Optimal. Leaf size=31 \[ e^x \left (5-\log \left (e^{-1/e} \left (x+\frac {3 e^{x^2} x}{\log (x)}\right )\right )\right ) \]
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Rubi [A] time = 5.57, antiderivative size = 34, normalized size of antiderivative = 1.10, number of steps used = 27, number of rules used = 7, integrand size = 113, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {6741, 6742, 2199, 2194, 2178, 2176, 2555} \begin {gather*} (1+5 e) e^{x-1}-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right ) \end {gather*}
Antiderivative was successfully verified.
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Rule 2176
Rule 2178
Rule 2194
Rule 2199
Rule 2555
Rule 6741
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 e^{x+x^2}+e^{x+x^2} \left (-3+15 x-6 x^2\right ) \log (x)+e^x (-1+5 x) \log ^2(x)+\left (-3 e^{x+x^2} x \log (x)-e^x x \log ^2(x)\right ) \log \left (\frac {e^{-1/e} \left (3 e^{x^2} x+x \log (x)\right )}{\log (x)}\right )}{x \log (x) \left (3 e^{x^2}+\log (x)\right )} \, dx\\ &=\int \left (\frac {e^x \left (-1+2 x^2 \log (x)\right )}{x \left (3 e^{x^2}+\log (x)\right )}+\frac {e^{-1+x} \left (e-e \log (x)+(1+5 e) x \log (x)-2 e x^2 \log (x)-e x \log (x) \log \left (x+\frac {3 e^{x^2} x}{\log (x)}\right )\right )}{x \log (x)}\right ) \, dx\\ &=\int \frac {e^x \left (-1+2 x^2 \log (x)\right )}{x \left (3 e^{x^2}+\log (x)\right )} \, dx+\int \frac {e^{-1+x} \left (e-e \log (x)+(1+5 e) x \log (x)-2 e x^2 \log (x)-e x \log (x) \log \left (x+\frac {3 e^{x^2} x}{\log (x)}\right )\right )}{x \log (x)} \, dx\\ &=\int \left (-\frac {e^x}{x \left (3 e^{x^2}+\log (x)\right )}+\frac {2 e^x x \log (x)}{3 e^{x^2}+\log (x)}\right ) \, dx+\int \left (\frac {e^{-1+x} \left (e-e \log (x)+(1+5 e) x \log (x)-2 e x^2 \log (x)\right )}{x \log (x)}-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )\right ) \, dx\\ &=2 \int \frac {e^x x \log (x)}{3 e^{x^2}+\log (x)} \, dx-\int \frac {e^x}{x \left (3 e^{x^2}+\log (x)\right )} \, dx+\int \frac {e^{-1+x} \left (e-e \log (x)+(1+5 e) x \log (x)-2 e x^2 \log (x)\right )}{x \log (x)} \, dx-\int e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right ) \, dx\\ &=-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )+2 \int \frac {e^x x \log (x)}{3 e^{x^2}+\log (x)} \, dx+\int \left (\frac {e^{-1+x} \left (-e+(1+5 e) x-2 e x^2\right )}{x}+\frac {e^x}{x \log (x)}\right ) \, dx-\int \frac {e^x}{x \left (3 e^{x^2}+\log (x)\right )} \, dx+\int \frac {e^x \left (-3 e^{x^2}+3 e^{x^2} \left (1+2 x^2\right ) \log (x)+\log ^2(x)\right )}{x \log (x) \left (3 e^{x^2}+\log (x)\right )} \, dx\\ &=-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )+2 \int \frac {e^x x \log (x)}{3 e^{x^2}+\log (x)} \, dx+\int \frac {e^{-1+x} \left (-e+(1+5 e) x-2 e x^2\right )}{x} \, dx+\int \frac {e^x}{x \log (x)} \, dx-\int \frac {e^x}{x \left (3 e^{x^2}+\log (x)\right )} \, dx+\int \left (-\frac {e^x \left (-1+2 x^2 \log (x)\right )}{x \left (3 e^{x^2}+\log (x)\right )}+\frac {e^x \left (-1+\log (x)+2 x^2 \log (x)\right )}{x \log (x)}\right ) \, dx\\ &=-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )+2 \int \frac {e^x x \log (x)}{3 e^{x^2}+\log (x)} \, dx+\int \left (e^{-1+x} (1+5 e)-\frac {e^x}{x}-2 e^x x\right ) \, dx+\int \frac {e^x}{x \log (x)} \, dx-\int \frac {e^x}{x \left (3 e^{x^2}+\log (x)\right )} \, dx-\int \frac {e^x \left (-1+2 x^2 \log (x)\right )}{x \left (3 e^{x^2}+\log (x)\right )} \, dx+\int \frac {e^x \left (-1+\log (x)+2 x^2 \log (x)\right )}{x \log (x)} \, dx\\ &=-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )-2 \int e^x x \, dx+2 \int \frac {e^x x \log (x)}{3 e^{x^2}+\log (x)} \, dx+(1+5 e) \int e^{-1+x} \, dx-\int \frac {e^x}{x} \, dx+\int \left (\frac {e^x \left (1+2 x^2\right )}{x}-\frac {e^x}{x \log (x)}\right ) \, dx+\int \frac {e^x}{x \log (x)} \, dx-\int \frac {e^x}{x \left (3 e^{x^2}+\log (x)\right )} \, dx-\int \left (-\frac {e^x}{x \left (3 e^{x^2}+\log (x)\right )}+\frac {2 e^x x \log (x)}{3 e^{x^2}+\log (x)}\right ) \, dx\\ &=e^{-1+x} (1+5 e)-2 e^x x-\text {Ei}(x)-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )+2 \int e^x \, dx+\int \frac {e^x \left (1+2 x^2\right )}{x} \, dx\\ &=2 e^x+e^{-1+x} (1+5 e)-2 e^x x-\text {Ei}(x)-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )+\int \left (\frac {e^x}{x}+2 e^x x\right ) \, dx\\ &=2 e^x+e^{-1+x} (1+5 e)-2 e^x x-\text {Ei}(x)-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )+2 \int e^x x \, dx+\int \frac {e^x}{x} \, dx\\ &=2 e^x+e^{-1+x} (1+5 e)-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )-2 \int e^x \, dx\\ &=e^{-1+x} (1+5 e)-e^x \log \left (\frac {x \left (3 e^{x^2}+\log (x)\right )}{\log (x)}\right )\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.22, size = 29, normalized size = 0.94 \begin {gather*} e^{-1+x} \left (1+5 e-e \log \left (x+\frac {3 e^{x^2} x}{\log (x)}\right )\right ) \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.93, size = 47, normalized size = 1.52 \begin {gather*} -{\left (e^{\left (x^{2} + x\right )} \log \left (\frac {{\left (3 \, x e^{\left (x^{2}\right )} + x \log \relax (x)\right )} e^{\left (-e^{\left (-1\right )}\right )}}{\log \relax (x)}\right ) - 5 \, e^{\left (x^{2} + x\right )}\right )} e^{\left (-x^{2}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.27, size = 46, normalized size = 1.48 \begin {gather*} -{\left (e^{\left (x + 1\right )} \log \relax (x) + e^{\left (x + 1\right )} \log \left (3 \, e^{\left (x^{2}\right )} + \log \relax (x)\right ) - e^{\left (x + 1\right )} \log \left (\log \relax (x)\right ) - 5 \, e^{\left (x + 1\right )} - e^{x}\right )} e^{\left (-1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [C] time = 0.32, size = 333, normalized size = 10.74
method | result | size |
risch | \(-{\mathrm e}^{x} \ln \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )+\frac {\left (i \pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right )^{3} {\mathrm e}+i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) {\mathrm e}-i \pi \,\mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i x \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right )^{2} {\mathrm e}+i \pi \mathrm {csgn}\left (\frac {i x \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right )^{3} {\mathrm e}-i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right )^{2} {\mathrm e}+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i x \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right ) {\mathrm e}-i \pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right ) {\mathrm e}-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x \left ({\mathrm e}^{x^{2}}+\frac {\ln \relax (x )}{3}\right )}{\ln \relax (x )}\right )^{2} {\mathrm e}+2-2 \ln \relax (3) {\mathrm e}-2 \,{\mathrm e} \ln \relax (x )+2 \ln \left (\ln \relax (x )\right ) {\mathrm e}+10 \,{\mathrm e}\right ) {\mathrm e}^{x -1}}{2}\) | \(333\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.44, size = 43, normalized size = 1.39 \begin {gather*} -{\left ({\left (e \log \relax (x) - 5 \, e - 1\right )} e^{x} + e^{\left (x + 1\right )} \log \left (3 \, e^{\left (x^{2}\right )} + \log \relax (x)\right ) - e^{\left (x + 1\right )} \log \left (\log \relax (x)\right )\right )} e^{\left (-1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.56, size = 29, normalized size = 0.94 \begin {gather*} -{\mathrm {e}}^x\,\left (\ln \left (\frac {{\mathrm {e}}^{-{\mathrm {e}}^{-1}}\,\left (3\,x\,{\mathrm {e}}^{x^2}+x\,\ln \relax (x)\right )}{\ln \relax (x)}\right )-5\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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