Integrand size = 23, antiderivative size = 89 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{2 x^2}+\frac {1}{4} \arctan \left (\sqrt {3}-2 x^2\right )-\frac {1}{4} \arctan \left (\sqrt {3}+2 x^2\right )-\frac {\log \left (1-\sqrt {3} x^2+x^4\right )}{8 \sqrt {3}}+\frac {\log \left (1+\sqrt {3} x^2+x^4\right )}{8 \sqrt {3}} \]
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Time = 0.06 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {1504, 1295, 1141, 1175, 632, 210, 1178, 642} \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=\frac {1}{4} \arctan \left (\sqrt {3}-2 x^2\right )-\frac {1}{4} \arctan \left (2 x^2+\sqrt {3}\right )-\frac {1}{2 x^2}-\frac {\log \left (x^4-\sqrt {3} x^2+1\right )}{8 \sqrt {3}}+\frac {\log \left (x^4+\sqrt {3} x^2+1\right )}{8 \sqrt {3}} \]
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Rule 210
Rule 632
Rule 642
Rule 1141
Rule 1175
Rule 1178
Rule 1295
Rule 1504
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1-x^2}{x^2 \left (1-x^2+x^4\right )} \, dx,x,x^2\right ) \\ & = -\frac {1}{2 x^2}-\frac {1}{2} \text {Subst}\left (\int \frac {x^2}{1-x^2+x^4} \, dx,x,x^2\right ) \\ & = -\frac {1}{2 x^2}+\frac {1}{4} \text {Subst}\left (\int \frac {1-x^2}{1-x^2+x^4} \, dx,x,x^2\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1+x^2}{1-x^2+x^4} \, dx,x,x^2\right ) \\ & = -\frac {1}{2 x^2}-\frac {1}{8} \text {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,x^2\right )-\frac {1}{8} \text {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,x^2\right )-\frac {\text {Subst}\left (\int \frac {\sqrt {3}+2 x}{-1-\sqrt {3} x-x^2} \, dx,x,x^2\right )}{8 \sqrt {3}}-\frac {\text {Subst}\left (\int \frac {\sqrt {3}-2 x}{-1+\sqrt {3} x-x^2} \, dx,x,x^2\right )}{8 \sqrt {3}} \\ & = -\frac {1}{2 x^2}-\frac {\log \left (1-\sqrt {3} x^2+x^4\right )}{8 \sqrt {3}}+\frac {\log \left (1+\sqrt {3} x^2+x^4\right )}{8 \sqrt {3}}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+2 x^2\right )+\frac {1}{4} \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+2 x^2\right ) \\ & = -\frac {1}{2 x^2}+\frac {1}{4} \tan ^{-1}\left (\sqrt {3}-2 x^2\right )-\frac {1}{4} \tan ^{-1}\left (\sqrt {3}+2 x^2\right )-\frac {\log \left (1-\sqrt {3} x^2+x^4\right )}{8 \sqrt {3}}+\frac {\log \left (1+\sqrt {3} x^2+x^4\right )}{8 \sqrt {3}} \\ \end{align*}
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.55 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{2 x^2}-\frac {1}{4} \text {RootSum}\left [1-\text {$\#$1}^4+\text {$\#$1}^8\&,\frac {\log (x-\text {$\#$1}) \text {$\#$1}^2}{-1+2 \text {$\#$1}^4}\&\right ] \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.45
method | result | size |
risch | \(-\frac {1}{2 x^{2}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (9 \textit {\_Z}^{4}+3 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-6 \textit {\_R}^{3}+x^{2}-\textit {\_R} \right )\right )}{4}\) | \(40\) |
default | \(-\frac {1}{2 x^{2}}+\frac {\sqrt {3}\, \left (-\frac {\ln \left (1+x^{4}-x^{2} \sqrt {3}\right )}{2}-\sqrt {3}\, \arctan \left (2 x^{2}-\sqrt {3}\right )\right )}{12}+\frac {\sqrt {3}\, \left (\frac {\ln \left (1+x^{4}+x^{2} \sqrt {3}\right )}{2}-\sqrt {3}\, \arctan \left (2 x^{2}+\sqrt {3}\right )\right )}{12}\) | \(82\) |
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Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.90 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=-\frac {\sqrt {6} x^{2} \sqrt {i \, \sqrt {3} - 1} \log \left (6 \, x^{2} + i \, \sqrt {6} \sqrt {3} \sqrt {i \, \sqrt {3} - 1}\right ) - \sqrt {6} x^{2} \sqrt {i \, \sqrt {3} - 1} \log \left (6 \, x^{2} - i \, \sqrt {6} \sqrt {3} \sqrt {i \, \sqrt {3} - 1}\right ) - \sqrt {6} x^{2} \sqrt {-i \, \sqrt {3} - 1} \log \left (6 \, x^{2} + i \, \sqrt {6} \sqrt {3} \sqrt {-i \, \sqrt {3} - 1}\right ) + \sqrt {6} x^{2} \sqrt {-i \, \sqrt {3} - 1} \log \left (6 \, x^{2} - i \, \sqrt {6} \sqrt {3} \sqrt {-i \, \sqrt {3} - 1}\right ) + 12}{24 \, x^{2}} \]
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Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.85 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=- \frac {\sqrt {3} \log {\left (x^{4} - \sqrt {3} x^{2} + 1 \right )}}{24} + \frac {\sqrt {3} \log {\left (x^{4} + \sqrt {3} x^{2} + 1 \right )}}{24} - \frac {\operatorname {atan}{\left (2 x^{2} - \sqrt {3} \right )}}{4} - \frac {\operatorname {atan}{\left (2 x^{2} + \sqrt {3} \right )}}{4} - \frac {1}{2 x^{2}} \]
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\[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=\int { -\frac {x^{4} - 1}{{\left (x^{8} - x^{4} + 1\right )} x^{3}} \,d x } \]
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none
Time = 0.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.91 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=-\frac {1}{24} \, \sqrt {3} x^{4} \log \left (x^{4} + \sqrt {3} x^{2} + 1\right ) + \frac {1}{24} \, \sqrt {3} x^{4} \log \left (x^{4} - \sqrt {3} x^{2} + 1\right ) - \frac {1}{4} \, x^{4} \arctan \left (2 \, x^{2} + \sqrt {3}\right ) - \frac {1}{4} \, x^{4} \arctan \left (2 \, x^{2} - \sqrt {3}\right ) - \frac {1}{2 \, x^{2}} \]
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Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.63 \[ \int \frac {1-x^4}{x^3 \left (1-x^4+x^8\right )} \, dx=\mathrm {atan}\left (\frac {2\,x^2}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\mathrm {atan}\left (\frac {2\,x^2}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )-\frac {1}{2\,x^2} \]
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