Integrand size = 16, antiderivative size = 53 \[ \int \frac {x^2}{1-x^2+x^4} \, dx=-\frac {1}{2} \arctan \left (\sqrt {3}-2 x\right )+\frac {1}{2} \arctan \left (\sqrt {3}+2 x\right )-\frac {\text {arctanh}\left (\frac {\sqrt {3} x}{1+x^2}\right )}{2 \sqrt {3}} \] Output:
1/2*arctan(-3^(1/2)+2*x)+1/2*arctan(3^(1/2)+2*x)-1/6*arctanh(3^(1/2)*x/(x^ 2+1))*3^(1/2)
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.77 \[ \int \frac {x^2}{1-x^2+x^4} \, dx=\frac {\sqrt {-1-i \sqrt {3}} \left (i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (1-i \sqrt {3}\right ) x\right )+\sqrt {-1+i \sqrt {3}} \left (-i+\sqrt {3}\right ) \arctan \left (\frac {1}{2} \left (1+i \sqrt {3}\right ) x\right )}{2 \sqrt {6}} \] Input:
Integrate[x^2/(1 - x^2 + x^4),x]
Output:
(Sqrt[-1 - I*Sqrt[3]]*(I + Sqrt[3])*ArcTan[((1 - I*Sqrt[3])*x)/2] + Sqrt[- 1 + I*Sqrt[3]]*(-I + Sqrt[3])*ArcTan[((1 + I*Sqrt[3])*x)/2])/(2*Sqrt[6])
Time = 0.44 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.47, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, 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 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^2-\sqrt {3} x+1}dx+\frac {1}{2} \int \frac {1}{x^2+\sqrt {3} x+1}dx\right )-\frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{2} \left (-\int \frac {1}{-\left (2 x-\sqrt {3}\right )^2-1}d\left (2 x-\sqrt {3}\right )-\int \frac {1}{-\left (2 x+\sqrt {3}\right )^2-1}d\left (2 x+\sqrt {3}\right )\right )-\frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{2} \left (\arctan \left (2 x+\sqrt {3}\right )-\arctan \left (\sqrt {3}-2 x\right )\right )-\frac {1}{2} \int \frac {1-x^2}{x^4-x^2+1}dx\) |
\(\Big \downarrow \) 1478 |
\(\displaystyle \frac {1}{2} \left (\frac {\int -\frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}+\frac {\int -\frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}\right )+\frac {1}{2} \left (\arctan \left (2 x+\sqrt {3}\right )-\arctan \left (\sqrt {3}-2 x\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \left (-\frac {\int \frac {\sqrt {3}-2 x}{x^2-\sqrt {3} x+1}dx}{2 \sqrt {3}}-\frac {\int \frac {2 x+\sqrt {3}}{x^2+\sqrt {3} x+1}dx}{2 \sqrt {3}}\right )+\frac {1}{2} \left (\arctan \left (2 x+\sqrt {3}\right )-\arctan \left (\sqrt {3}-2 x\right )\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{2} \left (\arctan \left (2 x+\sqrt {3}\right )-\arctan \left (\sqrt {3}-2 x\right )\right )+\frac {1}{2} \left (\frac {\log \left (x^2-\sqrt {3} x+1\right )}{2 \sqrt {3}}-\frac {\log \left (x^2+\sqrt {3} x+1\right )}{2 \sqrt {3}}\right )\) |
Input:
Int[x^2/(1 - x^2 + x^4),x]
Output:
(-ArcTan[Sqrt[3] - 2*x] + ArcTan[Sqrt[3] + 2*x])/2 + (Log[1 - Sqrt[3]*x + x^2]/(2*Sqrt[3]) - Log[1 + Sqrt[3]*x + x^2]/(2*Sqrt[3]))/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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.72
method | result | size |
risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}-\textit {\_Z}^{2}+1\right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (x -\textit {\_R} \right )}{2 \textit {\_R}^{3}-\textit {\_R}}\right )}{2}\) | \(38\) |
default | \(-\frac {\sqrt {3}\, \left (-\frac {\ln \left (x^{2}-\sqrt {3}\, x +1\right )}{2}-\sqrt {3}\, \arctan \left (2 x -\sqrt {3}\right )\right )}{6}-\frac {\sqrt {3}\, \left (\frac {\ln \left (x^{2}+\sqrt {3}\, x +1\right )}{2}-\sqrt {3}\, \arctan \left (2 x +\sqrt {3}\right )\right )}{6}\) | \(69\) |
Input:
int(x^2/(x^4-x^2+1),x,method=_RETURNVERBOSE)
Output:
1/2*sum(_R^2/(2*_R^3-_R)*ln(x-_R),_R=RootOf(_Z^4-_Z^2+1))
Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.02 \[ \int \frac {x^2}{1-x^2+x^4} \, dx=-\frac {1}{12} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{12} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + \frac {1}{2} \, \arctan \left (2 \, x + \sqrt {3}\right ) - \frac {1}{2} \, \arctan \left (-2 \, x + \sqrt {3}\right ) \] Input:
integrate(x^2/(x^4-x^2+1),x, algorithm="fricas")
Output:
-1/12*sqrt(3)*log(x^2 + sqrt(3)*x + 1) + 1/12*sqrt(3)*log(x^2 - sqrt(3)*x + 1) + 1/2*arctan(2*x + sqrt(3)) - 1/2*arctan(-2*x + sqrt(3))
Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.19 \[ \int \frac {x^2}{1-x^2+x^4} \, dx=\frac {\sqrt {3} \log {\left (x^{2} - \sqrt {3} x + 1 \right )}}{12} - \frac {\sqrt {3} \log {\left (x^{2} + \sqrt {3} x + 1 \right )}}{12} + \frac {\operatorname {atan}{\left (2 x - \sqrt {3} \right )}}{2} + \frac {\operatorname {atan}{\left (2 x + \sqrt {3} \right )}}{2} \] Input:
integrate(x**2/(x**4-x**2+1),x)
Output:
sqrt(3)*log(x**2 - sqrt(3)*x + 1)/12 - sqrt(3)*log(x**2 + sqrt(3)*x + 1)/1 2 + atan(2*x - sqrt(3))/2 + atan(2*x + sqrt(3))/2
\[ \int \frac {x^2}{1-x^2+x^4} \, dx=\int { \frac {x^{2}}{x^{4} - x^{2} + 1} \,d x } \] Input:
integrate(x^2/(x^4-x^2+1),x, algorithm="maxima")
Output:
integrate(x^2/(x^4 - x^2 + 1), x)
Time = 0.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06 \[ \int \frac {x^2}{1-x^2+x^4} \, dx=-\frac {1}{12} \, \sqrt {3} \log \left (x^{2} + \sqrt {3} x + 1\right ) + \frac {1}{12} \, \sqrt {3} \log \left (x^{2} - \sqrt {3} x + 1\right ) + \frac {1}{2} \, \arctan \left (2 \, x + \sqrt {3}\right ) + \frac {1}{2} \, \arctan \left (2 \, x - \sqrt {3}\right ) \] Input:
integrate(x^2/(x^4-x^2+1),x, algorithm="giac")
Output:
-1/12*sqrt(3)*log(x^2 + sqrt(3)*x + 1) + 1/12*sqrt(3)*log(x^2 - sqrt(3)*x + 1) + 1/2*arctan(2*x + sqrt(3)) + 1/2*arctan(2*x - sqrt(3))
Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int \frac {x^2}{1-x^2+x^4} \, dx=-\mathrm {atan}\left (\frac {x}{2}-\frac {\sqrt {3}\,x\,1{}\mathrm {i}}{2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right )+\mathrm {atan}\left (\frac {x}{2}+\frac {\sqrt {3}\,x\,1{}\mathrm {i}}{2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{6}\right ) \] Input:
int(x^2/(x^4 - x^2 + 1),x)
Output:
atan(x/2 + (3^(1/2)*x*1i)/2)*((3^(1/2)*1i)/6 + 1/2) - atan(x/2 - (3^(1/2)* x*1i)/2)*((3^(1/2)*1i)/6 - 1/2)
Time = 0.16 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91 \[ \int \frac {x^2}{1-x^2+x^4} \, dx=-\frac {\mathit {atan} \left (\sqrt {3}-2 x \right )}{2}+\frac {\mathit {atan} \left (\sqrt {3}+2 x \right )}{2}+\frac {\sqrt {3}\, \mathrm {log}\left (-\sqrt {3}\, x +x^{2}+1\right )}{12}-\frac {\sqrt {3}\, \mathrm {log}\left (\sqrt {3}\, x +x^{2}+1\right )}{12} \] Input:
int(x^2/(x^4-x^2+1),x)
Output:
( - 6*atan(sqrt(3) - 2*x) + 6*atan(sqrt(3) + 2*x) + sqrt(3)*log( - sqrt(3) *x + x**2 + 1) - sqrt(3)*log(sqrt(3)*x + x**2 + 1))/12