Integrand size = 16, antiderivative size = 113 \[ \int \frac {1}{x^6 \left (1+2 x^4+x^8\right )} \, dx=-\frac {9}{20 x^5}+\frac {9}{4 x}+\frac {1}{4 x^5 \left (1+x^4\right )}-\frac {9 \arctan \left (1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {9 \arctan \left (1+\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {9 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {9 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}} \]
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Time = 0.04 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {28, 296, 331, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {1}{x^6 \left (1+2 x^4+x^8\right )} \, dx=-\frac {9 \arctan \left (1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {9 \arctan \left (\sqrt {2} x+1\right )}{8 \sqrt {2}}-\frac {9}{20 x^5}+\frac {9 \log \left (x^2-\sqrt {2} x+1\right )}{16 \sqrt {2}}-\frac {9 \log \left (x^2+\sqrt {2} x+1\right )}{16 \sqrt {2}}+\frac {1}{4 x^5 \left (x^4+1\right )}+\frac {9}{4 x} \]
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Rule 28
Rule 210
Rule 296
Rule 303
Rule 331
Rule 631
Rule 642
Rule 1176
Rule 1179
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^6 \left (1+x^4\right )^2} \, dx \\ & = \frac {1}{4 x^5 \left (1+x^4\right )}+\frac {9}{4} \int \frac {1}{x^6 \left (1+x^4\right )} \, dx \\ & = -\frac {9}{20 x^5}+\frac {1}{4 x^5 \left (1+x^4\right )}-\frac {9}{4} \int \frac {1}{x^2 \left (1+x^4\right )} \, dx \\ & = -\frac {9}{20 x^5}+\frac {9}{4 x}+\frac {1}{4 x^5 \left (1+x^4\right )}+\frac {9}{4} \int \frac {x^2}{1+x^4} \, dx \\ & = -\frac {9}{20 x^5}+\frac {9}{4 x}+\frac {1}{4 x^5 \left (1+x^4\right )}-\frac {9}{8} \int \frac {1-x^2}{1+x^4} \, dx+\frac {9}{8} \int \frac {1+x^2}{1+x^4} \, dx \\ & = -\frac {9}{20 x^5}+\frac {9}{4 x}+\frac {1}{4 x^5 \left (1+x^4\right )}+\frac {9}{16} \int \frac {1}{1-\sqrt {2} x+x^2} \, dx+\frac {9}{16} \int \frac {1}{1+\sqrt {2} x+x^2} \, dx+\frac {9 \int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}}+\frac {9 \int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}} \\ & = -\frac {9}{20 x^5}+\frac {9}{4 x}+\frac {1}{4 x^5 \left (1+x^4\right )}+\frac {9 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {9 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}}+\frac {9 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {9 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} x\right )}{8 \sqrt {2}} \\ & = -\frac {9}{20 x^5}+\frac {9}{4 x}+\frac {1}{4 x^5 \left (1+x^4\right )}-\frac {9 \tan ^{-1}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {9 \tan ^{-1}\left (1+\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {9 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {9 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^6 \left (1+2 x^4+x^8\right )} \, dx=\frac {1}{160} \left (-\frac {32}{x^5}+\frac {320}{x}+\frac {40 x^3}{1+x^4}-90 \sqrt {2} \arctan \left (1-\sqrt {2} x\right )+90 \sqrt {2} \arctan \left (1+\sqrt {2} x\right )+45 \sqrt {2} \log \left (1-\sqrt {2} x+x^2\right )-45 \sqrt {2} \log \left (1+\sqrt {2} x+x^2\right )\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.39
method | result | size |
risch | \(\frac {\frac {9}{4} x^{8}+\frac {9}{5} x^{4}-\frac {1}{5}}{x^{5} \left (x^{4}+1\right )}+\frac {9 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (\textit {\_R}^{3}+x \right )\right )}{16}\) | \(44\) |
default | \(-\frac {1}{5 x^{5}}+\frac {2}{x}+\frac {x^{3}}{4 x^{4}+4}+\frac {9 \sqrt {2}\, \left (\ln \left (\frac {1+x^{2}-x \sqrt {2}}{1+x^{2}+x \sqrt {2}}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{32}\) | \(75\) |
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Result contains complex when optimal does not.
Time = 0.28 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.12 \[ \int \frac {1}{x^6 \left (1+2 x^4+x^8\right )} \, dx=\frac {360 \, x^{8} + 288 \, x^{4} - 45 \, \sqrt {2} {\left (-\left (i - 1\right ) \, x^{9} - \left (i - 1\right ) \, x^{5}\right )} \log \left (2 \, x + \left (i + 1\right ) \, \sqrt {2}\right ) - 45 \, \sqrt {2} {\left (\left (i + 1\right ) \, x^{9} + \left (i + 1\right ) \, x^{5}\right )} \log \left (2 \, x - \left (i - 1\right ) \, \sqrt {2}\right ) - 45 \, \sqrt {2} {\left (-\left (i + 1\right ) \, x^{9} - \left (i + 1\right ) \, x^{5}\right )} \log \left (2 \, x + \left (i - 1\right ) \, \sqrt {2}\right ) - 45 \, \sqrt {2} {\left (\left (i - 1\right ) \, x^{9} + \left (i - 1\right ) \, x^{5}\right )} \log \left (2 \, x - \left (i + 1\right ) \, \sqrt {2}\right ) - 32}{160 \, {\left (x^{9} + x^{5}\right )}} \]
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Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.90 \[ \int \frac {1}{x^6 \left (1+2 x^4+x^8\right )} \, dx=\frac {9 \sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{32} - \frac {9 \sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{32} + \frac {9 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{16} + \frac {9 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{16} + \frac {45 x^{8} + 36 x^{4} - 4}{20 x^{9} + 20 x^{5}} \]
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Time = 0.27 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.84 \[ \int \frac {1}{x^6 \left (1+2 x^4+x^8\right )} \, dx=\frac {9}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {9}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {9}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {9}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) + \frac {45 \, x^{8} + 36 \, x^{4} - 4}{20 \, {\left (x^{9} + x^{5}\right )}} \]
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Time = 0.28 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^6 \left (1+2 x^4+x^8\right )} \, dx=\frac {x^{3}}{4 \, {\left (x^{4} + 1\right )}} + \frac {9}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {9}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {9}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {9}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) + \frac {10 \, x^{4} - 1}{5 \, x^{5}} \]
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Time = 0.05 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.49 \[ \int \frac {1}{x^6 \left (1+2 x^4+x^8\right )} \, dx=\frac {\frac {9\,x^8}{4}+\frac {9\,x^4}{5}-\frac {1}{5}}{x^9+x^5}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {9}{16}-\frac {9}{16}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {9}{16}+\frac {9}{16}{}\mathrm {i}\right ) \]
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