Integrand size = 16, antiderivative size = 99 \[ \int \frac {x^2}{1+2 x^4+x^8} \, dx=\frac {x^3}{4 \left (1+x^4\right )}-\frac {\arctan \left (1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {\log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}} \]
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Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {28, 296, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {x^2}{1+2 x^4+x^8} \, dx=-\frac {\arctan \left (1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {\arctan \left (\sqrt {2} x+1\right )}{8 \sqrt {2}}+\frac {\log \left (x^2-\sqrt {2} x+1\right )}{16 \sqrt {2}}-\frac {\log \left (x^2+\sqrt {2} x+1\right )}{16 \sqrt {2}}+\frac {x^3}{4 \left (x^4+1\right )} \]
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Rule 28
Rule 210
Rule 296
Rule 303
Rule 631
Rule 642
Rule 1176
Rule 1179
Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2}{\left (1+x^4\right )^2} \, dx \\ & = \frac {x^3}{4 \left (1+x^4\right )}+\frac {1}{4} \int \frac {x^2}{1+x^4} \, dx \\ & = \frac {x^3}{4 \left (1+x^4\right )}-\frac {1}{8} \int \frac {1-x^2}{1+x^4} \, dx+\frac {1}{8} \int \frac {1+x^2}{1+x^4} \, dx \\ & = \frac {x^3}{4 \left (1+x^4\right )}+\frac {1}{16} \int \frac {1}{1-\sqrt {2} x+x^2} \, dx+\frac {1}{16} \int \frac {1}{1+\sqrt {2} x+x^2} \, dx+\frac {\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}}+\frac {\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}} \\ & = \frac {x^3}{4 \left (1+x^4\right )}+\frac {\log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} x\right )}{8 \sqrt {2}} \\ & = \frac {x^3}{4 \left (1+x^4\right )}-\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {\tan ^{-1}\left (1+\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {\log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}} \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.93 \[ \int \frac {x^2}{1+2 x^4+x^8} \, dx=\frac {1}{32} \left (\frac {8 x^3}{1+x^4}-2 \sqrt {2} \arctan \left (1-\sqrt {2} x\right )+2 \sqrt {2} \arctan \left (1+\sqrt {2} x\right )+\sqrt {2} \log \left (1-\sqrt {2} x+x^2\right )-\sqrt {2} \log \left (1+\sqrt {2} x+x^2\right )\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.05 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.35
method | result | size |
risch | \(\frac {x^{3}}{4 x^{4}+4}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{16}\) | \(35\) |
default | \(\frac {x^{3}}{4 x^{4}+4}+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x^{2}-x \sqrt {2}}{1+x^{2}+x \sqrt {2}}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{32}\) | \(65\) |
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Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.00 \[ \int \frac {x^2}{1+2 x^4+x^8} \, dx=\frac {8 \, x^{3} + \sqrt {2} {\left (\left (i - 1\right ) \, x^{4} + i - 1\right )} \log \left (2 \, x + \left (i + 1\right ) \, \sqrt {2}\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, x^{4} - i - 1\right )} \log \left (2 \, x - \left (i - 1\right ) \, \sqrt {2}\right ) + \sqrt {2} {\left (\left (i + 1\right ) \, x^{4} + i + 1\right )} \log \left (2 \, x + \left (i - 1\right ) \, \sqrt {2}\right ) + \sqrt {2} {\left (-\left (i - 1\right ) \, x^{4} - i + 1\right )} \log \left (2 \, x - \left (i + 1\right ) \, \sqrt {2}\right )}{32 \, {\left (x^{4} + 1\right )}} \]
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Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.84 \[ \int \frac {x^2}{1+2 x^4+x^8} \, dx=\frac {x^{3}}{4 x^{4} + 4} + \frac {\sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{32} - \frac {\sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{32} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{16} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{16} \]
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Time = 0.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.85 \[ \int \frac {x^2}{1+2 x^4+x^8} \, dx=\frac {x^{3}}{4 \, {\left (x^{4} + 1\right )}} + \frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {1}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \]
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Time = 0.30 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.85 \[ \int \frac {x^2}{1+2 x^4+x^8} \, dx=\frac {x^{3}}{4 \, {\left (x^{4} + 1\right )}} + \frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {1}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \]
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Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.46 \[ \int \frac {x^2}{1+2 x^4+x^8} \, dx=\frac {x^3}{4\,\left (x^4+1\right )}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{16}-\frac {1}{16}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{16}+\frac {1}{16}{}\mathrm {i}\right ) \]
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