Optimal. Leaf size=25 \[ e^{3+\frac {2 x^2}{5 \log (x) \left (\frac {2}{x}+\log (x)\right )}} \]
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Rubi [F]
time = 4.53, 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 {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{10 \log (x)+5 x \log ^2(x)}\right ) \left (-4 x^2+\left (12 x^2-4 x^3\right ) \log (x)+4 x^3 \log ^2(x)\right )}{20 \log ^2(x)+20 x \log ^3(x)+5 x^2 \log ^4(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 {4 \exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^2 \left (-1+3 \log (x)-x \log (x)+x \log ^2(x)\right )}{5 \log ^2(x) (2+x \log (x))^2} \, dx\\ &=\frac {4}{5} \int \frac {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^2 \left (-1+3 \log (x)-x \log (x)+x \log ^2(x)\right )}{\log ^2(x) (2+x \log (x))^2} \, dx\\ &=\frac {4}{5} \int \left (-\frac {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^2}{4 \log ^2(x)}+\frac {3 \exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^2}{4 \log (x)}+\frac {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) (-2+x) x^3}{4 (2+x \log (x))^2}-\frac {3 \exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^3}{4 (2+x \log (x))}\right ) \, dx\\ &=-\left (\frac {1}{5} \int \frac {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^2}{\log ^2(x)} \, dx\right )+\frac {1}{5} \int \frac {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) (-2+x) x^3}{(2+x \log (x))^2} \, dx+\frac {3}{5} \int \frac {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^2}{\log (x)} \, dx-\frac {3}{5} \int \frac {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^3}{2+x \log (x)} \, dx\\ &=-\left (\frac {1}{5} \int \frac {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^2}{\log ^2(x)} \, dx\right )+\frac {1}{5} \int \left (-\frac {2 \exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^3}{(2+x \log (x))^2}+\frac {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^4}{(2+x \log (x))^2}\right ) \, dx+\frac {3}{5} \int \frac {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^2}{\log (x)} \, dx-\frac {3}{5} \int \frac {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^3}{2+x \log (x)} \, dx\\ &=-\left (\frac {1}{5} \int \frac {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^2}{\log ^2(x)} \, dx\right )+\frac {1}{5} \int \frac {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^4}{(2+x \log (x))^2} \, dx-\frac {2}{5} \int \frac {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^3}{(2+x \log (x))^2} \, dx+\frac {3}{5} \int \frac {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^2}{\log (x)} \, dx-\frac {3}{5} \int \frac {\exp \left (\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{5 \log (x) (2+x \log (x))}\right ) x^3}{2+x \log (x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.30, size = 34, normalized size = 1.36 \begin {gather*} e^{\frac {2 x^3+30 \log (x)+15 x \log ^2(x)}{10 \log (x)+5 x \log ^2(x)}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.06, size = 33, normalized size = 1.32
method | result | size |
risch | \({\mathrm e}^{\frac {15 x \ln \left (x \right )^{2}+30 \ln \left (x \right )+2 x^{3}}{5 \ln \left (x \right ) \left (x \ln \left (x \right )+2\right )}}\) | \(33\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.43, size = 42, normalized size = 1.68 \begin {gather*} e^{\left (\frac {2 \, x^{2}}{5 \, \log \left (x\right )^{2}} - \frac {16}{5 \, {\left (x \log \left (x\right )^{5} + 2 \, \log \left (x\right )^{4}\right )}} - \frac {4 \, x}{5 \, \log \left (x\right )^{3}} + \frac {8}{5 \, \log \left (x\right )^{4}} + 3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.37, size = 33, normalized size = 1.32 \begin {gather*} e^{\left (\frac {2 \, x^{3} + 15 \, x \log \left (x\right )^{2} + 30 \, \log \left (x\right )}{5 \, {\left (x \log \left (x\right )^{2} + 2 \, \log \left (x\right )\right )}}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 0.22, size = 32, normalized size = 1.28 \begin {gather*} e^{\frac {2 x^{3} + 15 x \log {\left (x \right )}^{2} + 30 \log {\left (x \right )}}{5 x \log {\left (x \right )}^{2} + 10 \log {\left (x \right )}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 57 vs.
\(2 (22) = 44\).
time = 0.43, size = 57, normalized size = 2.28 \begin {gather*} e^{\left (\frac {2 \, x^{3}}{5 \, {\left (x \log \left (x\right )^{2} + 2 \, \log \left (x\right )\right )}} + \frac {3 \, x \log \left (x\right )^{2}}{x \log \left (x\right )^{2} + 2 \, \log \left (x\right )} + \frac {6 \, \log \left (x\right )}{x \log \left (x\right )^{2} + 2 \, \log \left (x\right )}\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 3.42, size = 51, normalized size = 2.04 \begin {gather*} x^{\frac {3\,x}{x\,\ln \left (x\right )+2}}\,x^{\frac {6}{x\,{\ln \left (x\right )}^2+2\,\ln \left (x\right )}}\,{\mathrm {e}}^{\frac {2\,x^3}{5\,x\,{\ln \left (x\right )}^2+10\,\ln \left (x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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