Integrand size = 26, antiderivative size = 123 \[ \int \frac {-1+x}{\left (-4-2 x+x^2\right ) \sqrt [3]{-2-2 x+x^2}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2^{2/3} \sqrt [3]{-2-2 x+x^2}}{\sqrt {3}}\right )}{2 \sqrt [3]{2}}+\frac {\log \left (-2+2^{2/3} \sqrt [3]{-2-2 x+x^2}\right )}{2 \sqrt [3]{2}}-\frac {\log \left (2+2^{2/3} \sqrt [3]{-2-2 x+x^2}+\sqrt [3]{2} \left (-2-2 x+x^2\right )^{2/3}\right )}{4 \sqrt [3]{2}} \]
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\[ \int \frac {-1+x}{\left (-4-2 x+x^2\right ) \sqrt [3]{-2-2 x+x^2}} \, dx=\int \frac {-1+x}{\left (-4-2 x+x^2\right ) \sqrt [3]{-2-2 x+x^2}} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {-1+x}{\left (-4-2 x+x^2\right ) \sqrt [3]{-2-2 x+x^2}} \, dx \\ \end{align*}
Time = 0.18 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.88 \[ \int \frac {-1+x}{\left (-4-2 x+x^2\right ) \sqrt [3]{-2-2 x+x^2}} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {1+2^{2/3} \sqrt [3]{-2-2 x+x^2}}{\sqrt {3}}\right )+2 \log \left (-2+2^{2/3} \sqrt [3]{-2-2 x+x^2}\right )-\log \left (2+2^{2/3} \sqrt [3]{-2-2 x+x^2}+\sqrt [3]{2} \left (-2-2 x+x^2\right )^{2/3}\right )}{4 \sqrt [3]{2}} \]
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Time = 6.82 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.68
method | result | size |
pseudoelliptic | \(\frac {2^{\frac {2}{3}} \left (2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (2^{\frac {2}{3}} \left (x^{2}-2 x -2\right )^{\frac {1}{3}}+1\right )}{3}\right )+2 \ln \left (\left (x^{2}-2 x -2\right )^{\frac {1}{3}}-2^{\frac {1}{3}}\right )-\ln \left (\left (x^{2}-2 x -2\right )^{\frac {2}{3}}+2^{\frac {1}{3}} \left (x^{2}-2 x -2\right )^{\frac {1}{3}}+2^{\frac {2}{3}}\right )\right )}{8}\) | \(84\) |
trager | \(\text {Expression too large to display}\) | \(1005\) |
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Time = 0.30 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.76 \[ \int \frac {-1+x}{\left (-4-2 x+x^2\right ) \sqrt [3]{-2-2 x+x^2}} \, dx=\frac {1}{4} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{6} \, \sqrt {3} 2^{\frac {1}{6}} {\left (2^{\frac {5}{6}} + 2 \, \sqrt {2} {\left (x^{2} - 2 \, x - 2\right )}^{\frac {1}{3}}\right )}\right ) - \frac {1}{8} \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (x^{2} - 2 \, x - 2\right )}^{\frac {1}{3}} + {\left (x^{2} - 2 \, x - 2\right )}^{\frac {2}{3}}\right ) + \frac {1}{4} \cdot 2^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} + {\left (x^{2} - 2 \, x - 2\right )}^{\frac {1}{3}}\right ) \]
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\[ \int \frac {-1+x}{\left (-4-2 x+x^2\right ) \sqrt [3]{-2-2 x+x^2}} \, dx=\int \frac {x - 1}{\left (x^{2} - 2 x - 4\right ) \sqrt [3]{x^{2} - 2 x - 2}}\, dx \]
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\[ \int \frac {-1+x}{\left (-4-2 x+x^2\right ) \sqrt [3]{-2-2 x+x^2}} \, dx=\int { \frac {x - 1}{{\left (x^{2} - 2 \, x - 2\right )}^{\frac {1}{3}} {\left (x^{2} - 2 \, x - 4\right )}} \,d x } \]
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Time = 0.32 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.68 \[ \int \frac {-1+x}{\left (-4-2 x+x^2\right ) \sqrt [3]{-2-2 x+x^2}} \, dx=\frac {1}{4} \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, {\left (\frac {1}{2} \, x^{2} - x - 1\right )}^{\frac {1}{3}} + 1\right )}\right ) - \frac {1}{8} \cdot 2^{\frac {2}{3}} \log \left ({\left (\frac {1}{2} \, x^{2} - x - 1\right )}^{\frac {2}{3}} + {\left (\frac {1}{2} \, x^{2} - x - 1\right )}^{\frac {1}{3}} + 1\right ) + \frac {1}{4} \cdot 2^{\frac {2}{3}} \log \left ({\left | {\left (\frac {1}{2} \, x^{2} - x - 1\right )}^{\frac {1}{3}} - 1 \right |}\right ) \]
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Time = 6.18 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.89 \[ \int \frac {-1+x}{\left (-4-2 x+x^2\right ) \sqrt [3]{-2-2 x+x^2}} \, dx=\frac {2^{2/3}\,\ln \left (\frac {9\,{\left (x^2-2\,x-2\right )}^{1/3}}{4}-\frac {9\,2^{1/3}}{4}\right )}{4}+\frac {2^{2/3}\,\ln \left (\frac {9\,{\left (x^2-2\,x-2\right )}^{1/3}}{4}-\frac {9\,2^{1/3}\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{8}-\frac {2^{2/3}\,\ln \left (\frac {9\,{\left (x^2-2\,x-2\right )}^{1/3}}{4}-\frac {9\,2^{1/3}\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2}{16}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{8} \]
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