Optimal. Leaf size=93 \[ \frac{\log \left (x^4-\sqrt{2} x^2+1\right )}{8 \sqrt{2}}-\frac{\log \left (x^4+\sqrt{2} x^2+1\right )}{8 \sqrt{2}}-\frac{\tan ^{-1}\left (1-\sqrt{2} x^2\right )}{4 \sqrt{2}}+\frac{\tan ^{-1}\left (\sqrt{2} x^2+1\right )}{4 \sqrt{2}} \]
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Rubi [A] time = 0.0630532, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.636, Rules used = {275, 297, 1162, 617, 204, 1165, 628} \[ \frac{\log \left (x^4-\sqrt{2} x^2+1\right )}{8 \sqrt{2}}-\frac{\log \left (x^4+\sqrt{2} x^2+1\right )}{8 \sqrt{2}}-\frac{\tan ^{-1}\left (1-\sqrt{2} x^2\right )}{4 \sqrt{2}}+\frac{\tan ^{-1}\left (\sqrt{2} x^2+1\right )}{4 \sqrt{2}} \]
Antiderivative was successfully verified.
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Rule 275
Rule 297
Rule 1162
Rule 617
Rule 204
Rule 1165
Rule 628
Rubi steps
\begin{align*} \int \frac{x^5}{1+x^8} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^2}{1+x^4} \, dx,x,x^2\right )\\ &=-\left (\frac{1}{4} \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,x^2\right )\right )+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,x^2\right )\\ &=\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,x^2\right )+\frac{1}{8} \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,x^2\right )+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,x^2\right )}{8 \sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,x^2\right )}{8 \sqrt{2}}\\ &=\frac{\log \left (1-\sqrt{2} x^2+x^4\right )}{8 \sqrt{2}}-\frac{\log \left (1+\sqrt{2} x^2+x^4\right )}{8 \sqrt{2}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} x^2\right )}{4 \sqrt{2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} x^2\right )}{4 \sqrt{2}}\\ &=-\frac{\tan ^{-1}\left (1-\sqrt{2} x^2\right )}{4 \sqrt{2}}+\frac{\tan ^{-1}\left (1+\sqrt{2} x^2\right )}{4 \sqrt{2}}+\frac{\log \left (1-\sqrt{2} x^2+x^4\right )}{8 \sqrt{2}}-\frac{\log \left (1+\sqrt{2} x^2+x^4\right )}{8 \sqrt{2}}\\ \end{align*}
Mathematica [A] time = 0.0388304, size = 149, normalized size = 1.6 \[ -\frac{\log \left (x^2-2 x \sin \left (\frac{\pi }{8}\right )+1\right )+\log \left (x^2+2 x \sin \left (\frac{\pi }{8}\right )+1\right )-\log \left (x^2-2 x \cos \left (\frac{\pi }{8}\right )+1\right )-\log \left (x^2+2 x \cos \left (\frac{\pi }{8}\right )+1\right )-2 \tan ^{-1}\left (x \sec \left (\frac{\pi }{8}\right )-\tan \left (\frac{\pi }{8}\right )\right )+2 \tan ^{-1}\left (\csc \left (\frac{\pi }{8}\right ) \left (x+\cos \left (\frac{\pi }{8}\right )\right )\right )+2 \tan ^{-1}\left (\cot \left (\frac{\pi }{8}\right )-x \csc \left (\frac{\pi }{8}\right )\right )+2 \tan ^{-1}\left (\sec \left (\frac{\pi }{8}\right ) \left (x+\sin \left (\frac{\pi }{8}\right )\right )\right )}{8 \sqrt{2}} \]
Antiderivative was successfully verified.
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Maple [A] time = 0., size = 66, normalized size = 0.7 \begin{align*}{\frac{\arctan \left ( 1+{x}^{2}\sqrt{2} \right ) \sqrt{2}}{8}}+{\frac{\arctan \left ( -1+{x}^{2}\sqrt{2} \right ) \sqrt{2}}{8}}+{\frac{\sqrt{2}}{16}\ln \left ({\frac{1+{x}^{4}-{x}^{2}\sqrt{2}}{1+{x}^{4}+{x}^{2}\sqrt{2}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A] time = 1.45548, size = 108, normalized size = 1.16 \begin{align*} \frac{1}{8} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x^{2} + \sqrt{2}\right )}\right ) + \frac{1}{8} \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (2 \, x^{2} - \sqrt{2}\right )}\right ) - \frac{1}{16} \, \sqrt{2} \log \left (x^{4} + \sqrt{2} x^{2} + 1\right ) + \frac{1}{16} \, \sqrt{2} \log \left (x^{4} - \sqrt{2} x^{2} + 1\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 1.30445, size = 321, normalized size = 3.45 \begin{align*} -\frac{1}{4} \, \sqrt{2} \arctan \left (-\sqrt{2} x^{2} + \sqrt{2} \sqrt{x^{4} + \sqrt{2} x^{2} + 1} - 1\right ) - \frac{1}{4} \, \sqrt{2} \arctan \left (-\sqrt{2} x^{2} + \sqrt{2} \sqrt{x^{4} - \sqrt{2} x^{2} + 1} + 1\right ) - \frac{1}{16} \, \sqrt{2} \log \left (x^{4} + \sqrt{2} x^{2} + 1\right ) + \frac{1}{16} \, \sqrt{2} \log \left (x^{4} - \sqrt{2} x^{2} + 1\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A] time = 0.150571, size = 80, normalized size = 0.86 \begin{align*} \frac{\sqrt{2} \log{\left (x^{4} - \sqrt{2} x^{2} + 1 \right )}}{16} - \frac{\sqrt{2} \log{\left (x^{4} + \sqrt{2} x^{2} + 1 \right )}}{16} + \frac{\sqrt{2} \operatorname{atan}{\left (\sqrt{2} x^{2} - 1 \right )}}{8} + \frac{\sqrt{2} \operatorname{atan}{\left (\sqrt{2} x^{2} + 1 \right )}}{8} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.26236, size = 269, normalized size = 2.89 \begin{align*} -\frac{1}{8} \, \sqrt{2} \arctan \left (\frac{2 \, x + \sqrt{-\sqrt{2} + 2}}{\sqrt{\sqrt{2} + 2}}\right ) + \frac{1}{8} \, \sqrt{2} \arctan \left (\frac{2 \, x - \sqrt{-\sqrt{2} + 2}}{\sqrt{\sqrt{2} + 2}}\right ) - \frac{1}{8} \, \sqrt{2} \arctan \left (\frac{2 \, x + \sqrt{\sqrt{2} + 2}}{\sqrt{-\sqrt{2} + 2}}\right ) + \frac{1}{8} \, \sqrt{2} \arctan \left (\frac{2 \, x - \sqrt{\sqrt{2} + 2}}{\sqrt{-\sqrt{2} + 2}}\right ) + \frac{1}{16} \, \sqrt{2} \log \left (x^{2} + x \sqrt{\sqrt{2} + 2} + 1\right ) + \frac{1}{16} \, \sqrt{2} \log \left (x^{2} - x \sqrt{\sqrt{2} + 2} + 1\right ) - \frac{1}{16} \, \sqrt{2} \log \left (x^{2} + x \sqrt{-\sqrt{2} + 2} + 1\right ) - \frac{1}{16} \, \sqrt{2} \log \left (x^{2} - x \sqrt{-\sqrt{2} + 2} + 1\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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