Integrand size = 42, antiderivative size = 54 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{x-x^4}}{1+x^2-2 x^3-x^4-x^5+x^6} \, dx=-\text {RootSum}\left [-1+\text {$\#$1}^3+\text {$\#$1}^6\&,\frac {-\log (x) \text {$\#$1}+\log \left (\sqrt [3]{x-x^4}-x \text {$\#$1}\right ) \text {$\#$1}}{1+2 \text {$\#$1}^3}\&\right ] \]
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\[ \int \frac {\left (2+x^3\right ) \sqrt [3]{x-x^4}}{1+x^2-2 x^3-x^4-x^5+x^6} \, dx=\int \frac {\left (2+x^3\right ) \sqrt [3]{x-x^4}}{1+x^2-2 x^3-x^4-x^5+x^6} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [3]{x-x^4} \int \frac {\sqrt [3]{x} \sqrt [3]{1-x^3} \left (2+x^3\right )}{1+x^2-2 x^3-x^4-x^5+x^6} \, dx}{\sqrt [3]{x} \sqrt [3]{1-x^3}} \\ & = \frac {\left (3 \sqrt [3]{x-x^4}\right ) \text {Subst}\left (\int \frac {x^3 \sqrt [3]{1-x^9} \left (2+x^9\right )}{1+x^6-2 x^9-x^{12}-x^{15}+x^{18}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x} \sqrt [3]{1-x^3}} \\ & = \frac {\left (3 \sqrt [3]{x-x^4}\right ) \text {Subst}\left (\int \left (\frac {2 x^3 \sqrt [3]{1-x^9}}{1+x^6-2 x^9-x^{12}-x^{15}+x^{18}}+\frac {x^{12} \sqrt [3]{1-x^9}}{1+x^6-2 x^9-x^{12}-x^{15}+x^{18}}\right ) \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x} \sqrt [3]{1-x^3}} \\ & = \frac {\left (3 \sqrt [3]{x-x^4}\right ) \text {Subst}\left (\int \frac {x^{12} \sqrt [3]{1-x^9}}{1+x^6-2 x^9-x^{12}-x^{15}+x^{18}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x} \sqrt [3]{1-x^3}}+\frac {\left (6 \sqrt [3]{x-x^4}\right ) \text {Subst}\left (\int \frac {x^3 \sqrt [3]{1-x^9}}{1+x^6-2 x^9-x^{12}-x^{15}+x^{18}} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{x} \sqrt [3]{1-x^3}} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{x-x^4}}{1+x^2-2 x^3-x^4-x^5+x^6} \, dx=-\text {RootSum}\left [-1+\text {$\#$1}^3+\text {$\#$1}^6\&,\frac {-\log (x) \text {$\#$1}+\log \left (\sqrt [3]{x-x^4}-x \text {$\#$1}\right ) \text {$\#$1}}{1+2 \text {$\#$1}^3}\&\right ] \]
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Time = 80.76 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83
method | result | size |
pseudoelliptic | \(-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}+\textit {\_Z}^{3}-1\right )}{\sum }\frac {\textit {\_R} \ln \left (\frac {-\textit {\_R} x +\left (-x^{4}+x \right )^{\frac {1}{3}}}{x}\right )}{2 \textit {\_R}^{3}+1}\right )\) | \(45\) |
trager | \(\text {Expression too large to display}\) | \(12058\) |
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Exception generated. \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{x-x^4}}{1+x^2-2 x^3-x^4-x^5+x^6} \, dx=\text {Exception raised: TypeError} \]
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Not integrable
Time = 1.84 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{x-x^4}}{1+x^2-2 x^3-x^4-x^5+x^6} \, dx=\int \frac {\sqrt [3]{- x \left (x - 1\right ) \left (x^{2} + x + 1\right )} \left (x^{3} + 2\right )}{x^{6} - x^{5} - x^{4} - 2 x^{3} + x^{2} + 1}\, dx \]
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Not integrable
Time = 0.32 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{x-x^4}}{1+x^2-2 x^3-x^4-x^5+x^6} \, dx=\int { \frac {{\left (-x^{4} + x\right )}^{\frac {1}{3}} {\left (x^{3} + 2\right )}}{x^{6} - x^{5} - x^{4} - 2 \, x^{3} + x^{2} + 1} \,d x } \]
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Not integrable
Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{x-x^4}}{1+x^2-2 x^3-x^4-x^5+x^6} \, dx=\int { \frac {{\left (-x^{4} + x\right )}^{\frac {1}{3}} {\left (x^{3} + 2\right )}}{x^{6} - x^{5} - x^{4} - 2 \, x^{3} + x^{2} + 1} \,d x } \]
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Not integrable
Time = 0.00 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.78 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{x-x^4}}{1+x^2-2 x^3-x^4-x^5+x^6} \, dx=\int \frac {{\left (x-x^4\right )}^{1/3}\,\left (x^3+2\right )}{x^6-x^5-x^4-2\,x^3+x^2+1} \,d x \]
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