Optimal. Leaf size=27 \[ \frac {1}{3} e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} (x-\log (x))^2 \]
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Rubi [F]
time = 2.81, 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 {e^{\frac {4 x+4 x^2+x^3+\left (4+4 x+x^2\right ) \log (x)}{x^2}} \left (4 x^2-2 x^3+3 x^4+x^5+\left (-8 x-6 x^2-8 x^3-2 x^4\right ) \log (x)+\left (4+16 x+9 x^2+x^3\right ) \log ^2(x)+(-8-4 x) \log ^3(x)\right )}{3 x^3} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{3} \int \frac {e^{\frac {4 x+4 x^2+x^3+\left (4+4 x+x^2\right ) \log (x)}{x^2}} \left (4 x^2-2 x^3+3 x^4+x^5+\left (-8 x-6 x^2-8 x^3-2 x^4\right ) \log (x)+\left (4+16 x+9 x^2+x^3\right ) \log ^2(x)+(-8-4 x) \log ^3(x)\right )}{x^3} \, dx\\ &=\frac {1}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \left (4 x^2-2 x^3+3 x^4+x^5+\left (-8 x-6 x^2-8 x^3-2 x^4\right ) \log (x)+\left (4+16 x+9 x^2+x^3\right ) \log ^2(x)+(-8-4 x) \log ^3(x)\right )}{x^3} \, dx\\ &=\frac {1}{3} \int \left (\frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \left (4-2 x+3 x^2+x^3\right )}{x}-\frac {2 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \left (4+3 x+4 x^2+x^3\right ) \log (x)}{x^2}+\frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} (2+x) \left (2+7 x+x^2\right ) \log ^2(x)}{x^3}-\frac {4 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} (2+x) \log ^3(x)}{x^3}\right ) \, dx\\ &=\frac {1}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \left (4-2 x+3 x^2+x^3\right )}{x} \, dx+\frac {1}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} (2+x) \left (2+7 x+x^2\right ) \log ^2(x)}{x^3} \, dx-\frac {2}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \left (4+3 x+4 x^2+x^3\right ) \log (x)}{x^2} \, dx-\frac {4}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} (2+x) \log ^3(x)}{x^3} \, dx\\ &=\frac {1}{3} \int \left (-2 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}}+\frac {4 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}}}{x}+3 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} x+e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} x^2\right ) \, dx+\frac {1}{3} \int \left (e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^2(x)+\frac {4 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^2(x)}{x^3}+\frac {16 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^2(x)}{x^2}+\frac {9 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^2(x)}{x}\right ) \, dx-\frac {2}{3} \int \left (4 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log (x)+\frac {4 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log (x)}{x^2}+\frac {3 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log (x)}{x}+e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} x \log (x)\right ) \, dx-\frac {4}{3} \int \left (\frac {2 e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^3(x)}{x^3}+\frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^3(x)}{x^2}\right ) \, dx\\ &=\frac {1}{3} \int e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} x^2 \, dx+\frac {1}{3} \int e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^2(x) \, dx-\frac {2}{3} \int e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \, dx-\frac {2}{3} \int e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} x \log (x) \, dx+\frac {4}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}}}{x} \, dx+\frac {4}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^2(x)}{x^3} \, dx-\frac {4}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^3(x)}{x^2} \, dx-2 \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log (x)}{x} \, dx-\frac {8}{3} \int e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log (x) \, dx-\frac {8}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log (x)}{x^2} \, dx-\frac {8}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^3(x)}{x^3} \, dx+3 \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^2(x)}{x} \, dx+\frac {16}{3} \int \frac {e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} \log ^2(x)}{x^2} \, dx+\int e^{\frac {(2+x)^2 (x+\log (x))}{x^2}} x \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.04, size = 33, normalized size = 1.22 \begin {gather*} \frac {1}{3} e^{4+\frac {4}{x}+x} x^{\frac {(2+x)^2}{x^2}} (x-\log (x))^2 \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.07, size = 30, normalized size = 1.11
method | result | size |
risch | \(\frac {\left (\ln \left (x \right )^{2}-2 x \ln \left (x \right )+x^{2}\right ) {\mathrm e}^{\frac {\left (2+x \right )^{2} \left (x +\ln \left (x \right )\right )}{x^{2}}}}{3}\) | \(30\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.34, size = 48, normalized size = 1.78 \begin {gather*} \frac {1}{3} \, {\left (x^{3} e^{4} - 2 \, x^{2} e^{4} \log \left (x\right ) + x e^{4} \log \left (x\right )^{2}\right )} e^{\left (x + \frac {4 \, \log \left (x\right )}{x} + \frac {4}{x} + \frac {4 \, \log \left (x\right )}{x^{2}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.40, size = 43, normalized size = 1.59 \begin {gather*} \frac {1}{3} \, {\left (x^{2} - 2 \, x \log \left (x\right ) + \log \left (x\right )^{2}\right )} e^{\left (\frac {x^{3} + 4 \, x^{2} + {\left (x^{2} + 4 \, x + 4\right )} \log \left (x\right ) + 4 \, x}{x^{2}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.27, size = 44, normalized size = 1.63 \begin {gather*} \frac {\left (x^{2} - 2 x \log {\left (x \right )} + \log {\left (x \right )}^{2}\right ) e^{\frac {x^{3} + 4 x^{2} + 4 x + \left (x^{2} + 4 x + 4\right ) \log {\left (x \right )}}{x^{2}}}}{3} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.74, size = 42, normalized size = 1.56 \begin {gather*} x\,x^{4/x}\,x^{\frac {4}{x^2}}\,{\mathrm {e}}^{x+\frac {4}{x}+4}\,\left (\frac {x^2}{3}-\frac {2\,x\,\ln \left (x\right )}{3}+\frac {{\ln \left (x\right )}^2}{3}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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