Integrand size = 14, antiderivative size = 89 \[ \int \frac {1}{x^7 \left (1+x^4+x^8\right )} \, dx=-\frac {1}{6 x^6}+\frac {1}{2 x^2}-\frac {\arctan \left (\frac {1-2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {\arctan \left (\frac {1+2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{8} \log \left (1-x^2+x^4\right )-\frac {1}{8} \log \left (1+x^2+x^4\right ) \]
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Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {1373, 1137, 1295, 12, 1141, 1175, 632, 210, 1178, 642} \[ \int \frac {1}{x^7 \left (1+x^4+x^8\right )} \, dx=-\frac {\arctan \left (\frac {1-2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {\arctan \left (\frac {2 x^2+1}{\sqrt {3}}\right )}{4 \sqrt {3}}-\frac {1}{6 x^6}+\frac {1}{2 x^2}+\frac {1}{8} \log \left (x^4-x^2+1\right )-\frac {1}{8} \log \left (x^4+x^2+1\right ) \]
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Rule 12
Rule 210
Rule 632
Rule 642
Rule 1137
Rule 1141
Rule 1175
Rule 1178
Rule 1295
Rule 1373
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{x^4 \left (1+x^2+x^4\right )} \, dx,x,x^2\right ) \\ & = -\frac {1}{6 x^6}+\frac {1}{6} \text {Subst}\left (\int \frac {-3-3 x^2}{x^2 \left (1+x^2+x^4\right )} \, dx,x,x^2\right ) \\ & = -\frac {1}{6 x^6}+\frac {1}{2 x^2}-\frac {1}{6} \text {Subst}\left (\int -\frac {3 x^2}{1+x^2+x^4} \, dx,x,x^2\right ) \\ & = -\frac {1}{6 x^6}+\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}{6 x^6}+\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}{6 x^6}+\frac {1}{2 x^2}+\frac {1}{8} \text {Subst}\left (\int \frac {1+2 x}{-1-x-x^2} \, dx,x,x^2\right )+\frac {1}{8} \text {Subst}\left (\int \frac {1-2 x}{-1+x-x^2} \, dx,x,x^2\right )+\frac {1}{8} \text {Subst}\left (\int \frac {1}{1-x+x^2} \, dx,x,x^2\right )+\frac {1}{8} \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,x^2\right ) \\ & = -\frac {1}{6 x^6}+\frac {1}{2 x^2}+\frac {1}{8} \log \left (1-x^2+x^4\right )-\frac {1}{8} \log \left (1+x^2+x^4\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,-1+2 x^2\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 x^2\right ) \\ & = -\frac {1}{6 x^6}+\frac {1}{2 x^2}-\frac {\tan ^{-1}\left (\frac {1-2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {\tan ^{-1}\left (\frac {1+2 x^2}{\sqrt {3}}\right )}{4 \sqrt {3}}+\frac {1}{8} \log \left (1-x^2+x^4\right )-\frac {1}{8} \log \left (1+x^2+x^4\right ) \\ \end{align*}
Time = 0.06 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.20 \[ \int \frac {1}{x^7 \left (1+x^4+x^8\right )} \, dx=\frac {1}{24} \left (-\frac {4}{x^6}+\frac {12}{x^2}-2 \sqrt {3} \arctan \left (\frac {1-2 x}{\sqrt {3}}\right )-2 \sqrt {3} \arctan \left (\frac {1+2 x}{\sqrt {3}}\right )-2 \sqrt {3} \arctan \left (\frac {1-2 x^2}{\sqrt {3}}\right )-3 \log \left (1-x+x^2\right )-3 \log \left (1+x+x^2\right )+3 \log \left (1-x^2+x^4\right )\right ) \]
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Time = 0.13 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.78
method | result | size |
risch | \(\frac {\frac {x^{4}}{2}-\frac {1}{6}}{x^{6}}-\frac {\ln \left (x^{4}+x^{2}+1\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x^{2}+\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{12}+\frac {\ln \left (x^{4}-x^{2}+1\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {2 \left (x^{2}-\frac {1}{2}\right ) \sqrt {3}}{3}\right )}{12}\) | \(69\) |
default | \(-\frac {1}{6 x^{6}}+\frac {1}{2 x^{2}}-\frac {\ln \left (x^{2}-x +1\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x -1\right ) \sqrt {3}}{3}\right )}{12}-\frac {\ln \left (x^{2}+x +1\right )}{8}-\frac {\arctan \left (\frac {\left (1+2 x \right ) \sqrt {3}}{3}\right ) \sqrt {3}}{12}+\frac {\ln \left (x^{4}-x^{2}+1\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 x^{2}-1\right ) \sqrt {3}}{3}\right )}{12}\) | \(95\) |
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Time = 0.25 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^7 \left (1+x^4+x^8\right )} \, dx=\frac {2 \, \sqrt {3} x^{6} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + 2 \, \sqrt {3} x^{6} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) - 3 \, x^{6} \log \left (x^{4} + x^{2} + 1\right ) + 3 \, x^{6} \log \left (x^{4} - x^{2} + 1\right ) + 12 \, x^{4} - 4}{24 \, x^{6}} \]
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Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^7 \left (1+x^4+x^8\right )} \, dx=\frac {\log {\left (x^{4} - x^{2} + 1 \right )}}{8} - \frac {\log {\left (x^{4} + x^{2} + 1 \right )}}{8} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} - \frac {\sqrt {3}}{3} \right )}}{12} + \frac {\sqrt {3} \operatorname {atan}{\left (\frac {2 \sqrt {3} x^{2}}{3} + \frac {\sqrt {3}}{3} \right )}}{12} + \frac {3 x^{4} - 1}{6 x^{6}} \]
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Time = 0.27 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^7 \left (1+x^4+x^8\right )} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) + \frac {3 \, x^{4} - 1}{6 \, x^{6}} - \frac {1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) + \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \]
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Time = 0.32 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^7 \left (1+x^4+x^8\right )} \, dx=\frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} + 1\right )}\right ) + \frac {1}{12} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{2} - 1\right )}\right ) + \frac {3 \, x^{4} - 1}{6 \, x^{6}} - \frac {1}{8} \, \log \left (x^{4} + x^{2} + 1\right ) + \frac {1}{8} \, \log \left (x^{4} - x^{2} + 1\right ) \]
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Time = 0.02 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.70 \[ \int \frac {1}{x^7 \left (1+x^4+x^8\right )} \, dx=\mathrm {atanh}\left (\frac {2\,x^2}{-1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\mathrm {atanh}\left (\frac {2\,x^2}{1+\sqrt {3}\,1{}\mathrm {i}}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{12}\right )+\frac {\frac {x^4}{2}-\frac {1}{6}}{x^6} \]
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