Integrand size = 81, antiderivative size = 212 \[ \int \frac {(-1+x) (-1+k x) (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (b-2 (b+b k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x-2 \sqrt [6]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{5/6}}-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [6]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{5/6}}-\frac {\text {arctanh}\left (\frac {x}{\sqrt [6]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{5/6}}-\frac {\text {arctanh}\left (\frac {\frac {x^2}{\sqrt [6]{b}}+\sqrt [6]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{x \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{5/6}} \]
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\[ \int \frac {(-1+x) (-1+k x) (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (b-2 (b+b k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx=\int \frac {(-1+x) (-1+k x) (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (b-2 (b+b k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx \]
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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 {(-1+x) (-1+k x) (-2+(1+k) x)}{\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \left (b-2 (b+b k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx}{\sqrt [3]{(1-x) x (1-k x)}} \\ & = -\frac {\left (\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {(1-x)^{2/3} (-1+k x) (-2+(1+k) x)}{\sqrt [3]{x} \sqrt [3]{1-k x} \left (b-2 (b+b k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx}{\sqrt [3]{(1-x) x (1-k x)}} \\ & = \frac {\left (\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {(1-x)^{2/3} (1-k x)^{2/3} (-2+(1+k) x)}{\sqrt [3]{x} \left (b-2 (b+b k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+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 \left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-2+(1+k) x^3\right )}{b-2 (b+b k) x^3+\left (b+4 b k+b k^2\right ) x^6-2 b k (1+k) x^9+\left (-1+b k^2\right ) x^{12}} \, 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 \frac {x \left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (2-(1+k) x^3\right )}{x^{12}-b \left (-1+x^3\right )^2 \left (-1+k x^3\right )^2} \, 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^4 \left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3}}{b-2 b (1+k) x^3+b (1+k (4+k)) x^6-2 b k (1+k) x^9-\left (1-b k^2\right ) x^{12}}+\frac {2 x \left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3}}{-b+2 b (1+k) x^3-b (1+k (4+k)) x^6+2 b k (1+k) x^9+\left (1-b k^2\right ) x^{12}}\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 \left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3}}{-b+2 b (1+k) x^3-b (1+k (4+k)) x^6+2 b k (1+k) x^9+\left (1-b k^2\right ) x^{12}} \, 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^4 \left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3}}{b-2 b (1+k) x^3+b (1+k (4+k)) x^6-2 b k (1+k) x^9-\left (1-b k^2\right ) x^{12}} \, 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 \left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3}}{x^{12}-b \left (-1+x^3\right )^2 \left (-1+k x^3\right )^2} \, 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^4 \left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3}}{-x^{12}+b \left (-1+x^3\right )^2 \left (-1+k x^3\right )^2} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}} \\ \end{align*}
Time = 12.53 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.75 \[ \int \frac {(-1+x) (-1+k x) (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (b-2 (b+b k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx=\frac {\sqrt {3} \left (\arctan \left (\frac {\sqrt {3} x}{x-2 \sqrt [6]{b} \sqrt [3]{(-1+x) x (-1+k x)}}\right )-\arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [6]{b} \sqrt [3]{(-1+x) x (-1+k x)}}\right )\right )-2 \text {arctanh}\left (\frac {x}{\sqrt [6]{b} \sqrt [3]{(-1+x) x (-1+k x)}}\right )-\text {arctanh}\left (\frac {x^2+\sqrt [3]{b} ((-1+x) x (-1+k x))^{2/3}}{\sqrt [6]{b} x \sqrt [3]{(-1+x) x (-1+k x)}}\right )}{2 b^{5/6}} \]
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Time = 1.37 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.16
method | result | size |
pseudoelliptic | \(-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x \left (\frac {1}{b}\right )^{\frac {1}{6}}-2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 x \left (\frac {1}{b}\right )^{\frac {1}{6}}}\right )-\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (x \left (\frac {1}{b}\right )^{\frac {1}{6}}+2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 x \left (\frac {1}{b}\right )^{\frac {1}{6}}}\right )+\ln \left (\frac {x \left (\frac {1}{b}\right )^{\frac {1}{6}}+\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {x \left (\frac {1}{b}\right )^{\frac {1}{6}}-\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right )+\frac {\ln \left (\frac {\left (\frac {1}{b}\right )^{\frac {1}{6}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x +\left (\frac {1}{b}\right )^{\frac {1}{3}} x^{2}+\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2}-\frac {\ln \left (\frac {\left (\frac {1}{b}\right )^{\frac {1}{6}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x -\left (\frac {1}{b}\right )^{\frac {1}{3}} x^{2}-\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2}}{2 \left (\frac {1}{b}\right )^{\frac {1}{6}} b}\) | \(246\) |
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Timed out. \[ \int \frac {(-1+x) (-1+k x) (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (b-2 (b+b k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(-1+x) (-1+k x) (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (b-2 (b+b k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {(-1+x) (-1+k x) (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (b-2 (b+b k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx=\int { -\frac {{\left ({\left (k + 1\right )} x - 2\right )} {\left (k x - 1\right )} {\left (x - 1\right )}}{{\left (2 \, b {\left (k + 1\right )} k x^{3} - {\left (b k^{2} - 1\right )} x^{4} - {\left (b k^{2} + 4 \, b k + b\right )} x^{2} + 2 \, {\left (b k + b\right )} x - b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]
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Time = 0.46 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.35 \[ \int \frac {(-1+x) (-1+k x) (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (b-2 (b+b k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx=-\left (-\frac {1}{b}\right )^{\frac {5}{6}} \arctan \left (\frac {{\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-\frac {1}{b}\right )^{\frac {1}{6}}}\right ) + \frac {\sqrt {3} \left (-b^{5}\right )^{\frac {5}{6}} \log \left (\sqrt {3} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} \left (-\frac {1}{b}\right )^{\frac {1}{6}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-\frac {1}{b}\right )^{\frac {1}{3}}\right )}{4 \, b^{5}} - \frac {\sqrt {3} \left (-b^{5}\right )^{\frac {5}{6}} \log \left (-\sqrt {3} {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} \left (-\frac {1}{b}\right )^{\frac {1}{6}} + {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \left (-\frac {1}{b}\right )^{\frac {1}{3}}\right )}{4 \, b^{5}} - \frac {\left (-b^{5}\right )^{\frac {5}{6}} \arctan \left (\frac {\sqrt {3} \left (-\frac {1}{b}\right )^{\frac {1}{6}} + 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-\frac {1}{b}\right )^{\frac {1}{6}}}\right )}{2 \, b^{5}} - \frac {\left (-b^{5}\right )^{\frac {5}{6}} \arctan \left (-\frac {\sqrt {3} \left (-\frac {1}{b}\right )^{\frac {1}{6}} - 2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{\left (-\frac {1}{b}\right )^{\frac {1}{6}}}\right )}{2 \, b^{5}} \]
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Timed out. \[ \int \frac {(-1+x) (-1+k x) (-2+(1+k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (b-2 (b+b k) x+\left (b+4 b k+b k^2\right ) x^2-2 b k (1+k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx=\int \frac {\left (x\,\left (k+1\right )-2\right )\,\left (k\,x-1\right )\,\left (x-1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (b+x^4\,\left (b\,k^2-1\right )+x^2\,\left (b\,k^2+4\,b\,k+b\right )-2\,x\,\left (b+b\,k\right )-2\,b\,k\,x^3\,\left (k+1\right )\right )} \,d x \]
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