Integrand size = 47, antiderivative size = 291 \[ \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b-(1+a d) x+d x^2\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{d} \sqrt [3]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}}{-2 b+2 x+\sqrt [3]{d} \sqrt [3]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}}\right )}{d^{2/3}}+\frac {\log \left (b-x+\sqrt [3]{d} \sqrt [3]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}\right )}{d^{2/3}}-\frac {\log \left (b^2-2 b x+x^2+\left (-b \sqrt [3]{d}+\sqrt [3]{d} x\right ) \sqrt [3]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}+d^{2/3} \left (-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4\right )^{2/3}\right )}{2 d^{2/3}} \]
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\[ \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b-(1+a d) x+d x^2\right )} \, dx=\int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b-(1+a d) x+d x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {a b-2 b x+x^2}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (b-(1+a d) x+d x^2\right )} \, dx}{\sqrt [3]{x (-a+x) (-b+x)^2}} \\ & = \frac {\left (\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \left (\frac {1}{d \sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}}-\frac {b-a b d-(1+a d-2 b d) x}{d \sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (b+(-1-a d) x+d x^2\right )}\right ) \, dx}{\sqrt [3]{x (-a+x) (-b+x)^2}} \\ & = \frac {\left (\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}} \, dx}{d \sqrt [3]{x (-a+x) (-b+x)^2}}-\frac {\left (\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {b-a b d-(1+a d-2 b d) x}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (b+(-1-a d) x+d x^2\right )} \, dx}{d \sqrt [3]{x (-a+x) (-b+x)^2}} \\ & = -\frac {\left (\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \left (\frac {-1-a d+2 b d-\sqrt {1+2 a d-4 b d+a^2 d^2}}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (-1-a d-\sqrt {1+2 a d-4 b d+a^2 d^2}+2 d x\right )}+\frac {-1-a d+2 b d+\sqrt {1+2 a d-4 b d+a^2 d^2}}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (-1-a d+\sqrt {1+2 a d-4 b d+a^2 d^2}+2 d x\right )}\right ) \, dx}{d \sqrt [3]{x (-a+x) (-b+x)^2}}+\frac {\left (\sqrt [3]{x} (-b+x)^{2/3} \sqrt [3]{1-\frac {x}{a}}\right ) \int \frac {1}{\sqrt [3]{x} (-b+x)^{2/3} \sqrt [3]{1-\frac {x}{a}}} \, dx}{d \sqrt [3]{x (-a+x) (-b+x)^2}} \\ & = -\frac {\left (\left (-1-a d+2 b d-\sqrt {1+2 a d-4 b d+a^2 d^2}\right ) \sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (-1-a d-\sqrt {1+2 a d-4 b d+a^2 d^2}+2 d x\right )} \, dx}{d \sqrt [3]{x (-a+x) (-b+x)^2}}-\frac {\left (\left (-1-a d+2 b d+\sqrt {1+2 a d-4 b d+a^2 d^2}\right ) \sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (-1-a d+\sqrt {1+2 a d-4 b d+a^2 d^2}+2 d x\right )} \, dx}{d \sqrt [3]{x (-a+x) (-b+x)^2}}+\frac {\left (\sqrt [3]{x} \sqrt [3]{1-\frac {x}{a}} \left (1-\frac {x}{b}\right )^{2/3}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{1-\frac {x}{a}} \left (1-\frac {x}{b}\right )^{2/3}} \, dx}{d \sqrt [3]{x (-a+x) (-b+x)^2}} \\ & = \frac {3 x \sqrt [3]{1-\frac {x}{a}} \left (1-\frac {x}{b}\right )^{2/3} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{3},\frac {2}{3},\frac {5}{3},\frac {x}{a},\frac {x}{b}\right )}{2 d \sqrt [3]{-\left ((a-x) (b-x)^2 x\right )}}-\frac {\left (\left (-1-a d+2 b d-\sqrt {1+2 a d-4 b d+a^2 d^2}\right ) \sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (-1-a d-\sqrt {1+2 a d-4 b d+a^2 d^2}+2 d x\right )} \, dx}{d \sqrt [3]{x (-a+x) (-b+x)^2}}-\frac {\left (\left (-1-a d+2 b d+\sqrt {1+2 a d-4 b d+a^2 d^2}\right ) \sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (-1-a d+\sqrt {1+2 a d-4 b d+a^2 d^2}+2 d x\right )} \, dx}{d \sqrt [3]{x (-a+x) (-b+x)^2}} \\ \end{align*}
Time = 15.53 (sec) , antiderivative size = 202, normalized size of antiderivative = 0.69 \[ \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b-(1+a d) x+d x^2\right )} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{d} \sqrt [3]{(b-x)^2 x (-a+x)}}{-2 b+2 x+\sqrt [3]{d} \sqrt [3]{x (-a+x) (-b+x)^2}}\right )+2 \log \left (-b+x-\sqrt [3]{d} \sqrt [3]{x (-a+x) (-b+x)^2}\right )-\log \left (b^2-2 b x+x^2-b \sqrt [3]{d} \sqrt [3]{x (-a+x) (-b+x)^2}+\sqrt [3]{d} x \sqrt [3]{x (-a+x) (-b+x)^2}+d^{2/3} \left (x (-a+x) (-b+x)^2\right )^{2/3}\right )}{2 d^{2/3}} \]
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\[\int \frac {a b -2 b x +x^{2}}{\left (x \left (-a +x \right ) \left (-b +x \right )^{2}\right )^{\frac {1}{3}} \left (b -\left (a d +1\right ) x +d \,x^{2}\right )}d x\]
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Timed out. \[ \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b-(1+a d) x+d x^2\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b-(1+a d) x+d x^2\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b-(1+a d) x+d x^2\right )} \, dx=\int { \frac {a b - 2 \, b x + x^{2}}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2} x\right )^{\frac {1}{3}} {\left (d x^{2} - {\left (a d + 1\right )} x + b\right )}} \,d x } \]
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\[ \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b-(1+a d) x+d x^2\right )} \, dx=\int { \frac {a b - 2 \, b x + x^{2}}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2} x\right )^{\frac {1}{3}} {\left (d x^{2} - {\left (a d + 1\right )} x + b\right )}} \,d x } \]
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Timed out. \[ \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b-(1+a d) x+d x^2\right )} \, dx=\int \frac {x^2-2\,b\,x+a\,b}{{\left (-x\,\left (a-x\right )\,{\left (b-x\right )}^2\right )}^{1/3}\,\left (d\,x^2+\left (-a\,d-1\right )\,x+b\right )} \,d x \]
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