Optimal. Leaf size=31 \[ x \left (-4+x-\log \left (-1+\frac {e^3}{5}-x+\frac {1}{2} (3+2 x)+\log (x)\right )\right ) \]
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Rubi [A] time = 0.41, antiderivative size = 26, normalized size of antiderivative = 0.84, number of steps used = 12, number of rules used = 7, integrand size = 64, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {6741, 6742, 6688, 2299, 2178, 2549, 12} \begin {gather*} x^2-4 x-x \log \left (\frac {1}{10} \left (10 \log (x)+2 e^3+5\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2178
Rule 2299
Rule 2549
Rule 6688
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-30+10 x+e^3 (-8+4 x)+(-40+20 x) \log (x)+\left (-5-2 e^3-10 \log (x)\right ) \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )}{5 \left (1+\frac {2 e^3}{5}\right )+10 \log (x)} \, dx\\ &=\int \left (\frac {2 \left (-15 \left (1+\frac {4 e^3}{15}\right )+5 \left (1+\frac {2 e^3}{5}\right ) x-20 \log (x)+10 x \log (x)\right )}{5 \left (1+\frac {2 e^3}{5}\right )+10 \log (x)}-\log \left (\frac {1}{10} \left (5 \left (1+\frac {2 e^3}{5}\right )+10 \log (x)\right )\right )\right ) \, dx\\ &=2 \int \frac {-15 \left (1+\frac {4 e^3}{15}\right )+5 \left (1+\frac {2 e^3}{5}\right ) x-20 \log (x)+10 x \log (x)}{5 \left (1+\frac {2 e^3}{5}\right )+10 \log (x)} \, dx-\int \log \left (\frac {1}{10} \left (5 \left (1+\frac {2 e^3}{5}\right )+10 \log (x)\right )\right ) \, dx\\ &=-x \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )+2 \int \frac {5 (-3+x)+2 e^3 (-2+x)+10 (-2+x) \log (x)}{5 \left (1+\frac {2 e^3}{5}\right )+10 \log (x)} \, dx+\int \frac {10}{5 \left (1+\frac {2 e^3}{5}\right )+10 \log (x)} \, dx\\ &=-x \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )+2 \int \left (-2+x+\frac {5}{-5 \left (1+\frac {2 e^3}{5}\right )-10 \log (x)}\right ) \, dx+10 \int \frac {1}{5 \left (1+\frac {2 e^3}{5}\right )+10 \log (x)} \, dx\\ &=-4 x+x^2-x \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )+10 \int \frac {1}{-5 \left (1+\frac {2 e^3}{5}\right )-10 \log (x)} \, dx+10 \operatorname {Subst}\left (\int \frac {e^x}{5 \left (1+\frac {2 e^3}{5}\right )+10 x} \, dx,x,\log (x)\right )\\ &=-4 x+x^2+e^{-\frac {1}{2}-\frac {e^3}{5}} \text {Ei}\left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )-x \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )+10 \operatorname {Subst}\left (\int \frac {e^x}{-5 \left (1+\frac {2 e^3}{5}\right )-10 x} \, dx,x,\log (x)\right )\\ &=-4 x+x^2-x \log \left (\frac {1}{10} \left (5+2 e^3+10 \log (x)\right )\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.18, size = 24, normalized size = 0.77 \begin {gather*} -4 x+x^2-x \log \left (\frac {1}{2}+\frac {e^3}{5}+\log (x)\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.16, size = 19, normalized size = 0.61 \begin {gather*} x^{2} - x \log \left (\frac {1}{5} \, e^{3} + \log \relax (x) + \frac {1}{2}\right ) - 4 \, x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.18, size = 25, normalized size = 0.81 \begin {gather*} x^{2} + x \log \left (10\right ) - x \log \left (2 \, e^{3} + 10 \, \log \relax (x) + 5\right ) - 4 \, x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.24, size = 20, normalized size = 0.65
method | result | size |
norman | \(x^{2}-4 x -x \ln \left (\ln \relax (x )+\frac {{\mathrm e}^{3}}{5}+\frac {1}{2}\right )\) | \(20\) |
risch | \(x^{2}-4 x -x \ln \left (\ln \relax (x )+\frac {{\mathrm e}^{3}}{5}+\frac {1}{2}\right )\) | \(20\) |
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*} 4 \, e^{\left (-\frac {1}{5} \, e^{3} - \frac {1}{2}\right )} E_{1}\left (-\frac {1}{5} \, e^{3} - \log \relax (x) - \frac {1}{2}\right ) \log \relax (x) - 2 \, e^{\left (-\frac {2}{5} \, e^{3} - 1\right )} E_{1}\left (-\frac {2}{5} \, e^{3} - 2 \, \log \relax (x) - 1\right ) \log \relax (x) + x {\left (\log \relax (5) + \log \relax (2)\right )} - 4 \, e^{\left (-\frac {1}{5} \, e^{3} - \frac {1}{2}\right )} E_{2}\left (-\frac {1}{5} \, e^{3} - \log \relax (x) - \frac {1}{2}\right ) + e^{\left (-\frac {2}{5} \, e^{3} - 1\right )} E_{2}\left (-\frac {2}{5} \, e^{3} - 2 \, \log \relax (x) - 1\right ) + \frac {4}{5} \, e^{\left (-\frac {1}{5} \, e^{3} + \frac {5}{2}\right )} E_{1}\left (-\frac {1}{5} \, e^{3} - \log \relax (x) - \frac {1}{2}\right ) + 3 \, e^{\left (-\frac {1}{5} \, e^{3} - \frac {1}{2}\right )} E_{1}\left (-\frac {1}{5} \, e^{3} - \log \relax (x) - \frac {1}{2}\right ) - \frac {2}{5} \, e^{\left (-\frac {2}{5} \, e^{3} + 2\right )} E_{1}\left (-\frac {2}{5} \, e^{3} - 2 \, \log \relax (x) - 1\right ) - e^{\left (-\frac {2}{5} \, e^{3} - 1\right )} E_{1}\left (-\frac {2}{5} \, e^{3} - 2 \, \log \relax (x) - 1\right ) - x \log \left (2 \, e^{3} + 10 \, \log \relax (x) + 5\right ) + 10 \, \int \frac {1}{2 \, e^{3} + 10 \, \log \relax (x) + 5}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.66, size = 17, normalized size = 0.55 \begin {gather*} -x\,\left (\ln \left (\frac {{\mathrm {e}}^3}{5}+\ln \relax (x)+\frac {1}{2}\right )-x+4\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.39, size = 20, normalized size = 0.65 \begin {gather*} x^{2} - x \log {\left (\log {\relax (x )} + \frac {1}{2} + \frac {e^{3}}{5} \right )} - 4 x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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