Integrand size = 71, antiderivative size = 125 \[ \int \frac {-2 a b x^2+(a+b) x^3}{(-a+x) (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (-a b d+(a+b) d x+(1-d) x^2\right )} \, dx=-\frac {4 \left (a b x^2+(-a-b) x^3+x^4\right )^{3/4}}{x (-a+x) (-b+x)}+2 \sqrt [4]{d} \arctan \left (\frac {x}{\sqrt [4]{d} \sqrt [4]{a b x^2+(-a-b) x^3+x^4}}\right )+2 \sqrt [4]{d} \text {arctanh}\left (\frac {x}{\sqrt [4]{d} \sqrt [4]{a b x^2+(-a-b) x^3+x^4}}\right ) \]
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\[ \int \frac {-2 a b x^2+(a+b) x^3}{(-a+x) (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (-a b d+(a+b) d x+(1-d) x^2\right )} \, dx=\int \frac {-2 a b x^2+(a+b) x^3}{(-a+x) (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (-a b d+(a+b) d x+(1-d) x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {x^2 (-2 a b+(a+b) x)}{(-a+x) (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (-a b d+(a+b) d x+(1-d) x^2\right )} \, dx \\ & = \frac {\left (\sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \int \frac {x^{3/2} (-2 a b+(a+b) x)}{(-a+x)^{5/4} (-b+x)^{5/4} \left (-a b d+(a+b) d x+(1-d) x^2\right )} \, dx}{\sqrt [4]{x^2 (-a+x) (-b+x)}} \\ & = \frac {\left (\sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \int \left (\frac {\left (a+b-\frac {\sqrt {4 a b+a^2 d-2 a b d+b^2 d}}{\sqrt {d}}\right ) x^{3/2}}{(-a+x)^{5/4} (-b+x)^{5/4} \left ((a+b) d-\sqrt {d} \sqrt {4 a b+a^2 d-2 a b d+b^2 d}+2 (1-d) x\right )}+\frac {\left (a+b+\frac {\sqrt {4 a b+a^2 d-2 a b d+b^2 d}}{\sqrt {d}}\right ) x^{3/2}}{(-a+x)^{5/4} (-b+x)^{5/4} \left ((a+b) d+\sqrt {d} \sqrt {4 a b+a^2 d-2 a b d+b^2 d}+2 (1-d) x\right )}\right ) \, dx}{\sqrt [4]{x^2 (-a+x) (-b+x)}} \\ & = \frac {\left (\left (a+b-\frac {\sqrt {2 a b (2-d)+a^2 d+b^2 d}}{\sqrt {d}}\right ) \sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \int \frac {x^{3/2}}{(-a+x)^{5/4} (-b+x)^{5/4} \left ((a+b) d-\sqrt {d} \sqrt {4 a b+a^2 d-2 a b d+b^2 d}+2 (1-d) x\right )} \, dx}{\sqrt [4]{x^2 (-a+x) (-b+x)}}+\frac {\left (\left (a+b+\frac {\sqrt {2 a b (2-d)+a^2 d+b^2 d}}{\sqrt {d}}\right ) \sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \int \frac {x^{3/2}}{(-a+x)^{5/4} (-b+x)^{5/4} \left ((a+b) d+\sqrt {d} \sqrt {4 a b+a^2 d-2 a b d+b^2 d}+2 (1-d) x\right )} \, dx}{\sqrt [4]{x^2 (-a+x) (-b+x)}} \\ \end{align*}
Time = 67.10 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.37 \[ \int \frac {-2 a b x^2+(a+b) x^3}{(-a+x) (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (-a b d+(a+b) d x+(1-d) x^2\right )} \, dx=-\frac {2 x \left (2 \sqrt {\frac {x}{-a+x}}+\sqrt [4]{d} \sqrt [4]{\frac {b-x}{a-x}} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{\frac {b-x}{a-x}}}{\sqrt {\frac {x}{-a+x}}}\right )-\sqrt [4]{d} \sqrt [4]{\frac {b-x}{a-x}} \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{\frac {b-x}{a-x}}}{\sqrt {\frac {x}{-a+x}}}\right )\right )}{\sqrt {\frac {x}{-a+x}} \sqrt [4]{x^2 (-a+x) (-b+x)}} \]
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Time = 0.94 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.08
method | result | size |
pseudoelliptic | \(-\frac {2 \left (2 x \left (\frac {1}{d}\right )^{\frac {1}{4}}+\frac {\left (2 \arctan \left (\frac {\left (x^{2} \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{4}}}{x \left (\frac {1}{d}\right )^{\frac {1}{4}}}\right )-\ln \left (\frac {x \left (\frac {1}{d}\right )^{\frac {1}{4}}+\left (x^{2} \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{4}}}{-x \left (\frac {1}{d}\right )^{\frac {1}{4}}+\left (x^{2} \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{4}}}\right )\right ) \left (x^{2} \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{4}}}{2}\right )}{\left (x^{2} \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{4}} \left (\frac {1}{d}\right )^{\frac {1}{4}}}\) | \(135\) |
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Timed out. \[ \int \frac {-2 a b x^2+(a+b) x^3}{(-a+x) (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (-a b d+(a+b) d x+(1-d) x^2\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {-2 a b x^2+(a+b) x^3}{(-a+x) (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (-a b d+(a+b) d x+(1-d) x^2\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {-2 a b x^2+(a+b) x^3}{(-a+x) (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (-a b d+(a+b) d x+(1-d) x^2\right )} \, dx=\int { \frac {2 \, a b x^{2} - {\left (a + b\right )} x^{3}}{\left ({\left (a - x\right )} {\left (b - x\right )} x^{2}\right )^{\frac {1}{4}} {\left (a b d - {\left (a + b\right )} d x + {\left (d - 1\right )} x^{2}\right )} {\left (a - x\right )} {\left (b - x\right )}} \,d x } \]
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\[ \int \frac {-2 a b x^2+(a+b) x^3}{(-a+x) (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (-a b d+(a+b) d x+(1-d) x^2\right )} \, dx=\int { \frac {2 \, a b x^{2} - {\left (a + b\right )} x^{3}}{\left ({\left (a - x\right )} {\left (b - x\right )} x^{2}\right )^{\frac {1}{4}} {\left (a b d - {\left (a + b\right )} d x + {\left (d - 1\right )} x^{2}\right )} {\left (a - x\right )} {\left (b - x\right )}} \,d x } \]
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Timed out. \[ \int \frac {-2 a b x^2+(a+b) x^3}{(-a+x) (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (-a b d+(a+b) d x+(1-d) x^2\right )} \, dx=\int -\frac {x^3\,\left (a+b\right )-2\,a\,b\,x^2}{\left (a-x\right )\,\left (b-x\right )\,\left (\left (d-1\right )\,x^2-d\,\left (a+b\right )\,x+a\,b\,d\right )\,{\left (x^2\,\left (a-x\right )\,\left (b-x\right )\right )}^{1/4}} \,d x \]
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