Integrand size = 52, antiderivative size = 217 \[ \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b) x+(b+k) x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{2 x-2 k x^2+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{b^{2/3}}+\frac {\log \left (-x+k x^2+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{b^{2/3}}-\frac {\log \left (x^2-2 k x^3+k^2 x^4+\left (\sqrt [3]{b} x-\sqrt [3]{b} k x^2\right ) \left (x+(-1-k) x^2+k x^3\right )^{2/3}+b^{2/3} \left (x+(-1-k) x^2+k x^3\right )^{4/3}\right )}{2 b^{2/3}} \]
[Out]
\[ \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b) x+(b+k) x^2\right )} \, dx=\int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b) x+(b+k) x^2\right )} \, dx \]
[In]
[Out]
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 k x+(1-2 k) x^2}{(1-x)^{2/3} x^{2/3} (1-k x)^{2/3} \left (b-(1+2 b) x+(b+k) 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 {\sqrt [3]{1-x} (-1+(-1+2 k) x)}{x^{2/3} (1-k x)^{2/3} \left (b-(1+2 b) x+(b+k) 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 (-1+2 k-\sqrt {1+4 b-4 b k}\right ) \sqrt [3]{1-x}}{x^{2/3} (1-k x)^{2/3} \left (-1-2 b-\sqrt {1+4 b-4 b k}+2 (b+k) x\right )}+\frac {\left (-1+2 k+\sqrt {1+4 b-4 b k}\right ) \sqrt [3]{1-x}}{x^{2/3} (1-k x)^{2/3} \left (-1-2 b+\sqrt {1+4 b-4 b k}+2 (b+k) x\right )}\right ) \, dx}{((1-x) x (1-k x))^{2/3}} \\ & = \frac {\left (\left (-1+2 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-x}}{x^{2/3} (1-k x)^{2/3} \left (-1-2 b-\sqrt {1+4 b-4 b k}+2 (b+k) x\right )} \, dx}{((1-x) x (1-k x))^{2/3}}+\frac {\left (\left (-1+2 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-x}}{x^{2/3} (1-k x)^{2/3} \left (-1-2 b+\sqrt {1+4 b-4 b k}+2 (b+k) x\right )} \, dx}{((1-x) x (1-k x))^{2/3}} \\ \end{align*}
\[ \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b) x+(b+k) x^2\right )} \, dx=\int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b) x+(b+k) x^2\right )} \, dx \]
[In]
[Out]
\[\int \frac {-1+2 k x +\left (1-2 k \right ) x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (b -\left (1+2 b \right ) x +\left (b +k \right ) x^{2}\right )}d x\]
[In]
[Out]
Timed out. \[ \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b) x+(b+k) x^2\right )} \, dx=\text {Timed out} \]
[In]
[Out]
Timed out. \[ \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b) x+(b+k) x^2\right )} \, dx=\text {Timed out} \]
[In]
[Out]
\[ \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b) x+(b+k) x^2\right )} \, dx=\int { -\frac {{\left (2 \, k - 1\right )} x^{2} - 2 \, k x + 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (b + k\right )} x^{2} - {\left (2 \, b + 1\right )} x + b\right )}} \,d x } \]
[In]
[Out]
\[ \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b) x+(b+k) x^2\right )} \, dx=\int { -\frac {{\left (2 \, k - 1\right )} x^{2} - 2 \, k x + 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (b + k\right )} x^{2} - {\left (2 \, b + 1\right )} x + b\right )}} \,d x } \]
[In]
[Out]
Timed out. \[ \int \frac {-1+2 k x+(1-2 k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-(1+2 b) x+(b+k) x^2\right )} \, dx=\int -\frac {\left (2\,k-1\right )\,x^2-2\,k\,x+1}{\left (\left (b+k\right )\,x^2+\left (-2\,b-1\right )\,x+b\right )\,{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}} \,d x \]
[In]
[Out]