Integrand size = 56, antiderivative size = 218 \[ \int \frac {-a b+(2 a-b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a^2+(2 a-b d) x+(-1+d) x^2\right )} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{d} \sqrt [3]{a b x+(-a-b) x^2+x^3}}{-2 a+2 x+\sqrt [3]{d} \sqrt [3]{a b x+(-a-b) x^2+x^3}}\right )}{d^{2/3}}-\frac {\log \left (a-x+\sqrt [3]{d} \sqrt [3]{a b x+(-a-b) x^2+x^3}\right )}{d^{2/3}}+\frac {\log \left (a^2-2 a x+x^2+\left (-a \sqrt [3]{d}+\sqrt [3]{d} x\right ) \sqrt [3]{a b x+(-a-b) x^2+x^3}+d^{2/3} \left (a b x+(-a-b) x^2+x^3\right )^{2/3}\right )}{2 d^{2/3}} \]
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\[ \int \frac {-a b+(2 a-b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a^2+(2 a-b d) x+(-1+d) x^2\right )} \, dx=\int \frac {-a b+(2 a-b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a^2+(2 a-b d) x+(-1+d) x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [3]{x} \sqrt [3]{-a+x} \sqrt [3]{-b+x}\right ) \int \frac {-a b+(2 a-b) x}{\sqrt [3]{x} \sqrt [3]{-a+x} \sqrt [3]{-b+x} \left (-a^2+(2 a-b d) x+(-1+d) x^2\right )} \, dx}{\sqrt [3]{x (-a+x) (-b+x)}} \\ & = \frac {\left (\sqrt [3]{x} \sqrt [3]{-a+x} \sqrt [3]{-b+x}\right ) \int \left (\frac {2 a-b-\frac {\sqrt {4 a^2-4 a b+b^2 d}}{\sqrt {d}}}{\sqrt [3]{x} \sqrt [3]{-a+x} \sqrt [3]{-b+x} \left (2 a-b d-\sqrt {d} \sqrt {4 a^2-4 a b+b^2 d}+2 (-1+d) x\right )}+\frac {2 a-b+\frac {\sqrt {4 a^2-4 a b+b^2 d}}{\sqrt {d}}}{\sqrt [3]{x} \sqrt [3]{-a+x} \sqrt [3]{-b+x} \left (2 a-b d+\sqrt {d} \sqrt {4 a^2-4 a b+b^2 d}+2 (-1+d) x\right )}\right ) \, dx}{\sqrt [3]{x (-a+x) (-b+x)}} \\ & = \frac {\left (\left (2 a-b-\frac {\sqrt {4 a^2-4 a b+b^2 d}}{\sqrt {d}}\right ) \sqrt [3]{x} \sqrt [3]{-a+x} \sqrt [3]{-b+x}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{-a+x} \sqrt [3]{-b+x} \left (2 a-b d-\sqrt {d} \sqrt {4 a^2-4 a b+b^2 d}+2 (-1+d) x\right )} \, dx}{\sqrt [3]{x (-a+x) (-b+x)}}+\frac {\left (\left (2 a-b+\frac {\sqrt {4 a^2-4 a b+b^2 d}}{\sqrt {d}}\right ) \sqrt [3]{x} \sqrt [3]{-a+x} \sqrt [3]{-b+x}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{-a+x} \sqrt [3]{-b+x} \left (2 a-b d+\sqrt {d} \sqrt {4 a^2-4 a b+b^2 d}+2 (-1+d) x\right )} \, dx}{\sqrt [3]{x (-a+x) (-b+x)}} \\ \end{align*}
\[ \int \frac {-a b+(2 a-b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a^2+(2 a-b d) x+(-1+d) x^2\right )} \, dx=\int \frac {-a b+(2 a-b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a^2+(2 a-b d) x+(-1+d) x^2\right )} \, dx \]
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\[\int \frac {-a b +\left (2 a -b \right ) x}{\left (x \left (-a +x \right ) \left (-b +x \right )\right )^{\frac {1}{3}} \left (-a^{2}+\left (-b d +2 a \right ) x +\left (-1+d \right ) x^{2}\right )}d x\]
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Timed out. \[ \int \frac {-a b+(2 a-b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a^2+(2 a-b d) x+(-1+d) x^2\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {-a b+(2 a-b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a^2+(2 a-b d) x+(-1+d) x^2\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {-a b+(2 a-b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a^2+(2 a-b d) x+(-1+d) x^2\right )} \, dx=\int { -\frac {a b - {\left (2 \, a - b\right )} x}{\left ({\left (a - x\right )} {\left (b - x\right )} x\right )^{\frac {1}{3}} {\left ({\left (d - 1\right )} x^{2} - a^{2} - {\left (b d - 2 \, a\right )} x\right )}} \,d x } \]
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\[ \int \frac {-a b+(2 a-b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a^2+(2 a-b d) x+(-1+d) x^2\right )} \, dx=\int { -\frac {a b - {\left (2 \, a - b\right )} x}{\left ({\left (a - x\right )} {\left (b - x\right )} x\right )^{\frac {1}{3}} {\left ({\left (d - 1\right )} x^{2} - a^{2} - {\left (b d - 2 \, a\right )} x\right )}} \,d x } \]
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Timed out. \[ \int \frac {-a b+(2 a-b) x}{\sqrt [3]{x (-a+x) (-b+x)} \left (-a^2+(2 a-b d) x+(-1+d) x^2\right )} \, dx=-\int \frac {a\,b-x\,\left (2\,a-b\right )}{{\left (x\,\left (a-x\right )\,\left (b-x\right )\right )}^{1/3}\,\left (x\,\left (2\,a-b\,d\right )-a^2+x^2\,\left (d-1\right )\right )} \,d x \]
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