Integrand size = 58, antiderivative size = 119 \[ \int \frac {-a b+(-a+2 b) x}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (-b^2+(2 b-a d) x+(-1+d) x^2\right )} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}}{b-x}\right )}{d^{3/4}}-\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}}{b-x}\right )}{d^{3/4}} \]
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\[ \int \frac {-a b+(-a+2 b) x}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (-b^2+(2 b-a d) x+(-1+d) x^2\right )} \, dx=\int \frac {-a b+(-a+2 b) x}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (-b^2+(2 b-a d) x+(-1+d) x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [4]{x} \sqrt [4]{-a+x} \sqrt {-b+x}\right ) \int \frac {-a b+(-a+2 b) x}{\sqrt [4]{x} \sqrt [4]{-a+x} \sqrt {-b+x} \left (-b^2+(2 b-a d) x+(-1+d) x^2\right )} \, dx}{\sqrt [4]{x (-a+x) (-b+x)^2}} \\ & = \frac {\left (\sqrt [4]{x} \sqrt [4]{-a+x} \sqrt {-b+x}\right ) \int \left (\frac {-a+2 b-\frac {\sqrt {-4 a b+4 b^2+a^2 d}}{\sqrt {d}}}{\sqrt [4]{x} \sqrt [4]{-a+x} \sqrt {-b+x} \left (2 b-a d-\sqrt {d} \sqrt {-4 a b+4 b^2+a^2 d}+2 (-1+d) x\right )}+\frac {-a+2 b+\frac {\sqrt {-4 a b+4 b^2+a^2 d}}{\sqrt {d}}}{\sqrt [4]{x} \sqrt [4]{-a+x} \sqrt {-b+x} \left (2 b-a d+\sqrt {d} \sqrt {-4 a b+4 b^2+a^2 d}+2 (-1+d) x\right )}\right ) \, dx}{\sqrt [4]{x (-a+x) (-b+x)^2}} \\ & = \frac {\left (\left (-a+2 b-\frac {\sqrt {-4 a b+4 b^2+a^2 d}}{\sqrt {d}}\right ) \sqrt [4]{x} \sqrt [4]{-a+x} \sqrt {-b+x}\right ) \int \frac {1}{\sqrt [4]{x} \sqrt [4]{-a+x} \sqrt {-b+x} \left (2 b-a d-\sqrt {d} \sqrt {-4 a b+4 b^2+a^2 d}+2 (-1+d) x\right )} \, dx}{\sqrt [4]{x (-a+x) (-b+x)^2}}+\frac {\left (\left (-a+2 b+\frac {\sqrt {-4 a b+4 b^2+a^2 d}}{\sqrt {d}}\right ) \sqrt [4]{x} \sqrt [4]{-a+x} \sqrt {-b+x}\right ) \int \frac {1}{\sqrt [4]{x} \sqrt [4]{-a+x} \sqrt {-b+x} \left (2 b-a d+\sqrt {d} \sqrt {-4 a b+4 b^2+a^2 d}+2 (-1+d) x\right )} \, dx}{\sqrt [4]{x (-a+x) (-b+x)^2}} \\ \end{align*}
Time = 15.36 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.61 \[ \int \frac {-a b+(-a+2 b) x}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (-b^2+(2 b-a d) x+(-1+d) x^2\right )} \, dx=\frac {2 \left (\arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{x (-a+x) (-b+x)^2}}{b-x}\right )+\text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{(b-x)^2 x (-a+x)}}{-b+x}\right )\right )}{d^{3/4}} \]
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\[\int \frac {-a b +\left (-a +2 b \right ) x}{\left (x \left (-a +x \right ) \left (-b +x \right )^{2}\right )^{\frac {1}{4}} \left (-b^{2}+\left (-a d +2 b \right ) x +\left (-1+d \right ) x^{2}\right )}d x\]
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Timed out. \[ \int \frac {-a b+(-a+2 b) x}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (-b^2+(2 b-a d) x+(-1+d) x^2\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {-a b+(-a+2 b) x}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (-b^2+(2 b-a d) x+(-1+d) x^2\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {-a b+(-a+2 b) x}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (-b^2+(2 b-a d) x+(-1+d) x^2\right )} \, dx=\int { -\frac {a b + {\left (a - 2 \, b\right )} x}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2} x\right )^{\frac {1}{4}} {\left ({\left (d - 1\right )} x^{2} - b^{2} - {\left (a d - 2 \, b\right )} x\right )}} \,d x } \]
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\[ \int \frac {-a b+(-a+2 b) x}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (-b^2+(2 b-a d) x+(-1+d) x^2\right )} \, dx=\int { -\frac {a b + {\left (a - 2 \, b\right )} x}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2} x\right )^{\frac {1}{4}} {\left ({\left (d - 1\right )} x^{2} - b^{2} - {\left (a d - 2 \, b\right )} x\right )}} \,d x } \]
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Timed out. \[ \int \frac {-a b+(-a+2 b) x}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (-b^2+(2 b-a d) x+(-1+d) x^2\right )} \, dx=-\int \frac {a\,b+x\,\left (a-2\,b\right )}{{\left (-x\,\left (a-x\right )\,{\left (b-x\right )}^2\right )}^{1/4}\,\left (x\,\left (2\,b-a\,d\right )-b^2+x^2\,\left (d-1\right )\right )} \,d x \]
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