Integrand size = 31, antiderivative size = 26 \[ \int \frac {-1-2 x^2+2 x^4}{\left (1+2 x^4\right ) \sqrt {1+x^6}} \, dx=-\arctan \left (\frac {x \sqrt {1+x^6}}{1-x^2+x^4}\right ) \]
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\[ \int \frac {-1-2 x^2+2 x^4}{\left (1+2 x^4\right ) \sqrt {1+x^6}} \, dx=\int \frac {-1-2 x^2+2 x^4}{\left (1+2 x^4\right ) \sqrt {1+x^6}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{\sqrt {1+x^6}}-\frac {2 \left (1+x^2\right )}{\left (1+2 x^4\right ) \sqrt {1+x^6}}\right ) \, dx \\ & = -\left (2 \int \frac {1+x^2}{\left (1+2 x^4\right ) \sqrt {1+x^6}} \, dx\right )+\int \frac {1}{\sqrt {1+x^6}} \, dx \\ & = \frac {x \left (1+x^2\right ) \sqrt {\frac {1-x^2+x^4}{\left (1+\left (1+\sqrt {3}\right ) x^2\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1+\left (1-\sqrt {3}\right ) x^2}{1+\left (1+\sqrt {3}\right ) x^2}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 \sqrt [4]{3} \sqrt {\frac {x^2 \left (1+x^2\right )}{\left (1+\left (1+\sqrt {3}\right ) x^2\right )^2}} \sqrt {1+x^6}}-2 \int \left (\frac {i \left (i+\sqrt {2}\right )}{2 \sqrt {2} \left (i-\sqrt {2} x^2\right ) \sqrt {1+x^6}}-\frac {i \left (i-\sqrt {2}\right )}{2 \sqrt {2} \left (i+\sqrt {2} x^2\right ) \sqrt {1+x^6}}\right ) \, dx \\ & = \frac {x \left (1+x^2\right ) \sqrt {\frac {1-x^2+x^4}{\left (1+\left (1+\sqrt {3}\right ) x^2\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1+\left (1-\sqrt {3}\right ) x^2}{1+\left (1+\sqrt {3}\right ) x^2}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 \sqrt [4]{3} \sqrt {\frac {x^2 \left (1+x^2\right )}{\left (1+\left (1+\sqrt {3}\right ) x^2\right )^2}} \sqrt {1+x^6}}-\frac {1}{2} \left (2 i-\sqrt {2}\right ) \int \frac {1}{\left (i-\sqrt {2} x^2\right ) \sqrt {1+x^6}} \, dx-\frac {1}{2} \left (2 i+\sqrt {2}\right ) \int \frac {1}{\left (i+\sqrt {2} x^2\right ) \sqrt {1+x^6}} \, dx \\ & = \frac {x \left (1+x^2\right ) \sqrt {\frac {1-x^2+x^4}{\left (1+\left (1+\sqrt {3}\right ) x^2\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1+\left (1-\sqrt {3}\right ) x^2}{1+\left (1+\sqrt {3}\right ) x^2}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 \sqrt [4]{3} \sqrt {\frac {x^2 \left (1+x^2\right )}{\left (1+\left (1+\sqrt {3}\right ) x^2\right )^2}} \sqrt {1+x^6}}-\frac {1}{2} \left (2 i-\sqrt {2}\right ) \int \left (-\frac {(-1)^{3/4}}{2 \left (\sqrt [4]{-1}-\sqrt [4]{2} x\right ) \sqrt {1+x^6}}-\frac {(-1)^{3/4}}{2 \left (\sqrt [4]{-1}+\sqrt [4]{2} x\right ) \sqrt {1+x^6}}\right ) \, dx-\frac {1}{2} \left (2 i+\sqrt {2}\right ) \int \left (-\frac {\sqrt [4]{-1}}{2 \left (-(-1)^{3/4}-\sqrt [4]{2} x\right ) \sqrt {1+x^6}}-\frac {\sqrt [4]{-1}}{2 \left (-(-1)^{3/4}+\sqrt [4]{2} x\right ) \sqrt {1+x^6}}\right ) \, dx \\ & = \frac {x \left (1+x^2\right ) \sqrt {\frac {1-x^2+x^4}{\left (1+\left (1+\sqrt {3}\right ) x^2\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {1+\left (1-\sqrt {3}\right ) x^2}{1+\left (1+\sqrt {3}\right ) x^2}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{2 \sqrt [4]{3} \sqrt {\frac {x^2 \left (1+x^2\right )}{\left (1+\left (1+\sqrt {3}\right ) x^2\right )^2}} \sqrt {1+x^6}}+\left (\left (\frac {1}{4}+\frac {i}{4}\right ) \left (1+i \sqrt {2}\right )\right ) \int \frac {1}{\left (-(-1)^{3/4}-\sqrt [4]{2} x\right ) \sqrt {1+x^6}} \, dx+\left (\left (\frac {1}{4}+\frac {i}{4}\right ) \left (1+i \sqrt {2}\right )\right ) \int \frac {1}{\left (-(-1)^{3/4}+\sqrt [4]{2} x\right ) \sqrt {1+x^6}} \, dx-\left (\left (\frac {1}{4}+\frac {i}{4}\right ) \left (i+\sqrt {2}\right )\right ) \int \frac {1}{\left (\sqrt [4]{-1}-\sqrt [4]{2} x\right ) \sqrt {1+x^6}} \, dx-\left (\left (\frac {1}{4}+\frac {i}{4}\right ) \left (i+\sqrt {2}\right )\right ) \int \frac {1}{\left (\sqrt [4]{-1}+\sqrt [4]{2} x\right ) \sqrt {1+x^6}} \, dx \\ \end{align*}
Time = 9.85 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {-1-2 x^2+2 x^4}{\left (1+2 x^4\right ) \sqrt {1+x^6}} \, dx=-\arctan \left (\frac {x \sqrt {1+x^6}}{1-x^2+x^4}\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.42 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.96
method | result | size |
trager | \(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-2 \sqrt {x^{6}+1}\, x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{2 x^{4}+1}\right )}{2}\) | \(51\) |
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Time = 0.28 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {-1-2 x^2+2 x^4}{\left (1+2 x^4\right ) \sqrt {1+x^6}} \, dx=\frac {1}{2} \, \arctan \left (\frac {2 \, \sqrt {x^{6} + 1} x}{2 \, x^{2} - 1}\right ) \]
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\[ \int \frac {-1-2 x^2+2 x^4}{\left (1+2 x^4\right ) \sqrt {1+x^6}} \, dx=\int \frac {2 x^{4} - 2 x^{2} - 1}{\sqrt {\left (x^{2} + 1\right ) \left (x^{4} - x^{2} + 1\right )} \left (2 x^{4} + 1\right )}\, dx \]
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\[ \int \frac {-1-2 x^2+2 x^4}{\left (1+2 x^4\right ) \sqrt {1+x^6}} \, dx=\int { \frac {2 \, x^{4} - 2 \, x^{2} - 1}{\sqrt {x^{6} + 1} {\left (2 \, x^{4} + 1\right )}} \,d x } \]
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\[ \int \frac {-1-2 x^2+2 x^4}{\left (1+2 x^4\right ) \sqrt {1+x^6}} \, dx=\int { \frac {2 \, x^{4} - 2 \, x^{2} - 1}{\sqrt {x^{6} + 1} {\left (2 \, x^{4} + 1\right )}} \,d x } \]
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Timed out. \[ \int \frac {-1-2 x^2+2 x^4}{\left (1+2 x^4\right ) \sqrt {1+x^6}} \, dx=\int -\frac {-2\,x^4+2\,x^2+1}{\sqrt {x^6+1}\,\left (2\,x^4+1\right )} \,d x \]
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