Integrand size = 16, antiderivative size = 106 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=-\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}+\frac {7 \arctan \left (1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {7 \arctan \left (1+\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {7 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {7 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}} \]
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Time = 0.03 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {28, 296, 331, 217, 1179, 642, 1176, 631, 210} \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=\frac {7 \arctan \left (1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {7 \arctan \left (\sqrt {2} x+1\right )}{8 \sqrt {2}}-\frac {7}{12 x^3}+\frac {7 \log \left (x^2-\sqrt {2} x+1\right )}{16 \sqrt {2}}-\frac {7 \log \left (x^2+\sqrt {2} x+1\right )}{16 \sqrt {2}}+\frac {1}{4 x^3 \left (x^4+1\right )} \]
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Rule 28
Rule 210
Rule 217
Rule 296
Rule 331
Rule 631
Rule 642
Rule 1176
Rule 1179
Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{x^4 \left (1+x^4\right )^2} \, dx \\ & = \frac {1}{4 x^3 \left (1+x^4\right )}+\frac {7}{4} \int \frac {1}{x^4 \left (1+x^4\right )} \, dx \\ & = -\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}-\frac {7}{4} \int \frac {1}{1+x^4} \, dx \\ & = -\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}-\frac {7}{8} \int \frac {1-x^2}{1+x^4} \, dx-\frac {7}{8} \int \frac {1+x^2}{1+x^4} \, dx \\ & = -\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}-\frac {7}{16} \int \frac {1}{1-\sqrt {2} x+x^2} \, dx-\frac {7}{16} \int \frac {1}{1+\sqrt {2} x+x^2} \, dx+\frac {7 \int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}}+\frac {7 \int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}} \\ & = -\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}+\frac {7 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {7 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {7 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {7 \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} x\right )}{8 \sqrt {2}} \\ & = -\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}+\frac {7 \tan ^{-1}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {7 \tan ^{-1}\left (1+\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {7 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {7 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=\frac {1}{96} \left (-\frac {32}{x^3}-\frac {24 x}{1+x^4}+42 \sqrt {2} \arctan \left (1-\sqrt {2} x\right )-42 \sqrt {2} \arctan \left (1+\sqrt {2} x\right )+21 \sqrt {2} \log \left (1-\sqrt {2} x+x^2\right )-21 \sqrt {2} \log \left (1+\sqrt {2} x+x^2\right )\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.37
method | result | size |
risch | \(\frac {-\frac {7 x^{4}}{12}-\frac {1}{3}}{x^{3} \left (x^{4}+1\right )}+\frac {7 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+1\right )}{\sum }\textit {\_R} \ln \left (x -\textit {\_R} \right )\right )}{16}\) | \(39\) |
default | \(-\frac {1}{3 x^{3}}-\frac {x}{4 \left (x^{4}+1\right )}-\frac {7 \sqrt {2}\, \left (\ln \left (\frac {1+x^{2}+x \sqrt {2}}{1+x^{2}-x \sqrt {2}}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{32}\) | \(68\) |
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Result contains complex when optimal does not.
Time = 0.25 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=-\frac {56 \, x^{4} + 21 \, \sqrt {2} {\left (\left (i + 1\right ) \, x^{7} + \left (i + 1\right ) \, x^{3}\right )} \log \left (2 \, x + \left (i + 1\right ) \, \sqrt {2}\right ) + 21 \, \sqrt {2} {\left (-\left (i - 1\right ) \, x^{7} - \left (i - 1\right ) \, x^{3}\right )} \log \left (2 \, x - \left (i - 1\right ) \, \sqrt {2}\right ) + 21 \, \sqrt {2} {\left (\left (i - 1\right ) \, x^{7} + \left (i - 1\right ) \, x^{3}\right )} \log \left (2 \, x + \left (i - 1\right ) \, \sqrt {2}\right ) + 21 \, \sqrt {2} {\left (-\left (i + 1\right ) \, x^{7} - \left (i + 1\right ) \, x^{3}\right )} \log \left (2 \, x - \left (i + 1\right ) \, \sqrt {2}\right ) + 32}{96 \, {\left (x^{7} + x^{3}\right )}} \]
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Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=\frac {- 7 x^{4} - 4}{12 x^{7} + 12 x^{3}} + \frac {7 \sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{32} - \frac {7 \sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{32} - \frac {7 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{16} - \frac {7 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{16} \]
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Time = 0.26 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.85 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=-\frac {7}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {7}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {7}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {7}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {7 \, x^{4} + 4}{12 \, {\left (x^{7} + x^{3}\right )}} \]
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Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=-\frac {7}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {7}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {7}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {7}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {x}{4 \, {\left (x^{4} + 1\right )}} - \frac {1}{3 \, x^{3}} \]
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Time = 8.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.48 \[ \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx=-\frac {\frac {7\,x^4}{12}+\frac {1}{3}}{x^7+x^3}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {7}{16}-\frac {7}{16}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {7}{16}+\frac {7}{16}{}\mathrm {i}\right ) \]
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