Integrand size = 79, antiderivative size = 95 \[ \int \frac {\left (a^2-2 a x+x^2\right ) \left (-a b+2 (a-b) x+x^2\right )}{(x (-a+x) (-b+x))^{3/4} \left (-a^3 d+\left (b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{d} \left (a b x+(-a-b) x^2+x^3\right )^{3/4}}{x (-b+x)}\right )}{d^{3/4}}-\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{d} \left (a b x+(-a-b) x^2+x^3\right )^{3/4}}{x (-b+x)}\right )}{d^{3/4}} \]
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\[ \int \frac {\left (a^2-2 a x+x^2\right ) \left (-a b+2 (a-b) x+x^2\right )}{(x (-a+x) (-b+x))^{3/4} \left (-a^3 d+\left (b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx=\int \frac {\left (a^2-2 a x+x^2\right ) \left (-a b+2 (a-b) x+x^2\right )}{(x (-a+x) (-b+x))^{3/4} \left (-a^3 d+\left (b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {(-a+x)^2 \left (-a b+2 (a-b) x+x^2\right )}{(x (-a+x) (-b+x))^{3/4} \left (-a^3 d+\left (b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx \\ & = \frac {\left (x^{3/4} (-a+x)^{3/4} (-b+x)^{3/4}\right ) \int \frac {(-a+x)^{5/4} \left (-a b+2 (a-b) x+x^2\right )}{x^{3/4} (-b+x)^{3/4} \left (-a^3 d+\left (b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx}{(x (-a+x) (-b+x))^{3/4}} \\ & = \frac {\left (4 x^{3/4} (-a+x)^{3/4} (-b+x)^{3/4}\right ) \text {Subst}\left (\int \frac {\left (-a+x^4\right )^{5/4} \left (-a b+2 (a-b) x^4+x^8\right )}{\left (-b+x^4\right )^{3/4} \left (-a^3 d+\left (b+3 a^2 d\right ) x^4-(1+3 a d) x^8+d x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{(x (-a+x) (-b+x))^{3/4}} \\ & = \frac {\left (4 x^{3/4} (-a+x)^{3/4} (-b+x)^{3/4}\right ) \text {Subst}\left (\int \left (\frac {a b \left (-a+x^4\right )^{5/4}}{\left (-b+x^4\right )^{3/4} \left (a^3 d-b \left (1+\frac {3 a^2 d}{b}\right ) x^4+(1+3 a d) x^8-d x^{12}\right )}+\frac {2 (-a+b) x^4 \left (-a+x^4\right )^{5/4}}{\left (-b+x^4\right )^{3/4} \left (a^3 d-b \left (1+\frac {3 a^2 d}{b}\right ) x^4+(1+3 a d) x^8-d x^{12}\right )}+\frac {x^8 \left (-a+x^4\right )^{5/4}}{\left (-b+x^4\right )^{3/4} \left (-a^3 d+b \left (1+\frac {3 a^2 d}{b}\right ) x^4-(1+3 a d) x^8+d x^{12}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{(x (-a+x) (-b+x))^{3/4}} \\ & = \frac {\left (4 x^{3/4} (-a+x)^{3/4} (-b+x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^8 \left (-a+x^4\right )^{5/4}}{\left (-b+x^4\right )^{3/4} \left (-a^3 d+b \left (1+\frac {3 a^2 d}{b}\right ) x^4-(1+3 a d) x^8+d x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{(x (-a+x) (-b+x))^{3/4}}-\frac {\left (8 (a-b) x^{3/4} (-a+x)^{3/4} (-b+x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^4 \left (-a+x^4\right )^{5/4}}{\left (-b+x^4\right )^{3/4} \left (a^3 d-b \left (1+\frac {3 a^2 d}{b}\right ) x^4+(1+3 a d) x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{(x (-a+x) (-b+x))^{3/4}}+\frac {\left (4 a b x^{3/4} (-a+x)^{3/4} (-b+x)^{3/4}\right ) \text {Subst}\left (\int \frac {\left (-a+x^4\right )^{5/4}}{\left (-b+x^4\right )^{3/4} \left (a^3 d-b \left (1+\frac {3 a^2 d}{b}\right ) x^4+(1+3 a d) x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{(x (-a+x) (-b+x))^{3/4}} \\ \end{align*}
\[ \int \frac {\left (a^2-2 a x+x^2\right ) \left (-a b+2 (a-b) x+x^2\right )}{(x (-a+x) (-b+x))^{3/4} \left (-a^3 d+\left (b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx=\int \frac {\left (a^2-2 a x+x^2\right ) \left (-a b+2 (a-b) x+x^2\right )}{(x (-a+x) (-b+x))^{3/4} \left (-a^3 d+\left (b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx \]
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\[\int \frac {\left (a^{2}-2 a x +x^{2}\right ) \left (-a b +2 \left (a -b \right ) x +x^{2}\right )}{\left (x \left (-a +x \right ) \left (-b +x \right )\right )^{\frac {3}{4}} \left (-d \,a^{3}+\left (3 a^{2} d +b \right ) x -\left (3 a d +1\right ) x^{2}+d \,x^{3}\right )}d x\]
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Timed out. \[ \int \frac {\left (a^2-2 a x+x^2\right ) \left (-a b+2 (a-b) x+x^2\right )}{(x (-a+x) (-b+x))^{3/4} \left (-a^3 d+\left (b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {\left (a^2-2 a x+x^2\right ) \left (-a b+2 (a-b) x+x^2\right )}{(x (-a+x) (-b+x))^{3/4} \left (-a^3 d+\left (b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {\left (a^2-2 a x+x^2\right ) \left (-a b+2 (a-b) x+x^2\right )}{(x (-a+x) (-b+x))^{3/4} \left (-a^3 d+\left (b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx=\int { \frac {{\left (a^{2} - 2 \, a x + x^{2}\right )} {\left (a b - 2 \, {\left (a - b\right )} x - x^{2}\right )}}{{\left (a^{3} d - d x^{3} + {\left (3 \, a d + 1\right )} x^{2} - {\left (3 \, a^{2} d + b\right )} x\right )} \left ({\left (a - x\right )} {\left (b - x\right )} x\right )^{\frac {3}{4}}} \,d x } \]
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\[ \int \frac {\left (a^2-2 a x+x^2\right ) \left (-a b+2 (a-b) x+x^2\right )}{(x (-a+x) (-b+x))^{3/4} \left (-a^3 d+\left (b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx=\int { \frac {{\left (a^{2} - 2 \, a x + x^{2}\right )} {\left (a b - 2 \, {\left (a - b\right )} x - x^{2}\right )}}{{\left (a^{3} d - d x^{3} + {\left (3 \, a d + 1\right )} x^{2} - {\left (3 \, a^{2} d + b\right )} x\right )} \left ({\left (a - x\right )} {\left (b - x\right )} x\right )^{\frac {3}{4}}} \,d x } \]
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Timed out. \[ \int \frac {\left (a^2-2 a x+x^2\right ) \left (-a b+2 (a-b) x+x^2\right )}{(x (-a+x) (-b+x))^{3/4} \left (-a^3 d+\left (b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx=\int \frac {\left (a^2-2\,a\,x+x^2\right )\,\left (2\,x\,\left (a-b\right )-a\,b+x^2\right )}{{\left (x\,\left (a-x\right )\,\left (b-x\right )\right )}^{3/4}\,\left (x\,\left (3\,d\,a^2+b\right )-a^3\,d+d\,x^3-x^2\,\left (3\,a\,d+1\right )\right )} \,d x \]
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