Integrand size = 77, antiderivative size = 276 \[ \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x^2}{\sqrt [3]{b} x^2+2 \left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{2 b^{2/3}}+\frac {\log \left (-\sqrt [6]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}+\frac {\log \left (\sqrt [6]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{2/3}}-\frac {\log \left (\sqrt [3]{b} x^2-\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{2/3}}-\frac {\log \left (\sqrt [3]{b} x^2+\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{2/3}} \]
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\[ \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {x^{5/3} (-2+(1+k) x)}{\sqrt [3]{1-x} \sqrt [3]{1-k x} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx}{\sqrt [3]{(1-x) x (1-k x)}} \\ & = \frac {\left (3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \text {Subst}\left (\int \frac {x^7 \left (-2+(1+k) x^3\right )}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (1-(2+2 k) x^3+\left (1+4 k+k^2\right ) x^6-\left (2 k+2 k^2\right ) x^9+\left (-b+k^2\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}} \\ & = \frac {\left (3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \text {Subst}\left (\int \left (\frac {(1+k) x^{10}}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (1-2 (1+k) x^3+(1+k (4+k)) x^6-2 k (1+k) x^9-b \left (1-\frac {k^2}{b}\right ) x^{12}\right )}+\frac {2 x^7}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (-1+2 (1+k) x^3-(1+k (4+k)) x^6+2 k (1+k) x^9+b \left (1-\frac {k^2}{b}\right ) x^{12}\right )}\right ) \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}} \\ & = \frac {\left (6 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \text {Subst}\left (\int \frac {x^7}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (-1+2 (1+k) x^3-(1+k (4+k)) x^6+2 k (1+k) x^9+b \left (1-\frac {k^2}{b}\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}+\frac {\left (3 (1+k) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \text {Subst}\left (\int \frac {x^{10}}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (1-2 (1+k) x^3+(1+k (4+k)) x^6-2 k (1+k) x^9-b \left (1-\frac {k^2}{b}\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}} \\ \end{align*}
Time = 12.43 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.70 \[ \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=-\frac {2 \sqrt {3} \arctan \left (\frac {1+\frac {2 ((-1+x) x (-1+k x))^{2/3}}{\sqrt [3]{b} x^2}}{\sqrt {3}}\right )-2 \log \left (-\sqrt [6]{b} x+\sqrt [3]{(-1+x) x (-1+k x)}\right )-2 \log \left (\sqrt [6]{b} x+\sqrt [3]{(-1+x) x (-1+k x)}\right )+\log \left (\sqrt [3]{b} x^2-\sqrt [6]{b} x \sqrt [3]{(-1+x) x (-1+k x)}+((-1+x) x (-1+k x))^{2/3}\right )+\log \left (\sqrt [3]{b} x^2+\sqrt [6]{b} x \sqrt [3]{(-1+x) x (-1+k x)}+((-1+x) x (-1+k x))^{2/3}\right )}{4 b^{2/3}} \]
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Time = 0.56 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.46
method | result | size |
pseudoelliptic | \(\frac {-2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x^{2}+2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}\right )}{3 b^{\frac {1}{3}} x^{2}}\right )+2 \ln \left (\frac {-b^{\frac {1}{3}} x^{2}+\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )-\ln \left (\frac {b^{\frac {1}{3}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}} x +\left (-1+x \right ) \left (k x -1\right ) \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}+b^{\frac {2}{3}} x^{3}}{x^{3}}\right )}{4 b^{\frac {2}{3}}}\) | \(127\) |
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Timed out. \[ \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\int { \frac {{\left ({\left (k + 1\right )} x - 2\right )} x^{2}}{{\left ({\left (k^{2} - b\right )} x^{4} - 2 \, {\left (k^{2} + k\right )} x^{3} + {\left (k^{2} + 4 \, k + 1\right )} x^{2} - 2 \, {\left (k + 1\right )} x + 1\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]
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Time = 0.41 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.45 \[ \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + b^{\frac {1}{3}}\right )}}{3 \, b^{\frac {1}{3}}}\right )}{2 \, b^{\frac {2}{3}}} - \frac {\log \left ({\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {4}{3}} + b^{\frac {1}{3}} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + b^{\frac {2}{3}}\right )}{4 \, b^{\frac {2}{3}}} + \frac {\log \left ({\left | {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} - b^{\frac {1}{3}} \right |}\right )}{2 \, b^{\frac {2}{3}}} \]
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Timed out. \[ \int \frac {x^2 (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (1-(2+2 k) x+\left (1+4 k+k^2\right ) x^2-\left (2 k+2 k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx=\int -\frac {x^2\,\left (x\,\left (k+1\right )-2\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (\left (b-k^2\right )\,x^4+\left (2\,k^2+2\,k\right )\,x^3+\left (-k^2-4\,k-1\right )\,x^2+\left (2\,k+2\right )\,x-1\right )} \,d x \]
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