Optimal. Leaf size=102 \[ -\frac {\log \left (x^2-\sqrt {2} x+1\right )}{8 \sqrt {2}}+\frac {\log \left (x^2+\sqrt {2} x+1\right )}{8 \sqrt {2}}-\frac {1}{x}-\frac {1}{4} \tan ^{-1}(x)+\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} x+1\right )}{4 \sqrt {2}}+\frac {1}{4} \tanh ^{-1}(x) \]
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Rubi [A] time = 0.06, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.846, Rules used = {325, 301, 297, 1162, 617, 204, 1165, 628, 298, 203, 206} \[ -\frac {\log \left (x^2-\sqrt {2} x+1\right )}{8 \sqrt {2}}+\frac {\log \left (x^2+\sqrt {2} x+1\right )}{8 \sqrt {2}}-\frac {1}{x}-\frac {1}{4} \tan ^{-1}(x)+\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {\tan ^{-1}\left (\sqrt {2} x+1\right )}{4 \sqrt {2}}+\frac {1}{4} \tanh ^{-1}(x) \]
Antiderivative was successfully verified.
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Rule 203
Rule 204
Rule 206
Rule 297
Rule 298
Rule 301
Rule 325
Rule 617
Rule 628
Rule 1162
Rule 1165
Rubi steps
\begin {align*} \int \frac {1}{x^2 \left (1-x^8\right )} \, dx &=-\frac {1}{x}+\int \frac {x^6}{1-x^8} \, dx\\ &=-\frac {1}{x}+\frac {1}{2} \int \frac {x^2}{1-x^4} \, dx-\frac {1}{2} \int \frac {x^2}{1+x^4} \, dx\\ &=-\frac {1}{x}+\frac {1}{4} \int \frac {1}{1-x^2} \, dx-\frac {1}{4} \int \frac {1}{1+x^2} \, dx+\frac {1}{4} \int \frac {1-x^2}{1+x^4} \, dx-\frac {1}{4} \int \frac {1+x^2}{1+x^4} \, dx\\ &=-\frac {1}{x}-\frac {1}{4} \tan ^{-1}(x)+\frac {1}{4} \tanh ^{-1}(x)-\frac {1}{8} \int \frac {1}{1-\sqrt {2} x+x^2} \, dx-\frac {1}{8} \int \frac {1}{1+\sqrt {2} x+x^2} \, dx-\frac {\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{8 \sqrt {2}}-\frac {\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{8 \sqrt {2}}\\ &=-\frac {1}{x}-\frac {1}{4} \tan ^{-1}(x)+\frac {1}{4} \tanh ^{-1}(x)-\frac {\log \left (1-\sqrt {2} x+x^2\right )}{8 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} x+x^2\right )}{8 \sqrt {2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{4 \sqrt {2}}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} x\right )}{4 \sqrt {2}}\\ &=-\frac {1}{x}-\frac {1}{4} \tan ^{-1}(x)+\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{4 \sqrt {2}}-\frac {\tan ^{-1}\left (1+\sqrt {2} x\right )}{4 \sqrt {2}}+\frac {1}{4} \tanh ^{-1}(x)-\frac {\log \left (1-\sqrt {2} x+x^2\right )}{8 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} x+x^2\right )}{8 \sqrt {2}}\\ \end {align*}
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Mathematica [A] time = 0.04, size = 109, normalized size = 1.07 \[ -\frac {\sqrt {2} x \log \left (x^2-\sqrt {2} x+1\right )-\sqrt {2} x \log \left (x^2+\sqrt {2} x+1\right )+2 x \log (1-x)-2 x \log (x+1)+4 x \tan ^{-1}(x)-2 \sqrt {2} x \tan ^{-1}\left (1-\sqrt {2} x\right )+2 \sqrt {2} x \tan ^{-1}\left (\sqrt {2} x+1\right )+16}{16 x} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.61, size = 123, normalized size = 1.21 \[ \frac {4 \, \sqrt {2} x \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} + \sqrt {2} x + 1} - 1\right ) + 4 \, \sqrt {2} x \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} - \sqrt {2} x + 1} + 1\right ) + \sqrt {2} x \log \left (x^{2} + \sqrt {2} x + 1\right ) - \sqrt {2} x \log \left (x^{2} - \sqrt {2} x + 1\right ) - 4 \, x \arctan \relax (x) + 2 \, x \log \left (x + 1\right ) - 2 \, x \log \left (x - 1\right ) - 16}{16 \, x} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.17, size = 95, normalized size = 0.93 \[ -\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {1}{x} - \frac {1}{4} \, \arctan \relax (x) + \frac {1}{8} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{8} \, \log \left ({\left | x - 1 \right |}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 79, normalized size = 0.77 \[ -\frac {\arctan \relax (x )}{4}-\frac {\sqrt {2}\, \arctan \left (\sqrt {2}\, x -1\right )}{8}-\frac {\sqrt {2}\, \arctan \left (\sqrt {2}\, x +1\right )}{8}-\frac {\sqrt {2}\, \ln \left (\frac {x^{2}-\sqrt {2}\, x +1}{x^{2}+\sqrt {2}\, x +1}\right )}{16}-\frac {\ln \left (x -1\right )}{8}+\frac {\ln \left (x +1\right )}{8}-\frac {1}{x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 2.35, size = 93, normalized size = 0.91 \[ -\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {1}{8} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) - \frac {1}{16} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {1}{x} - \frac {1}{4} \, \arctan \relax (x) + \frac {1}{8} \, \log \left (x + 1\right ) - \frac {1}{8} \, \log \left (x - 1\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.04, size = 50, normalized size = 0.49 \[ -\frac {\mathrm {atan}\relax (x)}{4}-\frac {1}{x}-\frac {\mathrm {atan}\left (x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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