Optimal. Leaf size=78 \[ \frac {1}{4} \sqrt {2 x^8+1} \left (x^4+1\right )+\frac {\log \left (\sqrt {2 x^8+1}+\sqrt {2} x^4\right )}{4 \sqrt {2}}-\frac {1}{2} \tanh ^{-1}\left (\sqrt {2 x^8+1}+\sqrt {2} x^4\right ) \]
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Rubi [A] time = 0.06, antiderivative size = 56, normalized size of antiderivative = 0.72, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1475, 815, 844, 215, 266, 63, 207} \begin {gather*} -\frac {1}{4} \tanh ^{-1}\left (\sqrt {2 x^8+1}\right )+\frac {\sinh ^{-1}\left (\sqrt {2} x^4\right )}{4 \sqrt {2}}+\frac {1}{4} \sqrt {2 x^8+1} \left (x^4+1\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 63
Rule 207
Rule 215
Rule 266
Rule 815
Rule 844
Rule 1475
Rubi steps
\begin {align*} \int \frac {\left (1+2 x^4\right ) \sqrt {1+2 x^8}}{x} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {(1+2 x) \sqrt {1+2 x^2}}{x} \, dx,x,x^4\right )\\ &=\frac {1}{4} \left (1+x^4\right ) \sqrt {1+2 x^8}+\frac {1}{16} \operatorname {Subst}\left (\int \frac {4+4 x}{x \sqrt {1+2 x^2}} \, dx,x,x^4\right )\\ &=\frac {1}{4} \left (1+x^4\right ) \sqrt {1+2 x^8}+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+2 x^2}} \, dx,x,x^4\right )+\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+2 x^2}} \, dx,x,x^4\right )\\ &=\frac {1}{4} \left (1+x^4\right ) \sqrt {1+2 x^8}+\frac {\sinh ^{-1}\left (\sqrt {2} x^4\right )}{4 \sqrt {2}}+\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+2 x}} \, dx,x,x^8\right )\\ &=\frac {1}{4} \left (1+x^4\right ) \sqrt {1+2 x^8}+\frac {\sinh ^{-1}\left (\sqrt {2} x^4\right )}{4 \sqrt {2}}+\frac {1}{8} \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{2}+\frac {x^2}{2}} \, dx,x,\sqrt {1+2 x^8}\right )\\ &=\frac {1}{4} \left (1+x^4\right ) \sqrt {1+2 x^8}+\frac {\sinh ^{-1}\left (\sqrt {2} x^4\right )}{4 \sqrt {2}}-\frac {1}{4} \tanh ^{-1}\left (\sqrt {1+2 x^8}\right )\\ \end {align*}
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Mathematica [A] time = 0.04, size = 53, normalized size = 0.68 \begin {gather*} \frac {1}{8} \left (-2 \tanh ^{-1}\left (\sqrt {2 x^8+1}\right )+\sqrt {2} \sinh ^{-1}\left (\sqrt {2} x^4\right )+2 \sqrt {2 x^8+1} \left (x^4+1\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.16, size = 81, normalized size = 1.04 \begin {gather*} \frac {1}{4} \left (1+x^4\right ) \sqrt {1+2 x^8}+\frac {1}{2} \tanh ^{-1}\left (\sqrt {2} x^4-\sqrt {1+2 x^8}\right )-\frac {\log \left (-\sqrt {2} x^4+\sqrt {1+2 x^8}\right )}{4 \sqrt {2}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 61, normalized size = 0.78 \begin {gather*} \frac {1}{4} \, \sqrt {2 \, x^{8} + 1} {\left (x^{4} + 1\right )} + \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} x^{4} - \sqrt {2 \, x^{8} + 1}\right ) + \frac {1}{4} \, \log \left (\frac {\sqrt {2 \, x^{8} + 1} - 1}{x^{4}}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.32, size = 86, normalized size = 1.10 \begin {gather*} \frac {1}{4} \, \sqrt {2 \, x^{8} + 1} {\left (x^{4} + 1\right )} - \frac {1}{8} \, \sqrt {2} \log \left (-\sqrt {2} x^{4} + \sqrt {2 \, x^{8} + 1}\right ) + \frac {1}{4} \, \log \left (\sqrt {2} x^{4} - \sqrt {2 \, x^{8} + 1} + 1\right ) - \frac {1}{4} \, \log \left (-\sqrt {2} x^{4} + \sqrt {2 \, x^{8} + 1} + 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.55, size = 68, normalized size = 0.87
method | result | size |
trager | \(\left (\frac {x^{4}}{4}+\frac {1}{4}\right ) \sqrt {2 x^{8}+1}-\frac {\RootOf \left (\textit {\_Z}^{2}-2\right ) \ln \left (\RootOf \left (\textit {\_Z}^{2}-2\right ) x^{4}-\sqrt {2 x^{8}+1}\right )}{8}+\frac {\ln \left (\frac {\sqrt {2 x^{8}+1}-1}{x^{4}}\right )}{4}\) | \(68\) |
meijerg | \(-\frac {4 \sqrt {\pi }-4 \sqrt {\pi }\, \sqrt {2 x^{8}+1}+4 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {2 x^{8}+1}}{2}\right )-2 \left (2-\ln \relax (2)+8 \ln \relax (x )\right ) \sqrt {\pi }}{16 \sqrt {\pi }}-\frac {\sqrt {2}\, \left (-2 \sqrt {\pi }\, \sqrt {2}\, x^{4} \sqrt {2 x^{8}+1}-2 \sqrt {\pi }\, \arcsinh \left (\sqrt {2}\, x^{4}\right )\right )}{16 \sqrt {\pi }}\) | \(103\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 114, normalized size = 1.46 \begin {gather*} -\frac {1}{16} \, \sqrt {2} \log \left (-\frac {\sqrt {2} - \frac {\sqrt {2 \, x^{8} + 1}}{x^{4}}}{\sqrt {2} + \frac {\sqrt {2 \, x^{8} + 1}}{x^{4}}}\right ) + \frac {1}{4} \, \sqrt {2 \, x^{8} + 1} + \frac {\sqrt {2 \, x^{8} + 1}}{4 \, x^{4} {\left (\frac {2 \, x^{8} + 1}{x^{8}} - 2\right )}} - \frac {1}{8} \, \log \left (\sqrt {2 \, x^{8} + 1} + 1\right ) + \frac {1}{8} \, \log \left (\sqrt {2 \, x^{8} + 1} - 1\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.24, size = 47, normalized size = 0.60 \begin {gather*} \frac {\sqrt {2}\,\mathrm {asinh}\left (\sqrt {2}\,x^4\right )}{8}-\frac {\mathrm {atanh}\left (\sqrt {2}\,\sqrt {x^8+\frac {1}{2}}\right )}{4}+\frac {\sqrt {2}\,\sqrt {x^8+\frac {1}{2}}\,\left (\frac {x^4}{2}+\frac {1}{2}\right )}{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 31.28, size = 65, normalized size = 0.83 \begin {gather*} \frac {x^{4} \sqrt {2 x^{8} + 1}}{4} + \frac {\sqrt {2 x^{8} + 1}}{4} + \frac {\log {\left (x^{8} \right )}}{8} - \frac {\log {\left (\sqrt {2 x^{8} + 1} + 1 \right )}}{4} + \frac {\sqrt {2} \operatorname {asinh}{\left (\sqrt {2} x^{4} \right )}}{8} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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