Optimal. Leaf size=250 \[ \frac {\log \left (a^2-\sqrt [3]{d} \left (x (-a-b)+a b+x^2\right )^{2/3}-2 a x+x^2\right )}{2 d^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{d} \left (x (-a-b)+a b+x^2\right )^{2/3}}{2 a^2+\sqrt [3]{d} \left (x (-a-b)+a b+x^2\right )^{2/3}-4 a x+2 x^2}\right )}{2 d^{2/3}}-\frac {\log \left (a^4-4 a^3 x+\left (x (-a-b)+a b+x^2\right )^{2/3} \left (a^2 \sqrt [3]{d}-2 a \sqrt [3]{d} x+\sqrt [3]{d} x^2\right )+6 a^2 x^2+d^{2/3} \left (x (-a-b)+a b+x^2\right )^{4/3}-4 a x^3+x^4\right )}{4 d^{2/3}} \]
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Rubi [F] time = 7.40, 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 {(-b+x) \left (-a (a-2 b)-2 b x+x^2\right )}{((-a+x) (-b+x))^{2/3} \left (a^4-b^2 d-2 \left (2 a^3-b d\right ) x+\left (6 a^2-d\right ) x^2-4 a x^3+x^4\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {(-b+x) \left (-a (a-2 b)-2 b x+x^2\right )}{((-a+x) (-b+x))^{2/3} \left (a^4-b^2 d-2 \left (2 a^3-b d\right ) x+\left (6 a^2-d\right ) x^2-4 a x^3+x^4\right )} \, dx &=\int \frac {\sqrt [3]{(a-x) (b-x)} (a-2 b+x)}{a^4-b^2 d-2 \left (2 a^3-b d\right ) x+\left (6 a^2-d\right ) x^2-4 a x^3+x^4} \, dx\\ &=\frac {\sqrt [3]{(a-x) (b-x)} \int \frac {\sqrt [3]{a-x} \sqrt [3]{b-x} (a-2 b+x)}{a^4-b^2 d-2 \left (2 a^3-b d\right ) x+\left (6 a^2-d\right ) x^2-4 a x^3+x^4} \, dx}{\sqrt [3]{a-x} \sqrt [3]{b-x}}\\ &=\frac {\sqrt [3]{(a-x) (b-x)} \int \left (\frac {2 \left (1-\frac {a}{2 b}\right ) b \sqrt [3]{a-x} \sqrt [3]{b-x}}{-a^4+b^2 d+2 \left (2 a^3-b d\right ) x-\left (6 a^2-d\right ) x^2+4 a x^3-x^4}+\frac {\sqrt [3]{a-x} \sqrt [3]{b-x} x}{a^4-b^2 d-2 \left (2 a^3-b d\right ) x+\left (6 a^2-d\right ) x^2-4 a x^3+x^4}\right ) \, dx}{\sqrt [3]{a-x} \sqrt [3]{b-x}}\\ &=\frac {\sqrt [3]{(a-x) (b-x)} \int \frac {\sqrt [3]{a-x} \sqrt [3]{b-x} x}{a^4-b^2 d-2 \left (2 a^3-b d\right ) x+\left (6 a^2-d\right ) x^2-4 a x^3+x^4} \, dx}{\sqrt [3]{a-x} \sqrt [3]{b-x}}+\frac {\left ((-a+2 b) \sqrt [3]{(a-x) (b-x)}\right ) \int \frac {\sqrt [3]{a-x} \sqrt [3]{b-x}}{-a^4+b^2 d+2 \left (2 a^3-b d\right ) x-\left (6 a^2-d\right ) x^2+4 a x^3-x^4} \, dx}{\sqrt [3]{a-x} \sqrt [3]{b-x}}\\ &=-\frac {\left (3 \sqrt [3]{(a-x) (b-x)}\right ) \operatorname {Subst}\left (\int \frac {x^3 \left (-a+x^3\right ) \sqrt [3]{-a+b+x^3}}{a^2 d+b^2 d+2 b d x^3-2 a d \left (b+x^3\right )+x^6 \left (d-x^6\right )} \, dx,x,\sqrt [3]{a-x}\right )}{\sqrt [3]{a-x} \sqrt [3]{b-x}}-\frac {\left (3 (-a+2 b) \sqrt [3]{(a-x) (b-x)}\right ) \operatorname {Subst}\left (\int \frac {x^3 \sqrt [3]{-a+b+x^3}}{a^2 d+b^2 d+2 b d x^3-2 a d \left (b+x^3\right )+x^6 \left (d-x^6\right )} \, dx,x,\sqrt [3]{a-x}\right )}{\sqrt [3]{a-x} \sqrt [3]{b-x}}\\ &=-\frac {\left (3 \sqrt [3]{(a-x) (b-x)}\right ) \operatorname {Subst}\left (\int \frac {x^3 \left (-a+x^3\right ) \sqrt [3]{-a+b+x^3}}{a^2 \left (1+\frac {b^2}{a^2}\right ) d+2 b d x^3-2 a d \left (b+x^3\right )+x^6 \left (d-x^6\right )} \, dx,x,\sqrt [3]{a-x}\right )}{\sqrt [3]{a-x} \sqrt [3]{b-x}}-\frac {\left (3 (-a+2 b) \sqrt [3]{(a-x) (b-x)}\right ) \operatorname {Subst}\left (\int \frac {x^3 \sqrt [3]{-a+b+x^3}}{a^2 \left (1+\frac {b^2}{a^2}\right ) d+2 b d x^3-2 a d \left (b+x^3\right )+x^6 \left (d-x^6\right )} \, dx,x,\sqrt [3]{a-x}\right )}{\sqrt [3]{a-x} \sqrt [3]{b-x}}\\ &=-\frac {\left (3 \sqrt [3]{(a-x) (b-x)}\right ) \operatorname {Subst}\left (\int \left (\frac {x^6 \sqrt [3]{-a+b+x^3}}{a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right ) d-2 a \left (1-\frac {b}{a}\right ) d x^3+d x^6-x^{12}}+\frac {a x^3 \sqrt [3]{-a+b+x^3}}{-a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right ) d+2 a \left (1-\frac {b}{a}\right ) d x^3-d x^6+x^{12}}\right ) \, dx,x,\sqrt [3]{a-x}\right )}{\sqrt [3]{a-x} \sqrt [3]{b-x}}-\frac {\left (3 (-a+2 b) \sqrt [3]{(a-x) (b-x)}\right ) \operatorname {Subst}\left (\int \frac {x^3 \sqrt [3]{-a+b+x^3}}{a^2 \left (1+\frac {b^2}{a^2}\right ) d+2 b d x^3-2 a d \left (b+x^3\right )+x^6 \left (d-x^6\right )} \, dx,x,\sqrt [3]{a-x}\right )}{\sqrt [3]{a-x} \sqrt [3]{b-x}}\\ &=-\frac {\left (3 \sqrt [3]{(a-x) (b-x)}\right ) \operatorname {Subst}\left (\int \frac {x^6 \sqrt [3]{-a+b+x^3}}{a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right ) d-2 a \left (1-\frac {b}{a}\right ) d x^3+d x^6-x^{12}} \, dx,x,\sqrt [3]{a-x}\right )}{\sqrt [3]{a-x} \sqrt [3]{b-x}}-\frac {\left (3 a \sqrt [3]{(a-x) (b-x)}\right ) \operatorname {Subst}\left (\int \frac {x^3 \sqrt [3]{-a+b+x^3}}{-a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right ) d+2 a \left (1-\frac {b}{a}\right ) d x^3-d x^6+x^{12}} \, dx,x,\sqrt [3]{a-x}\right )}{\sqrt [3]{a-x} \sqrt [3]{b-x}}-\frac {\left (3 (-a+2 b) \sqrt [3]{(a-x) (b-x)}\right ) \operatorname {Subst}\left (\int \frac {x^3 \sqrt [3]{-a+b+x^3}}{a^2 \left (1+\frac {b^2}{a^2}\right ) d+2 b d x^3-2 a d \left (b+x^3\right )+x^6 \left (d-x^6\right )} \, dx,x,\sqrt [3]{a-x}\right )}{\sqrt [3]{a-x} \sqrt [3]{b-x}}\\ \end {align*}
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Mathematica [F] time = 1.91, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-b+x) \left (-a (a-2 b)-2 b x+x^2\right )}{((-a+x) (-b+x))^{2/3} \left (a^4-b^2 d-2 \left (2 a^3-b d\right ) x+\left (6 a^2-d\right ) x^2-4 a x^3+x^4\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 1.05, size = 250, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{d} \left (a b+(-a-b) x+x^2\right )^{2/3}}{2 a^2-4 a x+2 x^2+\sqrt [3]{d} \left (a b+(-a-b) x+x^2\right )^{2/3}}\right )}{2 d^{2/3}}+\frac {\log \left (a^2-2 a x+x^2-\sqrt [3]{d} \left (a b+(-a-b) x+x^2\right )^{2/3}\right )}{2 d^{2/3}}-\frac {\log \left (a^4-4 a^3 x+6 a^2 x^2-4 a x^3+x^4+d^{2/3} \left (a b+(-a-b) x+x^2\right )^{4/3}+\left (a b+(-a-b) x+x^2\right )^{2/3} \left (a^2 \sqrt [3]{d}-2 a \sqrt [3]{d} x+\sqrt [3]{d} x^2\right )\right )}{4 d^{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 (a - 2 \, b\right )} a + 2 \, b x - x^{2}\right )} {\left (b - x\right )}}{{\left (a^{4} - 4 \, a x^{3} + x^{4} - b^{2} d + {\left (6 \, a^{2} - d\right )} x^{2} - 2 \, {\left (2 \, a^{3} - b d\right )} x\right )} \left ({\left (a - x\right )} {\left (b - x\right )}\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 = 0.12, size = 0, normalized size = 0.00 \[\int \frac {\left (-b +x \right ) \left (-a \left (a -2 b \right )-2 b x +x^{2}\right )}{\left (\left (-a +x \right ) \left (-b +x \right )\right )^{\frac {2}{3}} \left (a^{4}-b^{2} d -2 \left (2 a^{3}-b d \right ) x +\left (6 a^{2}-d \right ) x^{2}-4 a \,x^{3}+x^{4}\right )}\, dx\]
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*} \int \frac {{\left ({\left (a - 2 \, b\right )} a + 2 \, b x - x^{2}\right )} {\left (b - x\right )}}{{\left (a^{4} - 4 \, a x^{3} + x^{4} - b^{2} d + {\left (6 \, a^{2} - d\right )} x^{2} - 2 \, {\left (2 \, a^{3} - b d\right )} x\right )} \left ({\left (a - x\right )} {\left (b - x\right )}\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 = -0.00 \begin {gather*} \int -\frac {\left (b-x\right )\,\left (-x^2+2\,b\,x+a\,\left (a-2\,b\right )\right )}{{\left (\left (a-x\right )\,\left (b-x\right )\right )}^{2/3}\,\left (x^2\,\left (d-6\,a^2\right )-2\,x\,\left (b\,d-2\,a^3\right )+b^2\,d+4\,a\,x^3-a^4-x^4\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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