Optimal. Leaf size=27 \[ \frac {\log \left (\log \left (\left (e^3+2 x\right ) \left (\frac {x}{3}+4 \left (e^x+x\right )\right )\right )\right )}{x} \]
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Rubi [F]
time = 1.34, 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 {13 e^3 x+52 x^2+e^x \left (24 x+12 e^3 x+24 x^2\right )+\left (e^x \left (-12 e^3-24 x\right )-13 e^3 x-26 x^2\right ) \log \left (\frac {1}{3} \left (13 e^3 x+26 x^2+e^x \left (12 e^3+24 x\right )\right )\right ) \log \left (\log \left (\frac {1}{3} \left (13 e^3 x+26 x^2+e^x \left (12 e^3+24 x\right )\right )\right )\right )}{\left (13 e^3 x^3+26 x^4+e^x \left (12 e^3 x^2+24 x^3\right )\right ) \log \left (\frac {1}{3} \left (13 e^3 x+26 x^2+e^x \left (12 e^3+24 x\right )\right )\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\frac {x \left (13 e^3+12 e^{3+x}+52 x+24 e^x (1+x)\right )}{\left (e^3+2 x\right ) \left (12 e^x+13 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )}-\log \left (\log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )\right )}{x^2} \, dx\\ &=\int \left (-\frac {13 (-1+x)}{x \left (12 e^x+13 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )}+\frac {2 \left (1+\frac {e^3}{2}\right ) x+2 x^2-e^3 \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right ) \log \left (\log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )\right )-2 x \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right ) \log \left (\log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )\right )}{x^2 \left (e^3+2 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )}\right ) \, dx\\ &=-\left (13 \int \frac {-1+x}{x \left (12 e^x+13 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )} \, dx\right )+\int \frac {2 \left (1+\frac {e^3}{2}\right ) x+2 x^2-e^3 \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right ) \log \left (\log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )\right )-2 x \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right ) \log \left (\log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )\right )}{x^2 \left (e^3+2 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )} \, dx\\ &=-\left (13 \int \left (\frac {1}{\left (12 e^x+13 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )}-\frac {1}{x \left (12 e^x+13 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )}\right ) \, dx\right )+\int \frac {\frac {x \left (2+e^3+2 x\right )}{\left (e^3+2 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )}-\log \left (\log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )\right )}{x^2} \, dx\\ &=-\left (13 \int \frac {1}{\left (12 e^x+13 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )} \, dx\right )+13 \int \frac {1}{x \left (12 e^x+13 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )} \, dx+\int \left (\frac {2+e^3+2 x}{x \left (e^3+2 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )}-\frac {\log \left (\log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )\right )}{x^2}\right ) \, dx\\ &=-\left (13 \int \frac {1}{\left (12 e^x+13 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )} \, dx\right )+13 \int \frac {1}{x \left (12 e^x+13 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )} \, dx+\int \frac {2+e^3+2 x}{x \left (e^3+2 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )} \, dx-\int \frac {\log \left (\log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )\right )}{x^2} \, dx\\ &=-\left (13 \int \frac {1}{\left (12 e^x+13 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )} \, dx\right )+13 \int \frac {1}{x \left (12 e^x+13 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )} \, dx+\int \left (\frac {2+e^3}{e^3 x \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )}-\frac {4}{e^3 \left (e^3+2 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )}\right ) \, dx-\int \frac {\log \left (\log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )\right )}{x^2} \, dx\\ &=-\left (13 \int \frac {1}{\left (12 e^x+13 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )} \, dx\right )+13 \int \frac {1}{x \left (12 e^x+13 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )} \, dx-\frac {4 \int \frac {1}{\left (e^3+2 x\right ) \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )} \, dx}{e^3}+\frac {\left (2+e^3\right ) \int \frac {1}{x \log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )} \, dx}{e^3}-\int \frac {\log \left (\log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )\right )}{x^2} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.11, size = 26, normalized size = 0.96 \begin {gather*} \frac {\log \left (\log \left (\frac {1}{3} \left (e^3+2 x\right ) \left (12 e^x+13 x\right )\right )\right )}{x} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
3.
time = 0.27, size = 103, normalized size = 3.81
method | result | size |
risch | \(\frac {\ln \left (-\ln \left (3\right )+\ln \left (x +\frac {12 \,{\mathrm e}^{x}}{13}\right )+\ln \left (2 x +{\mathrm e}^{3}\right )-\frac {i \pi \,\mathrm {csgn}\left (i \left (x +\frac {12 \,{\mathrm e}^{x}}{13}\right ) \left (2 x +{\mathrm e}^{3}\right )\right ) \left (-\mathrm {csgn}\left (i \left (x +\frac {12 \,{\mathrm e}^{x}}{13}\right ) \left (2 x +{\mathrm e}^{3}\right )\right )+\mathrm {csgn}\left (i \left (x +\frac {12 \,{\mathrm e}^{x}}{13}\right )\right )\right ) \left (-\mathrm {csgn}\left (i \left (x +\frac {12 \,{\mathrm e}^{x}}{13}\right ) \left (2 x +{\mathrm e}^{3}\right )\right )+\mathrm {csgn}\left (i \left (2 x +{\mathrm e}^{3}\right )\right )\right )}{2}\right )}{x}\) | \(103\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.57, size = 26, normalized size = 0.96 \begin {gather*} \frac {\log \left (-\log \left (3\right ) + \log \left (13 \, x + 12 \, e^{x}\right ) + \log \left (2 \, x + e^{3}\right )\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.33, size = 27, normalized size = 1.00 \begin {gather*} \frac {\log \left (\log \left (\frac {26}{3} \, x^{2} + \frac {13}{3} \, x e^{3} + 4 \, {\left (2 \, x + e^{3}\right )} e^{x}\right )\right )}{x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 2.46, size = 31, normalized size = 1.15 \begin {gather*} \frac {\log {\left (\log {\left (\frac {26 x^{2}}{3} + \frac {13 x e^{3}}{3} + \left (8 x + 4 e^{3}\right ) e^{x} \right )} \right )}}{x} \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 = 1.32, size = 47, normalized size = 1.74 \begin {gather*} \frac {\ln \left (\ln \left (\frac {{\mathrm {e}}^x\,\left (24\,x+12\,{\mathrm {e}}^3\right )}{3}+\frac {13\,x\,{\mathrm {e}}^3}{3}+\frac {26\,x^2}{3}\right )\right )\,\left (2\,x^2+{\mathrm {e}}^3\,x\right )}{x^2\,\left (2\,x+{\mathrm {e}}^3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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