Integrand size = 16, antiderivative size = 112 \[ \int \frac {1+x^3+x^6}{x+x^5} \, dx=\frac {x^2}{2}-\frac {\arctan \left (x^2\right )}{2}-\frac {\arctan \left (1-\sqrt {2} x\right )}{2 \sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} x\right )}{2 \sqrt {2}}+\log (x)+\frac {\log \left (1-\sqrt {2} x+x^2\right )}{4 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} x+x^2\right )}{4 \sqrt {2}}-\frac {1}{4} \log \left (1+x^4\right ) \]
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Time = 0.07 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.812, Rules used = {1607, 1847, 303, 1176, 631, 210, 1179, 642, 1848, 1262, 649, 209, 266} \[ \int \frac {1+x^3+x^6}{x+x^5} \, dx=-\frac {\arctan \left (x^2\right )}{2}-\frac {\arctan \left (1-\sqrt {2} x\right )}{2 \sqrt {2}}+\frac {\arctan \left (\sqrt {2} x+1\right )}{2 \sqrt {2}}-\frac {1}{4} \log \left (x^4+1\right )+\frac {x^2}{2}+\frac {\log \left (x^2-\sqrt {2} x+1\right )}{4 \sqrt {2}}-\frac {\log \left (x^2+\sqrt {2} x+1\right )}{4 \sqrt {2}}+\log (x) \]
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Rule 209
Rule 210
Rule 266
Rule 303
Rule 631
Rule 642
Rule 649
Rule 1176
Rule 1179
Rule 1262
Rule 1607
Rule 1847
Rule 1848
Rubi steps \begin{align*} \text {integral}& = \int \frac {1+x^3+x^6}{x \left (1+x^4\right )} \, dx \\ & = \int \left (\frac {x^2}{1+x^4}+\frac {1+x^6}{x \left (1+x^4\right )}\right ) \, dx \\ & = \int \frac {x^2}{1+x^4} \, dx+\int \frac {1+x^6}{x \left (1+x^4\right )} \, dx \\ & = -\left (\frac {1}{2} \int \frac {1-x^2}{1+x^4} \, dx\right )+\frac {1}{2} \int \frac {1+x^2}{1+x^4} \, dx+\int \left (\frac {1}{x}+x+\frac {x \left (-1-x^2\right )}{1+x^4}\right ) \, dx \\ & = \frac {x^2}{2}+\log (x)+\frac {1}{4} \int \frac {1}{1-\sqrt {2} x+x^2} \, dx+\frac {1}{4} \int \frac {1}{1+\sqrt {2} x+x^2} \, dx+\frac {\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{4 \sqrt {2}}+\frac {\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{4 \sqrt {2}}+\int \frac {x \left (-1-x^2\right )}{1+x^4} \, dx \\ & = \frac {x^2}{2}+\log (x)+\frac {\log \left (1-\sqrt {2} x+x^2\right )}{4 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} x+x^2\right )}{4 \sqrt {2}}+\frac {1}{2} \text {Subst}\left (\int \frac {-1-x}{1+x^2} \, dx,x,x^2\right )+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{2 \sqrt {2}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} x\right )}{2 \sqrt {2}} \\ & = \frac {x^2}{2}-\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{2 \sqrt {2}}+\frac {\tan ^{-1}\left (1+\sqrt {2} x\right )}{2 \sqrt {2}}+\log (x)+\frac {\log \left (1-\sqrt {2} x+x^2\right )}{4 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} x+x^2\right )}{4 \sqrt {2}}-\frac {1}{2} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,x^2\right )-\frac {1}{2} \text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,x^2\right ) \\ & = \frac {x^2}{2}-\frac {1}{2} \tan ^{-1}\left (x^2\right )-\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{2 \sqrt {2}}+\frac {\tan ^{-1}\left (1+\sqrt {2} x\right )}{2 \sqrt {2}}+\log (x)+\frac {\log \left (1-\sqrt {2} x+x^2\right )}{4 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} x+x^2\right )}{4 \sqrt {2}}-\frac {1}{4} \log \left (1+x^4\right ) \\ \end{align*}
Time = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.90 \[ \int \frac {1+x^3+x^6}{x+x^5} \, dx=\frac {1}{8} \left (4 x^2-2 \left (-2+\sqrt {2}\right ) \arctan \left (1-\sqrt {2} x\right )+2 \left (2+\sqrt {2}\right ) \arctan \left (1+\sqrt {2} x\right )+8 \log (x)+\sqrt {2} \log \left (1-\sqrt {2} x+x^2\right )-\sqrt {2} \log \left (1+\sqrt {2} x+x^2\right )-2 \log \left (1+x^4\right )\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.80 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.48
method | result | size |
risch | \(\frac {x^{2}}{2}+\ln \left (x \right )+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{4}+4 \textit {\_Z}^{3}+8 \textit {\_Z}^{2}+4 \textit {\_Z} +1\right )}{\sum }\textit {\_R} \ln \left (-\textit {\_R}^{3}-5 \textit {\_R}^{2}-10 \textit {\_R} +3 x -5\right )\right )}{4}\) | \(54\) |
default | \(\frac {x^{2}}{2}+\ln \left (x \right )-\frac {\arctan \left (x^{2}\right )}{2}+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x^{2}-x \sqrt {2}}{1+x^{2}+x \sqrt {2}}\right )+2 \arctan \left (x \sqrt {2}+1\right )+2 \arctan \left (x \sqrt {2}-1\right )\right )}{8}-\frac {\ln \left (x^{4}+1\right )}{4}\) | \(74\) |
meijerg | \(\frac {x^{2}}{2}-\frac {\arctan \left (x^{2}\right )}{2}+\frac {x^{3} \sqrt {2}\, \ln \left (1-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {x^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2-\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {3}{4}}}-\frac {x^{3} \sqrt {2}\, \ln \left (1+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}+\sqrt {x^{4}}\right )}{8 \left (x^{4}\right )^{\frac {3}{4}}}+\frac {x^{3} \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}{2+\sqrt {2}\, \left (x^{4}\right )^{\frac {1}{4}}}\right )}{4 \left (x^{4}\right )^{\frac {3}{4}}}-\frac {\ln \left (x^{4}+1\right )}{4}+\ln \left (x \right )\) | \(160\) |
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Result contains complex when optimal does not.
Time = 0.95 (sec) , antiderivative size = 515, normalized size of antiderivative = 4.60 \[ \int \frac {1+x^3+x^6}{x+x^5} \, dx=\text {Too large to display} \]
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Time = 0.53 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.54 \[ \int \frac {1+x^3+x^6}{x+x^5} \, dx=\frac {x^{2}}{2} + \log {\left (x \right )} + \operatorname {RootSum} {\left (256 t^{4} + 256 t^{3} + 128 t^{2} + 16 t + 1, \left ( t \mapsto t \log {\left (\frac {1792 t^{4}}{73} + \frac {704 t^{3}}{219} - \frac {3152 t^{2}}{219} - \frac {2584 t}{219} + x - \frac {344}{219} \right )} \right )\right )} \]
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Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.88 \[ \int \frac {1+x^3+x^6}{x+x^5} \, dx=\frac {1}{4} \, \sqrt {2} {\left (\sqrt {2} + 1\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {1}{4} \, \sqrt {2} {\left (\sqrt {2} - 1\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {1}{8} \, \sqrt {2} {\left (\sqrt {2} + 1\right )} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{8} \, \sqrt {2} {\left (\sqrt {2} - 1\right )} \log \left (x^{2} - \sqrt {2} x + 1\right ) + \frac {1}{2} \, x^{2} + \log \left (x\right ) \]
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Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.82 \[ \int \frac {1+x^3+x^6}{x+x^5} \, dx=\frac {1}{2} \, x^{2} + \frac {1}{4} \, {\left (\sqrt {2} + 2\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{4} \, {\left (\sqrt {2} - 2\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{8} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {1}{4} \, \log \left (x^{4} + 1\right ) + \log \left ({\left | x \right |}\right ) \]
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Time = 9.35 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.52 \[ \int \frac {1+x^3+x^6}{x+x^5} \, dx=\ln \left (x\right )+\left (\sum _{k=1}^4\ln \left (\mathrm {root}\left (z^4+z^3+\frac {z^2}{2}+\frac {z}{16}+\frac {1}{256},z,k\right )\,\left (8\,\mathrm {root}\left (z^4+z^3+\frac {z^2}{2}+\frac {z}{16}+\frac {1}{256},z,k\right )+x+\mathrm {root}\left (z^4+z^3+\frac {z^2}{2}+\frac {z}{16}+\frac {1}{256},z,k\right )\,x\,96+{\mathrm {root}\left (z^4+z^3+\frac {z^2}{2}+\frac {z}{16}+\frac {1}{256},z,k\right )}^2\,x\,240+{\mathrm {root}\left (z^4+z^3+\frac {z^2}{2}+\frac {z}{16}+\frac {1}{256},z,k\right )}^3\,x\,320-16\,{\mathrm {root}\left (z^4+z^3+\frac {z^2}{2}+\frac {z}{16}+\frac {1}{256},z,k\right )}^2+8\right )\right )\,\mathrm {root}\left (z^4+z^3+\frac {z^2}{2}+\frac {z}{16}+\frac {1}{256},z,k\right )\right )+\frac {x^2}{2} \]
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