Integrand size = 58, antiderivative size = 208 \[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{2 x-2 x^2+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{b^{2/3}}+\frac {\log \left (-x+x^2+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{b^{2/3}}-\frac {\log \left (x^2-2 x^3+x^4+\left (\sqrt [3]{b} x-\sqrt [3]{b} x^2\right ) \left (x+(-1-k) x^2+k x^3\right )^{2/3}+b^{2/3} \left (x+(-1-k) x^2+k x^3\right )^{4/3}\right )}{2 b^{2/3}} \]
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\[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{(1-x)^{2/3} x^{2/3} (1-k x)^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx}{((1-x) x (1-k x))^{2/3}} \\ & = \frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {(-1+(2-k) x) \sqrt [3]{1-k x}}{(1-x)^{2/3} x^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx}{((1-x) x (1-k x))^{2/3}} \\ & = \frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \left (\frac {\left (2-k-k \sqrt {1-4 b+4 b k}\right ) \sqrt [3]{1-k x}}{(1-x)^{2/3} x^{2/3} \left (-1-2 b k-\sqrt {1-4 b+4 b k}+2 \left (1+b k^2\right ) x\right )}+\frac {\left (2-k+k \sqrt {1-4 b+4 b k}\right ) \sqrt [3]{1-k x}}{(1-x)^{2/3} x^{2/3} \left (-1-2 b k+\sqrt {1-4 b+4 b k}+2 \left (1+b k^2\right ) x\right )}\right ) \, dx}{((1-x) x (1-k x))^{2/3}} \\ & = \frac {\left (\left (2-\left (1-\sqrt {1-4 b (1-k)}\right ) k\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {\sqrt [3]{1-k x}}{(1-x)^{2/3} x^{2/3} \left (-1-2 b k+\sqrt {1-4 b+4 b k}+2 \left (1+b k^2\right ) x\right )} \, dx}{((1-x) x (1-k x))^{2/3}}+\frac {\left (\left (2-k-k \sqrt {1-4 b+4 b k}\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {\sqrt [3]{1-k x}}{(1-x)^{2/3} x^{2/3} \left (-1-2 b k-\sqrt {1-4 b+4 b k}+2 \left (1+b k^2\right ) x\right )} \, dx}{((1-x) x (1-k x))^{2/3}} \\ \end{align*}
Time = 15.66 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.79 \[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} ((-1+x) x (-1+k x))^{2/3}}{2 x-2 x^2+\sqrt [3]{b} ((-1+x) x (-1+k x))^{2/3}}\right )+2 \log \left (-x+x^2+\sqrt [3]{b} ((-1+x) x (-1+k x))^{2/3}\right )-\log \left (x^2-2 x^3+x^4-\sqrt [3]{b} (-1+x) x ((-1+x) x (-1+k x))^{2/3}+b^{2/3} ((-1+x) x (-1+k x))^{4/3}\right )}{2 b^{2/3}} \]
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\[\int \frac {-1+2 x +\left (k^{2}-2 k \right ) x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (b -\left (2 b k +1\right ) x +\left (b \,k^{2}+1\right ) x^{2}\right )}d x\]
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Timed out. \[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\int { \frac {{\left (k^{2} - 2 \, k\right )} x^{2} + 2 \, x - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (b k^{2} + 1\right )} x^{2} - {\left (2 \, b k + 1\right )} x + b\right )}} \,d x } \]
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\[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\int { \frac {{\left (k^{2} - 2 \, k\right )} x^{2} + 2 \, x - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (b k^{2} + 1\right )} x^{2} - {\left (2 \, b k + 1\right )} x + b\right )}} \,d x } \]
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Timed out. \[ \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b k) x+\left (1+b k^2\right ) x^2\right )} \, dx=\int -\frac {\left (2\,k-k^2\right )\,x^2-2\,x+1}{\left (\left (b\,k^2+1\right )\,x^2+\left (-2\,b\,k-1\right )\,x+b\right )\,{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}} \,d x \]
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