Optimal. Leaf size=185 \[ \frac {\log \left (x^2-\sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{b^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}}{\sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}+2 x^2}\right )}{b^{2/3}}-\frac {\log \left (b^{2/3} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+\sqrt [3]{b} x^2 \sqrt [3]{k x^3+(-k-1) x^2+x}+x^4\right )}{2 b^{2/3}} \]
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Rubi [F] time = 15.98, 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 \left (5-4 (1+k) x+3 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(b+b k) x-b k x^2+x^5\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {align*} \int \frac {x \left (5-4 (1+k) x+3 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(b+b k) x-b k x^2+x^5\right )} \, dx &=\frac {\left (\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {x^{2/3} \left (5-4 (1+k) x+3 k x^2\right )}{\sqrt [3]{1-x} \sqrt [3]{1-k x} \left (-b+(b+b k) x-b k x^2+x^5\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 {x^{2/3} \left (5-4 (1+k) x+3 k x^2\right )}{\sqrt [3]{1-x} \sqrt [3]{1-k x} \left (x^5-b (-1+x) (-1+k x)\right )} \, dx}{\sqrt [3]{(1-x) x (1-k x)}}\\ &=\frac {\left (3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (5-4 (1+k) x^3+3 k x^6\right )}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (x^{15}-b \left (-1+x^3\right ) \left (-1+k x^3\right )\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\\ &=\frac {\left (3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \left (\frac {4 (1+k) x^7}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (b-b (1+k) x^3+b k x^6-x^{15}\right )}+\frac {5 x^4}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (-b+b (1+k) x^3-b k x^6+x^{15}\right )}+\frac {3 k x^{10}}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (-b+b (1+k) x^3-b k x^6+x^{15}\right )}\right ) \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\\ &=\frac {\left (15 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (-b+b (1+k) x^3-b k x^6+x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}+\frac {\left (9 k \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^{10}}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (-b+b (1+k) x^3-b k x^6+x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}+\frac {\left (12 (1+k) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^7}{\sqrt [3]{1-x^3} \sqrt [3]{1-k x^3} \left (b-b (1+k) x^3+b k x^6-x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\\ \end {align*}
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Mathematica [F] time = 2.07, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \left (5-4 (1+k) x+3 k x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(b+b k) x-b k x^2+x^5\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 2.90, size = 185, normalized size = 1.00 \begin {gather*} \frac {\log \left (x^2-\sqrt [3]{b} \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]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}}{\sqrt [3]{b} \sqrt [3]{k x^3+(-k-1) x^2+x}+2 x^2}\right )}{b^{2/3}}-\frac {\log \left (b^{2/3} \left (k x^3+(-k-1) x^2+x\right )^{2/3}+\sqrt [3]{b} x^2 \sqrt [3]{k x^3+(-k-1) x^2+x}+x^4\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}{{\left (x^{5} - b k x^{2} + {\left (b k + b\right )} x - b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.06, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \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 {1}{3}} \left (-b +\left (b k +b \right ) x -b k \,x^{2}+x^{5}\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 (3 \, k x^{2} - 4 \, {\left (k + 1\right )} x + 5\right )} x}{{\left (x^{5} - b k x^{2} + {\left (b k + b\right )} x - b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{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\,\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 )}^{1/3}\,\left (-x^5+b\,k\,x^2+\left (-b-b\,k\right )\,x+b\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 \left (3 k x^{2} - 4 k x - 4 x + 5\right )}{\sqrt [3]{x \left (x - 1\right ) \left (k x - 1\right )} \left (- b k x^{2} + b k x + b x - b + x^{5}\right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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