Integrand size = 18, antiderivative size = 85 \[ \int \frac {1+x^4}{1+2 x^4+x^8} \, dx=-\frac {\arctan \left (1-\sqrt {2} x\right )}{2 \sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} x\right )}{2 \sqrt {2}}-\frac {\log \left (1-\sqrt {2} x+x^2\right )}{4 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} x+x^2\right )}{4 \sqrt {2}} \]
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Time = 0.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {28, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {1+x^4}{1+2 x^4+x^8} \, dx=-\frac {\arctan \left (1-\sqrt {2} x\right )}{2 \sqrt {2}}+\frac {\arctan \left (\sqrt {2} x+1\right )}{2 \sqrt {2}}-\frac {\log \left (x^2-\sqrt {2} x+1\right )}{4 \sqrt {2}}+\frac {\log \left (x^2+\sqrt {2} x+1\right )}{4 \sqrt {2}} \]
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Rule 28
Rule 210
Rule 217
Rule 631
Rule 642
Rule 1176
Rule 1179
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{1+x^4} \, dx \\ & = \frac {1}{2} \int \frac {1-x^2}{1+x^4} \, dx+\frac {1}{2} \int \frac {1+x^2}{1+x^4} \, dx \\ & = \frac {1}{4} \int \frac {1}{1-\sqrt {2} x+x^2} \, dx+\frac {1}{4} \int \frac {1}{1+\sqrt {2} x+x^2} \, dx-\frac {\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{4 \sqrt {2}}-\frac {\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{4 \sqrt {2}} \\ & = -\frac {\log \left (1-\sqrt {2} x+x^2\right )}{4 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} x+x^2\right )}{4 \sqrt {2}}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{2 \sqrt {2}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} x\right )}{2 \sqrt {2}} \\ & = -\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{2 \sqrt {2}}+\frac {\tan ^{-1}\left (1+\sqrt {2} x\right )}{2 \sqrt {2}}-\frac {\log \left (1-\sqrt {2} x+x^2\right )}{4 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} x+x^2\right )}{4 \sqrt {2}} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.75 \[ \int \frac {1+x^4}{1+2 x^4+x^8} \, dx=\frac {-2 \arctan \left (1-\sqrt {2} x\right )+2 \arctan \left (1+\sqrt {2} x\right )-\log \left (1-\sqrt {2} x+x^2\right )+\log \left (1+\sqrt {2} x+x^2\right )}{4 \sqrt {2}} \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.26
method | result | size |
risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{3}}\right )}{4}\) | \(22\) |
default | \(\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x^{2}+x \sqrt {2}}{1+x^{2}-x \sqrt {2}}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{8}\) | \(52\) |
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Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.72 \[ \int \frac {1+x^4}{1+2 x^4+x^8} \, dx=\left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (2 \, x + \left (i + 1\right ) \, \sqrt {2}\right ) - \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (2 \, x - \left (i - 1\right ) \, \sqrt {2}\right ) + \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \sqrt {2} \log \left (2 \, x + \left (i - 1\right ) \, \sqrt {2}\right ) - \left (\frac {1}{8} i + \frac {1}{8}\right ) \, \sqrt {2} \log \left (2 \, x - \left (i + 1\right ) \, \sqrt {2}\right ) \]
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Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86 \[ \int \frac {1+x^4}{1+2 x^4+x^8} \, dx=- \frac {\sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{8} + \frac {\sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{8} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{4} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{4} \]
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none
Time = 0.31 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.85 \[ \int \frac {1+x^4}{1+2 x^4+x^8} \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{8} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \]
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none
Time = 0.30 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.85 \[ \int \frac {1+x^4}{1+2 x^4+x^8} \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{8} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{8} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \]
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Time = 8.34 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.39 \[ \int \frac {1+x^4}{1+2 x^4+x^8} \, dx=\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{4}+\frac {1}{4}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{4}-\frac {1}{4}{}\mathrm {i}\right ) \]
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