Integrand size = 69, antiderivative size = 109 \[ \int \frac {(-b+x) (a-3 b+2 x)}{(-a+x) ((-a+x) (-b+x))^{3/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx=-\frac {4 (b-x)}{\left (a b-a x-b x+x^2\right )^{3/4}}-2 \sqrt [4]{d} \arctan \left (\frac {\sqrt [4]{d} \left (a b+(-a-b) x+x^2\right )^{3/4}}{b-x}\right )+2 \sqrt [4]{d} \text {arctanh}\left (\frac {\sqrt [4]{a b+(-a-b) x+x^2}}{\sqrt [4]{d} (a-x)}\right ) \]
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\[ \int \frac {(-b+x) (a-3 b+2 x)}{(-a+x) ((-a+x) (-b+x))^{3/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx=\int \frac {(-b+x) (a-3 b+2 x)}{(-a+x) ((-a+x) (-b+x))^{3/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \int \frac {(-b+x) (a-3 b+2 x)}{(-a+x) \left (a b+(-a-b) x+x^2\right )^{3/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx \\ & = \int \frac {(b-x)^2 (a-3 b+2 x)}{((a-x) (b-x))^{7/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx \\ & = \int \frac {(b-x)^2 (a-3 b+2 x)}{\left (a b+(-a-b) x+x^2\right )^{7/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx \\ & = \int \frac {(b-x)^2 (a-3 b+2 x)}{\left (a b-(a+b) x+x^2\right )^{7/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx \\ & = \int \left (\frac {2}{d \left (a b-(a+b) x+x^2\right )^{7/4}}-\frac {2 b-2 a^3 d-a b^2 d+3 b^3 d-2 \left (1-3 a^2 d-a b d+4 b^2 d\right ) x-7 (a-b) d x^2}{d \left (a b-(a+b) x+x^2\right )^{7/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )}\right ) \, dx \\ & = -\frac {\int \frac {2 b-2 a^3 d-a b^2 d+3 b^3 d-2 \left (1-3 a^2 d-a b d+4 b^2 d\right ) x-7 (a-b) d x^2}{\left (a b-(a+b) x+x^2\right )^{7/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx}{d}+\frac {2 \int \frac {1}{\left (a b-(a+b) x+x^2\right )^{7/4}} \, dx}{d} \\ & = \frac {8 (a+b-2 x)}{3 (a-b)^2 d \left (a b-(a+b) x+x^2\right )^{3/4}}-\frac {\int \left (\frac {2 a^3 d \left (1+\frac {b \left (-2+a b d-3 b^2 d\right )}{2 a^3 d}\right )}{\left (a b-(a+b) x+x^2\right )^{7/4} \left (-b+a^3 d+\left (1-3 a^2 d\right ) x+3 a d x^2-d x^3\right )}+\frac {2 \left (-1+3 a^2 d+a b d-4 b^2 d\right ) x}{\left (a b-(a+b) x+x^2\right )^{7/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )}+\frac {7 (-a+b) d x^2}{\left (a b-(a+b) x+x^2\right )^{7/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )}\right ) \, dx}{d}-\frac {8 \int \frac {1}{\left (a b+(-a-b) x+x^2\right )^{3/4}} \, dx}{3 (a-b)^2 d} \\ & = \frac {8 (a+b-2 x)}{3 (a-b)^2 d \left (a b-(a+b) x+x^2\right )^{3/4}}+(7 (a-b)) \int \frac {x^2}{\left (a b-(a+b) x+x^2\right )^{7/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx+\frac {\left (2 \left (1-3 a^2 d-a b d+4 b^2 d\right )\right ) \int \frac {x}{\left (a b-(a+b) x+x^2\right )^{7/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx}{d}-\frac {\left (-2 b+2 a^3 d+a b^2 d-3 b^3 d\right ) \int \frac {1}{\left (a b-(a+b) x+x^2\right )^{7/4} \left (-b+a^3 d+\left (1-3 a^2 d\right ) x+3 a d x^2-d x^3\right )} \, dx}{d}-\frac {\left (32 \sqrt {(-a-b+2 x)^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {(-a-b)^2-4 a b+4 x^4}} \, dx,x,\sqrt [4]{a b+(-a-b) x+x^2}\right )}{3 (a-b)^2 d (-a-b+2 x)} \\ & = \frac {8 (a+b-2 x)}{3 (a-b)^2 d \left (a b-(a+b) x+x^2\right )^{3/4}}+\frac {8 \sqrt {2} \left (1+\frac {2 \sqrt {(a-x) (b-x)}}{a-b}\right ) \sqrt {\frac {(a+b-2 x)^2}{(a-b)^2 \left (1+\frac {2 \sqrt {(a-x) (b-x)}}{a-b}\right )^2}} \sqrt {(-a-b+2 x)^2} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{a b-(a+b) x+x^2}}{\sqrt {a-b}}\right ),\frac {1}{2}\right )}{3 (a-b)^{3/2} d (a+b-2 x) \sqrt {(a+b-2 x)^2}}+(7 (a-b)) \int \frac {x^2}{\left (a b-(a+b) x+x^2\right )^{7/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx+\frac {\left (2 \left (1-3 a^2 d-a b d+4 b^2 d\right )\right ) \int \frac {x}{\left (a b-(a+b) x+x^2\right )^{7/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx}{d}-\frac {\left (-2 b+2 a^3 d+a b^2 d-3 b^3 d\right ) \int \frac {1}{\left (a b-(a+b) x+x^2\right )^{7/4} \left (-b+a^3 d+\left (1-3 a^2 d\right ) x+3 a d x^2-d x^3\right )} \, dx}{d} \\ \end{align*}
\[ \int \frac {(-b+x) (a-3 b+2 x)}{(-a+x) ((-a+x) (-b+x))^{3/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx=\int \frac {(-b+x) (a-3 b+2 x)}{(-a+x) ((-a+x) (-b+x))^{3/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx \]
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\[\int \frac {\left (-b +x \right ) \left (a -3 b +2 x \right )}{\left (-a +x \right ) \left (\left (-a +x \right ) \left (-b +x \right )\right )^{\frac {3}{4}} \left (b -a^{3} d -\left (-3 a^{2} d +1\right ) x -3 a d \,x^{2}+d \,x^{3}\right )}d x\]
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Timed out. \[ \int \frac {(-b+x) (a-3 b+2 x)}{(-a+x) ((-a+x) (-b+x))^{3/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(-b+x) (a-3 b+2 x)}{(-a+x) ((-a+x) (-b+x))^{3/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {(-b+x) (a-3 b+2 x)}{(-a+x) ((-a+x) (-b+x))^{3/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx=\int { -\frac {{\left (a - 3 \, b + 2 \, x\right )} {\left (b - x\right )}}{{\left (a^{3} d + 3 \, a d x^{2} - d x^{3} - {\left (3 \, a^{2} d - 1\right )} x - b\right )} \left ({\left (a - x\right )} {\left (b - x\right )}\right )^{\frac {3}{4}} {\left (a - x\right )}} \,d x } \]
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\[ \int \frac {(-b+x) (a-3 b+2 x)}{(-a+x) ((-a+x) (-b+x))^{3/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx=\int { -\frac {{\left (a - 3 \, b + 2 \, x\right )} {\left (b - x\right )}}{{\left (a^{3} d + 3 \, a d x^{2} - d x^{3} - {\left (3 \, a^{2} d - 1\right )} x - b\right )} \left ({\left (a - x\right )} {\left (b - x\right )}\right )^{\frac {3}{4}} {\left (a - x\right )}} \,d x } \]
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Timed out. \[ \int \frac {(-b+x) (a-3 b+2 x)}{(-a+x) ((-a+x) (-b+x))^{3/4} \left (b-a^3 d-\left (1-3 a^2 d\right ) x-3 a d x^2+d x^3\right )} \, dx=\int \frac {\left (b-x\right )\,\left (a-3\,b+2\,x\right )}{{\left (\left (a-x\right )\,\left (b-x\right )\right )}^{3/4}\,\left (a-x\right )\,\left (b-a^3\,d+d\,x^3+x\,\left (3\,a^2\,d-1\right )-3\,a\,d\,x^2\right )} \,d x \]
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