Integrand size = 13, antiderivative size = 347 \[ \int \frac {1+x^4}{1+x^8} \, dx=-\frac {1}{4} \sqrt {\frac {1}{2} \left (2-\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )-\frac {1}{4} \sqrt {\frac {1}{2} \left (2+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )+\frac {1}{4} \sqrt {\frac {1}{2} \left (2-\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2-\sqrt {2}}+2 x}{\sqrt {2+\sqrt {2}}}\right )+\frac {1}{4} \sqrt {\frac {1}{2} \left (2+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2+\sqrt {2}}+2 x}{\sqrt {2-\sqrt {2}}}\right )-\frac {\log \left (1-\sqrt {2-\sqrt {2}} x+x^2\right )}{8 \sqrt {2-\sqrt {2}}}+\frac {\log \left (1+\sqrt {2-\sqrt {2}} x+x^2\right )}{8 \sqrt {2-\sqrt {2}}}-\frac {\log \left (1-\sqrt {2+\sqrt {2}} x+x^2\right )}{8 \sqrt {2+\sqrt {2}}}+\frac {\log \left (1+\sqrt {2+\sqrt {2}} x+x^2\right )}{8 \sqrt {2+\sqrt {2}}} \]
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Time = 0.16 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {1427, 1108, 648, 632, 210, 642} \[ \int \frac {1+x^4}{1+x^8} \, dx=-\frac {1}{4} \sqrt {\frac {1}{2} \left (2-\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )-\frac {1}{4} \sqrt {\frac {1}{2} \left (2+\sqrt {2}\right )} \arctan \left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )+\frac {1}{4} \sqrt {\frac {1}{2} \left (2-\sqrt {2}\right )} \arctan \left (\frac {2 x+\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )+\frac {1}{4} \sqrt {\frac {1}{2} \left (2+\sqrt {2}\right )} \arctan \left (\frac {2 x+\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right )-\frac {\log \left (x^2-\sqrt {2-\sqrt {2}} x+1\right )}{8 \sqrt {2-\sqrt {2}}}+\frac {\log \left (x^2+\sqrt {2-\sqrt {2}} x+1\right )}{8 \sqrt {2-\sqrt {2}}}-\frac {\log \left (x^2-\sqrt {2+\sqrt {2}} x+1\right )}{8 \sqrt {2+\sqrt {2}}}+\frac {\log \left (x^2+\sqrt {2+\sqrt {2}} x+1\right )}{8 \sqrt {2+\sqrt {2}}} \]
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Rule 210
Rule 632
Rule 642
Rule 648
Rule 1108
Rule 1427
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {1}{1-\sqrt {2} x^2+x^4} \, dx+\frac {1}{2} \int \frac {1}{1+\sqrt {2} x^2+x^4} \, dx \\ & = \frac {\int \frac {\sqrt {2-\sqrt {2}}-x}{1-\sqrt {2-\sqrt {2}} x+x^2} \, dx}{4 \sqrt {2-\sqrt {2}}}+\frac {\int \frac {\sqrt {2-\sqrt {2}}+x}{1+\sqrt {2-\sqrt {2}} x+x^2} \, dx}{4 \sqrt {2-\sqrt {2}}}+\frac {\int \frac {\sqrt {2+\sqrt {2}}-x}{1-\sqrt {2+\sqrt {2}} x+x^2} \, dx}{4 \sqrt {2+\sqrt {2}}}+\frac {\int \frac {\sqrt {2+\sqrt {2}}+x}{1+\sqrt {2+\sqrt {2}} x+x^2} \, dx}{4 \sqrt {2+\sqrt {2}}} \\ & = \frac {1}{8} \int \frac {1}{1-\sqrt {2-\sqrt {2}} x+x^2} \, dx+\frac {1}{8} \int \frac {1}{1+\sqrt {2-\sqrt {2}} x+x^2} \, dx+\frac {1}{8} \int \frac {1}{1-\sqrt {2+\sqrt {2}} x+x^2} \, dx+\frac {1}{8} \int \frac {1}{1+\sqrt {2+\sqrt {2}} x+x^2} \, dx-\frac {\int \frac {-\sqrt {2-\sqrt {2}}+2 x}{1-\sqrt {2-\sqrt {2}} x+x^2} \, dx}{8 \sqrt {2-\sqrt {2}}}+\frac {\int \frac {\sqrt {2-\sqrt {2}}+2 x}{1+\sqrt {2-\sqrt {2}} x+x^2} \, dx}{8 \sqrt {2-\sqrt {2}}}-\frac {\int \frac {-\sqrt {2+\sqrt {2}}+2 x}{1-\sqrt {2+\sqrt {2}} x+x^2} \, dx}{8 \sqrt {2+\sqrt {2}}}+\frac {\int \frac {\sqrt {2+\sqrt {2}}+2 x}{1+\sqrt {2+\sqrt {2}} x+x^2} \, dx}{8 \sqrt {2+\sqrt {2}}} \\ & = -\frac {\log \left (1-\sqrt {2-\sqrt {2}} x+x^2\right )}{8 \sqrt {2-\sqrt {2}}}+\frac {\log \left (1+\sqrt {2-\sqrt {2}} x+x^2\right )}{8 \sqrt {2-\sqrt {2}}}-\frac {\log \left (1-\sqrt {2+\sqrt {2}} x+x^2\right )}{8 \sqrt {2+\sqrt {2}}}+\frac {\log \left (1+\sqrt {2+\sqrt {2}} x+x^2\right )}{8 \sqrt {2+\sqrt {2}}}-\frac {1}{4} \text {Subst}\left (\int \frac {1}{-2-\sqrt {2}-x^2} \, dx,x,-\sqrt {2-\sqrt {2}}+2 x\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{-2-\sqrt {2}-x^2} \, dx,x,\sqrt {2-\sqrt {2}}+2 x\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{-2+\sqrt {2}-x^2} \, dx,x,-\sqrt {2+\sqrt {2}}+2 x\right )-\frac {1}{4} \text {Subst}\left (\int \frac {1}{-2+\sqrt {2}-x^2} \, dx,x,\sqrt {2+\sqrt {2}}+2 x\right ) \\ & = -\frac {\tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2+\sqrt {2}}}-\frac {\tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2-\sqrt {2}}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}+2 x}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2+\sqrt {2}}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}+2 x}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2-\sqrt {2}}}-\frac {\log \left (1-\sqrt {2-\sqrt {2}} x+x^2\right )}{8 \sqrt {2-\sqrt {2}}}+\frac {\log \left (1+\sqrt {2-\sqrt {2}} x+x^2\right )}{8 \sqrt {2-\sqrt {2}}}-\frac {\log \left (1-\sqrt {2+\sqrt {2}} x+x^2\right )}{8 \sqrt {2+\sqrt {2}}}+\frac {\log \left (1+\sqrt {2+\sqrt {2}} x+x^2\right )}{8 \sqrt {2+\sqrt {2}}} \\ \end{align*}
Time = 0.13 (sec) , antiderivative size = 258, normalized size of antiderivative = 0.74 \[ \int \frac {1+x^4}{1+x^8} \, dx=\frac {1}{8} \left (2 \arctan \left (\sec \left (\frac {\pi }{8}\right ) \left (x+\sin \left (\frac {\pi }{8}\right )\right )\right ) \left (\cos \left (\frac {\pi }{8}\right )-\sin \left (\frac {\pi }{8}\right )\right )+2 \arctan \left (x \sec \left (\frac {\pi }{8}\right )-\tan \left (\frac {\pi }{8}\right )\right ) \left (\cos \left (\frac {\pi }{8}\right )-\sin \left (\frac {\pi }{8}\right )\right )+\log \left (1+x^2+2 x \cos \left (\frac {\pi }{8}\right )\right ) \left (\cos \left (\frac {\pi }{8}\right )-\sin \left (\frac {\pi }{8}\right )\right )+\log \left (1+x^2-2 x \cos \left (\frac {\pi }{8}\right )\right ) \left (-\cos \left (\frac {\pi }{8}\right )+\sin \left (\frac {\pi }{8}\right )\right )+2 \arctan \left (\left (x-\cos \left (\frac {\pi }{8}\right )\right ) \csc \left (\frac {\pi }{8}\right )\right ) \left (\cos \left (\frac {\pi }{8}\right )+\sin \left (\frac {\pi }{8}\right )\right )+2 \arctan \left (\left (x+\cos \left (\frac {\pi }{8}\right )\right ) \csc \left (\frac {\pi }{8}\right )\right ) \left (\cos \left (\frac {\pi }{8}\right )+\sin \left (\frac {\pi }{8}\right )\right )-\log \left (1+x^2-2 x \sin \left (\frac {\pi }{8}\right )\right ) \left (\cos \left (\frac {\pi }{8}\right )+\sin \left (\frac {\pi }{8}\right )\right )+\log \left (1+x^2+2 x \sin \left (\frac {\pi }{8}\right )\right ) \left (\cos \left (\frac {\pi }{8}\right )+\sin \left (\frac {\pi }{8}\right )\right )\right ) \]
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Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.62 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.08
method | result | size |
default | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )}{\sum }\frac {\left (\textit {\_R}^{4}+1\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{7}}\right )}{8}\) | \(27\) |
risch | \(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}+1\right )}{\sum }\frac {\left (\textit {\_R}^{4}+1\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{7}}\right )}{8}\) | \(27\) |
meijerg | \(\text {Expression too large to display}\) | \(566\) |
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Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.54 \[ \int \frac {1+x^4}{1+x^8} \, dx=\frac {1}{8} \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} \log \left (8 \, \sqrt {2} {\left (\left (-1\right )^{\frac {5}{8}} + \left (-1\right )^{\frac {1}{8}}\right )} + 16 \, x\right ) - \frac {1}{8} \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} \log \left (-8 \, \sqrt {2} {\left (\left (-1\right )^{\frac {5}{8}} + \left (-1\right )^{\frac {1}{8}}\right )} + 16 \, x\right ) - \frac {1}{8} i \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} \log \left (-8 \, \sqrt {2} {\left (i \, \left (-1\right )^{\frac {5}{8}} + i \, \left (-1\right )^{\frac {1}{8}}\right )} + 16 \, x\right ) + \frac {1}{8} i \, \sqrt {2} \left (-1\right )^{\frac {1}{8}} \log \left (-8 \, \sqrt {2} {\left (-i \, \left (-1\right )^{\frac {5}{8}} - i \, \left (-1\right )^{\frac {1}{8}}\right )} + 16 \, x\right ) - \left (\frac {1}{8} i + \frac {1}{8}\right ) \, \left (-1\right )^{\frac {1}{8}} \log \left (32 \, x + \left (16 i + 16\right ) \, \left (-1\right )^{\frac {5}{8}} - \left (16 i + 16\right ) \, \left (-1\right )^{\frac {1}{8}}\right ) + \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \left (-1\right )^{\frac {1}{8}} \log \left (32 \, x - \left (16 i - 16\right ) \, \left (-1\right )^{\frac {5}{8}} + \left (16 i - 16\right ) \, \left (-1\right )^{\frac {1}{8}}\right ) - \left (\frac {1}{8} i - \frac {1}{8}\right ) \, \left (-1\right )^{\frac {1}{8}} \log \left (32 \, x + \left (16 i - 16\right ) \, \left (-1\right )^{\frac {5}{8}} - \left (16 i - 16\right ) \, \left (-1\right )^{\frac {1}{8}}\right ) + \left (\frac {1}{8} i + \frac {1}{8}\right ) \, \left (-1\right )^{\frac {1}{8}} \log \left (32 \, x - \left (16 i + 16\right ) \, \left (-1\right )^{\frac {5}{8}} + \left (16 i + 16\right ) \, \left (-1\right )^{\frac {1}{8}}\right ) \]
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Time = 1.15 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.05 \[ \int \frac {1+x^4}{1+x^8} \, dx=\operatorname {RootSum} {\left (1048576 t^{8} + 1, \left ( t \mapsto t \log {\left (4096 t^{5} + 4 t + x \right )} \right )\right )} \]
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\[ \int \frac {1+x^4}{1+x^8} \, dx=\int { \frac {x^{4} + 1}{x^{8} + 1} \,d x } \]
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none
Time = 0.38 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.71 \[ \int \frac {1+x^4}{1+x^8} \, dx=\frac {1}{8} \, \sqrt {-2 \, \sqrt {2} + 4} \arctan \left (\frac {2 \, x + \sqrt {-\sqrt {2} + 2}}{\sqrt {\sqrt {2} + 2}}\right ) + \frac {1}{8} \, \sqrt {-2 \, \sqrt {2} + 4} \arctan \left (\frac {2 \, x - \sqrt {-\sqrt {2} + 2}}{\sqrt {\sqrt {2} + 2}}\right ) + \frac {1}{8} \, \sqrt {2 \, \sqrt {2} + 4} \arctan \left (\frac {2 \, x + \sqrt {\sqrt {2} + 2}}{\sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{8} \, \sqrt {2 \, \sqrt {2} + 4} \arctan \left (\frac {2 \, x - \sqrt {\sqrt {2} + 2}}{\sqrt {-\sqrt {2} + 2}}\right ) + \frac {1}{16} \, \sqrt {-2 \, \sqrt {2} + 4} \log \left (x^{2} + x \sqrt {\sqrt {2} + 2} + 1\right ) - \frac {1}{16} \, \sqrt {-2 \, \sqrt {2} + 4} \log \left (x^{2} - x \sqrt {\sqrt {2} + 2} + 1\right ) + \frac {1}{16} \, \sqrt {2 \, \sqrt {2} + 4} \log \left (x^{2} + x \sqrt {-\sqrt {2} + 2} + 1\right ) - \frac {1}{16} \, \sqrt {2 \, \sqrt {2} + 4} \log \left (x^{2} - x \sqrt {-\sqrt {2} + 2} + 1\right ) \]
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Time = 8.75 (sec) , antiderivative size = 311, normalized size of antiderivative = 0.90 \[ \int \frac {1+x^4}{1+x^8} \, dx=-\ln \left ({\left (\frac {\sqrt {-2\,\sqrt {2}-4}}{16}-\frac {\sqrt {4-2\,\sqrt {2}}}{16}\right )}^3\,\left (65536\,x-16384\,\sqrt {-2\,\sqrt {2}-4}+16384\,\sqrt {4-2\,\sqrt {2}}\right )+256\right )\,\left (\frac {\sqrt {-2\,\sqrt {2}-4}}{16}-\frac {\sqrt {4-2\,\sqrt {2}}}{16}\right )+\mathrm {atan}\left (\frac {x\,\sqrt {\sqrt {2}-2}\,1{}\mathrm {i}}{2}+\frac {x\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{2}+\frac {\sqrt {2}\,x\,\sqrt {\sqrt {2}-2}\,1{}\mathrm {i}}{2}-\frac {\sqrt {2}\,x\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {\sqrt {2}\,\sqrt {\sqrt {2}-2}\,1{}\mathrm {i}}{8}+\frac {\sqrt {2}\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}\right )-\frac {\mathrm {atan}\left (x\,{\left (\sqrt {2}+2\right )}^{3/2}\,\left (1-\frac {1}{2}{}\mathrm {i}\right )+\sqrt {2}\,x\,{\left (\sqrt {2}+2\right )}^{3/2}\,\left (-\frac {3}{4}+\frac {1}{4}{}\mathrm {i}\right )\right )\,\left (-2+\sqrt {2}\,\left (1-\mathrm {i}\right )\right )\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}+\frac {\mathrm {atan}\left (x\,{\left (\sqrt {2}+2\right )}^{3/2}\,\left (\frac {1}{2}+1{}\mathrm {i}\right )+\sqrt {2}\,x\,{\left (\sqrt {2}+2\right )}^{3/2}\,\left (-\frac {1}{4}-\frac {3}{4}{}\mathrm {i}\right )\right )\,\left (\sqrt {2}\,\left (1+1{}\mathrm {i}\right )-2{}\mathrm {i}\right )\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}+\sqrt {2}\,\ln \left (x+{\left (\sqrt {2}+2\right )}^{3/2}\,\left (-\frac {1}{2}-\mathrm {i}\right )+\sqrt {2}\,{\left (\sqrt {2}+2\right )}^{3/2}\,\left (\frac {1}{4}+\frac {3}{4}{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {\sqrt {2}-2}}{16}+\frac {\sqrt {\sqrt {2}+2}}{16}\right )\,1{}\mathrm {i} \]
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