Integrand size = 27, antiderivative size = 81 \[ \int \frac {3+x^4}{\sqrt [3]{-1+x^4} \left (-1-8 x^3+x^4\right )} \, dx=-\frac {1}{2} \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+\sqrt [3]{-1+x^4}}\right )+\frac {1}{2} \log \left (-2 x+\sqrt [3]{-1+x^4}\right )-\frac {1}{4} \log \left (4 x^2+2 x \sqrt [3]{-1+x^4}+\left (-1+x^4\right )^{2/3}\right ) \]
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\[ \int \frac {3+x^4}{\sqrt [3]{-1+x^4} \left (-1-8 x^3+x^4\right )} \, dx=\int \frac {3+x^4}{\sqrt [3]{-1+x^4} \left (-1-8 x^3+x^4\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{\sqrt [3]{-1+x^4}}+\frac {4 \left (1+2 x^3\right )}{\sqrt [3]{-1+x^4} \left (-1-8 x^3+x^4\right )}\right ) \, dx \\ & = 4 \int \frac {1+2 x^3}{\sqrt [3]{-1+x^4} \left (-1-8 x^3+x^4\right )} \, dx+\int \frac {1}{\sqrt [3]{-1+x^4}} \, dx \\ & = 4 \int \left (\frac {1}{\sqrt [3]{-1+x^4} \left (-1-8 x^3+x^4\right )}+\frac {2 x^3}{\sqrt [3]{-1+x^4} \left (-1-8 x^3+x^4\right )}\right ) \, dx+\frac {\sqrt [3]{1-x^4} \int \frac {1}{\sqrt [3]{1-x^4}} \, dx}{\sqrt [3]{-1+x^4}} \\ & = \frac {x \sqrt [3]{1-x^4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{3},\frac {5}{4},x^4\right )}{\sqrt [3]{-1+x^4}}+4 \int \frac {1}{\sqrt [3]{-1+x^4} \left (-1-8 x^3+x^4\right )} \, dx+8 \int \frac {x^3}{\sqrt [3]{-1+x^4} \left (-1-8 x^3+x^4\right )} \, dx \\ \end{align*}
Time = 1.18 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int \frac {3+x^4}{\sqrt [3]{-1+x^4} \left (-1-8 x^3+x^4\right )} \, dx=-\frac {1}{2} \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+\sqrt [3]{-1+x^4}}\right )+\frac {1}{2} \log \left (-2 x+\sqrt [3]{-1+x^4}\right )-\frac {1}{4} \log \left (4 x^2+2 x \sqrt [3]{-1+x^4}+\left (-1+x^4\right )^{2/3}\right ) \]
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Time = 4.69 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.90
method | result | size |
pseudoelliptic | \(-\frac {\ln \left (\frac {4 x^{2}+2 x \left (x^{4}-1\right )^{\frac {1}{3}}+\left (x^{4}-1\right )^{\frac {2}{3}}}{x^{2}}\right )}{4}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x +\left (x^{4}-1\right )^{\frac {1}{3}}\right )}{3 x}\right )}{2}+\frac {\ln \left (\frac {-2 x +\left (x^{4}-1\right )^{\frac {1}{3}}}{x}\right )}{2}\) | \(73\) |
trager | \(\frac {\ln \left (-\frac {32 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x^{3}+2 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) x^{4}-4 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{4}-1\right )^{\frac {2}{3}} x -8 \left (x^{4}-1\right )^{\frac {1}{3}} \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) x^{2}+32 x^{3} \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )+x^{4}-4 \left (x^{4}-1\right )^{\frac {2}{3}} x +4 \left (x^{4}-1\right )^{\frac {1}{3}} x^{2}+8 x^{3}-2 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )-1}{x^{4}-8 x^{3}-1}\right )}{2}+\operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \ln \left (\frac {32 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )^{2} x^{3}-2 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) x^{4}-4 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) \left (x^{4}-1\right )^{\frac {2}{3}} x +16 \left (x^{4}-1\right )^{\frac {1}{3}} \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right ) x^{2}+16 x^{3} \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )-x^{4}+2 \left (x^{4}-1\right )^{\frac {2}{3}} x +4 \left (x^{4}-1\right )^{\frac {1}{3}} x^{2}+2 \operatorname {RootOf}\left (4 \textit {\_Z}^{2}+2 \textit {\_Z} +1\right )+1}{x^{4}-8 x^{3}-1}\right )\) | \(320\) |
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Time = 1.37 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.38 \[ \int \frac {3+x^4}{\sqrt [3]{-1+x^4} \left (-1-8 x^3+x^4\right )} \, dx=-\frac {1}{2} \, \sqrt {3} \arctan \left (-\frac {8 \, \sqrt {3} {\left (x^{4} - 1\right )}^{\frac {1}{3}} x^{2} - 4 \, \sqrt {3} {\left (x^{4} - 1\right )}^{\frac {2}{3}} x + \sqrt {3} {\left (x^{4} - 8 \, x^{3} - 1\right )}}{3 \, {\left (x^{4} + 8 \, x^{3} - 1\right )}}\right ) + \frac {1}{4} \, \log \left (\frac {x^{4} - 8 \, x^{3} + 12 \, {\left (x^{4} - 1\right )}^{\frac {1}{3}} x^{2} - 6 \, {\left (x^{4} - 1\right )}^{\frac {2}{3}} x - 1}{x^{4} - 8 \, x^{3} - 1}\right ) \]
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Timed out. \[ \int \frac {3+x^4}{\sqrt [3]{-1+x^4} \left (-1-8 x^3+x^4\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {3+x^4}{\sqrt [3]{-1+x^4} \left (-1-8 x^3+x^4\right )} \, dx=\int { \frac {x^{4} + 3}{{\left (x^{4} - 8 \, x^{3} - 1\right )} {\left (x^{4} - 1\right )}^{\frac {1}{3}}} \,d x } \]
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\[ \int \frac {3+x^4}{\sqrt [3]{-1+x^4} \left (-1-8 x^3+x^4\right )} \, dx=\int { \frac {x^{4} + 3}{{\left (x^{4} - 8 \, x^{3} - 1\right )} {\left (x^{4} - 1\right )}^{\frac {1}{3}}} \,d x } \]
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Timed out. \[ \int \frac {3+x^4}{\sqrt [3]{-1+x^4} \left (-1-8 x^3+x^4\right )} \, dx=\int -\frac {x^4+3}{{\left (x^4-1\right )}^{1/3}\,\left (-x^4+8\,x^3+1\right )} \,d x \]
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