Optimal. Leaf size=34 \[ e^{\frac {3}{3+\log \left (\frac {(3-x)^2 x^3}{4 \left (-x+x^2\right )^2}\right )}} x \]
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Rubi [F]
time = 2.20, 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 {e^{\frac {3}{3+\log \left (\frac {9 x-6 x^2+x^3}{4-8 x+4 x^2}\right )}} \left (18-36 x+6 x^2+\left (18-24 x+6 x^2\right ) \log \left (\frac {9 x-6 x^2+x^3}{4-8 x+4 x^2}\right )+\left (3-4 x+x^2\right ) \log ^2\left (\frac {9 x-6 x^2+x^3}{4-8 x+4 x^2}\right )\right )}{27-36 x+9 x^2+\left (18-24 x+6 x^2\right ) \log \left (\frac {9 x-6 x^2+x^3}{4-8 x+4 x^2}\right )+\left (3-4 x+x^2\right ) \log ^2\left (\frac {9 x-6 x^2+x^3}{4-8 x+4 x^2}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {3}{3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )}} \left (6 \left (3-6 x+x^2\right )+6 \left (3-4 x+x^2\right ) \log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )+\left (3-4 x+x^2\right ) \log ^2\left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )\right )}{\left (3-4 x+x^2\right ) \left (3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )\right )^2} \, dx\\ &=\int \left (e^{\frac {3}{3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )}}-\frac {3 e^{\frac {3}{3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )}} \left (3+x^2\right )}{(-3+x) (-1+x) \left (3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )\right )^2}\right ) \, dx\\ &=-\left (3 \int \frac {e^{\frac {3}{3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )}} \left (3+x^2\right )}{(-3+x) (-1+x) \left (3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )\right )^2} \, dx\right )+\int e^{\frac {3}{3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )}} \, dx\\ &=-\left (3 \int \left (\frac {e^{\frac {3}{3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )}}}{\left (3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )\right )^2}+\frac {6 e^{\frac {3}{3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )}}}{(-3+x) \left (3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )\right )^2}-\frac {2 e^{\frac {3}{3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )}}}{(-1+x) \left (3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )\right )^2}\right ) \, dx\right )+\int e^{\frac {3}{3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )}} \, dx\\ &=-\left (3 \int \frac {e^{\frac {3}{3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )}}}{\left (3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )\right )^2} \, dx\right )+6 \int \frac {e^{\frac {3}{3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )}}}{(-1+x) \left (3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )\right )^2} \, dx-18 \int \frac {e^{\frac {3}{3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )}}}{(-3+x) \left (3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )\right )^2} \, dx+\int e^{\frac {3}{3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )}} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.08, size = 26, normalized size = 0.76 \begin {gather*} e^{\frac {3}{3+\log \left (\frac {(-3+x)^2 x}{4 (-1+x)^2}\right )}} x \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.23, size = 36, normalized size = 1.06
method | result | size |
risch | \(x \,{\mathrm e}^{\frac {3}{\ln \left (\frac {x^{3}-6 x^{2}+9 x}{4 x^{2}-8 x +4}\right )+3}}\) | \(36\) |
norman | \(\frac {\ln \left (\frac {x^{3}-6 x^{2}+9 x}{4 x^{2}-8 x +4}\right ) x \,{\mathrm e}^{\frac {3}{\ln \left (\frac {x^{3}-6 x^{2}+9 x}{4 x^{2}-8 x +4}\right )+3}}+3 x \,{\mathrm e}^{\frac {3}{\ln \left (\frac {x^{3}-6 x^{2}+9 x}{4 x^{2}-8 x +4}\right )+3}}}{\ln \left (\frac {x^{3}-6 x^{2}+9 x}{4 x^{2}-8 x +4}\right )+3}\) | \(130\) |
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*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 34, normalized size = 1.00 \begin {gather*} x e^{\left (\frac {3}{\log \left (\frac {x^{3} - 6 \, x^{2} + 9 \, x}{4 \, {\left (x^{2} - 2 \, x + 1\right )}}\right ) + 3}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 7.42, size = 29, normalized size = 0.85 \begin {gather*} x e^{\frac {3}{\log {\left (\frac {x^{3} - 6 x^{2} + 9 x}{4 x^{2} - 8 x + 4} \right )} + 3}} \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*} \text {could not integrate} \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 = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{\frac {3}{\ln \left (\frac {x^3-6\,x^2+9\,x}{4\,x^2-8\,x+4}\right )+3}}\,\left (\ln \left (\frac {x^3-6\,x^2+9\,x}{4\,x^2-8\,x+4}\right )\,\left (6\,x^2-24\,x+18\right )-36\,x+{\ln \left (\frac {x^3-6\,x^2+9\,x}{4\,x^2-8\,x+4}\right )}^2\,\left (x^2-4\,x+3\right )+6\,x^2+18\right )}{\ln \left (\frac {x^3-6\,x^2+9\,x}{4\,x^2-8\,x+4}\right )\,\left (6\,x^2-24\,x+18\right )-36\,x+{\ln \left (\frac {x^3-6\,x^2+9\,x}{4\,x^2-8\,x+4}\right )}^2\,\left (x^2-4\,x+3\right )+9\,x^2+27} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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