Optimal. Leaf size=32 \[ \left (15-\frac {e^{e^{5-\frac {5}{2 x}}}}{-\log (4)+\log \left (\frac {x}{5}\right )}\right )^2 \]
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Rubi [F]
time = 7.16, 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^{2 e^{\frac {-5+10 x}{2 x}}} \left (-2 x-5 e^{\frac {-5+10 x}{2 x}} \log (4)+5 e^{\frac {-5+10 x}{2 x}} \log \left (\frac {x}{5}\right )\right )+e^{e^{\frac {-5+10 x}{2 x}}} \left (-30 x \log (4)-75 e^{\frac {-5+10 x}{2 x}} \log ^2(4)+\left (30 x+150 e^{\frac {-5+10 x}{2 x}} \log (4)\right ) \log \left (\frac {x}{5}\right )-75 e^{\frac {-5+10 x}{2 x}} \log ^2\left (\frac {x}{5}\right )\right )}{-x^2 \log ^3(4)+3 x^2 \log ^2(4) \log \left (\frac {x}{5}\right )-3 x^2 \log (4) \log ^2\left (\frac {x}{5}\right )+x^2 \log ^3\left (\frac {x}{5}\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^{e^{5-\frac {5}{2 x}}-\frac {5}{2 x}} \left (e^{e^{5-\frac {5}{2 x}}}+15 \log (4)-15 \log \left (\frac {x}{5}\right )\right ) \left (-2 e^{\left .\frac {5}{2}\right /x} x-5 e^5 \log (4)+5 e^5 \log \left (\frac {x}{5}\right )\right )}{x^2 \log ^3\left (\frac {x}{20}\right )} \, dx\\ &=\int \left (-\frac {2 e^{e^{5-\frac {5}{2 x}}} \left (e^{e^{5-\frac {5}{2 x}}}+15 \log (4)-15 \log \left (\frac {x}{5}\right )\right )}{x \log ^3\left (\frac {x}{20}\right )}-\frac {5 e^{5+e^{5-\frac {5}{2 x}}-\frac {5}{2 x}} \left (e^{e^{5-\frac {5}{2 x}}}+15 \log (4)-15 \log \left (\frac {x}{5}\right )\right ) (\log (20)-\log (x))}{x^2 \log ^3\left (\frac {x}{20}\right )}\right ) \, dx\\ &=-\left (2 \int \frac {e^{e^{5-\frac {5}{2 x}}} \left (e^{e^{5-\frac {5}{2 x}}}+15 \log (4)-15 \log \left (\frac {x}{5}\right )\right )}{x \log ^3\left (\frac {x}{20}\right )} \, dx\right )-5 \int \frac {e^{5+e^{5-\frac {5}{2 x}}-\frac {5}{2 x}} \left (e^{e^{5-\frac {5}{2 x}}}+15 \log (4)-15 \log \left (\frac {x}{5}\right )\right ) (\log (20)-\log (x))}{x^2 \log ^3\left (\frac {x}{20}\right )} \, dx\\ &=-\left (2 \int \left (\frac {e^{2 e^{5-\frac {5}{2 x}}}}{x \log ^3\left (\frac {x}{20}\right )}-\frac {15 e^{e^{5-\frac {5}{2 x}}}}{x \log ^2\left (\frac {x}{20}\right )}\right ) \, dx\right )-5 \int \left (-\frac {e^{5+2 e^{5-\frac {5}{2 x}}-\frac {5}{2 x}}}{x^2 \log ^2\left (\frac {x}{20}\right )}+\frac {15 e^{5+e^{5-\frac {5}{2 x}}-\frac {5}{2 x}}}{x^2 \log \left (\frac {x}{20}\right )}\right ) \, dx\\ &=-\left (2 \int \frac {e^{2 e^{5-\frac {5}{2 x}}}}{x \log ^3\left (\frac {x}{20}\right )} \, dx\right )+5 \int \frac {e^{5+2 e^{5-\frac {5}{2 x}}-\frac {5}{2 x}}}{x^2 \log ^2\left (\frac {x}{20}\right )} \, dx+30 \int \frac {e^{e^{5-\frac {5}{2 x}}}}{x \log ^2\left (\frac {x}{20}\right )} \, dx-75 \int \frac {e^{5+e^{5-\frac {5}{2 x}}-\frac {5}{2 x}}}{x^2 \log \left (\frac {x}{20}\right )} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [F]
time = 2.62, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{2 e^{\frac {-5+10 x}{2 x}}} \left (-2 x-5 e^{\frac {-5+10 x}{2 x}} \log (4)+5 e^{\frac {-5+10 x}{2 x}} \log \left (\frac {x}{5}\right )\right )+e^{e^{\frac {-5+10 x}{2 x}}} \left (-30 x \log (4)-75 e^{\frac {-5+10 x}{2 x}} \log ^2(4)+\left (30 x+150 e^{\frac {-5+10 x}{2 x}} \log (4)\right ) \log \left (\frac {x}{5}\right )-75 e^{\frac {-5+10 x}{2 x}} \log ^2\left (\frac {x}{5}\right )\right )}{-x^2 \log ^3(4)+3 x^2 \log ^2(4) \log \left (\frac {x}{5}\right )-3 x^2 \log (4) \log ^2\left (\frac {x}{5}\right )+x^2 \log ^3\left (\frac {x}{5}\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.37, size = 0, normalized size = 0.00 \[\int \frac {\left (5 \,{\mathrm e}^{\frac {10 x -5}{2 x}} \ln \left (\frac {x}{5}\right )-10 \ln \left (2\right ) {\mathrm e}^{\frac {10 x -5}{2 x}}-2 x \right ) {\mathrm e}^{2 \,{\mathrm e}^{\frac {10 x -5}{2 x}}}+\left (-75 \,{\mathrm e}^{\frac {10 x -5}{2 x}} \ln \left (\frac {x}{5}\right )^{2}+\left (300 \ln \left (2\right ) {\mathrm e}^{\frac {10 x -5}{2 x}}+30 x \right ) \ln \left (\frac {x}{5}\right )-300 \ln \left (2\right )^{2} {\mathrm e}^{\frac {10 x -5}{2 x}}-60 x \ln \left (2\right )\right ) {\mathrm e}^{{\mathrm e}^{\frac {10 x -5}{2 x}}}}{x^{2} \ln \left (\frac {x}{5}\right )^{3}-6 x^{2} \ln \left (2\right ) \ln \left (\frac {x}{5}\right )^{2}+12 x^{2} \ln \left (2\right )^{2} \ln \left (\frac {x}{5}\right )-8 x^{2} \ln \left (2\right )^{3}}\, dx\]
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 [B] Leaf count of result is larger than twice the leaf count of optimal. 64 vs.
\(2 (27) = 54\).
time = 0.38, size = 64, normalized size = 2.00 \begin {gather*} \frac {30 \, {\left (2 \, \log \left (2\right ) - \log \left (\frac {1}{5} \, x\right )\right )} e^{\left (e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )}\right )} + e^{\left (2 \, e^{\left (\frac {5 \, {\left (2 \, x - 1\right )}}{2 \, x}\right )}\right )}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) \log \left (\frac {1}{5} \, x\right ) + \log \left (\frac {1}{5} \, x\right )^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 97 vs.
\(2 (22) = 44\).
time = 0.26, size = 97, normalized size = 3.03 \begin {gather*} \frac {\left (\log {\left (\frac {x}{5} \right )} - 2 \log {\left (2 \right )}\right ) e^{2 e^{\frac {5 x - \frac {5}{2}}{x}}} + \left (- 30 \log {\left (\frac {x}{5} \right )}^{2} + 120 \log {\left (2 \right )} \log {\left (\frac {x}{5} \right )} - 120 \log {\left (2 \right )}^{2}\right ) e^{e^{\frac {5 x - \frac {5}{2}}{x}}}}{\log {\left (\frac {x}{5} \right )}^{3} - 6 \log {\left (2 \right )} \log {\left (\frac {x}{5} \right )}^{2} + 12 \log {\left (2 \right )}^{2} \log {\left (\frac {x}{5} \right )} - 8 \log {\left (2 \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}}^{2\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}}\,\left (2\,x-5\,\ln \left (\frac {x}{5}\right )\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}+10\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}\,\ln \left (2\right )\right )+{\mathrm {e}}^{{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}}\,\left (75\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}\,{\ln \left (\frac {x}{5}\right )}^2+\left (-30\,x-300\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}\,\ln \left (2\right )\right )\,\ln \left (\frac {x}{5}\right )+60\,x\,\ln \left (2\right )+300\,{\mathrm {e}}^{\frac {5\,x-\frac {5}{2}}{x}}\,{\ln \left (2\right )}^2\right )}{-x^2\,{\ln \left (\frac {x}{5}\right )}^3+6\,\ln \left (2\right )\,x^2\,{\ln \left (\frac {x}{5}\right )}^2-12\,{\ln \left (2\right )}^2\,x^2\,\ln \left (\frac {x}{5}\right )+8\,{\ln \left (2\right )}^3\,x^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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