Optimal. Leaf size=30 \[ \log \left (4 \left (-2+e^{\frac {x+\frac {1}{\log (x)}}{5 \left (-5+e^x-x\right ) x}}\right )\right ) \]
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Rubi [F]
time = 56.39, 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 {1+x \log (x)}{\left (-25 x+5 e^x x-5 x^2\right ) \log (x)}\right ) \left (5-e^x+x+\left (5+e^x (-1-x)+2 x\right ) \log (x)+\left (x^2-e^x x^2\right ) \log ^2(x)\right )}{\exp \left (\frac {1+x \log (x)}{\left (-25 x+5 e^x x-5 x^2\right ) \log (x)}\right ) \left (125 x^2+5 e^{2 x} x^2+50 x^3+5 x^4+e^x \left (-50 x^2-10 x^3\right )\right ) \log ^2(x)+\left (-250 x^2-10 e^{2 x} x^2-100 x^3-10 x^4+e^x \left (100 x^2+20 x^3\right )\right ) \log ^2(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {5-e^x+x-\left (-5-2 x+e^x (1+x)\right ) \log (x)-\left (-1+e^x\right ) x^2 \log ^2(x)}{5 \left (1-2 e^{\frac {1+x \log (x)}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) x^2 \left (5-e^x+x\right )^2 \log ^2(x)} \, dx\\ &=\frac {1}{5} \int \frac {5-e^x+x-\left (-5-2 x+e^x (1+x)\right ) \log (x)-\left (-1+e^x\right ) x^2 \log ^2(x)}{\left (1-2 e^{\frac {1+x \log (x)}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) x^2 \left (5-e^x+x\right )^2 \log ^2(x)} \, dx\\ &=\frac {1}{5} \int \left (-\frac {1}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) \left (-5+e^x-x\right )^2}+\frac {e^x}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) \left (-5+e^x-x\right )^2}+\frac {e^x}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) \left (-5+e^x-x\right )^2 x^2 \log ^2(x)}-\frac {5}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) x^2 \left (5-e^x+x\right )^2 \log ^2(x)}-\frac {1}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) x \left (5-e^x+x\right )^2 \log ^2(x)}+\frac {e^x}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) \left (-5+e^x-x\right )^2 x^2 \log (x)}+\frac {e^x}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) \left (-5+e^x-x\right )^2 x \log (x)}-\frac {5}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) x^2 \left (5-e^x+x\right )^2 \log (x)}-\frac {2}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) x \left (5-e^x+x\right )^2 \log (x)}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {1}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) \left (-5+e^x-x\right )^2} \, dx\right )+\frac {1}{5} \int \frac {e^x}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) \left (-5+e^x-x\right )^2} \, dx+\frac {1}{5} \int \frac {e^x}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) \left (-5+e^x-x\right )^2 x^2 \log ^2(x)} \, dx-\frac {1}{5} \int \frac {1}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) x \left (5-e^x+x\right )^2 \log ^2(x)} \, dx+\frac {1}{5} \int \frac {e^x}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) \left (-5+e^x-x\right )^2 x^2 \log (x)} \, dx+\frac {1}{5} \int \frac {e^x}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) \left (-5+e^x-x\right )^2 x \log (x)} \, dx-\frac {2}{5} \int \frac {1}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) x \left (5-e^x+x\right )^2 \log (x)} \, dx-\int \frac {1}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) x^2 \left (5-e^x+x\right )^2 \log ^2(x)} \, dx-\int \frac {1}{\left (-1+2 e^{\frac {1}{5 \left (5-e^x+x\right )}+\frac {1}{5 x \left (5-e^x+x\right ) \log (x)}}\right ) x^2 \left (5-e^x+x\right )^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [F]
time = 0.81, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {1+x \log (x)}{\left (-25 x+5 e^x x-5 x^2\right ) \log (x)}} \left (5-e^x+x+\left (5+e^x (-1-x)+2 x\right ) \log (x)+\left (x^2-e^x x^2\right ) \log ^2(x)\right )}{e^{\frac {1+x \log (x)}{\left (-25 x+5 e^x x-5 x^2\right ) \log (x)}} \left (125 x^2+5 e^{2 x} x^2+50 x^3+5 x^4+e^x \left (-50 x^2-10 x^3\right )\right ) \log ^2(x)+\left (-250 x^2-10 e^{2 x} x^2-100 x^3-10 x^4+e^x \left (100 x^2+20 x^3\right )\right ) \log ^2(x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(86\) vs.
\(2(26)=52\).
time = 0.06, size = 87, normalized size = 2.90
method | result | size |
risch | \(-\frac {1}{5 \left (x -{\mathrm e}^{x}+5\right )}-\frac {1}{5 x \left (x -{\mathrm e}^{x}+5\right ) \ln \left (x \right )}-\frac {x \ln \left (x \right )+1}{\left (5 \,{\mathrm e}^{x} x -5 x^{2}-25 x \right ) \ln \left (x \right )}+\ln \left ({\mathrm e}^{\frac {x \ln \left (x \right )+1}{5 x \left ({\mathrm e}^{x}-5-x \right ) \ln \left (x \right )}}-2\right )\) | \(87\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 144 vs.
\(2 (26) = 52\).
time = 0.43, size = 144, normalized size = 4.80 \begin {gather*} \frac {{\left (x \log \left (x\right ) + 1\right )} e^{x} - 5 \, x \log \left (x\right ) - x - 5}{5 \, {\left (x e^{\left (2 \, x\right )} \log \left (x\right ) - {\left (x^{2} + 10 \, x\right )} e^{x} \log \left (x\right ) + 5 \, {\left (x^{2} + 5 \, x\right )} \log \left (x\right )\right )}} + \log \left ({\left (e^{\left (-\frac {1}{5 \, {\left (x - e^{x} + 5\right )}} - \frac {1}{5 \, {\left ({\left (x + 10\right )} e^{x} - 5 \, x - e^{\left (2 \, x\right )} - 25\right )} \log \left (x\right )} + \frac {1}{5 \, {\left (x e^{x} - 5 \, x\right )} \log \left (x\right )}\right )} - 2\right )} e^{\left (\frac {1}{5 \, {\left (x - e^{x} + 5\right )}} - \frac {1}{5 \, {\left (x e^{x} - 5 \, x\right )} \log \left (x\right )}\right )}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.36, size = 30, normalized size = 1.00 \begin {gather*} \log \left (e^{\left (-\frac {x \log \left (x\right ) + 1}{5 \, {\left (x^{2} - x e^{x} + 5 \, x\right )} \log \left (x\right )}\right )} - 2\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 1.94, size = 29, normalized size = 0.97 \begin {gather*} \log {\left (e^{\frac {x \log {\left (x \right )} + 1}{\left (- 5 x^{2} + 5 x e^{x} - 25 x\right ) \log {\left (x \right )}}} - 2 \right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.48, size = 43, normalized size = 1.43 \begin {gather*} \ln \left ({\mathrm {e}}^{-\frac {1}{5\,x-5\,{\mathrm {e}}^x+25}}\,{\mathrm {e}}^{-\frac {1}{5\,x^2\,\ln \left (x\right )+25\,x\,\ln \left (x\right )-5\,x\,{\mathrm {e}}^x\,\ln \left (x\right )}}-2\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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