Optimal. Leaf size=340 \[ \frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (x^2-\sqrt {2-\sqrt {2}} x+1\right )-\frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (x^2+\sqrt {2-\sqrt {2}} x+1\right )+\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (x^2-\sqrt {2+\sqrt {2}} x+1\right )-\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (x^2+\sqrt {2+\sqrt {2}} x+1\right )+x+\frac {\tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2 \left (2+\sqrt {2}\right )}}-\frac {\tan ^{-1}\left (\frac {2 x+\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {\tan ^{-1}\left (\frac {2 x+\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2 \left (2+\sqrt {2}\right )}} \]
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Rubi [A] time = 0.32, antiderivative size = 340, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 7, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {321, 213, 1169, 634, 618, 204, 628} \[ \frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (x^2-\sqrt {2-\sqrt {2}} x+1\right )-\frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (x^2+\sqrt {2-\sqrt {2}} x+1\right )+\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (x^2-\sqrt {2+\sqrt {2}} x+1\right )-\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (x^2+\sqrt {2+\sqrt {2}} x+1\right )+x+\frac {\tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2 \left (2+\sqrt {2}\right )}}-\frac {\tan ^{-1}\left (\frac {2 x+\sqrt {2-\sqrt {2}}}{\sqrt {2+\sqrt {2}}}\right )}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {\tan ^{-1}\left (\frac {2 x+\sqrt {2+\sqrt {2}}}{\sqrt {2-\sqrt {2}}}\right )}{4 \sqrt {2 \left (2+\sqrt {2}\right )}} \]
Antiderivative was successfully verified.
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Rule 204
Rule 213
Rule 321
Rule 618
Rule 628
Rule 634
Rule 1169
Rubi steps
\begin {align*} \int \frac {x^8}{1+x^8} \, dx &=x-\int \frac {1}{1+x^8} \, dx\\ &=x-\frac {\int \frac {\sqrt {2}-x^2}{1-\sqrt {2} x^2+x^4} \, dx}{2 \sqrt {2}}-\frac {\int \frac {\sqrt {2}+x^2}{1+\sqrt {2} x^2+x^4} \, dx}{2 \sqrt {2}}\\ &=x-\frac {\int \frac {\sqrt {2 \left (2-\sqrt {2}\right )}-\left (-1+\sqrt {2}\right ) x}{1-\sqrt {2-\sqrt {2}} x+x^2} \, dx}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {\int \frac {\sqrt {2 \left (2-\sqrt {2}\right )}+\left (-1+\sqrt {2}\right ) x}{1+\sqrt {2-\sqrt {2}} x+x^2} \, dx}{4 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {\int \frac {\sqrt {2 \left (2+\sqrt {2}\right )}-\left (1+\sqrt {2}\right ) x}{1-\sqrt {2+\sqrt {2}} x+x^2} \, dx}{4 \sqrt {2 \left (2+\sqrt {2}\right )}}-\frac {\int \frac {\sqrt {2 \left (2+\sqrt {2}\right )}+\left (1+\sqrt {2}\right ) x}{1+\sqrt {2+\sqrt {2}} x+x^2} \, dx}{4 \sqrt {2 \left (2+\sqrt {2}\right )}}\\ &=x-\frac {1}{8} \sqrt {\frac {1}{2} \left (3-2 \sqrt {2}\right )} \int \frac {1}{1-\sqrt {2+\sqrt {2}} x+x^2} \, dx-\frac {1}{8} \sqrt {\frac {1}{2} \left (3-2 \sqrt {2}\right )} \int \frac {1}{1+\sqrt {2+\sqrt {2}} x+x^2} \, dx+\frac {1}{16} \sqrt {2-\sqrt {2}} \int \frac {-\sqrt {2-\sqrt {2}}+2 x}{1-\sqrt {2-\sqrt {2}} x+x^2} \, dx-\frac {1}{16} \sqrt {2-\sqrt {2}} \int \frac {\sqrt {2-\sqrt {2}}+2 x}{1+\sqrt {2-\sqrt {2}} x+x^2} \, dx+\frac {1}{16} \sqrt {2+\sqrt {2}} \int \frac {-\sqrt {2+\sqrt {2}}+2 x}{1-\sqrt {2+\sqrt {2}} x+x^2} \, dx-\frac {1}{16} \sqrt {2+\sqrt {2}} \int \frac {\sqrt {2+\sqrt {2}}+2 x}{1+\sqrt {2+\sqrt {2}} x+x^2} \, dx-\frac {1}{8} \sqrt {\frac {1}{2} \left (3+2 \sqrt {2}\right )} \int \frac {1}{1-\sqrt {2-\sqrt {2}} x+x^2} \, dx-\frac {1}{8} \sqrt {\frac {1}{2} \left (3+2 \sqrt {2}\right )} \int \frac {1}{1+\sqrt {2-\sqrt {2}} x+x^2} \, dx\\ &=x+\frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (1-\sqrt {2-\sqrt {2}} x+x^2\right )-\frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (1+\sqrt {2-\sqrt {2}} x+x^2\right )+\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (1-\sqrt {2+\sqrt {2}} x+x^2\right )-\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (1+\sqrt {2+\sqrt {2}} x+x^2\right )+\frac {1}{4} \sqrt {\frac {1}{2} \left (3-2 \sqrt {2}\right )} \operatorname {Subst}\left (\int \frac {1}{-2+\sqrt {2}-x^2} \, dx,x,-\sqrt {2+\sqrt {2}}+2 x\right )+\frac {1}{4} \sqrt {\frac {1}{2} \left (3-2 \sqrt {2}\right )} \operatorname {Subst}\left (\int \frac {1}{-2+\sqrt {2}-x^2} \, dx,x,\sqrt {2+\sqrt {2}}+2 x\right )+\frac {1}{4} \sqrt {\frac {1}{2} \left (3+2 \sqrt {2}\right )} \operatorname {Subst}\left (\int \frac {1}{-2-\sqrt {2}-x^2} \, dx,x,-\sqrt {2-\sqrt {2}}+2 x\right )+\frac {1}{4} \sqrt {\frac {1}{2} \left (3+2 \sqrt {2}\right )} \operatorname {Subst}\left (\int \frac {1}{-2-\sqrt {2}-x^2} \, dx,x,\sqrt {2-\sqrt {2}}+2 x\right )\\ &=x+\frac {1}{8} \sqrt {2+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}-2 x}{\sqrt {2+\sqrt {2}}}\right )+\frac {1}{8} \sqrt {2-\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}-2 x}{\sqrt {2-\sqrt {2}}}\right )-\frac {1}{8} \sqrt {2+\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2-\sqrt {2}}+2 x}{\sqrt {2+\sqrt {2}}}\right )-\frac {1}{8} \sqrt {2-\sqrt {2}} \tan ^{-1}\left (\frac {\sqrt {2+\sqrt {2}}+2 x}{\sqrt {2-\sqrt {2}}}\right )+\frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (1-\sqrt {2-\sqrt {2}} x+x^2\right )-\frac {1}{16} \sqrt {2-\sqrt {2}} \log \left (1+\sqrt {2-\sqrt {2}} x+x^2\right )+\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (1-\sqrt {2+\sqrt {2}} x+x^2\right )-\frac {1}{16} \sqrt {2+\sqrt {2}} \log \left (1+\sqrt {2+\sqrt {2}} x+x^2\right )\\ \end {align*}
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Mathematica [A] time = 0.01, size = 210, normalized size = 0.62 \[ \frac {1}{8} \sin \left (\frac {\pi }{8}\right ) \log \left (x^2-2 x \sin \left (\frac {\pi }{8}\right )+1\right )-\frac {1}{8} \sin \left (\frac {\pi }{8}\right ) \log \left (x^2+2 x \sin \left (\frac {\pi }{8}\right )+1\right )+\frac {1}{8} \cos \left (\frac {\pi }{8}\right ) \log \left (x^2-2 x \cos \left (\frac {\pi }{8}\right )+1\right )-\frac {1}{8} \cos \left (\frac {\pi }{8}\right ) \log \left (x^2+2 x \cos \left (\frac {\pi }{8}\right )+1\right )+x-\frac {1}{4} \sin \left (\frac {\pi }{8}\right ) \tan ^{-1}\left (\csc \left (\frac {\pi }{8}\right ) \left (x-\cos \left (\frac {\pi }{8}\right )\right )\right )-\frac {1}{4} \sin \left (\frac {\pi }{8}\right ) \tan ^{-1}\left (\csc \left (\frac {\pi }{8}\right ) \left (x+\cos \left (\frac {\pi }{8}\right )\right )\right )-\frac {1}{4} \cos \left (\frac {\pi }{8}\right ) \tan ^{-1}\left (\sec \left (\frac {\pi }{8}\right ) \left (x-\sin \left (\frac {\pi }{8}\right )\right )\right )-\frac {1}{4} \cos \left (\frac {\pi }{8}\right ) \tan ^{-1}\left (\sec \left (\frac {\pi }{8}\right ) \left (x+\sin \left (\frac {\pi }{8}\right )\right )\right ) \]
Antiderivative was successfully verified.
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fricas [B] time = 0.93, size = 1028, normalized size = 3.02 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.20, size = 240, normalized size = 0.71 \[ -\frac {1}{8} \, \sqrt {\sqrt {2} + 2} \arctan \left (\frac {2 \, x + \sqrt {-\sqrt {2} + 2}}{\sqrt {\sqrt {2} + 2}}\right ) - \frac {1}{8} \, \sqrt {\sqrt {2} + 2} \arctan \left (\frac {2 \, x - \sqrt {-\sqrt {2} + 2}}{\sqrt {\sqrt {2} + 2}}\right ) - \frac {1}{8} \, \sqrt {-\sqrt {2} + 2} \arctan \left (\frac {2 \, x + \sqrt {\sqrt {2} + 2}}{\sqrt {-\sqrt {2} + 2}}\right ) - \frac {1}{8} \, \sqrt {-\sqrt {2} + 2} \arctan \left (\frac {2 \, x - \sqrt {\sqrt {2} + 2}}{\sqrt {-\sqrt {2} + 2}}\right ) - \frac {1}{16} \, \sqrt {\sqrt {2} + 2} \log \left (x^{2} + x \sqrt {\sqrt {2} + 2} + 1\right ) + \frac {1}{16} \, \sqrt {\sqrt {2} + 2} \log \left (x^{2} - x \sqrt {\sqrt {2} + 2} + 1\right ) - \frac {1}{16} \, \sqrt {-\sqrt {2} + 2} \log \left (x^{2} + x \sqrt {-\sqrt {2} + 2} + 1\right ) + \frac {1}{16} \, \sqrt {-\sqrt {2} + 2} \log \left (x^{2} - x \sqrt {-\sqrt {2} + 2} + 1\right ) + x \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.01, size = 24, normalized size = 0.07 \[ x -\frac {\ln \left (-\RootOf \left (\textit {\_Z}^{8}+1\right )+x \right )}{8 \RootOf \left (\textit {\_Z}^{8}+1\right )^{7}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ x - \int \frac {1}{x^{8} + 1}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.25, size = 289, normalized size = 0.85 \[ x-\mathrm {atan}\left (\frac {x\,\sqrt {-\sqrt {2}-2}\,1{}\mathrm {i}}{\sqrt {2-\sqrt {2}}\,\sqrt {-\sqrt {2}-2}+\sqrt {2}}-\frac {x\,\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{\sqrt {2-\sqrt {2}}\,\sqrt {-\sqrt {2}-2}+\sqrt {2}}\right )\,\left (\frac {\sqrt {-\sqrt {2}-2}\,1{}\mathrm {i}}{8}-\frac {\sqrt {2-\sqrt {2}}\,1{}\mathrm {i}}{8}\right )+\mathrm {atan}\left (\frac {x\,\sqrt {\sqrt {2}-2}\,1{}\mathrm {i}}{\sqrt {2}+\sqrt {\sqrt {2}-2}\,\sqrt {\sqrt {2}+2}}+\frac {x\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{\sqrt {2}+\sqrt {\sqrt {2}-2}\,\sqrt {\sqrt {2}+2}}\right )\,\left (\frac {\sqrt {\sqrt {2}-2}\,1{}\mathrm {i}}{8}+\frac {\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{8}\right )-\mathrm {atan}\left (-\frac {\sqrt {2}\,x\,\sqrt {\sqrt {2}+2}}{2}+x\,\sqrt {\sqrt {2}+2}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {\sqrt {2}\,1{}\mathrm {i}}{16}-\frac {1}{16}-\frac {1}{16}{}\mathrm {i}\right )\,\sqrt {\sqrt {2}+2}\,2{}\mathrm {i}+\mathrm {atan}\left (x\,\sqrt {\sqrt {2}+2}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )+\frac {\sqrt {2}\,x\,\sqrt {\sqrt {2}+2}\,1{}\mathrm {i}}{2}\right )\,\left (\frac {\sqrt {2}}{16}-\frac {1}{16}+\frac {1}{16}{}\mathrm {i}\right )\,\sqrt {\sqrt {2}+2}\,2{}\mathrm {i} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 3.15, size = 15, normalized size = 0.04 \[ x + \operatorname {RootSum} {\left (16777216 t^{8} + 1, \left (t \mapsto t \log {\left (- 8 t + x \right )} \right )\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
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