Optimal. Leaf size=29 \[ x^{1+\frac {e^{-2 e^{1+5 \left (-e^{1-x}+x\right )}}}{x^2}} \]
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Rubi [F] time = 5.45, 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 e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \left (1+e^{2 e^{1-5 e^{1-x}+5 x}} x^2-2 \log (x)+e^{-5 e^{1-x}+5 x} \left (-10 e x-10 e^{2-x} x\right ) \log (x)\right ) \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}}+x^{\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}}-2 e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \log (x)-10 e^{1-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+4 x} \left (e+e^x\right ) x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \log (x)\right ) \, dx\\ &=-\left (2 \int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \log (x) \, dx\right )-10 \int e^{1-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+4 x} \left (e+e^x\right ) x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \log (x) \, dx+\int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx+\int x^{\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx\\ &=2 \int \frac {\int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx}{x} \, dx+10 \int \frac {\int e^{2-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+4 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx+\int e^{1-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+5 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx}{x} \, dx-(2 \log (x)) \int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx-(10 \log (x)) \int e^{2-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+4 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx-(10 \log (x)) \int e^{1-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+5 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx+\int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx+\int x^{\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx\\ &=2 \int \frac {\int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx}{x} \, dx+10 \int \left (\frac {\int e^{2-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+4 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx}{x}+\frac {\int e^{1-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+5 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx}{x}\right ) \, dx-(2 \log (x)) \int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx-(10 \log (x)) \int e^{2-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+4 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx-(10 \log (x)) \int e^{1-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+5 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx+\int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx+\int x^{\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx\\ &=2 \int \frac {\int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx}{x} \, dx+10 \int \frac {\int e^{2-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+4 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx}{x} \, dx+10 \int \frac {\int e^{1-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+5 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx}{x} \, dx-(2 \log (x)) \int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx-(10 \log (x)) \int e^{2-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+4 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx-(10 \log (x)) \int e^{1-5 e^{1-x}-2 e^{1-5 e^{1-x}+5 x}+5 x} x^{-1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx+\int e^{-2 e^{1-5 e^{1-x}+5 x}} x^{-2+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx+\int x^{\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.30, size = 28, normalized size = 0.97 \begin {gather*} x^{1+\frac {e^{-2 e^{1-5 e^{1-x}+5 x}}}{x^2}} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.66, size = 32, normalized size = 1.10 \begin {gather*} x x^{\frac {e^{\left (-2 \, e^{\left ({\left ({\left (5 \, x + 1\right )} e - 5 \, e^{\left (-x + 2\right )}\right )} e^{\left (-1\right )}\right )}\right )}}{x^{2}}} \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 (x^{2} e^{\left (2 \, e^{\left (5 \, x - 5 \, e^{\left (-x + 1\right )} + 1\right )}\right )} - 10 \, {\left (x e + x e^{\left (-x + 2\right )}\right )} e^{\left (5 \, x - 5 \, e^{\left (-x + 1\right )}\right )} \log \relax (x) - 2 \, \log \relax (x) + 1\right )} x^{\frac {e^{\left (-2 \, e^{\left (5 \, x - 5 \, e^{\left (-x + 1\right )} + 1\right )}\right )}}{x^{2}}} e^{\left (-2 \, e^{\left (5 \, x - 5 \, e^{\left (-x + 1\right )} + 1\right )}\right )}}{x^{2}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.12, size = 26, normalized size = 0.90
method | result | size |
risch | \(x^{\frac {{\mathrm e}^{-2 \,{\mathrm e}^{1-5 \,{\mathrm e}^{1-x}+5 x}}}{x^{2}}} x\) | \(26\) |
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 {\left (x^{2} e^{\left (2 \, e^{\left (5 \, x - 5 \, e^{\left (-x + 1\right )} + 1\right )}\right )} - 10 \, {\left (x e + x e^{\left (-x + 2\right )}\right )} e^{\left (5 \, x - 5 \, e^{\left (-x + 1\right )}\right )} \log \relax (x) - 2 \, \log \relax (x) + 1\right )} x^{\frac {e^{\left (-2 \, e^{\left (5 \, x - 5 \, e^{\left (-x + 1\right )} + 1\right )}\right )}}{x^{2}} - 2} e^{\left (-2 \, e^{\left (5 \, x - 5 \, e^{\left (-x + 1\right )} + 1\right )}\right )}\,{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 = -0.03 \begin {gather*} \int -\frac {{\mathrm {e}}^{\frac {{\mathrm {e}}^{-2\,\mathrm {e}\,{\mathrm {e}}^{5\,x-5\,{\mathrm {e}}^{1-x}}}\,\ln \relax (x)}{x^2}}\,{\mathrm {e}}^{-2\,\mathrm {e}\,{\mathrm {e}}^{5\,x-5\,{\mathrm {e}}^{1-x}}}\,\left (2\,\ln \relax (x)-x^2\,{\mathrm {e}}^{2\,\mathrm {e}\,{\mathrm {e}}^{5\,x-5\,{\mathrm {e}}^{1-x}}}+{\mathrm {e}}^{5\,x-5\,{\mathrm {e}}^{1-x}}\,\ln \relax (x)\,\left (10\,x\,\mathrm {e}+10\,x\,\mathrm {e}\,{\mathrm {e}}^{1-x}\right )-1\right )}{x^2} \,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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