Optimal. Leaf size=343 \[ -\frac {\log \left (-\sqrt [6]{b} x+\sqrt [6]{b}+\sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{2 b^{2/3}}-\frac {\log \left (\sqrt [6]{b} x-\sqrt [6]{b}+\sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{2 b^{2/3}}+\frac {\log \left (\left (\sqrt [6]{b}-\sqrt [6]{b} x\right ) \sqrt [3]{k x^3+(-k-1) x^2+x}+\sqrt [3]{b} x^2-2 \sqrt [3]{b} x+\sqrt [3]{b}+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{4 b^{2/3}}+\frac {\log \left (\left (\sqrt [6]{b} x-\sqrt [6]{b}\right ) \sqrt [3]{k x^3+(-k-1) x^2+x}+\sqrt [3]{b} x^2-2 \sqrt [3]{b} x+\sqrt [3]{b}+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{4 b^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \left (k x^3+(-k-1) x^2+x\right )^{2/3}}{\sqrt {3} \sqrt [3]{b}}+\frac {x^2}{\sqrt {3}}-\frac {2 x}{\sqrt {3}}+\frac {1}{\sqrt {3}}}{(x-1)^2}\right )}{2 b^{2/3}} \]
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Rubi [F] time = 10.66, 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+(-1+2 k) x) \left (1-2 x+x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+4 b x+(1-6 b) x^2+(4 b-2 k) x^3+\left (-b+k^2\right ) x^4\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {align*} \int \frac {(-1+(-1+2 k) x) \left (1-2 x+x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+4 b x+(1-6 b) x^2+(4 b-2 k) x^3+\left (-b+k^2\right ) x^4\right )} \, dx &=\int \frac {(-1+x)^2 (-1+(-1+2 k) x)}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+4 b x+(1-6 b) x^2+(4 b-2 k) x^3+\left (-b+k^2\right ) x^4\right )} \, dx\\ &=\frac {\left (\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \int \frac {(-1+x)^2 (-1+(-1+2 k) x)}{\sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \left (-b+4 b x+(1-6 b) x^2+(4 b-2 k) x^3+\left (-b+k^2\right ) x^4\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 {(1-x)^{5/3} (-1+(-1+2 k) x)}{\sqrt [3]{x} \sqrt [3]{1-k x} \left (-b+4 b x+(1-6 b) x^2+(4 b-2 k) x^3+\left (-b+k^2\right ) x^4\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 \left (1-x^3\right )^{5/3} \left (-1+(-1+2 k) x^3\right )}{\sqrt [3]{1-k x^3} \left (-b+4 b x^3+(1-6 b) x^6+(4 b-2 k) x^9+\left (-b+k^2\right ) x^{12}\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 \frac {x \left (1-x^3\right )^{5/3} \left (1-(-1+2 k) x^3\right )}{\sqrt [3]{1-k x^3} \left (b \left (-1+x^3\right )^4-x^6 \left (-1+k x^3\right )^2\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 {x \left (1-x^3\right )^{5/3}}{\sqrt [3]{1-k x^3} \left (b-4 b x^3-(1-6 b) x^6-4 b \left (1-\frac {k}{2 b}\right ) x^9+b \left (1-\frac {k^2}{b}\right ) x^{12}\right )}+\frac {(1-2 k) x^4 \left (1-x^3\right )^{5/3}}{\sqrt [3]{1-k x^3} \left (b-4 b x^3-(1-6 b) x^6-4 b \left (1-\frac {k}{2 b}\right ) x^9+b \left (1-\frac {k^2}{b}\right ) x^{12}\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 \frac {x \left (1-x^3\right )^{5/3}}{\sqrt [3]{1-k x^3} \left (b-4 b x^3-(1-6 b) x^6-4 b \left (1-\frac {k}{2 b}\right ) x^9+b \left (1-\frac {k^2}{b}\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}+\frac {\left (3 (1-2 k) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (1-x^3\right )^{5/3}}{\sqrt [3]{1-k x^3} \left (b-4 b x^3-(1-6 b) x^6-4 b \left (1-\frac {k}{2 b}\right ) x^9+b \left (1-\frac {k^2}{b}\right ) x^{12}\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 \frac {x \left (1-x^3\right )^{5/3}}{\sqrt [3]{1-k x^3} \left (b \left (-1+x^3\right )^4-x^6 \left (-1+k x^3\right )^2\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}+\frac {\left (3 (1-2 k) \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x}\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (1-x^3\right )^{5/3}}{\sqrt [3]{1-k x^3} \left (b \left (-1+x^3\right )^4-x^6 \left (-1+k x^3\right )^2\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(1-x) x (1-k x)}}\\ \end {align*}
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Mathematica [F] time = 4.59, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-1+(-1+2 k) x) \left (1-2 x+x^2\right )}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+4 b x+(1-6 b) x^2+(4 b-2 k) x^3+\left (-b+k^2\right ) x^4\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 5.34, size = 276, normalized size = 0.80 \begin {gather*} -\frac {\log \left (-\sqrt [3]{b} x^2+2 \sqrt [3]{b} x-\sqrt [3]{b}+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{2 b^{2/3}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} x^2-2 \sqrt {3} \sqrt [3]{b} x+\sqrt {3} \sqrt [3]{b}}{\sqrt [3]{b} x^2-2 \sqrt [3]{b} x+\sqrt [3]{b}+2 \left (k x^3+(-k-1) x^2+x\right )^{2/3}}\right )}{2 b^{2/3}}+\frac {\log \left (b^{2/3} x^4-4 b^{2/3} x^3+6 b^{2/3} x^2-4 b^{2/3} x+b^{2/3}+\left (\sqrt [3]{b} x^2-2 \sqrt [3]{b} x+\sqrt [3]{b}\right ) \left (k x^3+(-k-1) x^2+x\right )^{2/3}+\left (k x^3+(-k-1) x^2+x\right )^{4/3}\right )}{4 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 ({\left (2 \, k - 1\right )} x - 1\right )} {\left (x^{2} - 2 \, x + 1\right )}}{{\left ({\left (k^{2} - b\right )} x^{4} + 2 \, {\left (2 \, b - k\right )} x^{3} - {\left (6 \, b - 1\right )} x^{2} + 4 \, b 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 = 0.09, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-1+\left (-1+2 k \right ) x \right ) \left (x^{2}-2 x +1\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (-b +4 b x +\left (1-6 b \right ) x^{2}+\left (4 b -2 k \right ) x^{3}+\left (k^{2}-b \right ) x^{4}\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 ({\left (2 \, k - 1\right )} x - 1\right )} {\left (x^{2} - 2 \, x + 1\right )}}{{\left ({\left (k^{2} - b\right )} x^{4} + 2 \, {\left (2 \, b - k\right )} x^{3} - {\left (6 \, b - 1\right )} x^{2} + 4 \, b 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 = -0.00 \begin {gather*} \int -\frac {\left (x\,\left (2\,k-1\right )-1\right )\,\left (x^2-2\,x+1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (\left (b-k^2\right )\,x^4+\left (2\,k-4\,b\right )\,x^3+\left (6\,b-1\right )\,x^2-4\,b\,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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