Integrand size = 85, antiderivative size = 119 \[ \int \frac {a b^2-2 (2 a-b) b x+(3 a-2 b) x^2}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (a^3+\left (-3 a^2+b^2 d\right ) x+(3 a-2 b d) x^2+(-1+d) x^3\right )} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}}{a-x}\right )}{d^{3/4}}-\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}}{a-x}\right )}{d^{3/4}} \]
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\[ \int \frac {a b^2-2 (2 a-b) b x+(3 a-2 b) x^2}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (a^3+\left (-3 a^2+b^2 d\right ) x+(3 a-2 b d) x^2+(-1+d) x^3\right )} \, dx=\int \frac {a b^2-2 (2 a-b) b x+(3 a-2 b) x^2}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (a^3+\left (-3 a^2+b^2 d\right ) x+(3 a-2 b d) x^2+(-1+d) x^3\right )} \, dx \]
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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^2-2 (2 a-b) b x+(3 a-2 b) x^2}{\sqrt [4]{x} \sqrt [4]{-a+x} \sqrt {-b+x} \left (a^3+\left (-3 a^2+b^2 d\right ) x+(3 a-2 b d) x^2+(-1+d) x^3\right )} \, dx}{\sqrt [4]{x (-a+x) (-b+x)^2}} \\ & = \frac {\left (\sqrt [4]{x} \sqrt [4]{-a+x} \sqrt {-b+x}\right ) \int \frac {\sqrt {-b+x} (-a b+(3 a-2 b) x)}{\sqrt [4]{x} \sqrt [4]{-a+x} \left (a^3+\left (-3 a^2+b^2 d\right ) x+(3 a-2 b d) x^2+(-1+d) x^3\right )} \, dx}{\sqrt [4]{x (-a+x) (-b+x)^2}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{-a+x} \sqrt {-b+x}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt {-b+x^4} \left (-a b+(3 a-2 b) x^4\right )}{\sqrt [4]{-a+x^4} \left (a^3+\left (-3 a^2+b^2 d\right ) x^4+(3 a-2 b d) x^8+(-1+d) x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x (-a+x) (-b+x)^2}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{-a+x} \sqrt {-b+x}\right ) \text {Subst}\left (\int \left (\frac {(3 a-2 b) x^6 \sqrt {-b+x^4}}{\sqrt [4]{-a+x^4} \left (a^3-3 a^2 \left (1-\frac {b^2 d}{3 a^2}\right ) x^4+3 a \left (1-\frac {2 b d}{3 a}\right ) x^8-(1-d) x^{12}\right )}+\frac {a b x^2 \sqrt {-b+x^4}}{\sqrt [4]{-a+x^4} \left (-a^3+3 a^2 \left (1-\frac {b^2 d}{3 a^2}\right ) x^4-3 a \left (1-\frac {2 b d}{3 a}\right ) x^8+(1-d) x^{12}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x (-a+x) (-b+x)^2}} \\ & = \frac {\left (4 (3 a-2 b) \sqrt [4]{x} \sqrt [4]{-a+x} \sqrt {-b+x}\right ) \text {Subst}\left (\int \frac {x^6 \sqrt {-b+x^4}}{\sqrt [4]{-a+x^4} \left (a^3-3 a^2 \left (1-\frac {b^2 d}{3 a^2}\right ) x^4+3 a \left (1-\frac {2 b d}{3 a}\right ) x^8-(1-d) x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x (-a+x) (-b+x)^2}}+\frac {\left (4 a b \sqrt [4]{x} \sqrt [4]{-a+x} \sqrt {-b+x}\right ) \text {Subst}\left (\int \frac {x^2 \sqrt {-b+x^4}}{\sqrt [4]{-a+x^4} \left (-a^3+3 a^2 \left (1-\frac {b^2 d}{3 a^2}\right ) x^4-3 a \left (1-\frac {2 b d}{3 a}\right ) x^8+(1-d) x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x (-a+x) (-b+x)^2}} \\ \end{align*}
Time = 10.76 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.61 \[ \int \frac {a b^2-2 (2 a-b) b x+(3 a-2 b) x^2}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (a^3+\left (-3 a^2+b^2 d\right ) x+(3 a-2 b d) x^2+(-1+d) x^3\right )} \, dx=\frac {2 \left (\arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{x (-a+x) (-b+x)^2}}{a-x}\right )+\text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{(b-x)^2 x (-a+x)}}{-a+x}\right )\right )}{d^{3/4}} \]
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\[\int \frac {a \,b^{2}-2 \left (2 a -b \right ) b x +\left (3 a -2 b \right ) x^{2}}{\left (x \left (-a +x \right ) \left (-b +x \right )^{2}\right )^{\frac {1}{4}} \left (a^{3}+\left (b^{2} d -3 a^{2}\right ) x +\left (-2 b d +3 a \right ) x^{2}+\left (-1+d \right ) x^{3}\right )}d x\]
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Timed out. \[ \int \frac {a b^2-2 (2 a-b) b x+(3 a-2 b) x^2}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (a^3+\left (-3 a^2+b^2 d\right ) x+(3 a-2 b d) x^2+(-1+d) x^3\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {a b^2-2 (2 a-b) b x+(3 a-2 b) x^2}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (a^3+\left (-3 a^2+b^2 d\right ) x+(3 a-2 b d) x^2+(-1+d) x^3\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {a b^2-2 (2 a-b) b x+(3 a-2 b) x^2}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (a^3+\left (-3 a^2+b^2 d\right ) x+(3 a-2 b d) x^2+(-1+d) x^3\right )} \, dx=\int { \frac {a b^{2} - 2 \, {\left (2 \, a - b\right )} b x + {\left (3 \, a - 2 \, b\right )} x^{2}}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2} x\right )^{\frac {1}{4}} {\left ({\left (d - 1\right )} x^{3} + a^{3} - {\left (2 \, b d - 3 \, a\right )} x^{2} + {\left (b^{2} d - 3 \, a^{2}\right )} x\right )}} \,d x } \]
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\[ \int \frac {a b^2-2 (2 a-b) b x+(3 a-2 b) x^2}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (a^3+\left (-3 a^2+b^2 d\right ) x+(3 a-2 b d) x^2+(-1+d) x^3\right )} \, dx=\int { \frac {a b^{2} - 2 \, {\left (2 \, a - b\right )} b x + {\left (3 \, a - 2 \, b\right )} x^{2}}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2} x\right )^{\frac {1}{4}} {\left ({\left (d - 1\right )} x^{3} + a^{3} - {\left (2 \, b d - 3 \, a\right )} x^{2} + {\left (b^{2} d - 3 \, a^{2}\right )} x\right )}} \,d x } \]
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Timed out. \[ \int \frac {a b^2-2 (2 a-b) b x+(3 a-2 b) x^2}{\sqrt [4]{x (-a+x) (-b+x)^2} \left (a^3+\left (-3 a^2+b^2 d\right ) x+(3 a-2 b d) x^2+(-1+d) x^3\right )} \, dx=\int \frac {x^2\,\left (3\,a-2\,b\right )+a\,b^2-2\,b\,x\,\left (2\,a-b\right )}{{\left (-x\,\left (a-x\right )\,{\left (b-x\right )}^2\right )}^{1/4}\,\left (x^2\,\left (3\,a-2\,b\,d\right )+a^3+x\,\left (b^2\,d-3\,a^2\right )+x^3\,\left (d-1\right )\right )} \,d x \]
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