Integrand size = 32, antiderivative size = 143 \[ \int \frac {\left (-4+x^2\right ) \sqrt [4]{2-x^2-2 x^4}}{x^2 \left (-2+x^2\right )} \, dx=-\frac {2 \sqrt [4]{2-x^2-2 x^4}}{x}+\frac {\arctan \left (\frac {2^{3/4} x \sqrt [4]{2-x^2-2 x^4}}{\sqrt {2} x^2-\sqrt {2-x^2-2 x^4}}\right )}{\sqrt [4]{2}}+\frac {\text {arctanh}\left (\frac {2 \sqrt [4]{2} x \sqrt [4]{2-x^2-2 x^4}}{2 x^2+\sqrt {2} \sqrt {2-x^2-2 x^4}}\right )}{\sqrt [4]{2}} \]
-2*(-2*x^4-x^2+2)^(1/4)/x+1/2*arctan(2^(3/4)*x*(-2*x^4-x^2+2)^(1/4)/(2^(1/ 2)*x^2-(-2*x^4-x^2+2)^(1/2)))*2^(3/4)+1/2*arctanh(2*2^(1/4)*x*(-2*x^4-x^2+ 2)^(1/4)/(2*x^2+2^(1/2)*(-2*x^4-x^2+2)^(1/2)))*2^(3/4)
Time = 0.30 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-4+x^2\right ) \sqrt [4]{2-x^2-2 x^4}}{x^2 \left (-2+x^2\right )} \, dx=-\frac {2 \sqrt [4]{2-x^2-2 x^4}}{x}+\frac {\arctan \left (\frac {2^{3/4} x \sqrt [4]{2-x^2-2 x^4}}{\sqrt {2} x^2-\sqrt {2-x^2-2 x^4}}\right )}{\sqrt [4]{2}}+\frac {\text {arctanh}\left (\frac {2 \sqrt [4]{2} x \sqrt [4]{2-x^2-2 x^4}}{2 x^2+\sqrt {2} \sqrt {2-x^2-2 x^4}}\right )}{\sqrt [4]{2}} \]
(-2*(2 - x^2 - 2*x^4)^(1/4))/x + ArcTan[(2^(3/4)*x*(2 - x^2 - 2*x^4)^(1/4) )/(Sqrt[2]*x^2 - Sqrt[2 - x^2 - 2*x^4])]/2^(1/4) + ArcTanh[(2*2^(1/4)*x*(2 - x^2 - 2*x^4)^(1/4))/(2*x^2 + Sqrt[2]*Sqrt[2 - x^2 - 2*x^4])]/2^(1/4)
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.
\(\displaystyle \int \frac {\left (x^2-4\right ) \sqrt [4]{-2 x^4-x^2+2}}{x^2 \left (x^2-2\right )} \, dx\) |
\(\Big \downarrow \) 2250 |
\(\displaystyle \int \frac {\left (x^2-4\right ) \sqrt [4]{-2 x^4-x^2+2}}{x^2 \left (x^2-2\right )}dx\) |
3.21.14.3.1 Defintions of rubi rules used
Int[(Px_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_) ^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Unintegrable[Px*(f*x)^m*(d + e*x^2)^ q*(a + b*x^2 + c*x^4)^p, x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x] && Pol yQ[Px, x]
Time = 201.35 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.25
method | result | size |
pseudoelliptic | \(\frac {\ln \left (\frac {\left (-2 x^{4}-x^{2}+2\right )^{\frac {1}{4}} 2^{\frac {3}{4}} x +\sqrt {2}\, x^{2}+\sqrt {-2 x^{4}-x^{2}+2}}{\sqrt {2}\, x^{2}-\left (-2 x^{4}-x^{2}+2\right )^{\frac {1}{4}} 2^{\frac {3}{4}} x +\sqrt {-2 x^{4}-x^{2}+2}}\right ) 2^{\frac {3}{4}} x +2 \arctan \left (\frac {2^{\frac {1}{4}} \left (-2 x^{4}-x^{2}+2\right )^{\frac {1}{4}}+x}{x}\right ) 2^{\frac {3}{4}} x +2 \arctan \left (\frac {2^{\frac {1}{4}} \left (-2 x^{4}-x^{2}+2\right )^{\frac {1}{4}}-x}{x}\right ) 2^{\frac {3}{4}} x -8 \left (-2 x^{4}-x^{2}+2\right )^{\frac {1}{4}}}{4 x}\) | \(179\) |
trager | \(-\frac {2 \left (-2 x^{4}-x^{2}+2\right )^{\frac {1}{4}}}{x}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right ) \ln \left (-\frac {-4 \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{3} x^{4}+4 \left (-2 x^{4}-x^{2}+2\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2} x^{3}-4 \sqrt {-2 x^{4}-x^{2}+2}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right ) x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{3} x^{2}+4 \left (-2 x^{4}-x^{2}+2\right )^{\frac {3}{4}} x +2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{3}}{x^{2}-2}\right )}{2}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2}\right ) \ln \left (-\frac {-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2} x^{4}-4 \left (-2 x^{4}-x^{2}+2\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2} x^{3}+4 \sqrt {-2 x^{4}-x^{2}+2}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2}\right ) x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2} x^{2}+4 \left (-2 x^{4}-x^{2}+2\right )^{\frac {3}{4}} x +2 \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2}\right )}{x^{2}-2}\right )}{2}\) | \(322\) |
risch | \(\text {Expression too large to display}\) | \(1169\) |
1/4*(ln(((-2*x^4-x^2+2)^(1/4)*2^(3/4)*x+2^(1/2)*x^2+(-2*x^4-x^2+2)^(1/2))/ (2^(1/2)*x^2-(-2*x^4-x^2+2)^(1/4)*2^(3/4)*x+(-2*x^4-x^2+2)^(1/2)))*2^(3/4) *x+2*arctan((2^(1/4)*(-2*x^4-x^2+2)^(1/4)+x)/x)*2^(3/4)*x+2*arctan((2^(1/4 )*(-2*x^4-x^2+2)^(1/4)-x)/x)*2^(3/4)*x-8*(-2*x^4-x^2+2)^(1/4))/x
Exception generated. \[ \int \frac {\left (-4+x^2\right ) \sqrt [4]{2-x^2-2 x^4}}{x^2 \left (-2+x^2\right )} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (residue poly has multiple non-linear fac tors)
\[ \int \frac {\left (-4+x^2\right ) \sqrt [4]{2-x^2-2 x^4}}{x^2 \left (-2+x^2\right )} \, dx=\int \frac {\left (x - 2\right ) \left (x + 2\right ) \sqrt [4]{- 2 x^{4} - x^{2} + 2}}{x^{2} \left (x^{2} - 2\right )}\, dx \]
\[ \int \frac {\left (-4+x^2\right ) \sqrt [4]{2-x^2-2 x^4}}{x^2 \left (-2+x^2\right )} \, dx=\int { \frac {{\left (-2 \, x^{4} - x^{2} + 2\right )}^{\frac {1}{4}} {\left (x^{2} - 4\right )}}{{\left (x^{2} - 2\right )} x^{2}} \,d x } \]
\[ \int \frac {\left (-4+x^2\right ) \sqrt [4]{2-x^2-2 x^4}}{x^2 \left (-2+x^2\right )} \, dx=\int { \frac {{\left (-2 \, x^{4} - x^{2} + 2\right )}^{\frac {1}{4}} {\left (x^{2} - 4\right )}}{{\left (x^{2} - 2\right )} x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (-4+x^2\right ) \sqrt [4]{2-x^2-2 x^4}}{x^2 \left (-2+x^2\right )} \, dx=\int \frac {\left (x^2-4\right )\,{\left (-2\,x^4-x^2+2\right )}^{1/4}}{x^2\,\left (x^2-2\right )} \,d x \]