Optimal. Leaf size=30 \[ e^{\frac {-5+x}{x+e^{-e^3} \left (\frac {x}{5}-\log ^2(x)\right )}} \]
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Rubi [F]
time = 15.44, 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 {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}\right ) \left (25 e^{e^3} x+125 e^{2 e^3} x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x)\right )}{x^3+10 e^{e^3} x^3+25 e^{2 e^3} x^3+\left (-10 x^2-50 e^{e^3} x^2\right ) \log ^2(x)+25 x \log ^4(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 {\exp \left (\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}\right ) \left (\left (25 e^{e^3}+125 e^{2 e^3}\right ) x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x)\right )}{x^3+10 e^{e^3} x^3+25 e^{2 e^3} x^3+\left (-10 x^2-50 e^{e^3} x^2\right ) \log ^2(x)+25 x \log ^4(x)} \, dx\\ &=\int \frac {\exp \left (\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}\right ) \left (\left (25 e^{e^3}+125 e^{2 e^3}\right ) x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x)\right )}{25 e^{2 e^3} x^3+\left (1+10 e^{e^3}\right ) x^3+\left (-10 x^2-50 e^{e^3} x^2\right ) \log ^2(x)+25 x \log ^4(x)} \, dx\\ &=\int \frac {\exp \left (\frac {e^{e^3} (25-5 x)}{-x-5 e^{e^3} x+5 \log ^2(x)}\right ) \left (\left (25 e^{e^3}+125 e^{2 e^3}\right ) x+e^{e^3} (-250+50 x) \log (x)-25 e^{e^3} x \log ^2(x)\right )}{\left (1+10 e^{e^3}+25 e^{2 e^3}\right ) x^3+\left (-10 x^2-50 e^{e^3} x^2\right ) \log ^2(x)+25 x \log ^4(x)} \, dx\\ &=\int \frac {25 \exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \left (\left (1+5 e^{e^3}\right ) x+2 (-5+x) \log (x)-x \log ^2(x)\right )}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx\\ &=25 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \left (\left (1+5 e^{e^3}\right ) x+2 (-5+x) \log (x)-x \log ^2(x)\right )}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx\\ &=25 \int \left (\frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) (5-x) \left (\left (1+5 e^{e^3}\right ) x-10 \log (x)\right )}{5 x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}+\frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right )}{5 \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )}\right ) \, dx\\ &=5 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) (5-x) \left (\left (1+5 e^{e^3}\right ) x-10 \log (x)\right )}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx+5 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right )}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)} \, dx\\ &=5 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right )}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)} \, dx+5 \int \left (\frac {5 \exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \left (\left (1+5 e^{e^3}\right ) x-10 \log (x)\right )}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}+\frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \left (-\left (\left (1+5 e^{e^3}\right ) x\right )+10 \log (x)\right )}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}\right ) \, dx\\ &=5 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \left (-\left (\left (1+5 e^{e^3}\right ) x\right )+10 \log (x)\right )}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx+5 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right )}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)} \, dx+25 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \left (\left (1+5 e^{e^3}\right ) x-10 \log (x)\right )}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx\\ &=5 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right )}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)} \, dx+5 \int \left (\frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \left (-1-5 e^{e^3}\right ) x}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}+\frac {10 \exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \log (x)}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}\right ) \, dx+25 \int \left (\frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \left (1+5 e^{e^3}\right )}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}-\frac {10 \exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \log (x)}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2}\right ) \, dx\\ &=5 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right )}{\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)} \, dx+50 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \log (x)}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx-250 \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) \log (x)}{x \left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx-\left (5 \left (1+5 e^{e^3}\right )\right ) \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right ) x}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx+\left (25 \left (1+5 e^{e^3}\right )\right ) \int \frac {\exp \left (e^3+\frac {5 e^{e^3} (-5+x)}{x+5 e^{e^3} x-5 \log ^2(x)}\right )}{\left (\left (1+5 e^{e^3}\right ) x-5 \log ^2(x)\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.13, size = 32, normalized size = 1.07 \begin {gather*} e^{-\frac {5 e^{e^3} (-5+x)}{-x-5 e^{e^3} x+5 \log ^2(x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.22, size = 28, normalized size = 0.93
method | result | size |
risch | \({\mathrm e}^{-\frac {5 \left (x -5\right ) {\mathrm e}^{{\mathrm e}^{3}}}{5 \ln \left (x \right )^{2}-5 x \,{\mathrm e}^{{\mathrm e}^{3}}-x}}\) | \(28\) |
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. 81 vs.
\(2 (28) = 56\).
time = 0.46, size = 81, normalized size = 2.70 \begin {gather*} e^{\left (-\frac {25 \, e^{\left (e^{3}\right )} \log \left (x\right )^{2}}{5 \, {\left (5 \, e^{\left (e^{3}\right )} + 1\right )} \log \left (x\right )^{2} - x {\left (25 \, e^{\left (2 \, e^{3}\right )} + 10 \, e^{\left (e^{3}\right )} + 1\right )}} - \frac {25 \, e^{\left (e^{3}\right )}}{x {\left (5 \, e^{\left (e^{3}\right )} + 1\right )} - 5 \, \log \left (x\right )^{2}} + \frac {5 \, e^{\left (e^{3}\right )}}{5 \, e^{\left (e^{3}\right )} + 1}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.44, size = 25, normalized size = 0.83 \begin {gather*} e^{\left (\frac {5 \, {\left (x - 5\right )} e^{\left (e^{3}\right )}}{5 \, x e^{\left (e^{3}\right )} - 5 \, \log \left (x\right )^{2} + x}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.34, size = 27, normalized size = 0.90 \begin {gather*} e^{\frac {\left (25 - 5 x\right ) e^{e^{3}}}{- 5 x e^{e^{3}} - x + 5 \log {\left (x \right )}^{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*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 2.72, size = 31, normalized size = 1.03 \begin {gather*} {\mathrm {e}}^{-\frac {25\,{\mathrm {e}}^{{\mathrm {e}}^3}-5\,x\,{\mathrm {e}}^{{\mathrm {e}}^3}}{-5\,{\ln \left (x\right )}^2+x+5\,x\,{\mathrm {e}}^{{\mathrm {e}}^3}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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