Integrand size = 65, antiderivative size = 89 \[ \int \frac {-a b+2 (a-b) x+x^2}{\sqrt [4]{x (-a+x) (-b+x)} \left (-a^3+\left (3 a^2+b d\right ) x-(3 a+d) x^2+x^3\right )} \, dx=-\frac {2 \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a b x+(-a-b) x^2+x^3}}{a-x}\right )}{d^{3/4}}+\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a b x+(-a-b) x^2+x^3}}{a-x}\right )}{d^{3/4}} \]
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\[ \int \frac {-a b+2 (a-b) x+x^2}{\sqrt [4]{x (-a+x) (-b+x)} \left (-a^3+\left (3 a^2+b d\right ) x-(3 a+d) x^2+x^3\right )} \, dx=\int \frac {-a b+2 (a-b) x+x^2}{\sqrt [4]{x (-a+x) (-b+x)} \left (-a^3+\left (3 a^2+b d\right ) x-(3 a+d) x^2+x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [4]{x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \int \frac {-a b+2 (a-b) x+x^2}{\sqrt [4]{x} \sqrt [4]{-a+x} \sqrt [4]{-b+x} \left (-a^3+\left (3 a^2+b d\right ) x-(3 a+d) x^2+x^3\right )} \, dx}{\sqrt [4]{x (-a+x) (-b+x)}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \frac {x^2 \left (-a b+2 (a-b) x^4+x^8\right )}{\sqrt [4]{-a+x^4} \sqrt [4]{-b+x^4} \left (-a^3+\left (3 a^2+b d\right ) x^4-(3 a+d) x^8+x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x (-a+x) (-b+x)}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \left (\frac {a b x^2}{\sqrt [4]{-a+x^4} \sqrt [4]{-b+x^4} \left (a^3-3 a^2 \left (1+\frac {b d}{3 a^2}\right ) x^4+3 a \left (1+\frac {d}{3 a}\right ) x^8-x^{12}\right )}+\frac {2 (-a+b) x^6}{\sqrt [4]{-a+x^4} \sqrt [4]{-b+x^4} \left (a^3-3 a^2 \left (1+\frac {b d}{3 a^2}\right ) x^4+3 a \left (1+\frac {d}{3 a}\right ) x^8-x^{12}\right )}+\frac {x^{10}}{\sqrt [4]{-a+x^4} \sqrt [4]{-b+x^4} \left (-a^3+3 a^2 \left (1+\frac {b d}{3 a^2}\right ) x^4-3 a \left (1+\frac {d}{3 a}\right ) x^8+x^{12}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x (-a+x) (-b+x)}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \frac {x^{10}}{\sqrt [4]{-a+x^4} \sqrt [4]{-b+x^4} \left (-a^3+3 a^2 \left (1+\frac {b d}{3 a^2}\right ) x^4-3 a \left (1+\frac {d}{3 a}\right ) x^8+x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x (-a+x) (-b+x)}}-\frac {\left (8 (a-b) \sqrt [4]{x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \frac {x^6}{\sqrt [4]{-a+x^4} \sqrt [4]{-b+x^4} \left (a^3-3 a^2 \left (1+\frac {b d}{3 a^2}\right ) x^4+3 a \left (1+\frac {d}{3 a}\right ) x^8-x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x (-a+x) (-b+x)}}+\frac {\left (4 a b \sqrt [4]{x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt [4]{-a+x^4} \sqrt [4]{-b+x^4} \left (a^3-3 a^2 \left (1+\frac {b d}{3 a^2}\right ) x^4+3 a \left (1+\frac {d}{3 a}\right ) x^8-x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x (-a+x) (-b+x)}} \\ \end{align*}
Time = 11.72 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.75 \[ \int \frac {-a b+2 (a-b) x+x^2}{\sqrt [4]{x (-a+x) (-b+x)} \left (-a^3+\left (3 a^2+b d\right ) x-(3 a+d) x^2+x^3\right )} \, dx=\frac {2 \left (\arctan \left (\frac {\sqrt [4]{d} x (-b+x)}{(x (-a+x) (-b+x))^{3/4}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{x (-a+x) (-b+x)}}{a-x}\right )\right )}{d^{3/4}} \]
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\[\int \frac {-a b +2 \left (a -b \right ) x +x^{2}}{\left (x \left (-a +x \right ) \left (-b +x \right )\right )^{\frac {1}{4}} \left (-a^{3}+\left (3 a^{2}+d b \right ) x -\left (3 a +d \right ) x^{2}+x^{3}\right )}d x\]
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Timed out. \[ \int \frac {-a b+2 (a-b) x+x^2}{\sqrt [4]{x (-a+x) (-b+x)} \left (-a^3+\left (3 a^2+b d\right ) x-(3 a+d) x^2+x^3\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {-a b+2 (a-b) x+x^2}{\sqrt [4]{x (-a+x) (-b+x)} \left (-a^3+\left (3 a^2+b d\right ) x-(3 a+d) x^2+x^3\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {-a b+2 (a-b) x+x^2}{\sqrt [4]{x (-a+x) (-b+x)} \left (-a^3+\left (3 a^2+b d\right ) x-(3 a+d) x^2+x^3\right )} \, dx=\int { \frac {a b - 2 \, {\left (a - b\right )} x - x^{2}}{{\left (a^{3} + {\left (3 \, a + d\right )} x^{2} - x^{3} - {\left (3 \, a^{2} + b d\right )} x\right )} \left ({\left (a - x\right )} {\left (b - x\right )} x\right )^{\frac {1}{4}}} \,d x } \]
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\[ \int \frac {-a b+2 (a-b) x+x^2}{\sqrt [4]{x (-a+x) (-b+x)} \left (-a^3+\left (3 a^2+b d\right ) x-(3 a+d) x^2+x^3\right )} \, dx=\int { \frac {a b - 2 \, {\left (a - b\right )} x - x^{2}}{{\left (a^{3} + {\left (3 \, a + d\right )} x^{2} - x^{3} - {\left (3 \, a^{2} + b d\right )} x\right )} \left ({\left (a - x\right )} {\left (b - x\right )} x\right )^{\frac {1}{4}}} \,d x } \]
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Timed out. \[ \int \frac {-a b+2 (a-b) x+x^2}{\sqrt [4]{x (-a+x) (-b+x)} \left (-a^3+\left (3 a^2+b d\right ) x-(3 a+d) x^2+x^3\right )} \, dx=\int \frac {2\,x\,\left (a-b\right )-a\,b+x^2}{{\left (x\,\left (a-x\right )\,\left (b-x\right )\right )}^{1/4}\,\left (x\,\left (3\,a^2+b\,d\right )-x^2\,\left (3\,a+d\right )-a^3+x^3\right )} \,d x \]
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