Integrand size = 77, antiderivative size = 116 \[ \int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx=\frac {4 \left (-x+x^2\right )}{\left (x-x^2-k x^2+k x^3\right )^{3/4}}+2 \sqrt [4]{d} \arctan \left (\frac {\sqrt [4]{d} \left (x+(-1-k) x^2+k x^3\right )^{3/4}}{(-1+x) x}\right )-2 \sqrt [4]{d} \text {arctanh}\left (\frac {\sqrt [4]{d} \left (x+(-1-k) x^2+k x^3\right )^{3/4}}{(-1+x) x}\right ) \]
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\[ \int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx=\int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left ((1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \int \frac {(-1+x) \sqrt [4]{x} \left (-1-2 (-1+k) x+k x^2\right )}{(1-x)^{3/4} (1-k x)^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx}{((1-x) x (1-k x))^{3/4}} \\ & = -\frac {\left ((1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \int \frac {\sqrt [4]{1-x} \sqrt [4]{x} \left (-1-2 (-1+k) x+k x^2\right )}{(1-k x)^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx}{((1-x) x (1-k x))^{3/4}} \\ & = \frac {\left ((1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \int \frac {\sqrt [4]{1-x} \sqrt [4]{x} \left (-1-2 (-1+k) x+k x^2\right )}{(1-k x)^{7/4} \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx}{((1-x) x (1-k x))^{3/4}} \\ & = \frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^4 \sqrt [4]{1-x^4} \left (-1-2 (-1+k) x^4+k x^8\right )}{\left (1-k x^4\right )^{7/4} \left (-d+(1+3 d k) x^4-\left (1+3 d k^2\right ) x^8+d k^3 x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}} \\ & = \frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^4 \sqrt [4]{1-x^4} \left (-1-2 (-1+k) x^4+k x^8\right )}{\left (1-k x^4\right )^{7/4} \left (x^4-x^8+d \left (-1+k x^4\right )^3\right )} \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}} \\ & = \frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \text {Subst}\left (\int \left (\frac {\sqrt [4]{1-x^4}}{d k^2 \left (1-k x^4\right )^{7/4}}+\frac {\sqrt [4]{1-x^4} \left (d-(1+d k (3+k)) x^4+\left (1+d (5-2 k) k^2\right ) x^8\right )}{d k^2 \left (1-k x^4\right )^{7/4} \left (x^4-x^8+d \left (-1+k x^4\right )^3\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{((1-x) x (1-k x))^{3/4}} \\ & = \frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \text {Subst}\left (\int \frac {\sqrt [4]{1-x^4}}{\left (1-k x^4\right )^{7/4}} \, dx,x,\sqrt [4]{x}\right )}{d k^2 ((1-x) x (1-k x))^{3/4}}+\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \text {Subst}\left (\int \frac {\sqrt [4]{1-x^4} \left (d-(1+d k (3+k)) x^4+\left (1+d (5-2 k) k^2\right ) x^8\right )}{\left (1-k x^4\right )^{7/4} \left (x^4-x^8+d \left (-1+k x^4\right )^3\right )} \, dx,x,\sqrt [4]{x}\right )}{d k^2 ((1-x) x (1-k x))^{3/4}} \\ & = \frac {4 (1-x)^{3/4} x (1-k x)^{3/4} \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{4},\frac {7}{4},\frac {5}{4},x,k x\right )}{d k^2 ((1-x) x (1-k x))^{3/4}}+\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \text {Subst}\left (\int \left (\frac {(1+d k (3+k)) x^4 \sqrt [4]{1-x^4}}{\left (1-k x^4\right )^{7/4} \left (d-(1+3 d k) x^4+\left (1+3 d k^2\right ) x^8-d k^3 x^{12}\right )}+\frac {\left (-1-d (5-2 k) k^2\right ) x^8 \sqrt [4]{1-x^4}}{\left (1-k x^4\right )^{7/4} \left (d-(1+3 d k) x^4+\left (1+3 d k^2\right ) x^8-d k^3 x^{12}\right )}+\frac {d \sqrt [4]{1-x^4}}{\left (1-k x^4\right )^{7/4} \left (-d+(1+3 d k) x^4-\left (1+3 d k^2\right ) x^8+d k^3 x^{12}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{d k^2 ((1-x) x (1-k x))^{3/4}} \\ & = \frac {4 (1-x)^{3/4} x (1-k x)^{3/4} \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{4},\frac {7}{4},\frac {5}{4},x,k x\right )}{d k^2 ((1-x) x (1-k x))^{3/4}}+\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \text {Subst}\left (\int \frac {\sqrt [4]{1-x^4}}{\left (1-k x^4\right )^{7/4} \left (-d+(1+3 d k) x^4-\left (1+3 d k^2\right ) x^8+d k^3 x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{k^2 ((1-x) x (1-k x))^{3/4}}+\frac {\left (4 \left (-1-d (5-2 k) k^2\right ) (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^8 \sqrt [4]{1-x^4}}{\left (1-k x^4\right )^{7/4} \left (d-(1+3 d k) x^4+\left (1+3 d k^2\right ) x^8-d k^3 x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{d k^2 ((1-x) x (1-k x))^{3/4}}+\frac {\left (4 (1+d k (3+k)) (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^4 \sqrt [4]{1-x^4}}{\left (1-k x^4\right )^{7/4} \left (d-(1+3 d k) x^4+\left (1+3 d k^2\right ) x^8-d k^3 x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{d k^2 ((1-x) x (1-k x))^{3/4}} \\ & = \frac {4 (1-x)^{3/4} x (1-k x)^{3/4} \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{4},\frac {7}{4},\frac {5}{4},x,k x\right )}{d k^2 ((1-x) x (1-k x))^{3/4}}+\frac {\left (4 (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \text {Subst}\left (\int \frac {\sqrt [4]{1-x^4}}{\left (1-k x^4\right )^{7/4} \left (x^4-x^8+d \left (-1+k x^4\right )^3\right )} \, dx,x,\sqrt [4]{x}\right )}{k^2 ((1-x) x (1-k x))^{3/4}}+\frac {\left (4 \left (-1-d (5-2 k) k^2\right ) (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^8 \sqrt [4]{1-x^4}}{\left (1-k x^4\right )^{7/4} \left (-x^4+x^8-d \left (-1+k x^4\right )^3\right )} \, dx,x,\sqrt [4]{x}\right )}{d k^2 ((1-x) x (1-k x))^{3/4}}+\frac {\left (4 (1+d k (3+k)) (1-x)^{3/4} x^{3/4} (1-k x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^4 \sqrt [4]{1-x^4}}{\left (1-k x^4\right )^{7/4} \left (-x^4+x^8-d \left (-1+k x^4\right )^3\right )} \, dx,x,\sqrt [4]{x}\right )}{d k^2 ((1-x) x (1-k x))^{3/4}} \\ \end{align*}
\[ \int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx=\int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx \]
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\[\int \frac {\left (-1+x \right ) x \left (-1-2 \left (-1+k \right ) x +k \,x^{2}\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {3}{4}} \left (k x -1\right ) \left (-d +\left (3 d k +1\right ) x -\left (3 d \,k^{2}+1\right ) x^{2}+d \,k^{3} x^{3}\right )}d x\]
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Timed out. \[ \int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx=\int { \frac {{\left (k x^{2} - 2 \, {\left (k - 1\right )} x - 1\right )} {\left (x - 1\right )} x}{{\left (d k^{3} x^{3} - {\left (3 \, d k^{2} + 1\right )} x^{2} + {\left (3 \, d k + 1\right )} x - d\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {3}{4}} {\left (k x - 1\right )}} \,d x } \]
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\[ \int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx=\int { \frac {{\left (k x^{2} - 2 \, {\left (k - 1\right )} x - 1\right )} {\left (x - 1\right )} x}{{\left (d k^{3} x^{3} - {\left (3 \, d k^{2} + 1\right )} x^{2} + {\left (3 \, d k + 1\right )} x - d\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {3}{4}} {\left (k x - 1\right )}} \,d x } \]
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Timed out. \[ \int \frac {(-1+x) x \left (-1-2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} (-1+k x) \left (-d+(1+3 d k) x-\left (1+3 d k^2\right ) x^2+d k^3 x^3\right )} \, dx=\int \frac {x\,\left (x-1\right )\,\left (2\,x\,\left (k-1\right )-k\,x^2+1\right )}{\left (k\,x-1\right )\,{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{3/4}\,\left (d+x^2\,\left (3\,d\,k^2+1\right )-x\,\left (3\,d\,k+1\right )-d\,k^3\,x^3\right )} \,d x \]
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