Integrand size = 13, antiderivative size = 87 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x} \, dx=\frac {3}{8} \left (1+x^4\right )^{2/3}+\frac {1}{4} \sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{1+x^4}}{\sqrt {3}}\right )+\frac {1}{4} \log \left (-1+\sqrt [3]{1+x^4}\right )-\frac {1}{8} \log \left (1+\sqrt [3]{1+x^4}+\left (1+x^4\right )^{2/3}\right ) \]
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Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.77, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {272, 52, 57, 632, 210, 31} \[ \int \frac {\left (1+x^4\right )^{2/3}}{x} \, dx=\frac {1}{4} \sqrt {3} \arctan \left (\frac {2 \sqrt [3]{x^4+1}+1}{\sqrt {3}}\right )+\frac {3}{8} \left (x^4+1\right )^{2/3}+\frac {3}{8} \log \left (1-\sqrt [3]{x^4+1}\right )-\frac {\log (x)}{2} \]
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Rule 31
Rule 52
Rule 57
Rule 210
Rule 272
Rule 632
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {(1+x)^{2/3}}{x} \, dx,x,x^4\right ) \\ & = \frac {3}{8} \left (1+x^4\right )^{2/3}+\frac {1}{4} \text {Subst}\left (\int \frac {1}{x \sqrt [3]{1+x}} \, dx,x,x^4\right ) \\ & = \frac {3}{8} \left (1+x^4\right )^{2/3}-\frac {\log (x)}{2}-\frac {3}{8} \text {Subst}\left (\int \frac {1}{1-x} \, dx,x,\sqrt [3]{1+x^4}\right )+\frac {3}{8} \text {Subst}\left (\int \frac {1}{1+x+x^2} \, dx,x,\sqrt [3]{1+x^4}\right ) \\ & = \frac {3}{8} \left (1+x^4\right )^{2/3}-\frac {\log (x)}{2}+\frac {3}{8} \log \left (1-\sqrt [3]{1+x^4}\right )-\frac {3}{4} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1+2 \sqrt [3]{1+x^4}\right ) \\ & = \frac {3}{8} \left (1+x^4\right )^{2/3}+\frac {1}{4} \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{1+x^4}}{\sqrt {3}}\right )-\frac {\log (x)}{2}+\frac {3}{8} \log \left (1-\sqrt [3]{1+x^4}\right ) \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x} \, dx=\frac {1}{8} \left (3 \left (1+x^4\right )^{2/3}+2 \sqrt {3} \arctan \left (\frac {1+2 \sqrt [3]{1+x^4}}{\sqrt {3}}\right )+2 \log \left (-1+\sqrt [3]{1+x^4}\right )-\log \left (1+\sqrt [3]{1+x^4}+\left (1+x^4\right )^{2/3}\right )\right ) \]
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Result contains higher order function than in optimal. Order 5 vs. order 3.
Time = 5.38 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74
method | result | size |
meijerg | \(-\frac {\sqrt {3}\, \Gamma \left (\frac {2}{3}\right ) \left (-\frac {2 \pi \sqrt {3}\, x^{4} \operatorname {hypergeom}\left (\left [\frac {1}{3}, 1, 1\right ], \left [2, 2\right ], -x^{4}\right )}{3 \Gamma \left (\frac {2}{3}\right )}-\frac {\left (\frac {3}{2}-\frac {\pi \sqrt {3}}{6}-\frac {3 \ln \left (3\right )}{2}+4 \ln \left (x \right )\right ) \pi \sqrt {3}}{\Gamma \left (\frac {2}{3}\right )}\right )}{12 \pi }\) | \(64\) |
pseudoelliptic | \(\frac {3 \left (x^{4}+1\right )^{\frac {2}{3}}}{8}+\frac {\ln \left (-1+\left (x^{4}+1\right )^{\frac {1}{3}}\right )}{4}-\frac {\ln \left (1+\left (x^{4}+1\right )^{\frac {1}{3}}+\left (x^{4}+1\right )^{\frac {2}{3}}\right )}{8}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (2 \left (x^{4}+1\right )^{\frac {1}{3}}+1\right ) \sqrt {3}}{3}\right )}{4}\) | \(64\) |
trager | \(\frac {3 \left (x^{4}+1\right )^{\frac {2}{3}}}{8}+\frac {\ln \left (\frac {-333 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{4}-393 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{4}-60 x^{4}+351 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {2}{3}}-48 \left (x^{4}+1\right )^{\frac {2}{3}}+144 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {1}{3}}+333 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}+165 \left (x^{4}+1\right )^{\frac {1}{3}}-384 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-80}{x^{4}}\right )}{4}-\frac {\ln \left (-\frac {-153 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{4}+162 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{4}-40 x^{4}+351 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {2}{3}}+165 \left (x^{4}+1\right )^{\frac {2}{3}}-495 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {1}{3}}+153 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}-48 \left (x^{4}+1\right )^{\frac {1}{3}}+195 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-100}{x^{4}}\right )}{4}-\frac {3 \ln \left (-\frac {-153 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2} x^{4}+162 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) x^{4}-40 x^{4}+351 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {2}{3}}+165 \left (x^{4}+1\right )^{\frac {2}{3}}-495 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right ) \left (x^{4}+1\right )^{\frac {1}{3}}+153 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )^{2}-48 \left (x^{4}+1\right )^{\frac {1}{3}}+195 \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )-100}{x^{4}}\right ) \operatorname {RootOf}\left (9 \textit {\_Z}^{2}+3 \textit {\_Z} +1\right )}{4}\) | \(426\) |
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Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.75 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x} \, dx=\frac {1}{4} \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} {\left (x^{4} + 1\right )}^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) + \frac {3}{8} \, {\left (x^{4} + 1\right )}^{\frac {2}{3}} - \frac {1}{8} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {2}{3}} + {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{3}} - 1\right ) \]
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Result contains complex when optimal does not.
Time = 0.56 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.43 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x} \, dx=- \frac {x^{\frac {8}{3}} \Gamma \left (- \frac {2}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {2}{3}, - \frac {2}{3} \\ \frac {1}{3} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{4}}} \right )}}{4 \Gamma \left (\frac {1}{3}\right )} \]
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Time = 0.27 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x} \, dx=\frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {3}{8} \, {\left (x^{4} + 1\right )}^{\frac {2}{3}} - \frac {1}{8} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {2}{3}} + {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{3}} - 1\right ) \]
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Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x} \, dx=\frac {1}{4} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {3}{8} \, {\left (x^{4} + 1\right )}^{\frac {2}{3}} - \frac {1}{8} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {2}{3}} + {\left (x^{4} + 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{4} \, \log \left ({\left (x^{4} + 1\right )}^{\frac {1}{3}} - 1\right ) \]
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Time = 5.52 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.02 \[ \int \frac {\left (1+x^4\right )^{2/3}}{x} \, dx=\frac {\ln \left (\frac {9\,{\left (x^4+1\right )}^{1/3}}{16}-\frac {9}{16}\right )}{4}+\ln \left (\frac {9\,{\left (x^4+1\right )}^{1/3}}{16}-9\,{\left (-\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}^2\right )\,\left (-\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )-\ln \left (\frac {9\,{\left (x^4+1\right )}^{1/3}}{16}-9\,{\left (\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}^2\right )\,\left (\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )+\frac {3\,{\left (x^4+1\right )}^{2/3}}{8} \]
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