Integrand size = 43, antiderivative size = 59 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{x+x^3-x^4}}{1+x^2-2 x^3+x^4-x^5+x^6} \, dx=-\text {RootSum}\left [1-\text {$\#$1}^3+\text {$\#$1}^6\&,\frac {-\log (x) \text {$\#$1}+\log \left (\sqrt [3]{x+x^3-x^4}-x \text {$\#$1}\right ) \text {$\#$1}}{-1+2 \text {$\#$1}^3}\&\right ] \]
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\[ \int \frac {\left (2+x^3\right ) \sqrt [3]{x+x^3-x^4}}{1+x^2-2 x^3+x^4-x^5+x^6} \, dx=\int \frac {\left (2+x^3\right ) \sqrt [3]{x+x^3-x^4}}{1+x^2-2 x^3+x^4-x^5+x^6} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [3]{x+x^3-x^4} \int \frac {\sqrt [3]{x} \sqrt [3]{1+x^2-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^2-x^3}} \\ & = \frac {\left (3 \sqrt [3]{x+x^3-x^4}\right ) \text {Subst}\left (\int \frac {x^3 \sqrt [3]{1+x^6-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^2-x^3}} \\ & = \frac {\left (3 \sqrt [3]{x+x^3-x^4}\right ) \text {Subst}\left (\int \left (\frac {2 x^3 \sqrt [3]{1+x^6-x^9}}{1+x^6-2 x^9+x^{12}-x^{15}+x^{18}}+\frac {x^{12} \sqrt [3]{1+x^6-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^2-x^3}} \\ & = \frac {\left (3 \sqrt [3]{x+x^3-x^4}\right ) \text {Subst}\left (\int \frac {x^{12} \sqrt [3]{1+x^6-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^2-x^3}}+\frac {\left (6 \sqrt [3]{x+x^3-x^4}\right ) \text {Subst}\left (\int \frac {x^3 \sqrt [3]{1+x^6-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^2-x^3}} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.64 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{x+x^3-x^4}}{1+x^2-2 x^3+x^4-x^5+x^6} \, dx=\frac {\sqrt [3]{x+x^3-x^4} \text {RootSum}\left [1+\text {$\#$1}^3+\text {$\#$1}^6\&,\frac {-2 \log \left (\sqrt [3]{x}\right ) \text {$\#$1}+\log \left (\sqrt [3]{-1-x^2+x^3}-x^{2/3} \text {$\#$1}\right ) \text {$\#$1}}{1+2 \text {$\#$1}^3}\&\right ]}{\sqrt [3]{x} \sqrt [3]{-1-x^2+x^3}} \]
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Time = 9.70 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90
method | result | size |
pseudoelliptic | \(-\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{6}-\textit {\_Z}^{3}+1\right )}{\sum }\frac {\textit {\_R} \ln \left (\frac {-\textit {\_R} x +{\left (-x \left (x^{3}-x^{2}-1\right )\right )}^{\frac {1}{3}}}{x}\right )}{2 \textit {\_R}^{3}-1}\right )\) | \(53\) |
trager | \(\text {Expression too large to display}\) | \(2091\) |
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Exception generated. \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{x+x^3-x^4}}{1+x^2-2 x^3+x^4-x^5+x^6} \, dx=\text {Exception raised: TypeError} \]
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Not integrable
Time = 1.88 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.69 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{x+x^3-x^4}}{1+x^2-2 x^3+x^4-x^5+x^6} \, dx=\int \frac {\sqrt [3]{- x \left (x^{3} - x^{2} - 1\right )} \left (x^{3} + 2\right )}{x^{6} - x^{5} + x^{4} - 2 x^{3} + x^{2} + 1}\, dx \]
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Not integrable
Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.73 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{x+x^3-x^4}}{1+x^2-2 x^3+x^4-x^5+x^6} \, dx=\int { \frac {{\left (-x^{4} + x^{3} + x\right )}^{\frac {1}{3}} {\left (x^{3} + 2\right )}}{x^{6} - x^{5} + x^{4} - 2 \, x^{3} + x^{2} + 1} \,d x } \]
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Not integrable
Time = 0.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.73 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{x+x^3-x^4}}{1+x^2-2 x^3+x^4-x^5+x^6} \, dx=\int { \frac {{\left (-x^{4} + x^{3} + x\right )}^{\frac {1}{3}} {\left (x^{3} + 2\right )}}{x^{6} - x^{5} + x^{4} - 2 \, x^{3} + x^{2} + 1} \,d x } \]
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Not integrable
Time = 0.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.73 \[ \int \frac {\left (2+x^3\right ) \sqrt [3]{x+x^3-x^4}}{1+x^2-2 x^3+x^4-x^5+x^6} \, dx=\int \frac {\left (x^3+2\right )\,{\left (-x^4+x^3+x\right )}^{1/3}}{x^6-x^5+x^4-2\,x^3+x^2+1} \,d x \]
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