Optimal. Leaf size=26 \[ \left (e^3+e^{\frac {e^2}{x}}+x+x \left (e^x+x\right )\right )^2 \log (5) \]
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Rubi [B] time = 1.53, antiderivative size = 191, normalized size of antiderivative = 7.35, number of steps used = 38, number of rules used = 12, integrand size = 177, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.068, Rules used = {14, 2196, 2176, 2194, 6688, 6742, 2288, 2209, 629, 2206, 2210, 2214} \begin {gather*} 2 e^x x^3 \log (5)+2 e^{\frac {e^2}{x}} x^2 \log (5)+2 e^x x^2 \log (5)+e^{2 x} x^2 \log (5)+\left (x^2+x+e^3\right )^2 \log (5)-\frac {2 e^{x+\frac {e^2}{x}} \left (e^2-x^2\right ) \log (5)}{\left (1-\frac {e^2}{x^2}\right ) x}+2 e^{\frac {e^2}{x}} x \log (5)-4 e^x x \log (5)+2 \left (2+e^3\right ) e^x x \log (5)+2 e^{\frac {e^2}{x}+3} \log (5)+e^{\frac {2 e^2}{x}} \log (5)+4 e^x \log (5)+2 e^{x+3} \log (5)-2 \left (2+e^3\right ) e^x \log (5) \end {gather*}
Antiderivative was successfully verified.
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Rule 14
Rule 629
Rule 2176
Rule 2194
Rule 2196
Rule 2206
Rule 2209
Rule 2210
Rule 2214
Rule 2288
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2 e^{2 x} x (1+x) \log (5)+\frac {2 e^x \left (-e^{2+\frac {e^2}{x}}+e^3 x+e^{\frac {e^2}{x}} x+e^{\frac {e^2}{x}} x^2+2 \left (1+\frac {e^3}{2}\right ) x^2+4 x^3+x^4\right ) \log (5)}{x}+\frac {2 \left (-e^{5+\frac {e^2}{x}}-e^{2+\frac {2 e^2}{x}}-e^{2+\frac {e^2}{x}} x+e^3 x^2-\left (1-\frac {1}{e^2}\right ) e^{2+\frac {e^2}{x}} x^2+2 e^{\frac {e^2}{x}} x^3+\left (1+2 e^3\right ) x^3+3 x^4+2 x^5\right ) \log (5)}{x^2}\right ) \, dx\\ &=(2 \log (5)) \int e^{2 x} x (1+x) \, dx+(2 \log (5)) \int \frac {e^x \left (-e^{2+\frac {e^2}{x}}+e^3 x+e^{\frac {e^2}{x}} x+e^{\frac {e^2}{x}} x^2+2 \left (1+\frac {e^3}{2}\right ) x^2+4 x^3+x^4\right )}{x} \, dx+(2 \log (5)) \int \frac {-e^{5+\frac {e^2}{x}}-e^{2+\frac {2 e^2}{x}}-e^{2+\frac {e^2}{x}} x+e^3 x^2-\left (1-\frac {1}{e^2}\right ) e^{2+\frac {e^2}{x}} x^2+2 e^{\frac {e^2}{x}} x^3+\left (1+2 e^3\right ) x^3+3 x^4+2 x^5}{x^2} \, dx\\ &=(2 \log (5)) \int \left (e^{2 x} x+e^{2 x} x^2\right ) \, dx+(2 \log (5)) \int \frac {e^x \left (-e^{2+\frac {e^2}{x}}+e^3 x (1+x)+e^{\frac {e^2}{x}} x (1+x)+x^2 \left (2+4 x+x^2\right )\right )}{x} \, dx+(2 \log (5)) \int \left (-\frac {e^{2+\frac {2 e^2}{x}}}{x^2}+(1+2 x) \left (e^3+x+x^2\right )+\frac {e^{\frac {e^2}{x}} \left (-e^5-e^2 x+\left (1-e^2\right ) x^2+2 x^3\right )}{x^2}\right ) \, dx\\ &=-\left ((2 \log (5)) \int \frac {e^{2+\frac {2 e^2}{x}}}{x^2} \, dx\right )+(2 \log (5)) \int e^{2 x} x \, dx+(2 \log (5)) \int e^{2 x} x^2 \, dx+(2 \log (5)) \int (1+2 x) \left (e^3+x+x^2\right ) \, dx+(2 \log (5)) \int \frac {e^{\frac {e^2}{x}} \left (-e^5-e^2 x+\left (1-e^2\right ) x^2+2 x^3\right )}{x^2} \, dx+(2 \log (5)) \int \left (e^{3+x}+2 e^x \left (1+\frac {e^3}{2}\right ) x+4 e^x x^2+e^x x^3+\frac {e^{\frac {e^2}{x}+x} \left (-e^2+x+x^2\right )}{x}\right ) \, dx\\ &=e^{\frac {2 e^2}{x}} \log (5)+e^{2 x} x \log (5)+e^{2 x} x^2 \log (5)+\left (e^3+x+x^2\right )^2 \log (5)-\log (5) \int e^{2 x} \, dx+(2 \log (5)) \int e^{3+x} \, dx-(2 \log (5)) \int e^{2 x} x \, dx+(2 \log (5)) \int e^x x^3 \, dx+(2 \log (5)) \int \left (-e^{2+\frac {e^2}{x}}+e^{\frac {e^2}{x}}-\frac {e^{5+\frac {e^2}{x}}}{x^2}-\frac {e^{2+\frac {e^2}{x}}}{x}+2 e^{\frac {e^2}{x}} x\right ) \, dx+(2 \log (5)) \int \frac {e^{\frac {e^2}{x}+x} \left (-e^2+x+x^2\right )}{x} \, dx+(8 \log (5)) \int e^x x^2 \, dx+\left (2 \left (2+e^3\right ) \log (5)\right ) \int e^x x \, dx\\ &=e^{\frac {2 e^2}{x}} \log (5)-\frac {1}{2} e^{2 x} \log (5)+2 e^{3+x} \log (5)+2 e^x \left (2+e^3\right ) x \log (5)+8 e^x x^2 \log (5)+e^{2 x} x^2 \log (5)+2 e^x x^3 \log (5)-\frac {2 e^{\frac {e^2}{x}+x} \left (e^2-x^2\right ) \log (5)}{\left (1-\frac {e^2}{x^2}\right ) x}+\left (e^3+x+x^2\right )^2 \log (5)+\log (5) \int e^{2 x} \, dx-(2 \log (5)) \int e^{2+\frac {e^2}{x}} \, dx+(2 \log (5)) \int e^{\frac {e^2}{x}} \, dx-(2 \log (5)) \int \frac {e^{5+\frac {e^2}{x}}}{x^2} \, dx-(2 \log (5)) \int \frac {e^{2+\frac {e^2}{x}}}{x} \, dx+(4 \log (5)) \int e^{\frac {e^2}{x}} x \, dx-(6 \log (5)) \int e^x x^2 \, dx-(16 \log (5)) \int e^x x \, dx-\left (2 \left (2+e^3\right ) \log (5)\right ) \int e^x \, dx\\ &=2 e^{3+\frac {e^2}{x}} \log (5)+e^{\frac {2 e^2}{x}} \log (5)+2 e^{3+x} \log (5)-2 e^x \left (2+e^3\right ) \log (5)-2 e^{2+\frac {e^2}{x}} x \log (5)+2 e^{\frac {e^2}{x}} x \log (5)-16 e^x x \log (5)+2 e^x \left (2+e^3\right ) x \log (5)+2 e^{\frac {e^2}{x}} x^2 \log (5)+2 e^x x^2 \log (5)+e^{2 x} x^2 \log (5)+2 e^x x^3 \log (5)-\frac {2 e^{\frac {e^2}{x}+x} \left (e^2-x^2\right ) \log (5)}{\left (1-\frac {e^2}{x^2}\right ) x}+\left (e^3+x+x^2\right )^2 \log (5)+2 e^2 \text {Ei}\left (\frac {e^2}{x}\right ) \log (5)+(12 \log (5)) \int e^x x \, dx+(16 \log (5)) \int e^x \, dx+\left (2 e^2 \log (5)\right ) \int e^{\frac {e^2}{x}} \, dx-\left (2 e^2 \log (5)\right ) \int \frac {e^{2+\frac {e^2}{x}}}{x} \, dx+\left (2 e^2 \log (5)\right ) \int \frac {e^{\frac {e^2}{x}}}{x} \, dx\\ &=2 e^{3+\frac {e^2}{x}} \log (5)+e^{\frac {2 e^2}{x}} \log (5)+16 e^x \log (5)+2 e^{3+x} \log (5)-2 e^x \left (2+e^3\right ) \log (5)+2 e^{\frac {e^2}{x}} x \log (5)-4 e^x x \log (5)+2 e^x \left (2+e^3\right ) x \log (5)+2 e^{\frac {e^2}{x}} x^2 \log (5)+2 e^x x^2 \log (5)+e^{2 x} x^2 \log (5)+2 e^x x^3 \log (5)-\frac {2 e^{\frac {e^2}{x}+x} \left (e^2-x^2\right ) \log (5)}{\left (1-\frac {e^2}{x^2}\right ) x}+\left (e^3+x+x^2\right )^2 \log (5)+2 e^4 \text {Ei}\left (\frac {e^2}{x}\right ) \log (5)-(12 \log (5)) \int e^x \, dx+\left (2 e^4 \log (5)\right ) \int \frac {e^{\frac {e^2}{x}}}{x} \, dx\\ &=2 e^{3+\frac {e^2}{x}} \log (5)+e^{\frac {2 e^2}{x}} \log (5)+4 e^x \log (5)+2 e^{3+x} \log (5)-2 e^x \left (2+e^3\right ) \log (5)+2 e^{\frac {e^2}{x}} x \log (5)-4 e^x x \log (5)+2 e^x \left (2+e^3\right ) x \log (5)+2 e^{\frac {e^2}{x}} x^2 \log (5)+2 e^x x^2 \log (5)+e^{2 x} x^2 \log (5)+2 e^x x^3 \log (5)-\frac {2 e^{\frac {e^2}{x}+x} \left (e^2-x^2\right ) \log (5)}{\left (1-\frac {e^2}{x^2}\right ) x}+\left (e^3+x+x^2\right )^2 \log (5)\\ \end {aligned} \end {gather*}
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Mathematica [B] time = 0.19, size = 106, normalized size = 4.08 \begin {gather*} 2 \left (e^{3+\frac {e^2}{x}}+\frac {1}{2} e^{\frac {2 e^2}{x}}+e^{3+x} x+e^{\frac {e^2}{x}+x} x+\frac {1}{2} e^{2 x} x^2+e^3 x (1+x)+e^{\frac {e^2}{x}} x (1+x)+e^x x^2 (1+x)+\frac {1}{2} x^2 (1+x)^2\right ) \log (5) \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.67, size = 89, normalized size = 3.42 \begin {gather*} x^{2} e^{\left (2 \, x\right )} \log \relax (5) + 2 \, {\left (x^{3} + x^{2} + x e^{3}\right )} e^{x} \log \relax (5) + 2 \, {\left (x e^{x} \log \relax (5) + {\left (x^{2} + x + e^{3}\right )} \log \relax (5)\right )} e^{\left (\frac {e^{2}}{x}\right )} + {\left (x^{4} + 2 \, x^{3} + x^{2} + 2 \, {\left (x^{2} + x\right )} e^{3}\right )} \log \relax (5) + e^{\left (\frac {2 \, e^{2}}{x}\right )} \log \relax (5) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [C] time = 0.28, size = 238, normalized size = 9.15 \begin {gather*} x^{4} \log \relax (5) - 2 \, x^{2} {\left (\frac {{\rm Ei}\left (\frac {e^{2}}{x}\right ) e^{10}}{x^{2}} - \frac {e^{\left (\frac {e^{2}}{x} + 8\right )}}{x} - e^{\left (\frac {e^{2}}{x} + 6\right )}\right )} e^{\left (-6\right )} \log \relax (5) + 2 \, x^{3} e^{x} \log \relax (5) + 2 \, x^{3} \log \relax (5) + 2 \, x^{2} e^{3} \log \relax (5) + 2 \, x {\left (\frac {{\rm Ei}\left (\frac {e^{2}}{x}\right ) e^{8}}{x} - e^{\left (\frac {e^{2}}{x} + 6\right )}\right )} e^{\left (-4\right )} \log \relax (5) - 2 \, x {\left (\frac {{\rm Ei}\left (\frac {e^{2}}{x}\right ) e^{6}}{x} - e^{\left (\frac {e^{2}}{x} + 4\right )}\right )} e^{\left (-4\right )} \log \relax (5) + x^{2} e^{\left (2 \, x\right )} \log \relax (5) + 2 \, x^{2} e^{x} \log \relax (5) + x^{2} \log \relax (5) + 2 \, x e^{3} \log \relax (5) + 2 \, {\rm Ei}\left (\frac {e^{2}}{x}\right ) e^{2} \log \relax (5) + 2 \, x e^{\left (x + 3\right )} \log \relax (5) + 2 \, x e^{\left (\frac {x^{2} + e^{2}}{x}\right )} \log \relax (5) + e^{\left (\frac {2 \, e^{2}}{x}\right )} \log \relax (5) + 2 \, e^{\left (\frac {e^{2}}{x} + 3\right )} \log \relax (5) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.13, size = 117, normalized size = 4.50
method | result | size |
risch | \(x^{4} \ln \relax (5)+2 x^{2} {\mathrm e}^{3} \ln \relax (5)+2 x^{3} \ln \relax (5)+2 x \,{\mathrm e}^{3} \ln \relax (5)+x^{2} \ln \relax (5)+x^{2} \ln \relax (5) {\mathrm e}^{2 x}+\left (2 x \,{\mathrm e}^{3} \ln \relax (5)+2 x^{3} \ln \relax (5)+2 x^{2} \ln \relax (5)\right ) {\mathrm e}^{x}+\ln \relax (5) {\mathrm e}^{\frac {2 \,{\mathrm e}^{2}}{x}}+\left (2 \,{\mathrm e}^{3} \ln \relax (5)+2 x^{2} \ln \relax (5)+2 x \,{\mathrm e}^{x} \ln \relax (5)+2 x \ln \relax (5)\right ) {\mathrm e}^{\frac {{\mathrm e}^{2}}{x}}\) | \(117\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.53, size = 228, normalized size = 8.77 \begin {gather*} x^{4} \log \relax (5) + 2 \, x^{3} \log \relax (5) + 2 \, x^{2} e^{3} \log \relax (5) + x^{2} \log \relax (5) + 2 \, x e^{3} \log \relax (5) + 2 \, {\rm Ei}\left (\frac {e^{2}}{x}\right ) e^{2} \log \relax (5) + \frac {1}{2} \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} \log \relax (5) + \frac {1}{2} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \log \relax (5) + 2 \, x e^{\left (x + \frac {e^{2}}{x}\right )} \log \relax (5) + 2 \, {\left (x^{3} - 3 \, x^{2} + 6 \, x - 6\right )} e^{x} \log \relax (5) + 8 \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} \log \relax (5) + 2 \, {\left (x e^{3} - e^{3}\right )} e^{x} \log \relax (5) + 4 \, {\left (x - 1\right )} e^{x} \log \relax (5) + 2 \, e^{4} \Gamma \left (-1, -\frac {e^{2}}{x}\right ) \log \relax (5) - 2 \, e^{2} \Gamma \left (-1, -\frac {e^{2}}{x}\right ) \log \relax (5) + 4 \, e^{4} \Gamma \left (-2, -\frac {e^{2}}{x}\right ) \log \relax (5) + 2 \, e^{\left (x + 3\right )} \log \relax (5) + e^{\left (\frac {2 \, e^{2}}{x}\right )} \log \relax (5) + 2 \, e^{\left (\frac {e^{2}}{x} + 3\right )} \log \relax (5) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.26, size = 91, normalized size = 3.50 \begin {gather*} x^2\,\left (\ln \relax (5)+2\,{\mathrm {e}}^3\,\ln \relax (5)\right )+2\,x^3\,\ln \relax (5)+x^4\,\ln \relax (5)+{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^2}{x}}\,\ln \relax (5)+2\,{\mathrm {e}}^{\frac {{\mathrm {e}}^2}{x}}\,\ln \relax (5)\,\left (x+{\mathrm {e}}^3+x\,{\mathrm {e}}^x+x^2\right )+2\,x\,{\mathrm {e}}^3\,\ln \relax (5)+x^2\,{\mathrm {e}}^{2\,x}\,\ln \relax (5)+2\,x\,{\mathrm {e}}^x\,\ln \relax (5)\,\left (x^2+x+{\mathrm {e}}^3\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 10.46, size = 134, normalized size = 5.15 \begin {gather*} x^{4} \log {\relax (5 )} + 2 x^{3} \log {\relax (5 )} + x^{2} e^{2 x} \log {\relax (5 )} + x^{2} \left (\log {\relax (5 )} + 2 e^{3} \log {\relax (5 )}\right ) + 2 x e^{3} \log {\relax (5 )} + \left (2 x^{3} \log {\relax (5 )} + 2 x^{2} \log {\relax (5 )} + 2 x e^{3} \log {\relax (5 )}\right ) e^{x} + \left (2 x^{2} \log {\relax (5 )} + 2 x e^{x} \log {\relax (5 )} + 2 x \log {\relax (5 )} + 2 e^{3} \log {\relax (5 )}\right ) e^{\frac {e^{2}}{x}} + e^{\frac {2 e^{2}}{x}} \log {\relax (5 )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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