Optimal. Leaf size=106 \[ -\frac {7}{12 x^3}+\frac {7 \log \left (x^2-\sqrt {2} x+1\right )}{16 \sqrt {2}}-\frac {7 \log \left (x^2+\sqrt {2} x+1\right )}{16 \sqrt {2}}+\frac {1}{4 x^3 \left (x^4+1\right )}+\frac {7 \tan ^{-1}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {7 \tan ^{-1}\left (\sqrt {2} x+1\right )}{8 \sqrt {2}} \]
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Rubi [A] time = 0.05, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {28, 290, 325, 211, 1165, 628, 1162, 617, 204} \[ \frac {1}{4 x^3 \left (x^4+1\right )}-\frac {7}{12 x^3}+\frac {7 \log \left (x^2-\sqrt {2} x+1\right )}{16 \sqrt {2}}-\frac {7 \log \left (x^2+\sqrt {2} x+1\right )}{16 \sqrt {2}}+\frac {7 \tan ^{-1}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {7 \tan ^{-1}\left (\sqrt {2} x+1\right )}{8 \sqrt {2}} \]
Antiderivative was successfully verified.
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Rule 28
Rule 204
Rule 211
Rule 290
Rule 325
Rule 617
Rule 628
Rule 1162
Rule 1165
Rubi steps
\begin {align*} \int \frac {1}{x^4 \left (1+2 x^4+x^8\right )} \, dx &=\int \frac {1}{x^4 \left (1+x^4\right )^2} \, dx\\ &=\frac {1}{4 x^3 \left (1+x^4\right )}+\frac {7}{4} \int \frac {1}{x^4 \left (1+x^4\right )} \, dx\\ &=-\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}-\frac {7}{4} \int \frac {1}{1+x^4} \, dx\\ &=-\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}-\frac {7}{8} \int \frac {1-x^2}{1+x^4} \, dx-\frac {7}{8} \int \frac {1+x^2}{1+x^4} \, dx\\ &=-\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}-\frac {7}{16} \int \frac {1}{1-\sqrt {2} x+x^2} \, dx-\frac {7}{16} \int \frac {1}{1+\sqrt {2} x+x^2} \, dx+\frac {7 \int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}}+\frac {7 \int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}}\\ &=-\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}+\frac {7 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {7 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {7 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {7 \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} x\right )}{8 \sqrt {2}}\\ &=-\frac {7}{12 x^3}+\frac {1}{4 x^3 \left (1+x^4\right )}+\frac {7 \tan ^{-1}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {7 \tan ^{-1}\left (1+\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {7 \log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {7 \log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}}\\ \end {align*}
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Mathematica [A] time = 0.07, size = 96, normalized size = 0.91 \[ \frac {1}{96} \left (-\frac {24 x}{x^4+1}-\frac {32}{x^3}+21 \sqrt {2} \log \left (x^2-\sqrt {2} x+1\right )-21 \sqrt {2} \log \left (x^2+\sqrt {2} x+1\right )+42 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} x\right )-42 \sqrt {2} \tan ^{-1}\left (\sqrt {2} x+1\right )\right ) \]
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 140, normalized size = 1.32 \[ -\frac {56 \, x^{4} - 84 \, \sqrt {2} {\left (x^{7} + x^{3}\right )} \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} + \sqrt {2} x + 1} - 1\right ) - 84 \, \sqrt {2} {\left (x^{7} + x^{3}\right )} \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} - \sqrt {2} x + 1} + 1\right ) + 21 \, \sqrt {2} {\left (x^{7} + x^{3}\right )} \log \left (x^{2} + \sqrt {2} x + 1\right ) - 21 \, \sqrt {2} {\left (x^{7} + x^{3}\right )} \log \left (x^{2} - \sqrt {2} x + 1\right ) + 32}{96 \, {\left (x^{7} + x^{3}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.44, size = 87, normalized size = 0.82 \[ -\frac {7}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {7}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {7}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {7}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {x}{4 \, {\left (x^{4} + 1\right )}} - \frac {1}{3 \, x^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 73, normalized size = 0.69 \[ -\frac {x}{4 \left (x^{4}+1\right )}-\frac {7 \sqrt {2}\, \arctan \left (\sqrt {2}\, x -1\right )}{16}-\frac {7 \sqrt {2}\, \arctan \left (\sqrt {2}\, x +1\right )}{16}-\frac {7 \sqrt {2}\, \ln \left (\frac {x^{2}+\sqrt {2}\, x +1}{x^{2}-\sqrt {2}\, x +1}\right )}{32}-\frac {1}{3 x^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 2.20, size = 90, normalized size = 0.85 \[ -\frac {7}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) - \frac {7}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {7}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {7}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) - \frac {7 \, x^{4} + 4}{12 \, {\left (x^{7} + x^{3}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.36, size = 51, normalized size = 0.48 \[ -\frac {\frac {7\,x^4}{12}+\frac {1}{3}}{x^7+x^3}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {7}{16}-\frac {7}{16}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {7}{16}+\frac {7}{16}{}\mathrm {i}\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.22, size = 99, normalized size = 0.93 \[ \frac {- 7 x^{4} - 4}{12 x^{7} + 12 x^{3}} + \frac {7 \sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{32} - \frac {7 \sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{32} - \frac {7 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{16} - \frac {7 \sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{16} \]
Verification of antiderivative is not currently implemented for this CAS.
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