Integrand size = 95, antiderivative size = 130 \[ \int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx=-\frac {4 \left (a b x^2+(-a-b) x^3+x^4\right )^{3/4}}{x^2 (-b+x)}+2 \sqrt [4]{d} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a b x^2+(-a-b) x^3+x^4}}{a-x}\right )-2 \sqrt [4]{d} \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a b x^2+(-a-b) x^3+x^4}}{a-x}\right ) \]
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\[ \int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx=\int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x^{3/2} \sqrt [4]{-a+x} (-b+x)^{5/4} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx}{\sqrt [4]{x^2 (-a+x) (-b+x)}} \\ & = \frac {\left (\sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \int \frac {(-a+x)^{3/4} (-2 a b+(3 a-b) x) \left (a^2-2 a x+x^2\right )}{x^{3/2} (-b+x)^{5/4} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx}{\sqrt [4]{x^2 (-a+x) (-b+x)}} \\ & = \frac {\left (\sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \int \frac {(-a+x)^{11/4} (-2 a b+(3 a-b) x)}{x^{3/2} (-b+x)^{5/4} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx}{\sqrt [4]{x^2 (-a+x) (-b+x)}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \frac {\left (-a+x^2\right )^{11/4} \left (-2 a b+(3 a-b) x^2\right )}{x^2 \left (-b+x^2\right )^{5/4} \left (a^3-3 a^2 x^2+(3 a-b d) x^4+(-1+d) x^6\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2 (-a+x) (-b+x)}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \left (-\frac {2 b \left (-a+x^2\right )^{11/4}}{a^2 x^2 \left (-b+x^2\right )^{5/4}}+\frac {\left (-a+x^2\right )^{11/4} \left (a^2 (3 a-7 b)+2 b (3 a-b d) x^2-2 b (1-d) x^4\right )}{a^2 \left (-b+x^2\right )^{5/4} \left (a^3-3 a^2 x^2+3 a \left (1-\frac {b d}{3 a}\right ) x^4-(1-d) x^6\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2 (-a+x) (-b+x)}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \frac {\left (-a+x^2\right )^{11/4} \left (a^2 (3 a-7 b)+2 b (3 a-b d) x^2-2 b (1-d) x^4\right )}{\left (-b+x^2\right )^{5/4} \left (a^3-3 a^2 x^2+3 a \left (1-\frac {b d}{3 a}\right ) x^4-(1-d) x^6\right )} \, dx,x,\sqrt {x}\right )}{a^2 \sqrt [4]{x^2 (-a+x) (-b+x)}}-\frac {\left (4 b \sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \frac {\left (-a+x^2\right )^{11/4}}{x^2 \left (-b+x^2\right )^{5/4}} \, dx,x,\sqrt {x}\right )}{a^2 \sqrt [4]{x^2 (-a+x) (-b+x)}} \\ & = \frac {\left (2 \sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \left (\frac {a^2 (3 a-7 b) \left (-a+x^2\right )^{11/4}}{\left (-b+x^2\right )^{5/4} \left (a^3-3 a^2 x^2+3 a \left (1-\frac {b d}{3 a}\right ) x^4-(1-d) x^6\right )}+\frac {2 b (3 a-b d) x^2 \left (-a+x^2\right )^{11/4}}{\left (-b+x^2\right )^{5/4} \left (a^3-3 a^2 x^2+3 a \left (1-\frac {b d}{3 a}\right ) x^4-(1-d) x^6\right )}+\frac {2 b (-1+d) x^4 \left (-a+x^2\right )^{11/4}}{\left (-b+x^2\right )^{5/4} \left (a^3-3 a^2 x^2+3 a \left (1-\frac {b d}{3 a}\right ) x^4-(1-d) x^6\right )}\right ) \, dx,x,\sqrt {x}\right )}{a^2 \sqrt [4]{x^2 (-a+x) (-b+x)}}-\frac {\left (4 b \sqrt {x} (-a+x) \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \frac {\left (1-\frac {x^2}{a}\right )^{11/4}}{x^2 \left (-b+x^2\right )^{5/4}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2 (-a+x) (-b+x)} \left (1-\frac {x}{a}\right )^{3/4}} \\ & = \frac {\left (2 (3 a-7 b) \sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \frac {\left (-a+x^2\right )^{11/4}}{\left (-b+x^2\right )^{5/4} \left (a^3-3 a^2 x^2+3 a \left (1-\frac {b d}{3 a}\right ) x^4-(1-d) x^6\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2 (-a+x) (-b+x)}}-\frac {\left (4 b (1-d) \sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \frac {x^4 \left (-a+x^2\right )^{11/4}}{\left (-b+x^2\right )^{5/4} \left (a^3-3 a^2 x^2+3 a \left (1-\frac {b d}{3 a}\right ) x^4-(1-d) x^6\right )} \, dx,x,\sqrt {x}\right )}{a^2 \sqrt [4]{x^2 (-a+x) (-b+x)}}+\frac {\left (4 b (3 a-b d) \sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \frac {x^2 \left (-a+x^2\right )^{11/4}}{\left (-b+x^2\right )^{5/4} \left (a^3-3 a^2 x^2+3 a \left (1-\frac {b d}{3 a}\right ) x^4-(1-d) x^6\right )} \, dx,x,\sqrt {x}\right )}{a^2 \sqrt [4]{x^2 (-a+x) (-b+x)}}+\frac {\left (4 \sqrt {x} (-a+x) \sqrt [4]{1-\frac {x}{b}}\right ) \text {Subst}\left (\int \frac {\left (1-\frac {x^2}{a}\right )^{11/4}}{x^2 \left (1-\frac {x^2}{b}\right )^{5/4}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2 (-a+x) (-b+x)} \left (1-\frac {x}{a}\right )^{3/4}} \\ & = \frac {4 (a-x) \sqrt [4]{1-\frac {x}{b}} \operatorname {AppellF1}\left (-\frac {1}{2},-\frac {11}{4},\frac {5}{4},\frac {1}{2},\frac {x}{a},\frac {x}{b}\right )}{\sqrt [4]{(a-x) (b-x) x^2} \left (1-\frac {x}{a}\right )^{3/4}}+\frac {\left (2 (3 a-7 b) \sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \frac {\left (-a+x^2\right )^{11/4}}{\left (-b+x^2\right )^{5/4} \left (a^3-3 a^2 x^2+3 a \left (1-\frac {b d}{3 a}\right ) x^4-(1-d) x^6\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2 (-a+x) (-b+x)}}-\frac {\left (4 b (1-d) \sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \frac {x^4 \left (-a+x^2\right )^{11/4}}{\left (-b+x^2\right )^{5/4} \left (a^3-3 a^2 x^2+3 a \left (1-\frac {b d}{3 a}\right ) x^4-(1-d) x^6\right )} \, dx,x,\sqrt {x}\right )}{a^2 \sqrt [4]{x^2 (-a+x) (-b+x)}}+\frac {\left (4 b (3 a-b d) \sqrt {x} \sqrt [4]{-a+x} \sqrt [4]{-b+x}\right ) \text {Subst}\left (\int \frac {x^2 \left (-a+x^2\right )^{11/4}}{\left (-b+x^2\right )^{5/4} \left (a^3-3 a^2 x^2+3 a \left (1-\frac {b d}{3 a}\right ) x^4-(1-d) x^6\right )} \, dx,x,\sqrt {x}\right )}{a^2 \sqrt [4]{x^2 (-a+x) (-b+x)}} \\ \end{align*}
Time = 46.26 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.82 \[ \int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx=\frac {4 (a-x)}{\sqrt [4]{x^2 (-a+x) (-b+x)}}+\frac {\sqrt [4]{d} x \sqrt [4]{\frac {-b+x}{a-x}} \left (\arctan \left (\frac {a-x \left (1+\sqrt {d} \sqrt {\frac {-b+x}{a-x}}\right )}{\sqrt {2} \sqrt [4]{d} \sqrt {\frac {x}{a-x}} (-a+x) \sqrt [4]{\frac {-b+x}{a-x}}}\right )-\text {arctanh}\left (\frac {\sqrt [4]{d} x \sqrt [4]{\frac {-b+x}{a-x}}}{\sqrt {\frac {x}{2 a-2 x}} \left (a+x \left (-1+\sqrt {d} \sqrt {\frac {-b+x}{a-x}}\right )\right )}\right )\right )}{\sqrt {\frac {x}{2 a-2 x}} \sqrt [4]{x^2 (-a+x) (-b+x)}} \]
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\[\int \frac {\left (-2 a b +\left (3 a -b \right ) x \right ) \left (-a^{3}+3 a^{2} x -3 a \,x^{2}+x^{3}\right )}{x \left (-b +x \right ) \left (x^{2} \left (-a +x \right ) \left (-b +x \right )\right )^{\frac {1}{4}} \left (a^{3}-3 a^{2} x +\left (-b d +3 a \right ) x^{2}+\left (-1+d \right ) x^{3}\right )}d x\]
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Timed out. \[ \int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx=\int { -\frac {{\left (a^{3} - 3 \, a^{2} x + 3 \, a x^{2} - x^{3}\right )} {\left (2 \, a b - {\left (3 \, a - b\right )} x\right )}}{\left ({\left (a - x\right )} {\left (b - x\right )} x^{2}\right )^{\frac {1}{4}} {\left ({\left (d - 1\right )} x^{3} + a^{3} - 3 \, a^{2} x - {\left (b d - 3 \, a\right )} x^{2}\right )} {\left (b - x\right )} x} \,d x } \]
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\[ \int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx=\int { -\frac {{\left (a^{3} - 3 \, a^{2} x + 3 \, a x^{2} - x^{3}\right )} {\left (2 \, a b - {\left (3 \, a - b\right )} x\right )}}{\left ({\left (a - x\right )} {\left (b - x\right )} x^{2}\right )^{\frac {1}{4}} {\left ({\left (d - 1\right )} x^{3} + a^{3} - 3 \, a^{2} x - {\left (b d - 3 \, a\right )} x^{2}\right )} {\left (b - x\right )} x} \,d x } \]
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Timed out. \[ \int \frac {(-2 a b+(3 a-b) x) \left (-a^3+3 a^2 x-3 a x^2+x^3\right )}{x (-b+x) \sqrt [4]{x^2 (-a+x) (-b+x)} \left (a^3-3 a^2 x+(3 a-b d) x^2+(-1+d) x^3\right )} \, dx=-\int \frac {\left (2\,a\,b-x\,\left (3\,a-b\right )\right )\,\left (a^3-3\,a^2\,x+3\,a\,x^2-x^3\right )}{x\,\left (b-x\right )\,{\left (x^2\,\left (a-x\right )\,\left (b-x\right )\right )}^{1/4}\,\left (x^2\,\left (3\,a-b\,d\right )-3\,a^2\,x+a^3+x^3\,\left (d-1\right )\right )} \,d x \]
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