Integrand size = 30, antiderivative size = 58 \[ \int \frac {-2-x+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=-2 \arctan \left (\frac {x}{\sqrt {1+x^2+x^4}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {2 x^2}{1+2 x^2+x^4+\left (-1+x^2\right ) \sqrt {1+x^2+x^4}}\right ) \]
Time = 0.46 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \frac {-2-x+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=-2 \arctan \left (\frac {x}{\sqrt {1+x^2+x^4}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {2 x^2}{1+2 x^2+x^4+\left (-1+x^2\right ) \sqrt {1+x^2+x^4}}\right ) \]
-2*ArcTan[x/Sqrt[1 + x^2 + x^4]] + ArcTanh[(2*x^2)/(1 + 2*x^2 + x^4 + (-1 + x^2)*Sqrt[1 + x^2 + x^4])]/2
Time = 0.34 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.79, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, Rules used = {2254, 25, 1576, 1154, 219, 2212, 216}
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.
\(\displaystyle \int \frac {2 x^2-x-2}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}} \, dx\) |
\(\Big \downarrow \) 2254 |
\(\displaystyle \int -\frac {x}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx+\int \frac {2 x^2-2}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {2 x^2-2}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx-\int \frac {x}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx\) |
\(\Big \downarrow \) 1576 |
\(\displaystyle \int \frac {2 x^2-2}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx-\frac {1}{2} \int \frac {1}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx^2\) |
\(\Big \downarrow \) 1154 |
\(\displaystyle \int \frac {1}{4-x^4}d\frac {1-x^2}{\sqrt {x^4+x^2+1}}+\int \frac {2 x^2-2}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \int \frac {2 x^2-2}{\left (x^2+1\right ) \sqrt {x^4+x^2+1}}dx+\frac {1}{2} \text {arctanh}\left (\frac {1-x^2}{2 \sqrt {x^4+x^2+1}}\right )\) |
\(\Big \downarrow \) 2212 |
\(\displaystyle \frac {1}{2} \text {arctanh}\left (\frac {1-x^2}{2 \sqrt {x^4+x^2+1}}\right )-2 \int \frac {1}{\frac {x^2}{x^4+x^2+1}+1}d\frac {x}{\sqrt {x^4+x^2+1}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{2} \text {arctanh}\left (\frac {1-x^2}{2 \sqrt {x^4+x^2+1}}\right )-2 \arctan \left (\frac {x}{\sqrt {x^4+x^2+1}}\right )\) |
3.8.50.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
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[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> Simp[A Subst[Int[1/(d - (b*d - 2*a*e)*x^2), x], x, x/Sqrt[a + b*x^2 + c*x^4]], x] /; FreeQ[{a, b, c, d, e, A, B}, x] & & EqQ[c*d^2 - a*e^2, 0] && EqQ[B*d + A*e, 0]
Int[(Pr_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^ (p_), x_Symbol] :> Module[{r = Expon[Pr, x], k}, Int[Sum[Coeff[Pr, x, 2*k]* x^(2*k), {k, 0, r/2}]*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x] + Int[x*Sum[C oeff[Pr, x, 2*k + 1]*x^(2*k), {k, 0, (r - 1)/2}]*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && PolyQ[Pr, x] && !Poly Q[Pr, x^2]
Time = 4.62 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.67
method | result | size |
default | \(2 \arctan \left (\frac {\sqrt {x^{4}+x^{2}+1}}{x}\right )-\frac {\operatorname {arctanh}\left (\frac {x^{2}-1}{2 \sqrt {x^{4}+x^{2}+1}}\right )}{2}\) | \(39\) |
pseudoelliptic | \(2 \arctan \left (\frac {\sqrt {x^{4}+x^{2}+1}}{x}\right )-\frac {\operatorname {arctanh}\left (\frac {x^{2}-1}{2 \sqrt {x^{4}+x^{2}+1}}\right )}{2}\) | \(39\) |
elliptic | \(\frac {\operatorname {arctanh}\left (\frac {-x^{2}+1}{2 \sqrt {\left (x^{2}+1\right )^{2}-x^{2}}}\right )}{2}+2 \arctan \left (\frac {\sqrt {x^{4}+x^{2}+1}}{x}\right )\) | \(46\) |
trager | \(\frac {\operatorname {RootOf}\left (4 \textit {\_Z}^{2}-4 \textit {\_Z} +17\right ) \ln \left (-\frac {-6 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-4 \textit {\_Z} +17\right ) x -8 x^{2}+4 \sqrt {x^{4}+x^{2}+1}+3 x +8}{{\left (2 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-4 \textit {\_Z} +17\right ) x -x -4\right )}^{2}}\right )}{2}-\frac {\ln \left (-\frac {6 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-4 \textit {\_Z} +17\right ) x -8 x^{2}+4 \sqrt {x^{4}+x^{2}+1}-3 x +8}{{\left (2 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-4 \textit {\_Z} +17\right ) x -x +4\right )}^{2}}\right ) \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-4 \textit {\_Z} +17\right )}{2}+\frac {\ln \left (-\frac {6 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-4 \textit {\_Z} +17\right ) x -8 x^{2}+4 \sqrt {x^{4}+x^{2}+1}-3 x +8}{{\left (2 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-4 \textit {\_Z} +17\right ) x -x +4\right )}^{2}}\right )}{2}\) | \(210\) |
Time = 0.31 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.09 \[ \int \frac {-2-x+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=2 \, \arctan \left (\frac {\sqrt {x^{4} + x^{2} + 1}}{x}\right ) + \frac {1}{4} \, \log \left (\frac {5 \, x^{4} + 2 \, x^{2} - 4 \, \sqrt {x^{4} + x^{2} + 1} {\left (x^{2} - 1\right )} + 5}{x^{4} + 2 \, x^{2} + 1}\right ) \]
2*arctan(sqrt(x^4 + x^2 + 1)/x) + 1/4*log((5*x^4 + 2*x^2 - 4*sqrt(x^4 + x^ 2 + 1)*(x^2 - 1) + 5)/(x^4 + 2*x^2 + 1))
\[ \int \frac {-2-x+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=\int \frac {2 x^{2} - x - 2}{\sqrt {\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )} \left (x^{2} + 1\right )}\, dx \]
\[ \int \frac {-2-x+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=\int { \frac {2 \, x^{2} - x - 2}{\sqrt {x^{4} + x^{2} + 1} {\left (x^{2} + 1\right )}} \,d x } \]
\[ \int \frac {-2-x+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=\int { \frac {2 \, x^{2} - x - 2}{\sqrt {x^{4} + x^{2} + 1} {\left (x^{2} + 1\right )}} \,d x } \]
Timed out. \[ \int \frac {-2-x+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=\int -\frac {-2\,x^2+x+2}{\left (x^2+1\right )\,\sqrt {x^4+x^2+1}} \,d x \]