Optimal. Leaf size=29 \[ x+\frac {5}{-5+e-x-5 \left (-\log (5)+\frac {3}{\log \left (\frac {3}{x}\right )}\right )} \]
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Rubi [F] time = 3.77, 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 {75+225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (30 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{x}\right )}{225 x+\left (150 x-30 e x+30 x^2-150 x \log (5)\right ) \log \left (\frac {3}{x}\right )+\left (25 x+e^2 x+10 x^2+x^3+e \left (-10 x-2 x^2\right )+\left (-50 x+10 e x-10 x^2\right ) \log (5)+25 x \log ^2(5)\right ) \log ^2\left (\frac {3}{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 \frac {75 (1+3 x)+30 x (5-e+x-5 \log (5)) \log \left (\frac {3}{x}\right )+x \left (e^2+x^2-2 e (5+x-5 \log (5))-10 x (-1+\log (5))+5 \left (6-10 \log (5)+5 \log ^2(5)\right )\right ) \log ^2\left (\frac {3}{x}\right )}{x \left (15+(5-e+x-5 \log (5)) \log \left (\frac {3}{x}\right )\right )^2} \, dx\\ &=\int \left (\frac {30-10 e+e^2+x^2+2 x (5-e-5 \log (5))-50 \log (5)+10 e \log (5)+25 \log ^2(5)}{(5-e+x-5 \log (5))^2}+\frac {150}{(5-e+x-5 \log (5))^2 \left (-15-x \log \left (\frac {3}{x}\right )-5 \left (1-\frac {e}{5}-\log (5)\right ) \log \left (\frac {3}{x}\right )\right )}+\frac {75 \left (x^2+(-5+e+5 \log (5))^2+x (25-2 e-5 \log (25))\right )}{x (5-e+x-5 \log (5))^2 \left (15+x \log \left (\frac {3}{x}\right )+5 \left (1-\frac {e}{5}-\log (5)\right ) \log \left (\frac {3}{x}\right )\right )^2}\right ) \, dx\\ &=75 \int \frac {x^2+(-5+e+5 \log (5))^2+x (25-2 e-5 \log (25))}{x (5-e+x-5 \log (5))^2 \left (15+x \log \left (\frac {3}{x}\right )+5 \left (1-\frac {e}{5}-\log (5)\right ) \log \left (\frac {3}{x}\right )\right )^2} \, dx+150 \int \frac {1}{(5-e+x-5 \log (5))^2 \left (-15-x \log \left (\frac {3}{x}\right )-5 \left (1-\frac {e}{5}-\log (5)\right ) \log \left (\frac {3}{x}\right )\right )} \, dx+\int \frac {30-10 e+e^2+x^2+2 x (5-e-5 \log (5))-50 \log (5)+10 e \log (5)+25 \log ^2(5)}{(5-e+x-5 \log (5))^2} \, dx\\ &=75 \int \frac {x^2+(-5+e+5 \log (5))^2+x (25-2 e-5 \log (25))}{x (5-e+x-5 \log (5))^2 \left (15+(5-e+x-5 \log (5)) \log \left (\frac {3}{x}\right )\right )^2} \, dx+150 \int \frac {1}{(5-e+x-5 \log (5))^2 \left (-15+(-5+e-x+5 \log (5)) \log \left (\frac {3}{x}\right )\right )} \, dx+\int \left (1+\frac {5}{(-5+e-x+5 \log (5))^2}\right ) \, dx\\ &=x-\frac {5}{5-e+x-5 \log (5)}+75 \int \left (\frac {1}{x \left (15+x \log \left (\frac {3}{x}\right )+5 \left (1-\frac {e}{5}-\log (5)\right ) \log \left (\frac {3}{x}\right )\right )^2}+\frac {15}{(5-e+x-5 \log (5))^2 \left (15+x \log \left (\frac {3}{x}\right )+5 \left (1-\frac {e}{5}-\log (5)\right ) \log \left (\frac {3}{x}\right )\right )^2}\right ) \, dx+150 \int \frac {1}{(5-e+x-5 \log (5))^2 \left (-15+(-5+e-x+5 \log (5)) \log \left (\frac {3}{x}\right )\right )} \, dx\\ &=x-\frac {5}{5-e+x-5 \log (5)}+75 \int \frac {1}{x \left (15+x \log \left (\frac {3}{x}\right )+5 \left (1-\frac {e}{5}-\log (5)\right ) \log \left (\frac {3}{x}\right )\right )^2} \, dx+150 \int \frac {1}{(5-e+x-5 \log (5))^2 \left (-15+(-5+e-x+5 \log (5)) \log \left (\frac {3}{x}\right )\right )} \, dx+1125 \int \frac {1}{(5-e+x-5 \log (5))^2 \left (15+x \log \left (\frac {3}{x}\right )+5 \left (1-\frac {e}{5}-\log (5)\right ) \log \left (\frac {3}{x}\right )\right )^2} \, dx\\ &=x-\frac {5}{5-e+x-5 \log (5)}+75 \int \frac {1}{x \left (15+(5-e+x-5 \log (5)) \log \left (\frac {3}{x}\right )\right )^2} \, dx+150 \int \frac {1}{(5-e+x-5 \log (5))^2 \left (-15+(-5+e-x+5 \log (5)) \log \left (\frac {3}{x}\right )\right )} \, dx+1125 \int \frac {1}{(5-e+x-5 \log (5))^2 \left (15-(-5+e-x+5 \log (5)) \log \left (\frac {3}{x}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.10, size = 48, normalized size = 1.66 \begin {gather*} \frac {15 x+\left (-5+x^2-x (-5+e+5 \log (5))\right ) \log \left (\frac {3}{x}\right )}{15+(5-e+x-5 \log (5)) \log \left (\frac {3}{x}\right )} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.48, size = 52, normalized size = 1.79 \begin {gather*} \frac {{\left (x^{2} - x e - 5 \, x \log \relax (5) + 5 \, x - 5\right )} \log \left (\frac {3}{x}\right ) + 15 \, x}{{\left (x - e - 5 \, \log \relax (5) + 5\right )} \log \left (\frac {3}{x}\right ) + 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.92, size = 271, normalized size = 9.34 \begin {gather*} \frac {e \log \left (\frac {3}{x}\right ) - \frac {10 \, e \log \relax (5) \log \left (\frac {3}{x}\right )}{x} + 5 \, \log \relax (5) \log \left (\frac {3}{x}\right ) - \frac {25 \, \log \relax (5)^{2} \log \left (\frac {3}{x}\right )}{x} - \frac {e^{2} \log \left (\frac {3}{x}\right )}{x} + \frac {10 \, e \log \left (\frac {3}{x}\right )}{x} + \frac {50 \, \log \relax (5) \log \left (\frac {3}{x}\right )}{x} + \frac {15 \, e}{x} + \frac {75 \, \log \relax (5)}{x} - \frac {30 \, \log \left (\frac {3}{x}\right )}{x} - \frac {75}{x} - \frac {75}{x^{2}} - 5 \, \log \left (\frac {3}{x}\right )}{\frac {e \log \left (\frac {3}{x}\right )}{x} + \frac {5 \, \log \relax (5) \log \left (\frac {3}{x}\right )}{x} - \frac {10 \, e \log \relax (5) \log \left (\frac {3}{x}\right )}{x^{2}} - \frac {25 \, \log \relax (5)^{2} \log \left (\frac {3}{x}\right )}{x^{2}} - \frac {5 \, \log \left (\frac {3}{x}\right )}{x} - \frac {e^{2} \log \left (\frac {3}{x}\right )}{x^{2}} + \frac {10 \, e \log \left (\frac {3}{x}\right )}{x^{2}} + \frac {50 \, \log \relax (5) \log \left (\frac {3}{x}\right )}{x^{2}} + \frac {15 \, e}{x^{2}} + \frac {75 \, \log \relax (5)}{x^{2}} - \frac {25 \, \log \left (\frac {3}{x}\right )}{x^{2}} - \frac {75}{x^{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.28, size = 90, normalized size = 3.10
method | result | size |
risch | \(\frac {5 x \ln \relax (5)+x \,{\mathrm e}-x^{2}-5 x +5}{5 \ln \relax (5)+{\mathrm e}-x -5}+\frac {75}{\left (5 \ln \relax (5)+{\mathrm e}-x -5\right ) \left (5 \ln \left (\frac {3}{x}\right ) \ln \relax (5)+{\mathrm e} \ln \left (\frac {3}{x}\right )-x \ln \left (\frac {3}{x}\right )-5 \ln \left (\frac {3}{x}\right )-15\right )}\) | \(90\) |
norman | \(\frac {\left (25 \ln \relax (5)^{2}+10 \,{\mathrm e} \ln \relax (5)+{\mathrm e}^{2}-50 \ln \relax (5)-10 \,{\mathrm e}+30\right ) \ln \left (\frac {3}{x}\right )+75-15 x -x^{2} \ln \left (\frac {3}{x}\right )-15 \,{\mathrm e}-75 \ln \relax (5)}{5 \ln \left (\frac {3}{x}\right ) \ln \relax (5)+{\mathrm e} \ln \left (\frac {3}{x}\right )-x \ln \left (\frac {3}{x}\right )-5 \ln \left (\frac {3}{x}\right )-15}\) | \(99\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.52, size = 87, normalized size = 3.00 \begin {gather*} \frac {x^{2} \log \relax (3) - {\left (5 \, {\left (\log \relax (5) - 1\right )} \log \relax (3) + e \log \relax (3) - 15\right )} x - {\left (x^{2} - x {\left (e + 5 \, \log \relax (5) - 5\right )} - 5\right )} \log \relax (x) - 5 \, \log \relax (3)}{x \log \relax (3) - 5 \, {\left (\log \relax (5) - 1\right )} \log \relax (3) - e \log \relax (3) - {\left (x - e - 5 \, \log \relax (5) + 5\right )} \log \relax (x) + 15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 13.77, size = 94, normalized size = 3.24 \begin {gather*} \frac {15\,x-5\,\ln \left (\frac {3}{x}\right )+5\,x\,\ln \left (\frac {3}{x}\right )+x^2\,\ln \left (\frac {3}{x}\right )-x\,\mathrm {e}\,\ln \left (\frac {3}{x}\right )-5\,x\,\ln \relax (5)\,\ln \left (\frac {3}{x}\right )}{5\,\ln \left (\frac {3}{x}\right )-\mathrm {e}\,\ln \left (\frac {3}{x}\right )-5\,\ln \relax (5)\,\ln \left (\frac {3}{x}\right )+x\,\ln \left (\frac {3}{x}\right )+15} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.61, size = 87, normalized size = 3.00 \begin {gather*} x + \frac {75}{15 x + \left (x^{2} - 10 x \log {\relax (5 )} - 2 e x + 10 x - 50 \log {\relax (5 )} - 10 e + e^{2} + 25 + 10 e \log {\relax (5 )} + 25 \log {\relax (5 )}^{2}\right ) \log {\left (\frac {3}{x} \right )} - 75 \log {\relax (5 )} - 15 e + 75} - \frac {5}{x - 5 \log {\relax (5 )} - e + 5} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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