Integrand size = 16, antiderivative size = 41 \[ \int \frac {1}{x \left (1-x^4+x^8\right )} \, dx=-\frac {\arctan \left (\frac {1-2 x^4}{\sqrt {3}}\right )}{4 \sqrt {3}}+\log (x)-\frac {1}{8} \log \left (1-x^4+x^8\right ) \]
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Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {1371, 719, 29, 648, 632, 210, 642} \[ \int \frac {1}{x \left (1-x^4+x^8\right )} \, dx=-\frac {\arctan \left (\frac {1-2 x^4}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{8} \log \left (x^8-x^4+1\right )+\log (x) \]
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Rule 29
Rule 210
Rule 632
Rule 642
Rule 648
Rule 719
Rule 1371
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {1}{x \left (1-x+x^2\right )} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^4\right )+\frac {1}{4} \text {Subst}\left (\int \frac {1-x}{1-x+x^2} \, dx,x,x^4\right ) \\ & = \log (x)+\frac {1}{8} \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,x^4\right )-\frac {1}{8} \text {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,x^4\right ) \\ & = \log (x)-\frac {1}{8} \log \left (1-x^4+x^8\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x^4\right ) \\ & = -\frac {\tan ^{-1}\left (\frac {1-2 x^4}{\sqrt {3}}\right )}{4 \sqrt {3}}+\log (x)-\frac {1}{8} \log \left (1-x^4+x^8\right ) \\ \end{align*}
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.34 \[ \int \frac {1}{x \left (1-x^4+x^8\right )} \, dx=\log (x)-\frac {1}{4} \text {RootSum}\left [1-\text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x-\text {$\#$1})+\log (x-\text {$\#$1}) \text {$\#$1}^4}{-1+2 \text {$\#$1}^4}\&\right ] \]
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Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.80
method | result | size |
risch | \(\ln \left (x \right )-\frac {\ln \left (x^{8}-x^{4}+1\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x^{4}-\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{12}\) | \(33\) |
default | \(\ln \left (x \right )-\frac {\ln \left (x^{8}-x^{4}+1\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{4}-1\right ) \sqrt {3}}{3}\right )}{12}\) | \(35\) |
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Time = 0.23 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.83 \[ \int \frac {1}{x \left (1-x^4+x^8\right )} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} - 1\right )}\right ) - \frac {1}{8} \, \log \left (x^{8} - x^{4} + 1\right ) + \log \left (x\right ) \]
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Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x \left (1-x^4+x^8\right )} \, dx=\log {\left (x \right )} - \frac {\log {\left (x^{8} - x^{4} + 1 \right )}}{8} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{4}}{3} - \frac {\sqrt {3}}{3} \right )}}{12} \]
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Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x \left (1-x^4+x^8\right )} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} - 1\right )}\right ) - \frac {1}{8} \, \log \left (x^{8} - x^{4} + 1\right ) + \frac {1}{4} \, \log \left (x^{4}\right ) \]
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Time = 0.31 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x \left (1-x^4+x^8\right )} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{4} - 1\right )}\right ) - \frac {1}{8} \, \log \left (x^{8} - x^{4} + 1\right ) + \frac {1}{4} \, \log \left (x^{4}\right ) \]
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Time = 8.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x \left (1-x^4+x^8\right )} \, dx=\ln \left (x\right )-\frac {\ln \left (x^8-x^4+1\right )}{8}-\frac {\sqrt {3}\,\mathrm {atan}\left (\frac {\sqrt {3}}{3}-\frac {2\,\sqrt {3}\,x^4}{3}\right )}{12} \]
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