Optimal. Leaf size=28 \[ e^{-3+x-e^{2 x} \left (5+2 x-\frac {\log ((-3+x) x)}{x}\right )} \]
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Rubi [F] time = 18.76, 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 {\exp \left (\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{-3 x^2+x^3} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {-3 x+x^2+e^{2 x} \left (-5 x-2 x^2\right )+e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \left (-3 x^2+x^3+e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )+e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{(-3+x) x^2} \, dx\\ &=\int \frac {\exp \left (-3+x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \left (3 x^2-x^3-e^{2 x} \left (-3+2 x+36 x^2-4 x^4\right )-e^{2 x} \left (3-7 x+2 x^2\right ) \log \left (-3 x+x^2\right )\right )}{(3-x) x^2} \, dx\\ &=\int \left (\exp \left (-3+x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right )-\frac {\exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \left (3-2 x-36 x^2+4 x^4-3 \log ((-3+x) x)+7 x \log ((-3+x) x)-2 x^2 \log ((-3+x) x)\right )}{(-3+x) x^2}\right ) \, dx\\ &=\int \exp \left (-3+x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \, dx-\int \frac {\exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \left (3-2 x-36 x^2+4 x^4-3 \log ((-3+x) x)+7 x \log ((-3+x) x)-2 x^2 \log ((-3+x) x)\right )}{(-3+x) x^2} \, dx\\ &=\int \exp \left (-3+x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \, dx-\int \left (\frac {\exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \left (3-2 x-36 x^2+4 x^4\right )}{(-3+x) x^2}-\frac {\exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) (-1+2 x) \log ((-3+x) x)}{x^2}\right ) \, dx\\ &=\int \exp \left (-3+x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \, dx-\int \frac {\exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \left (3-2 x-36 x^2+4 x^4\right )}{(-3+x) x^2} \, dx+\int \frac {\exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) (-1+2 x) \log ((-3+x) x)}{x^2} \, dx\\ &=\int \exp \left (-3+x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \, dx-\int \left (12 \exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right )-\frac {\exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right )}{3 (-3+x)}-\frac {\exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right )}{x^2}+\frac {\exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right )}{3 x}+4 \exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) x\right ) \, dx+\int \left (-\frac {\exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \log ((-3+x) x)}{x^2}+\frac {2 \exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \log ((-3+x) x)}{x}\right ) \, dx\\ &=\frac {1}{3} \int \frac {\exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right )}{-3+x} \, dx-\frac {1}{3} \int \frac {\exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right )}{x} \, dx+2 \int \frac {\exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \log ((-3+x) x)}{x} \, dx-4 \int \exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) x \, dx-12 \int \exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \, dx+\int \exp \left (-3+x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \, dx+\int \frac {\exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right )}{x^2} \, dx-\int \frac {\exp \left (-3+3 x+\frac {e^{2 x} \left (-5 x-2 x^2\right )}{x}+\frac {e^{2 x} \log \left (-3 x+x^2\right )}{x}\right ) \log ((-3+x) x)}{x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.17, size = 33, normalized size = 1.18 \begin {gather*} e^{-3+x-e^{2 x} (5+2 x)} ((-3+x) x)^{\frac {e^{2 x}}{x}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.63, size = 40, normalized size = 1.43 \begin {gather*} e^{\left (\frac {x^{2} - {\left (2 \, x^{2} + 5 \, x\right )} e^{\left (2 \, x\right )} + e^{\left (2 \, x\right )} \log \left (x^{2} - 3 \, x\right ) - 3 \, x}{x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.37, size = 33, normalized size = 1.18 \begin {gather*} e^{\left (-2 \, x e^{\left (2 \, x\right )} + x + \frac {e^{\left (2 \, x\right )} \log \left (x^{2} - 3 \, x\right )}{x} - 5 \, e^{\left (2 \, x\right )} - 3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.19, size = 149, normalized size = 5.32
method | result | size |
risch | \(x^{\frac {{\mathrm e}^{2 x}}{x}} \left (x -3\right )^{\frac {{\mathrm e}^{2 x}}{x}} {\mathrm e}^{-\frac {i {\mathrm e}^{2 x} \pi \mathrm {csgn}\left (i x \left (x -3\right )\right )^{3}-i {\mathrm e}^{2 x} \pi \mathrm {csgn}\left (i x \left (x -3\right )\right )^{2} \mathrm {csgn}\left (i x \right )-i {\mathrm e}^{2 x} \pi \mathrm {csgn}\left (i x \left (x -3\right )\right )^{2} \mathrm {csgn}\left (i \left (x -3\right )\right )+i {\mathrm e}^{2 x} \pi \,\mathrm {csgn}\left (i x \left (x -3\right )\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x -3\right )\right )+4 \,{\mathrm e}^{2 x} x^{2}+10 x \,{\mathrm e}^{2 x}-2 x^{2}+6 x}{2 x}}\) | \(149\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.54, size = 39, normalized size = 1.39 \begin {gather*} e^{\left (-2 \, x e^{\left (2 \, x\right )} + x + \frac {e^{\left (2 \, x\right )} \log \left (x - 3\right )}{x} + \frac {e^{\left (2 \, x\right )} \log \relax (x)}{x} - 5 \, e^{\left (2 \, x\right )} - 3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.36, size = 36, normalized size = 1.29 \begin {gather*} {\mathrm {e}}^{-5\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^{-2\,x\,{\mathrm {e}}^{2\,x}}\,{\mathrm {e}}^x\,{\left (x^2-3\,x\right )}^{\frac {{\mathrm {e}}^{2\,x}}{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.96, size = 37, normalized size = 1.32 \begin {gather*} e^{\frac {x^{2} - 3 x + \left (- 2 x^{2} - 5 x\right ) e^{2 x} + e^{2 x} \log {\left (x^{2} - 3 x \right )}}{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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