Optimal. Leaf size=30 \[ 5-\log ^2\left (5 e^{1-\frac {2 x}{5}}+\frac {1}{4} (1+x) (-1+\log (x))\right ) \]
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Rubi [F] time = 141.03, 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 {\left (-2+16 e^{\frac {1}{5} (5-2 x)} x-2 x \log (x)\right ) \log \left (\frac {1}{4} \left (-1+20 e^{\frac {1}{5} (5-2 x)}-x+(1+x) \log (x)\right )\right )}{-x+20 e^{\frac {1}{5} (5-2 x)} x-x^2+\left (x+x^2\right ) \log (x)} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {2 (1+x \log (x)) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{x (1+x) (-1+\log (x))}+\frac {8 e \left (5-2 x-2 x^2+7 x \log (x)+2 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{x (1+x) (-1+\log (x)) \left (20 e-e^{2 x/5}-e^{2 x/5} x+e^{2 x/5} \log (x)+e^{2 x/5} x \log (x)\right )}\right ) \, dx\\ &=-\left (2 \int \frac {(1+x \log (x)) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{x (1+x) (-1+\log (x))} \, dx\right )+(8 e) \int \frac {\left (5-2 x-2 x^2+7 x \log (x)+2 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{x (1+x) (-1+\log (x)) \left (20 e-e^{2 x/5}-e^{2 x/5} x+e^{2 x/5} \log (x)+e^{2 x/5} x \log (x)\right )} \, dx\\ &=-\left (2 \int \left (\frac {(1+x \log (x)) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{x (-1+\log (x))}-\frac {(1+x \log (x)) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{(1+x) (-1+\log (x))}\right ) \, dx\right )+(8 e) \int \left (\frac {\left (5-2 x-2 x^2+7 x \log (x)+2 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{x (-1+\log (x)) \left (20 e-e^{2 x/5}-e^{2 x/5} x+e^{2 x/5} \log (x)+e^{2 x/5} x \log (x)\right )}-\frac {\left (5-2 x-2 x^2+7 x \log (x)+2 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{(1+x) (-1+\log (x)) \left (20 e-e^{2 x/5}-e^{2 x/5} x+e^{2 x/5} \log (x)+e^{2 x/5} x \log (x)\right )}\right ) \, dx\\ &=-\left (2 \int \frac {(1+x \log (x)) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{x (-1+\log (x))} \, dx\right )+2 \int \frac {(1+x \log (x)) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{(1+x) (-1+\log (x))} \, dx+(8 e) \int \frac {\left (5-2 x-2 x^2+7 x \log (x)+2 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{x (-1+\log (x)) \left (20 e-e^{2 x/5}-e^{2 x/5} x+e^{2 x/5} \log (x)+e^{2 x/5} x \log (x)\right )} \, dx-(8 e) \int \frac {\left (5-2 x-2 x^2+7 x \log (x)+2 x^2 \log (x)\right ) \log \left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right )}{(1+x) (-1+\log (x)) \left (20 e-e^{2 x/5}-e^{2 x/5} x+e^{2 x/5} \log (x)+e^{2 x/5} x \log (x)\right )} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}
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Mathematica [A] time = 0.08, size = 31, normalized size = 1.03 \begin {gather*} -\log ^2\left (\frac {1}{4} \left (-1+20 e^{1-\frac {2 x}{5}}-x+(1+x) \log (x)\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.59, size = 25, normalized size = 0.83 \begin {gather*} -\log \left (\frac {1}{4} \, {\left (x + 1\right )} \log \relax (x) - \frac {1}{4} \, x + 5 \, e^{\left (-\frac {2}{5} \, x + 1\right )} - \frac {1}{4}\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*} \int -\frac {2 \, {\left (8 \, x e^{\left (-\frac {2}{5} \, x + 1\right )} - x \log \relax (x) - 1\right )} \log \left (\frac {1}{4} \, {\left (x + 1\right )} \log \relax (x) - \frac {1}{4} \, x + 5 \, e^{\left (-\frac {2}{5} \, x + 1\right )} - \frac {1}{4}\right )}{x^{2} - 20 \, x e^{\left (-\frac {2}{5} \, x + 1\right )} - {\left (x^{2} + x\right )} \log \relax (x) + x}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (-2 x \ln \relax (x )+16 x \,{\mathrm e}^{-\frac {2 x}{5}+1}-2\right ) \ln \left (\frac {\ln \relax (x ) \left (x +1\right )}{4}+5 \,{\mathrm e}^{-\frac {2 x}{5}+1}-\frac {x}{4}-\frac {1}{4}\right )}{\left (x^{2}+x \right ) \ln \relax (x )+20 x \,{\mathrm e}^{-\frac {2 x}{5}+1}-x^{2}-x}\, dx\]
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*} -2 \, \int \frac {{\left (8 \, x e^{\left (-\frac {2}{5} \, x + 1\right )} - x \log \relax (x) - 1\right )} \log \left (\frac {1}{4} \, {\left (x + 1\right )} \log \relax (x) - \frac {1}{4} \, x + 5 \, e^{\left (-\frac {2}{5} \, x + 1\right )} - \frac {1}{4}\right )}{x^{2} - 20 \, x e^{\left (-\frac {2}{5} \, x + 1\right )} - {\left (x^{2} + x\right )} \log \relax (x) + x}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.39, size = 26, normalized size = 0.87 \begin {gather*} -{\ln \left (5\,{\mathrm {e}}^{-\frac {2\,x}{5}}\,\mathrm {e}-\frac {x}{4}+\ln \left (x^{1/4}\right )\,\left (x+1\right )-\frac {1}{4}\right )}^2 \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 1.26, size = 29, normalized size = 0.97 \begin {gather*} - \log {\left (- \frac {x}{4} + \frac {\left (x + 1\right ) \log {\relax (x )}}{4} + 5 e^{1 - \frac {2 x}{5}} - \frac {1}{4} \right )}^{2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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