Optimal. Leaf size=99 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \left (x^3 (-a-b)+a b x^2+x^4\right )^{3/4}}{x^2 (x-b)}\right )}{d^{3/4}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \left (x^3 (-a-b)+a b x^2+x^4\right )^{3/4}}{x^2 (x-b)}\right )}{d^{3/4}} \]
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Rubi [F] time = 13.12, 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 {\left (a^2-2 a x+x^2\right ) \left (-2 a b x+(3 a-b) x^2\right )}{\left (x^2 (-a+x) (-b+x)\right )^{3/4} \left (a^3 d-3 a^2 d x+(-b+3 a d) x^2+(1-d) x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {\left (a^2-2 a x+x^2\right ) \left (-2 a b x+(3 a-b) x^2\right )}{\left (x^2 (-a+x) (-b+x)\right )^{3/4} \left (a^3 d-3 a^2 d x+(-b+3 a d) x^2+(1-d) x^3\right )} \, dx &=\int \frac {(-a+x)^2 \left (-2 a b x+(3 a-b) x^2\right )}{\left (x^2 (-a+x) (-b+x)\right )^{3/4} \left (a^3 d-3 a^2 d x+(-b+3 a d) x^2+(1-d) x^3\right )} \, dx\\ &=\int \frac {x (-a+x)^2 (-2 a b+(3 a-b) x)}{\left (x^2 (-a+x) (-b+x)\right )^{3/4} \left (a^3 d-3 a^2 d x+(-b+3 a d) x^2+(1-d) x^3\right )} \, dx\\ &=\frac {\left (x^{3/2} (-a+x)^{3/4} (-b+x)^{3/4}\right ) \int \frac {(-a+x)^{5/4} (-2 a b+(3 a-b) x)}{\sqrt {x} (-b+x)^{3/4} \left (a^3 d-3 a^2 d x+(-b+3 a d) x^2+(1-d) x^3\right )} \, dx}{\left (x^2 (-a+x) (-b+x)\right )^{3/4}}\\ &=\frac {\left (2 x^{3/2} (-a+x)^{3/4} (-b+x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {\left (-a+x^2\right )^{5/4} \left (-2 a b+(3 a-b) x^2\right )}{\left (-b+x^2\right )^{3/4} \left (a^3 d-3 a^2 d x^2+(-b+3 a d) x^4+(1-d) x^6\right )} \, dx,x,\sqrt {x}\right )}{\left (x^2 (-a+x) (-b+x)\right )^{3/4}}\\ &=\frac {\left (2 x^{3/2} (-a+x)^{3/4} (-b+x)^{3/4}\right ) \operatorname {Subst}\left (\int \left (\frac {2 a b \left (-a+x^2\right )^{5/4}}{\left (-b+x^2\right )^{3/4} \left (-a^3 d+3 a^2 d x^2+b \left (1-\frac {3 a d}{b}\right ) x^4-(1-d) x^6\right )}+\frac {(3 a-b) x^2 \left (-a+x^2\right )^{5/4}}{\left (-b+x^2\right )^{3/4} \left (a^3 d-3 a^2 d x^2-b \left (1-\frac {3 a d}{b}\right ) x^4+(1-d) x^6\right )}\right ) \, dx,x,\sqrt {x}\right )}{\left (x^2 (-a+x) (-b+x)\right )^{3/4}}\\ &=\frac {\left (2 (3 a-b) x^{3/2} (-a+x)^{3/4} (-b+x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (-a+x^2\right )^{5/4}}{\left (-b+x^2\right )^{3/4} \left (a^3 d-3 a^2 d x^2-b \left (1-\frac {3 a d}{b}\right ) x^4+(1-d) x^6\right )} \, dx,x,\sqrt {x}\right )}{\left (x^2 (-a+x) (-b+x)\right )^{3/4}}+\frac {\left (4 a b x^{3/2} (-a+x)^{3/4} (-b+x)^{3/4}\right ) \operatorname {Subst}\left (\int \frac {\left (-a+x^2\right )^{5/4}}{\left (-b+x^2\right )^{3/4} \left (-a^3 d+3 a^2 d x^2+b \left (1-\frac {3 a d}{b}\right ) x^4-(1-d) x^6\right )} \, dx,x,\sqrt {x}\right )}{\left (x^2 (-a+x) (-b+x)\right )^{3/4}}\\ \end {align*}
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Mathematica [F] time = 4.24, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a^2-2 a x+x^2\right ) \left (-2 a b x+(3 a-b) x^2\right )}{\left (x^2 (-a+x) (-b+x)\right )^{3/4} \left (a^3 d-3 a^2 d x+(-b+3 a d) x^2+(1-d) x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 5.79, size = 99, normalized size = 1.00 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \left (a b x^2+(-a-b) x^3+x^4\right )^{3/4}}{x^2 (-b+x)}\right )}{d^{3/4}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \left (a b x^2+(-a-b) x^3+x^4\right )^{3/4}}{x^2 (-b+x)}\right )}{d^{3/4}} \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 (2 \, a b x - {\left (3 \, a - b\right )} x^{2}\right )} {\left (a^{2} - 2 \, a x + x^{2}\right )}}{{\left (a^{3} d - 3 \, a^{2} d x - {\left (d - 1\right )} x^{3} + {\left (3 \, a d - b\right )} x^{2}\right )} \left ({\left (a - x\right )} {\left (b - x\right )} x^{2}\right )^{\frac {3}{4}}}\,{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 \[\int \frac {\left (a^{2}-2 a x +x^{2}\right ) \left (-2 a b x +\left (3 a -b \right ) x^{2}\right )}{\left (x^{2} \left (-a +x \right ) \left (-b +x \right )\right )^{\frac {3}{4}} \left (a^{3} d -3 a^{2} d x +\left (3 a d -b \right ) x^{2}+\left (1-d \right ) x^{3}\right )}\, dx\]
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 (2 \, a b x - {\left (3 \, a - b\right )} x^{2}\right )} {\left (a^{2} - 2 \, a x + x^{2}\right )}}{{\left (a^{3} d - 3 \, a^{2} d x - {\left (d - 1\right )} x^{3} + {\left (3 \, a d - b\right )} x^{2}\right )} \left ({\left (a - x\right )} {\left (b - x\right )} x^{2}\right )^{\frac {3}{4}}}\,{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 {\left (x^2\,\left (3\,a-b\right )-2\,a\,b\,x\right )\,\left (a^2-2\,a\,x+x^2\right )}{{\left (x^2\,\left (a-x\right )\,\left (b-x\right )\right )}^{3/4}\,\left (x^2\,\left (b-3\,a\,d\right )-a^3\,d+x^3\,\left (d-1\right )+3\,a^2\,d\,x\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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