Optimal. Leaf size=154 \[ -\frac {\log \left (b^{2/3} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+\sqrt [3]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+x^2\right )}{2 \sqrt [3]{b}}+\frac {\log \left (x-\sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{\sqrt [3]{b}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}+x}\right )}{\sqrt [3]{b}} \]
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Rubi [F] time = 4.80, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx &=\int \frac {x (-2+(1+k) x)}{((1-x) x (1-k x))^{2/3} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx\\ &=\frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {\sqrt [3]{x} (-2+(1+k) x)}{(1-x)^{2/3} (1-k x)^{2/3} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \left (\frac {\left (1+k+\frac {\sqrt {4+b-2 b k+b k^2}}{\sqrt {b}}\right ) \sqrt [3]{x}}{(1-x)^{2/3} (1-k x)^{2/3} \left (-b (1+k)-\sqrt {b} \sqrt {4+b-2 b k+b k^2}+2 (-1+b k) x\right )}+\frac {\left (1+k-\frac {\sqrt {4+b-2 b k+b k^2}}{\sqrt {b}}\right ) \sqrt [3]{x}}{(1-x)^{2/3} (1-k x)^{2/3} \left (-b (1+k)+\sqrt {b} \sqrt {4+b-2 b k+b k^2}+2 (-1+b k) x\right )}\right ) \, dx}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left (\left (1-\frac {\sqrt {4+b (1-k)^2}}{\sqrt {b}}+k\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {\sqrt [3]{x}}{(1-x)^{2/3} (1-k x)^{2/3} \left (-b (1+k)+\sqrt {b} \sqrt {4+b-2 b k+b k^2}+2 (-1+b k) x\right )} \, dx}{((1-x) x (1-k x))^{2/3}}+\frac {\left (\left (1+\frac {\sqrt {4+b (1-k)^2}}{\sqrt {b}}+k\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {\sqrt [3]{x}}{(1-x)^{2/3} (1-k x)^{2/3} \left (-b (1+k)-\sqrt {b} \sqrt {4+b-2 b k+b k^2}+2 (-1+b k) x\right )} \, dx}{((1-x) x (1-k x))^{2/3}}\\ \end {align*}
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Mathematica [F] time = 7.18, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-2 x+(1+k) x^2}{((1-x) x (1-k x))^{2/3} \left (b-b (1+k) x+(-1+b k) x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 0.39, size = 154, normalized size = 1.00 \begin {gather*} -\frac {\log \left (b^{2/3} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+\sqrt [3]{b} x \sqrt [3]{k x^3+(-k-1) x^2+x}+x^2\right )}{2 \sqrt [3]{b}}+\frac {\log \left (x-\sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{\sqrt [3]{b}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}+x}\right )}{\sqrt [3]{b}} \end {gather*}
Antiderivative was successfully verified.
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fricas [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.58, size = 130, normalized size = 0.84 \begin {gather*} -\frac {\sqrt {3} {\left | b \right |}^{\frac {2}{3}} \arctan \left (\frac {1}{3} \, \sqrt {3} b^{\frac {1}{3}} {\left (2 \, {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} + \frac {1}{b^{\frac {1}{3}}}\right )}\right )}{b} - \frac {{\left | b \right |}^{\frac {2}{3}} \log \left ({\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {2}{3}} + \frac {{\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}}}{b^{\frac {1}{3}}} + \frac {1}{b^{\frac {2}{3}}}\right )}{2 \, b} + \frac {\log \left ({\left | {\left (k - \frac {k}{x} - \frac {1}{x} + \frac {1}{x^{2}}\right )}^{\frac {1}{3}} - \frac {1}{b^{\frac {1}{3}}} \right |}\right )}{b^{\frac {1}{3}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.64, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-2 x +\left (1+k \right ) x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (b -b \left (1+k \right ) x +\left (b k -1\right ) x^{2}\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left (k + 1\right )} x^{2} - 2 \, x}{{\left (b {\left (k + 1\right )} x - {\left (b k - 1\right )} x^{2} - b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {2\,x-x^2\,\left (k+1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (\left (b\,k-1\right )\,x^2-b\,\left (k+1\right )\,x+b\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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