Integrand size = 62, antiderivative size = 209 \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x^2 \sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\frac {3 \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{2 x^2}+\frac {\left (-\sqrt {3} a-\sqrt {3} b\right ) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{\sqrt [3]{b}}+\frac {(a+b) \log \left (-\sqrt [3]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{\sqrt [3]{b}}+\frac {(-a-b) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 \sqrt [3]{b}} \]
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\[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x^2 \sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x^2 \sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{\sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx}{\sqrt [3]{(1-x) x (1-k x)}} \\ & = \frac {\left (\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \left (-\frac {2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )}{(b-k)^2 \sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x}}-\frac {(1+k) (a+k)}{(b-k) \sqrt [3]{1-x} x^{4/3} \sqrt [3]{1-k x}}-\frac {(a+b) \left (1+2 b+k^2\right )-(a+b) (1+k) \left (1+3 b-k+k^2\right ) x}{(b-k)^2 \sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x} \left (1+(-1-k) x+(-b+k) x^2\right )}\right ) \, dx}{\sqrt [3]{(1-x) x (1-k x)}} \\ & = -\frac {\left (\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {(a+b) \left (1+2 b+k^2\right )-(a+b) (1+k) \left (1+3 b-k+k^2\right ) x}{\sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x} \left (1+(-1-k) x+(-b+k) x^2\right )} \, dx}{(b-k)^2 \sqrt [3]{(1-x) x (1-k x)}}-\frac {\left ((1+k) (a+k) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {1}{\sqrt [3]{1-x} x^{4/3} \sqrt [3]{1-k x}} \, dx}{(b-k) \sqrt [3]{(1-x) x (1-k x)}}-\frac {\left (\left (2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )\right ) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {1}{\sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x}} \, dx}{(b-k)^2 \sqrt [3]{(1-x) x (1-k x)}} \\ & = \frac {3 \left (2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )\right ) (1-x) (1-k x)}{4 (b-k)^2 x \sqrt [3]{(1-x) x (1-k x)}}+\frac {3 (1+k) (a+k) (1-x) \left (\frac {(1-k) x}{1-k x}\right )^{4/3} (1-k x) \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {5}{3},\frac {1-x}{1-k x}\right )}{2 (1-k) (b-k) x \sqrt [3]{(1-x) x (1-k x)}}-\frac {\left (\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \left (\frac {-\left ((a+b) (1+k) \left (1+3 b-k+k^2\right )\right )-\frac {(a+b) \left (1+5 b+4 b^2-k+2 b k+5 b k^2-k^3+k^4\right )}{\sqrt {1+4 b-2 k+k^2}}}{\sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x} \left (-1-k-\sqrt {1+4 b-2 k+k^2}+2 (-b+k) x\right )}+\frac {-\left ((a+b) (1+k) \left (1+3 b-k+k^2\right )\right )+\frac {(a+b) \left (1+5 b+4 b^2-k+2 b k+5 b k^2-k^3+k^4\right )}{\sqrt {1+4 b-2 k+k^2}}}{\sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x} \left (-1-k+\sqrt {1+4 b-2 k+k^2}+2 (-b+k) x\right )}\right ) \, dx}{(b-k)^2 \sqrt [3]{(1-x) x (1-k x)}}-\frac {\left ((1+k) \left (2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )\right ) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {1}{\sqrt [3]{1-x} x^{4/3} \sqrt [3]{1-k x}} \, dx}{2 (b-k)^2 \sqrt [3]{(1-x) x (1-k x)}} \\ & = \frac {3 \left (2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )\right ) (1-x) (1-k x)}{4 (b-k)^2 x \sqrt [3]{(1-x) x (1-k x)}}+\frac {3 (1+k) (a+k) (1-x) \left (\frac {(1-k) x}{1-k x}\right )^{4/3} (1-k x) \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {5}{3},\frac {1-x}{1-k x}\right )}{2 (1-k) (b-k) x \sqrt [3]{(1-x) x (1-k x)}}+\frac {3 (1+k) \left (2 k^2-a \left (1+2 b+k^2\right )-b \left (1+4 k+k^2\right )\right ) (1-x) \left (\frac {(1-k) x}{1-k x}\right )^{4/3} (1-k x) \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {4}{3},\frac {5}{3},\frac {1-x}{1-k x}\right )}{4 (1-k) (b-k)^2 x \sqrt [3]{(1-x) x (1-k x)}}-\frac {\left (\left (-\left ((a+b) (1+k) \left (1+3 b-k+k^2\right )\right )+\frac {(a+b) \left (1+5 b+4 b^2-k+2 b k+5 b k^2-k^3+k^4\right )}{\sqrt {1+4 b-2 k+k^2}}\right ) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {1}{\sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x} \left (-1-k+\sqrt {1+4 b-2 k+k^2}+2 (-b+k) x\right )} \, dx}{(b-k)^2 \sqrt [3]{(1-x) x (1-k x)}}+\frac {\left ((a+b) \left (1+k^3+3 b (1+k)+\frac {4 b^2+(1-k)^2 \left (1+k+k^2\right )+b \left (5+2 k+5 k^2\right )}{\sqrt {4 b+(-1+k)^2}}\right ) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {1}{\sqrt [3]{1-x} x^{7/3} \sqrt [3]{1-k x} \left (-1-k-\sqrt {1+4 b-2 k+k^2}+2 (-b+k) x\right )} \, dx}{(b-k)^2 \sqrt [3]{(1-x) x (1-k x)}} \\ \end{align*}
Time = 15.41 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.80 \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x^2 \sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\frac {3 ((-1+x) x (-1+k x))^{2/3}}{2 x^2}-\frac {\sqrt {3} (a+b) \arctan \left (\frac {\sqrt {3} \sqrt [3]{b} x}{\sqrt [3]{b} x+2 \sqrt [3]{(-1+x) x (-1+k x)}}\right )}{\sqrt [3]{b}}+\frac {(a+b) \log \left (-\sqrt [3]{b} x+\sqrt [3]{(-1+x) x (-1+k x)}\right )}{\sqrt [3]{b}}-\frac {(a+b) \log \left (b^{2/3} x^2+\sqrt [3]{b} x \sqrt [3]{(-1+x) x (-1+k x)}+((-1+x) x (-1+k x))^{2/3}\right )}{2 \sqrt [3]{b}} \]
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Time = 0.79 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.66
method | result | size |
pseudoelliptic | \(\frac {\frac {3 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}} b^{\frac {1}{3}}}{2}+\left (\arctan \left (\frac {\sqrt {3}\, \left (b^{\frac {1}{3}} x +2 \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}\right )}{3 b^{\frac {1}{3}} x}\right ) \sqrt {3}+\ln \left (\frac {-b^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}}}{x}\right )-\frac {\ln \left (\frac {b^{\frac {2}{3}} x^{2}+b^{\frac {1}{3}} \left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {1}{3}} x +\left (\left (-1+x \right ) x \left (k x -1\right )\right )^{\frac {2}{3}}}{x^{2}}\right )}{2}\right ) x^{2} \left (a +b \right )}{b^{\frac {1}{3}} x^{2}}\) | \(138\) |
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Timed out. \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x^2 \sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x^2 \sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x^2 \sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\int { -\frac {{\left ({\left (a + k\right )} x^{2} - {\left (k + 1\right )} x + 1\right )} {\left ({\left (k + 1\right )} x - 2\right )}}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}} {\left ({\left (b - k\right )} x^{2} + {\left (k + 1\right )} x - 1\right )} x^{2}} \,d x } \]
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\[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x^2 \sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\int { -\frac {{\left ({\left (a + k\right )} x^{2} - {\left (k + 1\right )} x + 1\right )} {\left ({\left (k + 1\right )} x - 2\right )}}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}} {\left ({\left (b - k\right )} x^{2} + {\left (k + 1\right )} x - 1\right )} x^{2}} \,d x } \]
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Timed out. \[ \int \frac {(-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{x^2 \sqrt [3]{(1-x) x (1-k x)} \left (1-(1+k) x+(-b+k) x^2\right )} \, dx=\int -\frac {\left (x\,\left (k+1\right )-2\right )\,\left (\left (a+k\right )\,x^2+\left (-k-1\right )\,x+1\right )}{x^2\,{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (\left (b-k\right )\,x^2+\left (k+1\right )\,x-1\right )} \,d x \]
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