Optimal. Leaf size=31 \[ \frac {4}{\left (-5-\frac {e^{2 x}}{5}+\frac {1}{x}-x\right ) \left (-\frac {x}{2}+\log (x)\right )} \]
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Rubi [F] time = 4.18, 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 {-400+2000 x-600 x^2-400 x^3+e^{2 x} \left (80 x-40 x^2-80 x^3\right )+\left (400+400 x^2+160 e^{2 x} x^2\right ) \log (x)}{25 x^2-250 x^3+575 x^4+e^{4 x} x^4+250 x^5+25 x^6+e^{2 x} \left (-10 x^3+50 x^4+10 x^5\right )+\left (-100 x+1000 x^2-2300 x^3-4 e^{4 x} x^3-1000 x^4-100 x^5+e^{2 x} \left (40 x^2-200 x^3-40 x^4\right )\right ) \log (x)+\left (100-1000 x+2300 x^2+4 e^{4 x} x^2+1000 x^3+100 x^4+e^{2 x} \left (-40 x+200 x^2+40 x^3\right )\right ) \log ^2(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 {40 \left (-10+2 \left (25+e^{2 x}\right ) x-\left (15+e^{2 x}\right ) x^2-2 \left (5+e^{2 x}\right ) x^3+2 \left (5+\left (5+2 e^{2 x}\right ) x^2\right ) \log (x)\right )}{\left (5-\left (25+e^{2 x}\right ) x-5 x^2\right )^2 (x-2 \log (x))^2} \, dx\\ &=40 \int \frac {-10+2 \left (25+e^{2 x}\right ) x-\left (15+e^{2 x}\right ) x^2-2 \left (5+e^{2 x}\right ) x^3+2 \left (5+\left (5+2 e^{2 x}\right ) x^2\right ) \log (x)}{\left (5-\left (25+e^{2 x}\right ) x-5 x^2\right )^2 (x-2 \log (x))^2} \, dx\\ &=40 \int \left (\frac {5 \left (-1-2 x+9 x^2+2 x^3\right )}{\left (-5+25 x+e^{2 x} x+5 x^2\right )^2 (x-2 \log (x))}-\frac {-2+x+2 x^2-4 x \log (x)}{\left (-5+25 x+e^{2 x} x+5 x^2\right ) (x-2 \log (x))^2}\right ) \, dx\\ &=-\left (40 \int \frac {-2+x+2 x^2-4 x \log (x)}{\left (-5+25 x+e^{2 x} x+5 x^2\right ) (x-2 \log (x))^2} \, dx\right )+200 \int \frac {-1-2 x+9 x^2+2 x^3}{\left (-5+25 x+e^{2 x} x+5 x^2\right )^2 (x-2 \log (x))} \, dx\\ &=-\left (40 \int \left (-\frac {2}{\left (-5+25 x+e^{2 x} x+5 x^2\right ) (x-2 \log (x))^2}+\frac {x}{\left (-5+25 x+e^{2 x} x+5 x^2\right ) (x-2 \log (x))^2}+\frac {2 x^2}{\left (-5+25 x+e^{2 x} x+5 x^2\right ) (x-2 \log (x))^2}-\frac {4 x \log (x)}{\left (-5+25 x+e^{2 x} x+5 x^2\right ) (x-2 \log (x))^2}\right ) \, dx\right )+200 \int \left (-\frac {1}{\left (-5+25 x+e^{2 x} x+5 x^2\right )^2 (x-2 \log (x))}-\frac {2 x}{\left (-5+25 x+e^{2 x} x+5 x^2\right )^2 (x-2 \log (x))}+\frac {9 x^2}{\left (-5+25 x+e^{2 x} x+5 x^2\right )^2 (x-2 \log (x))}+\frac {2 x^3}{\left (-5+25 x+e^{2 x} x+5 x^2\right )^2 (x-2 \log (x))}\right ) \, dx\\ &=-\left (40 \int \frac {x}{\left (-5+25 x+e^{2 x} x+5 x^2\right ) (x-2 \log (x))^2} \, dx\right )+80 \int \frac {1}{\left (-5+25 x+e^{2 x} x+5 x^2\right ) (x-2 \log (x))^2} \, dx-80 \int \frac {x^2}{\left (-5+25 x+e^{2 x} x+5 x^2\right ) (x-2 \log (x))^2} \, dx+160 \int \frac {x \log (x)}{\left (-5+25 x+e^{2 x} x+5 x^2\right ) (x-2 \log (x))^2} \, dx-200 \int \frac {1}{\left (-5+25 x+e^{2 x} x+5 x^2\right )^2 (x-2 \log (x))} \, dx-400 \int \frac {x}{\left (-5+25 x+e^{2 x} x+5 x^2\right )^2 (x-2 \log (x))} \, dx+400 \int \frac {x^3}{\left (-5+25 x+e^{2 x} x+5 x^2\right )^2 (x-2 \log (x))} \, dx+1800 \int \frac {x^2}{\left (-5+25 x+e^{2 x} x+5 x^2\right )^2 (x-2 \log (x))} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.09, size = 32, normalized size = 1.03 \begin {gather*} -\frac {40 x}{\left (-5+25 x+e^{2 x} x+5 x^2\right ) (-x+2 \log (x))} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.76, size = 47, normalized size = 1.52 \begin {gather*} \frac {40 \, x}{5 \, x^{3} + x^{2} e^{\left (2 \, x\right )} + 25 \, x^{2} - 2 \, {\left (5 \, x^{2} + x e^{\left (2 \, x\right )} + 25 \, x - 5\right )} \log \relax (x) - 5 \, x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.32, size = 52, normalized size = 1.68 \begin {gather*} \frac {40 \, x}{5 \, x^{3} + x^{2} e^{\left (2 \, x\right )} - 10 \, x^{2} \log \relax (x) - 2 \, x e^{\left (2 \, x\right )} \log \relax (x) + 25 \, x^{2} - 50 \, x \log \relax (x) - 5 \, x + 10 \, \log \relax (x)} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.05, size = 30, normalized size = 0.97
method | result | size |
risch | \(\frac {40 x}{\left (5 x^{2}+x \,{\mathrm e}^{2 x}+25 x -5\right ) \left (x -2 \ln \relax (x )\right )}\) | \(30\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.53, size = 45, normalized size = 1.45 \begin {gather*} \frac {40 \, x}{5 \, x^{3} + 25 \, x^{2} + {\left (x^{2} - 2 \, x \log \relax (x)\right )} e^{\left (2 \, x\right )} - 10 \, {\left (x^{2} + 5 \, x - 1\right )} \log \relax (x) - 5 \, x} \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\,x}\,\left (80\,x^3+40\,x^2-80\,x\right )-2000\,x-\ln \relax (x)\,\left (160\,x^2\,{\mathrm {e}}^{2\,x}+400\,x^2+400\right )+600\,x^2+400\,x^3+400}{{\ln \relax (x)}^2\,\left ({\mathrm {e}}^{2\,x}\,\left (40\,x^3+200\,x^2-40\,x\right )-1000\,x+4\,x^2\,{\mathrm {e}}^{4\,x}+2300\,x^2+1000\,x^3+100\,x^4+100\right )-\ln \relax (x)\,\left (100\,x+4\,x^3\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{2\,x}\,\left (40\,x^4+200\,x^3-40\,x^2\right )-1000\,x^2+2300\,x^3+1000\,x^4+100\,x^5\right )+x^4\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{2\,x}\,\left (10\,x^5+50\,x^4-10\,x^3\right )+25\,x^2-250\,x^3+575\,x^4+250\,x^5+25\,x^6} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.47, size = 51, normalized size = 1.65 \begin {gather*} \frac {40 x}{5 x^{3} - 10 x^{2} \log {\relax (x )} + 25 x^{2} - 50 x \log {\relax (x )} - 5 x + \left (x^{2} - 2 x \log {\relax (x )}\right ) e^{2 x} + 10 \log {\relax (x )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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