Optimal. Leaf size=228 \[ -\frac {\sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{2-4 x+2 x^2+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{2 b^{2/3}}-\frac {\log \left (1-2 x+x^2-\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 b^{2/3}}+\frac {\log \left (1-4 x+6 x^2-4 x^3+x^4+\left (\sqrt [3]{b}-2 \sqrt [3]{b} x+\sqrt [3]{b} x^2\right ) \left (x+(-1-k) x^2+k x^3\right )^{2/3}+b^{2/3} \left (x+(-1-k) x^2+k x^3\right )^{4/3}\right )}{4 b^{2/3}} \]
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Rubi [F]
time = 6.95, 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 {x (-1+k x) \left (1-2 k x+(-1+2 k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (-1+4 x+(-6+b) x^2+(4-2 b k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {x (-1+k x) \left (1-2 k x+(-1+2 k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (-1+4 x+(-6+b) x^2+(4-2 b k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx &=\frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {\sqrt [3]{x} (-1+k x) \left (1-2 k x+(-1+2 k) x^2\right )}{(1-x)^{2/3} (1-k x)^{2/3} \left (-1+4 x+(-6+b) x^2+(4-2 b k) x^3+\left (-1+b k^2\right ) x^4\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 \frac {\sqrt [3]{x} \sqrt [3]{1-k x} \left (1-2 k x+(-1+2 k) x^2\right )}{(1-x)^{2/3} \left (-1+4 x+(-6+b) x^2+(4-2 b k) x^3+\left (-1+b k^2\right ) x^4\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 \frac {\sqrt [3]{1-x} \sqrt [3]{x} (1+(1-2 k) x) \sqrt [3]{1-k x}}{-1+4 x+(-6+b) x^2+(4-2 b k) x^3+\left (-1+b k^2\right ) x^4} \, dx}{((1-x) x (1-k x))^{2/3}}\\ &=-\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \text {Subst}\left (\int \frac {x^3 \sqrt [3]{1-x^3} \left (1+(1-2 k) x^3\right ) \sqrt [3]{1-k x^3}}{-1+4 x^3+(-6+b) x^6+(4-2 b k) x^9+\left (-1+b k^2\right ) x^{12}} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\\ &=-\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \text {Subst}\left (\int \left (\frac {x^3 \sqrt [3]{1-x^3} \sqrt [3]{1-k x^3}}{-1+4 x^3-6 \left (1-\frac {b}{6}\right ) x^6+4 \left (1-\frac {b k}{2}\right ) x^9-\left (1-b k^2\right ) x^{12}}+\frac {(-1+2 k) x^6 \sqrt [3]{1-x^3} \sqrt [3]{1-k x^3}}{1-4 x^3+6 \left (1-\frac {b}{6}\right ) x^6-4 \left (1-\frac {b k}{2}\right ) x^9+\left (1-b k^2\right ) x^{12}}\right ) \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\\ &=-\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \text {Subst}\left (\int \frac {x^3 \sqrt [3]{1-x^3} \sqrt [3]{1-k x^3}}{-1+4 x^3-6 \left (1-\frac {b}{6}\right ) x^6+4 \left (1-\frac {b k}{2}\right ) x^9-\left (1-b k^2\right ) x^{12}} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}-\frac {\left (3 (-1+2 k) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \text {Subst}\left (\int \frac {x^6 \sqrt [3]{1-x^3} \sqrt [3]{1-k x^3}}{1-4 x^3+6 \left (1-\frac {b}{6}\right ) x^6-4 \left (1-\frac {b k}{2}\right ) x^9+\left (1-b k^2\right ) x^{12}} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\\ \end {align*}
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Mathematica [F]
time = 24.15, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x (-1+k x) \left (1-2 k x+(-1+2 k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (-1+4 x+(-6+b) x^2+(4-2 b k) x^3+\left (-1+b k^2\right ) x^4\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {x \left (k x -1\right ) \left (1-2 k x +\left (-1+2 k \right ) x^{2}\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (-1+4 x +\left (-6+b \right ) x^{2}+\left (-2 b k +4\right ) x^{3}+\left (b \,k^{2}-1\right ) x^{4}\right )}\, dx\]
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*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F(-1)] Timed out
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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Sympy [F(-1)] Timed out
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 [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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 = -0.00 \begin {gather*} \int \frac {x\,\left (k\,x-1\right )\,\left (\left (2\,k-1\right )\,x^2-2\,k\,x+1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (\left (b\,k^2-1\right )\,x^4+\left (4-2\,b\,k\right )\,x^3+\left (b-6\right )\,x^2+4\,x-1\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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