Optimal. Leaf size=223 \[ -\frac {\log \left (\left (b^{2/3} x-b^{2/3} x^2\right ) \left (k x^3+(-k-1) x^2+x\right )^{2/3}+\sqrt [3]{b} \left (k x^3+(-k-1) x^2+x\right )^{4/3}+b x^4-2 b x^3+b x^2\right )}{2 \sqrt [3]{b}}+\frac {\log \left (\sqrt [6]{b} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+\sqrt {b} x^2-\sqrt {b} x\right )}{\sqrt [3]{b}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \left (k x^3+(-k-1) x^2+x\right )^{2/3}}{-2 \sqrt [3]{b} x^2+2 \sqrt [3]{b} x+\left (k x^3+(-k-1) x^2+x\right )^{2/3}}\right )}{\sqrt [3]{b}} \]
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Rubi [F] time = 3.60, 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 {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(b+2 k) x+\left (b+k^2\right ) x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {align*} \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(b+2 k) x+\left (b+k^2\right ) x^2\right )} \, dx &=\frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{(1-x)^{2/3} x^{2/3} (1-k x)^{2/3} \left (1-(b+2 k) x+\left (b+k^2\right ) 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 \frac {(-1+(2-k) x) \sqrt [3]{1-k x}}{(1-x)^{2/3} x^{2/3} \left (1-(b+2 k) x+\left (b+k^2\right ) 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 (2-k-\frac {k \sqrt {-4+b+4 k}}{\sqrt {b}}\right ) \sqrt [3]{1-k x}}{(1-x)^{2/3} x^{2/3} \left (-b-2 k-\sqrt {b} \sqrt {-4+b+4 k}+2 \left (b+k^2\right ) x\right )}+\frac {\left (2-k+\frac {k \sqrt {-4+b+4 k}}{\sqrt {b}}\right ) \sqrt [3]{1-k x}}{(1-x)^{2/3} x^{2/3} \left (-b-2 k+\sqrt {b} \sqrt {-4+b+4 k}+2 \left (b+k^2\right ) x\right )}\right ) \, dx}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left (\left (2-k \left (1-\frac {\sqrt {-4+b+4 k}}{\sqrt {b}}\right )\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {\sqrt [3]{1-k x}}{(1-x)^{2/3} x^{2/3} \left (-b-2 k+\sqrt {b} \sqrt {-4+b+4 k}+2 \left (b+k^2\right ) x\right )} \, dx}{((1-x) x (1-k x))^{2/3}}+\frac {\left (\left (2-k \left (1+\frac {\sqrt {-4+b+4 k}}{\sqrt {b}}\right )\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {\sqrt [3]{1-k x}}{(1-x)^{2/3} x^{2/3} \left (-b-2 k-\sqrt {b} \sqrt {-4+b+4 k}+2 \left (b+k^2\right ) x\right )} \, dx}{((1-x) x (1-k x))^{2/3}}\\ \end {align*}
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Mathematica [F] time = 4.67, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-1+2 x+\left (-2 k+k^2\right ) x^2}{((1-x) x (1-k x))^{2/3} \left (1-(b+2 k) x+\left (b+k^2\right ) x^2\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 0.95, size = 223, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \left (x+(-1-k) x^2+k x^3\right )^{2/3}}{2 \sqrt [3]{b} x-2 \sqrt [3]{b} x^2+\left (x+(-1-k) x^2+k x^3\right )^{2/3}}\right )}{\sqrt [3]{b}}+\frac {\log \left (-\sqrt {b} x+\sqrt {b} x^2+\sqrt [6]{b} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{\sqrt [3]{b}}-\frac {\log \left (b x^2-2 b x^3+b x^4+\left (b^{2/3} x-b^{2/3} x^2\right ) \left (x+(-1-k) x^2+k x^3\right )^{2/3}+\sqrt [3]{b} \left (x+(-1-k) x^2+k x^3\right )^{4/3}\right )}{2 \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 [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (k^{2} - 2 \, k\right )} x^{2} + 2 \, x - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (k^{2} + b\right )} x^{2} - {\left (b + 2 \, k\right )} x + 1\right )}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.18, size = 0, normalized size = 0.00 \[\int \frac {-1+2 x +\left (k^{2}-2 k \right ) x^{2}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (1-\left (b +2 k \right ) x +\left (k^{2}+b \right ) x^{2}\right )}\, dx\]
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*} \int \frac {{\left (k^{2} - 2 \, k\right )} x^{2} + 2 \, x - 1}{\left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {2}{3}} {\left ({\left (k^{2} + b\right )} x^{2} - {\left (b + 2 \, k\right )} x + 1\right )}}\,{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 = -0.00 \begin {gather*} \int -\frac {\left (2\,k-k^2\right )\,x^2-2\,x+1}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (\left (k^2+b\right )\,x^2+\left (-b-2\,k\right )\,x+1\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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