Integrand size = 30, antiderivative size = 58 \[ \int \frac {-2-x+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=-2 \arctan \left (\frac {x}{\sqrt {1+x^2+x^4}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {2 x^2}{1+2 x^2+x^4+\left (-1+x^2\right ) \sqrt {1+x^2+x^4}}\right ) \]
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Time = 0.07 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.79, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1701, 1712, 209, 1261, 738, 212} \[ \int \frac {-2-x+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=\frac {1}{2} \text {arctanh}\left (\frac {1-x^2}{2 \sqrt {x^4+x^2+1}}\right )-2 \arctan \left (\frac {x}{\sqrt {x^4+x^2+1}}\right ) \]
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Rule 209
Rule 212
Rule 738
Rule 1261
Rule 1701
Rule 1712
Rubi steps \begin{align*} \text {integral}& = -\int \frac {x}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx+\int \frac {-2+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {1+x+x^2}} \, dx,x,x^2\right )\right )-2 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {x}{\sqrt {1+x^2+x^4}}\right ) \\ & = -2 \arctan \left (\frac {x}{\sqrt {1+x^2+x^4}}\right )+\text {Subst}\left (\int \frac {1}{4-x^2} \, dx,x,\frac {1-x^2}{\sqrt {1+x^2+x^4}}\right ) \\ & = -2 \arctan \left (\frac {x}{\sqrt {1+x^2+x^4}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {1-x^2}{2 \sqrt {1+x^2+x^4}}\right ) \\ \end{align*}
Time = 0.46 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \frac {-2-x+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=-2 \arctan \left (\frac {x}{\sqrt {1+x^2+x^4}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {2 x^2}{1+2 x^2+x^4+\left (-1+x^2\right ) \sqrt {1+x^2+x^4}}\right ) \]
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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 | \(\frac {\operatorname {RootOf}\left (4 \textit {\_Z}^{2}-4 \textit {\_Z} +17\right ) \ln \left (-\frac {-6 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-4 \textit {\_Z} +17\right ) x -8 x^{2}+4 \sqrt {x^{4}+x^{2}+1}+3 x +8}{{\left (2 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-4 \textit {\_Z} +17\right ) x -x -4\right )}^{2}}\right )}{2}-\frac {\ln \left (-\frac {6 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-4 \textit {\_Z} +17\right ) x -8 x^{2}+4 \sqrt {x^{4}+x^{2}+1}-3 x +8}{{\left (2 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-4 \textit {\_Z} +17\right ) x -x +4\right )}^{2}}\right ) \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-4 \textit {\_Z} +17\right )}{2}+\frac {\ln \left (-\frac {6 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-4 \textit {\_Z} +17\right ) x -8 x^{2}+4 \sqrt {x^{4}+x^{2}+1}-3 x +8}{{\left (2 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}-4 \textit {\_Z} +17\right ) x -x +4\right )}^{2}}\right )}{2}\) | \(210\) |
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Time = 0.31 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.09 \[ \int \frac {-2-x+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=2 \, \arctan \left (\frac {\sqrt {x^{4} + x^{2} + 1}}{x}\right ) + \frac {1}{4} \, \log \left (\frac {5 \, x^{4} + 2 \, x^{2} - 4 \, \sqrt {x^{4} + x^{2} + 1} {\left (x^{2} - 1\right )} + 5}{x^{4} + 2 \, x^{2} + 1}\right ) \]
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\[ \int \frac {-2-x+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=\int \frac {2 x^{2} - x - 2}{\sqrt {\left (x^{2} - x + 1\right ) \left (x^{2} + x + 1\right )} \left (x^{2} + 1\right )}\, dx \]
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\[ \int \frac {-2-x+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=\int { \frac {2 \, x^{2} - x - 2}{\sqrt {x^{4} + x^{2} + 1} {\left (x^{2} + 1\right )}} \,d x } \]
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\[ \int \frac {-2-x+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=\int { \frac {2 \, x^{2} - x - 2}{\sqrt {x^{4} + x^{2} + 1} {\left (x^{2} + 1\right )}} \,d x } \]
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Timed out. \[ \int \frac {-2-x+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^2+x^4}} \, dx=\int -\frac {-2\,x^2+x+2}{\left (x^2+1\right )\,\sqrt {x^4+x^2+1}} \,d x \]
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