Optimal. Leaf size=23 \[ e^{-x} \left (e^x x+\frac {5 x}{1+\log (x)}\right )^2 \]
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Rubi [C] time = 1.80, antiderivative size = 96, normalized size of antiderivative = 4.17, number of steps used = 18, number of rules used = 11, integrand size = 133, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {6688, 6742, 2196, 2176, 2194, 2306, 2309, 2178, 2366, 6482, 2288} \begin {gather*} \frac {20 \text {Ei}(2 (\log (x)+1))}{e^2}-\frac {40 (\log (x)+1) \text {Ei}(2 (\log (x)+1))}{e^2}+\frac {20 (2 \log (x)+1) \text {Ei}(2 (\log (x)+1))}{e^2}+e^x x^2+20 x^2-\frac {10 x^2 (2 \log (x)+1)}{\log (x)+1}+\frac {25 e^{-x} x (x+x \log (x))}{(\log (x)+1)^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 2176
Rule 2178
Rule 2194
Rule 2196
Rule 2288
Rule 2306
Rule 2309
Rule 2366
Rule 6482
Rule 6688
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} x \left (5+e^x+e^x \log (x)\right ) \left (-5 x+e^x (2+x)+\left (-5 (-2+x)+2 e^x (2+x)\right ) \log (x)+e^x (2+x) \log ^2(x)\right )}{(1+\log (x))^3} \, dx\\ &=\int \left (e^x x (2+x)+\frac {10 x (1+2 \log (x))}{(1+\log (x))^2}-\frac {25 e^{-x} x (x-2 \log (x)+x \log (x))}{(1+\log (x))^3}\right ) \, dx\\ &=10 \int \frac {x (1+2 \log (x))}{(1+\log (x))^2} \, dx-25 \int \frac {e^{-x} x (x-2 \log (x)+x \log (x))}{(1+\log (x))^3} \, dx+\int e^x x (2+x) \, dx\\ &=\frac {20 \text {Ei}(2 (1+\log (x))) (1+2 \log (x))}{e^2}-\frac {10 x^2 (1+2 \log (x))}{1+\log (x)}+\frac {25 e^{-x} x (x+x \log (x))}{(1+\log (x))^3}-20 \int \left (\frac {2 \text {Ei}(2 (1+\log (x)))}{e^2 x}-\frac {x}{1+\log (x)}\right ) \, dx+\int \left (2 e^x x+e^x x^2\right ) \, dx\\ &=\frac {20 \text {Ei}(2 (1+\log (x))) (1+2 \log (x))}{e^2}-\frac {10 x^2 (1+2 \log (x))}{1+\log (x)}+\frac {25 e^{-x} x (x+x \log (x))}{(1+\log (x))^3}+2 \int e^x x \, dx+20 \int \frac {x}{1+\log (x)} \, dx-\frac {40 \int \frac {\text {Ei}(2 (1+\log (x)))}{x} \, dx}{e^2}+\int e^x x^2 \, dx\\ &=2 e^x x+e^x x^2+\frac {20 \text {Ei}(2 (1+\log (x))) (1+2 \log (x))}{e^2}-\frac {10 x^2 (1+2 \log (x))}{1+\log (x)}+\frac {25 e^{-x} x (x+x \log (x))}{(1+\log (x))^3}-2 \int e^x \, dx-2 \int e^x x \, dx+20 \operatorname {Subst}\left (\int \frac {e^{2 x}}{1+x} \, dx,x,\log (x)\right )-\frac {40 \operatorname {Subst}(\int \text {Ei}(2 (1+x)) \, dx,x,\log (x))}{e^2}\\ &=-2 e^x+e^x x^2+\frac {20 \text {Ei}(2 (1+\log (x)))}{e^2}+\frac {20 \text {Ei}(2 (1+\log (x))) (1+2 \log (x))}{e^2}-\frac {10 x^2 (1+2 \log (x))}{1+\log (x)}+\frac {25 e^{-x} x (x+x \log (x))}{(1+\log (x))^3}+2 \int e^x \, dx-\frac {20 \operatorname {Subst}(\int \text {Ei}(x) \, dx,x,2+2 \log (x))}{e^2}\\ &=20 x^2+e^x x^2+\frac {20 \text {Ei}(2 (1+\log (x)))}{e^2}-\frac {40 \text {Ei}(2+2 \log (x)) (1+\log (x))}{e^2}+\frac {20 \text {Ei}(2 (1+\log (x))) (1+2 \log (x))}{e^2}-\frac {10 x^2 (1+2 \log (x))}{1+\log (x)}+\frac {25 e^{-x} x (x+x \log (x))}{(1+\log (x))^3}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.80, size = 28, normalized size = 1.22 \begin {gather*} \frac {e^{-x} x^2 \left (5+e^x+e^x \log (x)\right )^2}{(1+\log (x))^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [B] time = 0.54, size = 72, normalized size = 3.13 \begin {gather*} \frac {x^{2} e^{\left (2 \, x\right )} \log \relax (x)^{2} + x^{2} e^{\left (2 \, x\right )} + 10 \, x^{2} e^{x} + 25 \, x^{2} + 2 \, {\left (x^{2} e^{\left (2 \, x\right )} + 5 \, x^{2} e^{x}\right )} \log \relax (x)}{e^{x} \log \relax (x)^{2} + 2 \, e^{x} \log \relax (x) + e^{x}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.18, size = 60, normalized size = 2.61 \begin {gather*} \frac {x^{2} e^{x} \log \relax (x)^{2} + 2 \, x^{2} e^{x} \log \relax (x) + 25 \, x^{2} e^{\left (-x\right )} + x^{2} e^{x} + 10 \, x^{2} \log \relax (x) + 10 \, x^{2}}{\log \relax (x)^{2} + 2 \, \log \relax (x) + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 35, normalized size = 1.52
method | result | size |
risch | \({\mathrm e}^{x} x^{2}+\frac {5 x^{2} \left (2 \,{\mathrm e}^{x} \ln \relax (x )+2 \,{\mathrm e}^{x}+5\right ) {\mathrm e}^{-x}}{\left (\ln \relax (x )+1\right )^{2}}\) | \(35\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.44, size = 57, normalized size = 2.48 \begin {gather*} \frac {25 \, x^{2} e^{\left (-x\right )} + 10 \, x^{2} \log \relax (x) + 10 \, x^{2} + {\left (x^{2} \log \relax (x)^{2} + 2 \, x^{2} \log \relax (x) + x^{2}\right )} e^{x}}{\log \relax (x)^{2} + 2 \, \log \relax (x) + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.66, size = 169, normalized size = 7.35 \begin {gather*} x^2\,{\mathrm {e}}^x+20\,x^2\,\ln \relax (x)-\frac {10\,x^2\,{\ln \relax (x)}^2-\frac {5\,x^2\,{\mathrm {e}}^{-x}\,\left (5\,x-2\,{\mathrm {e}}^x\right )}{2}+\frac {5\,x^2\,{\mathrm {e}}^{-x}\,\ln \relax (x)\,\left (6\,{\mathrm {e}}^x-5\,x+10\right )}{2}}{{\ln \relax (x)}^2+2\,\ln \relax (x)+1}+{\mathrm {e}}^{-x}\,\left (\frac {25\,x^4}{2}-\frac {125\,x^3}{2}+50\,x^2\right )+40\,x^2-\frac {20\,x^2\,{\ln \relax (x)}^2+\frac {25\,x\,{\mathrm {e}}^{-x}\,\left (2\,x+2\,x\,{\mathrm {e}}^x-4\,x^2+x^3\right )}{2}+\frac {25\,x\,{\mathrm {e}}^{-x}\,\ln \relax (x)\,\left (4\,x+4\,x\,{\mathrm {e}}^x-5\,x^2+x^3\right )}{2}}{\ln \relax (x)+1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.38, size = 53, normalized size = 2.30 \begin {gather*} \frac {10 x^{2}}{\log {\relax (x )} + 1} + \frac {25 x^{2} e^{- x} + \left (x^{2} \log {\relax (x )}^{2} + 2 x^{2} \log {\relax (x )} + x^{2}\right ) e^{x}}{\log {\relax (x )}^{2} + 2 \log {\relax (x )} + 1} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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