Optimal. Leaf size=186 \[ \frac {\log \left (b x^2-b^{2/3} \sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{b^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{k x^3+(-k-1) x^2+x}}{2 \sqrt [3]{b} x^2+\sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{b^{2/3}}-\frac {\log \left (b^{5/3} x^2 \sqrt [3]{k x^3+(-k-1) x^2+x}+b^{4/3} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+b^2 x^4\right )}{2 b^{2/3}} \]
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Rubi [F] time = 18.27, 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^3 \left (5-4 (1+k) x+3 k x^2\right )}{((1-x) x (1-k x))^{2/3} \left (-1+(1+k) x-k x^2+b x^5\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {x^3 \left (5-4 (1+k) x+3 k x^2\right )}{((1-x) x (1-k x))^{2/3} \left (-1+(1+k) x-k x^2+b x^5\right )} \, dx &=\frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {x^{7/3} \left (5-4 (1+k) x+3 k x^2\right )}{(1-x)^{2/3} (1-k x)^{2/3} \left (-1+(1+k) x-k x^2+b x^5\right )} \, 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 ) \operatorname {Subst}\left (\int \frac {x^9 \left (5-4 (1+k) x^3+3 k x^6\right )}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+b x^{15}\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 ) \operatorname {Subst}\left (\int \left (\frac {3 k}{b \left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3}}+\frac {3 k-3 k (1+k) x^3+3 k^2 x^6+5 b x^9-4 b (1+k) x^{12}}{b \left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )}\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 ) \operatorname {Subst}\left (\int \frac {3 k-3 k (1+k) x^3+3 k^2 x^6+5 b x^9-4 b (1+k) x^{12}}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{b ((1-x) x (1-k x))^{2/3}}+\frac {\left (9 k (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3}} \, dx,x,\sqrt [3]{x}\right )}{b ((1-x) x (1-k x))^{2/3}}\\ &=\frac {9 k x \left (\frac {1-x}{1-k x}\right )^{2/3} (1-k x) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};\frac {(1-k) x}{1-k x}\right )}{b ((1-x) x (1-k x))^{2/3}}+\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \left (\frac {3 k (1+k) x^3}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-(1+k) x^3+k x^6-b x^{15}\right )}+\frac {4 b (1+k) x^{12}}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-(1+k) x^3+k x^6-b x^{15}\right )}+\frac {3 k}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )}+\frac {3 k^2 x^6}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )}+\frac {5 b x^9}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )}\right ) \, dx,x,\sqrt [3]{x}\right )}{b ((1-x) x (1-k x))^{2/3}}\\ &=\frac {9 k x \left (\frac {1-x}{1-k x}\right )^{2/3} (1-k x) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};\frac {(1-k) x}{1-k x}\right )}{b ((1-x) x (1-k x))^{2/3}}+\frac {\left (15 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^9}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}+\frac {\left (9 k (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{b ((1-x) x (1-k x))^{2/3}}+\frac {\left (9 k^2 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^6}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (-1+(1+k) x^3-k x^6+b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{b ((1-x) x (1-k x))^{2/3}}+\frac {\left (12 (1+k) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^{12}}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-(1+k) x^3+k x^6-b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}+\frac {\left (9 k (1+k) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \operatorname {Subst}\left (\int \frac {x^3}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-(1+k) x^3+k x^6-b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{b ((1-x) x (1-k x))^{2/3}}\\ \end {align*}
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Mathematica [F] time = 1.17, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^3 \left (5-4 (1+k) x+3 k x^2\right )}{((1-x) x (1-k x))^{2/3} \left (-1+(1+k) x-k x^2+b x^5\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 8.71, size = 186, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x+(-1-k) x^2+k x^3}}{2 \sqrt [3]{b} x^2+\sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{b^{2/3}}+\frac {\log \left (b x^2-b^{2/3} \sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{b^{2/3}}-\frac {\log \left (b^2 x^4+b^{5/3} x^2 \sqrt [3]{x+(-1-k) x^2+k x^3}+b^{4/3} \left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 b^{2/3}} \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 (3 \, k x^{2} - 4 \, {\left (k + 1\right )} x + 5\right )} x^{3}}{{\left (b x^{5} - k x^{2} + {\left (k + 1\right )} x - 1\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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maple [F] time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{3} \left (5-4 \left (1+k \right ) x +3 k \,x^{2}\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (-1+\left (1+k \right ) x -k \,x^{2}+b \,x^{5}\right )}\, dx\]
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 (3 \, k x^{2} - 4 \, {\left (k + 1\right )} x + 5\right )} x^{3}}{{\left (b x^{5} - k x^{2} + {\left (k + 1\right )} x - 1\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 {x^3\,\left (3\,k\,x^2-4\,x\,\left (k+1\right )+5\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (b\,x^5-k\,x^2+\left (k+1\right )\,x-1\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \left (3 k x^{2} - 4 k x - 4 x + 5\right )}{\left (x \left (x - 1\right ) \left (k x - 1\right )\right )^{\frac {2}{3}} \left (b x^{5} - k x^{2} + k x + x - 1\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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