Integrand size = 29, antiderivative size = 97 \[ \int \frac {-3+5 x^8}{\left (1+x^8\right ) \sqrt [3]{1-x^3+x^8}} \, dx=-\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{-x+2 \sqrt [3]{1-x^3+x^8}}\right )-\log \left (x+\sqrt [3]{1-x^3+x^8}\right )+\frac {1}{2} \log \left (x^2-x \sqrt [3]{1-x^3+x^8}+\left (1-x^3+x^8\right )^{2/3}\right ) \]
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\[ \int \frac {-3+5 x^8}{\left (1+x^8\right ) \sqrt [3]{1-x^3+x^8}} \, dx=\int \frac {-3+5 x^8}{\left (1+x^8\right ) \sqrt [3]{1-x^3+x^8}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {5}{\sqrt [3]{1-x^3+x^8}}-\frac {8}{\left (1+x^8\right ) \sqrt [3]{1-x^3+x^8}}\right ) \, dx \\ & = 5 \int \frac {1}{\sqrt [3]{1-x^3+x^8}} \, dx-8 \int \frac {1}{\left (1+x^8\right ) \sqrt [3]{1-x^3+x^8}} \, dx \\ & = 5 \int \frac {1}{\sqrt [3]{1-x^3+x^8}} \, dx-8 \int \left (\frac {i}{2 \left (i-x^4\right ) \sqrt [3]{1-x^3+x^8}}+\frac {i}{2 \left (i+x^4\right ) \sqrt [3]{1-x^3+x^8}}\right ) \, dx \\ & = -\left (4 i \int \frac {1}{\left (i-x^4\right ) \sqrt [3]{1-x^3+x^8}} \, dx\right )-4 i \int \frac {1}{\left (i+x^4\right ) \sqrt [3]{1-x^3+x^8}} \, dx+5 \int \frac {1}{\sqrt [3]{1-x^3+x^8}} \, dx \\ & = -\left (4 i \int \left (-\frac {(-1)^{3/4}}{2 \left (\sqrt [4]{-1}-x^2\right ) \sqrt [3]{1-x^3+x^8}}-\frac {(-1)^{3/4}}{2 \left (\sqrt [4]{-1}+x^2\right ) \sqrt [3]{1-x^3+x^8}}\right ) \, dx\right )-4 i \int \left (-\frac {\sqrt [4]{-1}}{2 \left (-(-1)^{3/4}-x^2\right ) \sqrt [3]{1-x^3+x^8}}-\frac {\sqrt [4]{-1}}{2 \left (-(-1)^{3/4}+x^2\right ) \sqrt [3]{1-x^3+x^8}}\right ) \, dx+5 \int \frac {1}{\sqrt [3]{1-x^3+x^8}} \, dx \\ & = 5 \int \frac {1}{\sqrt [3]{1-x^3+x^8}} \, dx-\left (2 \sqrt [4]{-1}\right ) \int \frac {1}{\left (\sqrt [4]{-1}-x^2\right ) \sqrt [3]{1-x^3+x^8}} \, dx-\left (2 \sqrt [4]{-1}\right ) \int \frac {1}{\left (\sqrt [4]{-1}+x^2\right ) \sqrt [3]{1-x^3+x^8}} \, dx+\left (2 (-1)^{3/4}\right ) \int \frac {1}{\left (-(-1)^{3/4}-x^2\right ) \sqrt [3]{1-x^3+x^8}} \, dx+\left (2 (-1)^{3/4}\right ) \int \frac {1}{\left (-(-1)^{3/4}+x^2\right ) \sqrt [3]{1-x^3+x^8}} \, dx \\ & = 5 \int \frac {1}{\sqrt [3]{1-x^3+x^8}} \, dx-\left (2 \sqrt [4]{-1}\right ) \int \left (-\frac {(-1)^{7/8}}{2 \left (\sqrt [8]{-1}-x\right ) \sqrt [3]{1-x^3+x^8}}-\frac {(-1)^{7/8}}{2 \left (\sqrt [8]{-1}+x\right ) \sqrt [3]{1-x^3+x^8}}\right ) \, dx-\left (2 \sqrt [4]{-1}\right ) \int \left (-\frac {(-1)^{3/8}}{2 \left (-(-1)^{5/8}-x\right ) \sqrt [3]{1-x^3+x^8}}-\frac {(-1)^{3/8}}{2 \left (-(-1)^{5/8}+x\right ) \sqrt [3]{1-x^3+x^8}}\right ) \, dx+\left (2 (-1)^{3/4}\right ) \int \left (\frac {(-1)^{5/8}}{2 \left ((-1)^{3/8}-x\right ) \sqrt [3]{1-x^3+x^8}}+\frac {(-1)^{5/8}}{2 \left ((-1)^{3/8}+x\right ) \sqrt [3]{1-x^3+x^8}}\right ) \, dx+\left (2 (-1)^{3/4}\right ) \int \left (\frac {\sqrt [8]{-1}}{2 \left (-(-1)^{7/8}-x\right ) \sqrt [3]{1-x^3+x^8}}+\frac {\sqrt [8]{-1}}{2 \left (-(-1)^{7/8}+x\right ) \sqrt [3]{1-x^3+x^8}}\right ) \, dx \\ & = 5 \int \frac {1}{\sqrt [3]{1-x^3+x^8}} \, dx-\sqrt [8]{-1} \int \frac {1}{\left (\sqrt [8]{-1}-x\right ) \sqrt [3]{1-x^3+x^8}} \, dx-\sqrt [8]{-1} \int \frac {1}{\left (\sqrt [8]{-1}+x\right ) \sqrt [3]{1-x^3+x^8}} \, dx-(-1)^{3/8} \int \frac {1}{\left ((-1)^{3/8}-x\right ) \sqrt [3]{1-x^3+x^8}} \, dx-(-1)^{3/8} \int \frac {1}{\left ((-1)^{3/8}+x\right ) \sqrt [3]{1-x^3+x^8}} \, dx+(-1)^{5/8} \int \frac {1}{\left (-(-1)^{5/8}-x\right ) \sqrt [3]{1-x^3+x^8}} \, dx+(-1)^{5/8} \int \frac {1}{\left (-(-1)^{5/8}+x\right ) \sqrt [3]{1-x^3+x^8}} \, dx+(-1)^{7/8} \int \frac {1}{\left (-(-1)^{7/8}-x\right ) \sqrt [3]{1-x^3+x^8}} \, dx+(-1)^{7/8} \int \frac {1}{\left (-(-1)^{7/8}+x\right ) \sqrt [3]{1-x^3+x^8}} \, dx \\ \end{align*}
Time = 1.50 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.97 \[ \int \frac {-3+5 x^8}{\left (1+x^8\right ) \sqrt [3]{1-x^3+x^8}} \, dx=\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x-2 \sqrt [3]{1-x^3+x^8}}\right )-\log \left (x+\sqrt [3]{1-x^3+x^8}\right )+\frac {1}{2} \log \left (x^2-x \sqrt [3]{1-x^3+x^8}+\left (1-x^3+x^8\right )^{2/3}\right ) \]
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Time = 4.35 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.94
method | result | size |
pseudoelliptic | \(-\ln \left (\frac {x +\left (x^{8}-x^{3}+1\right )^{\frac {1}{3}}}{x}\right )+\frac {\ln \left (\frac {x^{2}-x \left (x^{8}-x^{3}+1\right )^{\frac {1}{3}}+\left (x^{8}-x^{3}+1\right )^{\frac {2}{3}}}{x^{2}}\right )}{2}-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x -2 \left (x^{8}-x^{3}+1\right )^{\frac {1}{3}}\right )}{3 x}\right )\) | \(91\) |
trager | \(\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \ln \left (\frac {x^{8}+\left (x^{8}-x^{3}+1\right )^{\frac {2}{3}} \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x -\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{8}-x^{3}+1\right )^{\frac {1}{3}} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{3}-2 \left (x^{8}-x^{3}+1\right )^{\frac {2}{3}} x +2 \left (x^{8}-x^{3}+1\right )^{\frac {1}{3}} x^{2}-2 x^{3}+1}{x^{8}+1}\right )-\ln \left (\frac {-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{8}+x^{8}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{3}+3 \left (x^{8}-x^{3}+1\right )^{\frac {2}{3}} \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x -3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{8}-x^{3}+1\right )^{\frac {1}{3}} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{3}-3 \left (x^{8}-x^{3}+1\right )^{\frac {2}{3}} x +3 \left (x^{8}-x^{3}+1\right )^{\frac {1}{3}} x^{2}-x^{3}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+1}{x^{8}+1}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+\ln \left (\frac {-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{8}+x^{8}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )^{2} x^{3}+3 \left (x^{8}-x^{3}+1\right )^{\frac {2}{3}} \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x -3 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) \left (x^{8}-x^{3}+1\right )^{\frac {1}{3}} x^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right ) x^{3}-3 \left (x^{8}-x^{3}+1\right )^{\frac {2}{3}} x +3 \left (x^{8}-x^{3}+1\right )^{\frac {1}{3}} x^{2}-x^{3}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}-\textit {\_Z} +1\right )+1}{x^{8}+1}\right )\) | \(446\) |
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Time = 1.21 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.25 \[ \int \frac {-3+5 x^8}{\left (1+x^8\right ) \sqrt [3]{1-x^3+x^8}} \, dx=-\sqrt {3} \arctan \left (\frac {4 \, \sqrt {3} {\left (x^{8} - x^{3} + 1\right )}^{\frac {1}{3}} x^{2} + 2 \, \sqrt {3} {\left (x^{8} - x^{3} + 1\right )}^{\frac {2}{3}} x + \sqrt {3} {\left (x^{8} - x^{3} + 1\right )}}{x^{8} - 9 \, x^{3} + 1}\right ) - \frac {1}{2} \, \log \left (\frac {x^{8} + 3 \, {\left (x^{8} - x^{3} + 1\right )}^{\frac {1}{3}} x^{2} + 3 \, {\left (x^{8} - x^{3} + 1\right )}^{\frac {2}{3}} x + 1}{x^{8} + 1}\right ) \]
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\[ \int \frac {-3+5 x^8}{\left (1+x^8\right ) \sqrt [3]{1-x^3+x^8}} \, dx=\int \frac {5 x^{8} - 3}{\left (x^{8} + 1\right ) \sqrt [3]{x^{8} - x^{3} + 1}}\, dx \]
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\[ \int \frac {-3+5 x^8}{\left (1+x^8\right ) \sqrt [3]{1-x^3+x^8}} \, dx=\int { \frac {5 \, x^{8} - 3}{{\left (x^{8} - x^{3} + 1\right )}^{\frac {1}{3}} {\left (x^{8} + 1\right )}} \,d x } \]
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\[ \int \frac {-3+5 x^8}{\left (1+x^8\right ) \sqrt [3]{1-x^3+x^8}} \, dx=\int { \frac {5 \, x^{8} - 3}{{\left (x^{8} - x^{3} + 1\right )}^{\frac {1}{3}} {\left (x^{8} + 1\right )}} \,d x } \]
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Timed out. \[ \int \frac {-3+5 x^8}{\left (1+x^8\right ) \sqrt [3]{1-x^3+x^8}} \, dx=\int \frac {5\,x^8-3}{\left (x^8+1\right )\,{\left (x^8-x^3+1\right )}^{1/3}} \,d x \]
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