ[{a, b, c}, x] && LtQ[b ^2 - 4*a*c, 0] && PosQ[a*c]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[2*(d/e) - b/c, 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^ 2, x], x], x] + Simp[e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; F reeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[2*(d/e) - b/c, 0] || ( !LtQ[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2] , 0]))
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[-2*(d/e) - b/c, 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ [c*d^2 - a*e^2, 0] && !GtQ[b^2 - 4*a*c, 0]
Time = 0.06 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.95
method | result | size |
default | \(-\frac {\ln \left (x^{2}+x +1\right )}{4}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}+\frac {\ln \left (x^{2}-x +1\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}\) | \(54\) |
risch | \(-\frac {\ln \left (4 x^{2}+4 x +4\right )}{4}+\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{6}+\frac {\ln \left (4 x^{2}-4 x +4\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{6}\) | \(60\) |
Input:
int(x^2/(x^4+x^2+1),x,method=_RETURNVERBOSE)
Output:
-1/4*ln(x^2+x+1)+1/6*arctan(1/3*(1+2*x)*3^(1/2))*3^(1/2)+1/4*ln(x^2-x+1)+1 /6*3^(1/2)*arctan(1/3*(2*x-1)*3^(1/2))
Time = 0.07 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93 \[ \int \frac {x^2}{1+x^2+x^4} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{4} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{4} \, \log \left (x^{2} - x + 1\right ) \] Input:
integrate(x^2/(x^4+x^2+1),x, algorithm="fricas")
Output:
1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3) *(2*x - 1)) - 1/4*log(x^2 + x + 1) + 1/4*log(x^2 - x + 1)
Time = 0.08 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.23 \[ \int \frac {x^2}{1+x^2+x^4} \, dx=\frac {\log {\left (x^{2} - x + 1 \right )}}{4} - \frac {\log {\left (x^{2} + x + 1 \right )}}{4} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} - \frac {\sqrt {3}}{3} \right )}}{6} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3} + \frac {\sqrt {3}}{3} \right )}}{6} \] Input:
integrate(x**2/(x**4+x**2+1),x)
Output:
log(x**2 - x + 1)/4 - log(x**2 + x + 1)/4 + sqrt(3)*atan(2*sqrt(3)*x/3 - s qrt(3)/3)/6 + sqrt(3)*atan(2*sqrt(3)*x/3 + sqrt(3)/3)/6
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93 \[ \int \frac {x^2}{1+x^2+x^4} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{4} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{4} \, \log \left (x^{2} - x + 1\right ) \] Input:
integrate(x^2/(x^4+x^2+1),x, algorithm="maxima")
Output:
1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3) *(2*x - 1)) - 1/4*log(x^2 + x + 1) + 1/4*log(x^2 - x + 1)
Time = 0.11 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.93 \[ \int \frac {x^2}{1+x^2+x^4} \, dx=\frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x + 1\right )}\right ) + \frac {1}{6} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x - 1\right )}\right ) - \frac {1}{4} \, \log \left (x^{2} + x + 1\right ) + \frac {1}{4} \, \log \left (x^{2} - x + 1\right ) \] Input:
integrate(x^2/(x^4+x^2+1),x, algorithm="giac")
Output:
1/6*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + 1)) + 1/6*sqrt(3)*arctan(1/3*sqrt(3) *(2*x - 1)) - 1/4*log(x^2 + x + 1) + 1/4*log(x^2 - x + 1)
Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77 \[ \int \frac {x^2}{1+x^2+x^4} \, dx=-\mathrm {atan}\left (-\frac {\sqrt {3}\,x}{2}+\frac {x\,1{}\mathrm {i}}{2}\right )\,\left (\frac {\sqrt {3}}{6}-\frac {1}{2}{}\mathrm {i}\right )+\mathrm {atan}\left (\frac {\sqrt {3}\,x}{2}+\frac {x\,1{}\mathrm {i}}{2}\right )\,\left (\frac {\sqrt {3}}{6}+\frac {1}{2}{}\mathrm {i}\right ) \] Input:
int(x^2/(x^2 + x^4 + 1),x)
Output:
atan((x*1i)/2 + (3^(1/2)*x)/2)*(3^(1/2)/6 + 1i/2) - atan((x*1i)/2 - (3^(1/ 2)*x)/2)*(3^(1/2)/6 - 1i/2)
Time = 0.19 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.89 \[ \int \frac {x^2}{1+x^2+x^4} \, dx=\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x -1}{\sqrt {3}}\right )}{6}+\frac {\sqrt {3}\, \mathit {atan} \left (\frac {2 x +1}{\sqrt {3}}\right )}{6}+\frac {\mathrm {log}\left (x^{2}-x +1\right )}{4}-\frac {\mathrm {log}\left (x^{2}+x +1\right )}{4} \] Input:
int(x^2/(x^4+x^2+1),x)
Output:
(2*sqrt(3)*atan((2*x - 1)/sqrt(3)) + 2*sqrt(3)*atan((2*x + 1)/sqrt(3)) + 3 *log(x**2 - x + 1) - 3*log(x**2 + x + 1))/12