Integrand size = 45, antiderivative size = 155 \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {\log \left (x-\sqrt [3]{b} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}-\frac {\log \left (x^2+\sqrt [3]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+b^{2/3} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 b^{2/3}} \]
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\[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx \]
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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 {-2+(1+k) x}{\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \left (b-b (1+k) x+(-1+b k) x^2\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 \left (\frac {1+k+\frac {\sqrt {4+b-2 b k+b k^2}}{\sqrt {b}}}{\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \left (-b (1+k)-\sqrt {b} \sqrt {4+b-2 b k+b k^2}+2 (-1+b k) x\right )}+\frac {1+k-\frac {\sqrt {4+b-2 b k+b k^2}}{\sqrt {b}}}{\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \left (-b (1+k)+\sqrt {b} \sqrt {4+b-2 b k+b k^2}+2 (-1+b k) x\right )}\right ) \, dx}{\sqrt [3]{(1-x) x (1-k x)}} \\ & = \frac {\left (\left (1-\frac {\sqrt {4+b (1-k)^2}}{\sqrt {b}}+k\right ) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \left (-b (1+k)+\sqrt {b} \sqrt {4+b-2 b k+b k^2}+2 (-1+b k) x\right )} \, dx}{\sqrt [3]{(1-x) x (1-k x)}}+\frac {\left (\left (1+\frac {\sqrt {4+b (1-k)^2}}{\sqrt {b}}+k\right ) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \left (-b (1+k)-\sqrt {b} \sqrt {4+b-2 b k+b k^2}+2 (-1+b k) x\right )} \, dx}{\sqrt [3]{(1-x) x (1-k x)}} \\ \end{align*}
Time = 15.27 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.79 \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=-\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} x}{x+2 \sqrt [3]{b} \sqrt [3]{(-1+x) x (-1+k x)}}\right )-2 \log \left (x-\sqrt [3]{b} \sqrt [3]{(-1+x) x (-1+k x)}\right )+\log \left (x^2+\sqrt [3]{b} x \sqrt [3]{(-1+x) x (-1+k x)}+b^{2/3} ((-1+x) x (-1+k x))^{2/3}\right )}{2 b^{2/3}} \]
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Time = 0.58 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.83
method | result | size |
pseudoelliptic | \(\frac {2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\left (\frac {1}{b}\right )^{\frac {1}{3}} x +2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 \left (\frac {1}{b}\right )^{\frac {1}{3}} x}\right )+2 \ln \left (\frac {-\left (\frac {1}{b}\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right )-\ln \left (\frac {\left (\frac {1}{b}\right )^{\frac {2}{3}} x^{2}+\left (\frac {1}{b}\right )^{\frac {1}{3}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2 \left (\frac {1}{b}\right )^{\frac {1}{3}} b}\) | \(129\) |
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Timed out. \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\int { -\frac {{\left (k + 1\right )} x - 2}{{\left (b {\left (k + 1\right )} x - {\left (b k - 1\right )} x^{2} - b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]
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Time = 0.35 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.83 \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\frac {\sqrt {3} {\left | b \right |}^{\frac {4}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} b^{\frac {1}{3}} {\left (2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} + \frac {1}{b^{\frac {1}{3}}}\right )}\right )}{b^{2}} - \frac {{\left | b \right |}^{\frac {4}{3}} \log \left ({\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \frac {{\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{b^{\frac {1}{3}}} + \frac {1}{b^{\frac {2}{3}}}\right )}{2 \, b^{2}} + \frac {\log \left ({\left | {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} - \frac {1}{b^{\frac {1}{3}}} \right |}\right )}{b^{\frac {2}{3}}} \]
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Timed out. \[ \int \frac {-2+(1+k) x}{\sqrt [3]{(1-x) x (1-k x)} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx=\int \frac {x\,\left (k+1\right )-2}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (\left (b\,k-1\right )\,x^2-b\,\left (k+1\right )\,x+b\right )} \,d x \]
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