Optimal. Leaf size=26 \[ \frac {1}{3 \left (1+e^x+\frac {3 x}{e^x+\log (4)}\right ) \log (x)} \]
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Rubi [F]
time = 4.58, 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^{3 x}+e^{2 x} (-1-2 \log (4))-3 x \log (4)-\log ^2(4)+e^x \left (-3 x-2 \log (4)-\log ^2(4)\right )+\left (-e^{3 x} x-3 x \log (4)-2 e^{2 x} x \log (4)+e^x \left (-3 x+3 x^2-x \log ^2(4)\right )\right ) \log (x)}{3 \left (e^{4 x} x+9 x^3+6 x^2 \log (4)+x \log ^2(4)+e^{3 x} (2 x+2 x \log (4))+e^{2 x} \left (x+6 x^2+4 x \log (4)+x \log ^2(4)\right )+e^x \left (6 x^2+\left (2 x+6 x^2\right ) \log (4)+2 x \log ^2(4)\right )\right ) \log ^2(x)} \, 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^{3 x}+e^{2 x} (-1-2 \log (4))-3 x \log (4)-\log ^2(4)+e^x \left (-3 x-2 \log (4)-\log ^2(4)\right )+\left (-e^{3 x} x-3 x \log (4)-2 e^{2 x} x \log (4)+e^x \left (-3 x+3 x^2-x \log ^2(4)\right )\right ) \log (x)}{\left (e^{4 x} x+9 x^3+6 x^2 \log (4)+x \log ^2(4)+e^{3 x} (2 x+2 x \log (4))+e^{2 x} \left (x+6 x^2+4 x \log (4)+x \log ^2(4)\right )+e^x \left (6 x^2+\left (2 x+6 x^2\right ) \log (4)+2 x \log ^2(4)\right )\right ) \log ^2(x)} \, dx\\ &=\frac {1}{3} \int \frac {-\left (\left (e^x+\log (4)\right ) \left (e^{2 x}+3 x+\log (4)+e^x (1+\log (4))\right )\right )-x \left (e^{3 x}+e^x \left (3-3 x+\log ^2(4)\right )+e^{2 x} \log (16)+\log (64)\right ) \log (x)}{x \left (e^{2 x}+3 x+\log (4)+e^x (1+\log (4))\right )^2 \log ^2(x)} \, dx\\ &=\frac {1}{3} \int \left (\frac {6 e^x x-4 e^x \left (1-\frac {\log (2)}{2}\right )-3 x (1-\log (4))-\log ^2(4) \left (1+\frac {-\log (4) \log (16)+\log (256)}{\log ^2(4)}\right )}{\left (e^{2 x}+3 x+\log (4)+e^x (1+\log (4))\right )^2 \log (x)}+\frac {-e^x-\log (4)-e^x x \log (x)+x (1-\log (4)) \log (x)}{x \left (e^{2 x}+3 x+\log (4)+e^x (1+\log (4))\right ) \log ^2(x)}\right ) \, dx\\ &=\frac {1}{3} \int \frac {6 e^x x-4 e^x \left (1-\frac {\log (2)}{2}\right )-3 x (1-\log (4))-\log ^2(4) \left (1+\frac {-\log (4) \log (16)+\log (256)}{\log ^2(4)}\right )}{\left (e^{2 x}+3 x+\log (4)+e^x (1+\log (4))\right )^2 \log (x)} \, dx+\frac {1}{3} \int \frac {-e^x-\log (4)-e^x x \log (x)+x (1-\log (4)) \log (x)}{x \left (e^{2 x}+3 x+\log (4)+e^x (1+\log (4))\right ) \log ^2(x)} \, dx\\ &=\frac {1}{3} \int \left (\frac {e^x}{x \left (-e^{2 x}-3 x-\log (4)-e^x (1+\log (4))\right ) \log ^2(x)}+\frac {\log (4)}{x \left (-e^{2 x}-3 x-\log (4)-e^x (1+\log (4))\right ) \log ^2(x)}+\frac {e^x}{\left (-e^{2 x}-3 x-\log (4)-e^x (1+\log (4))\right ) \log (x)}+\frac {1-\log (4)}{\left (e^{2 x}+3 x+\log (4)+e^x (1+\log (4))\right ) \log (x)}\right ) \, dx+\frac {1}{3} \int \left (\frac {6 e^x x}{\left (e^{2 x}+3 x+\log (4)+e^x (1+\log (4))\right )^2 \log (x)}+\frac {2 e^x (-2+\log (2))}{\left (e^{2 x}+3 x+\log (4)+e^x (1+\log (4))\right )^2 \log (x)}+\frac {3 x (-1+\log (4))}{\left (e^{2 x}+3 x+\log (4)+e^x (1+\log (4))\right )^2 \log (x)}+\frac {-\log ^2(4)+\log (4) \log (16)-\log (256)}{\left (e^{2 x}+3 x+\log (4)+e^x (1+\log (4))\right )^2 \log (x)}\right ) \, dx\\ &=\frac {1}{3} \int \frac {e^x}{x \left (-e^{2 x}-3 x-\log (4)-e^x (1+\log (4))\right ) \log ^2(x)} \, dx+\frac {1}{3} \int \frac {e^x}{\left (-e^{2 x}-3 x-\log (4)-e^x (1+\log (4))\right ) \log (x)} \, dx+2 \int \frac {e^x x}{\left (e^{2 x}+3 x+\log (4)+e^x (1+\log (4))\right )^2 \log (x)} \, dx-\frac {1}{3} (2 (2-\log (2))) \int \frac {e^x}{\left (e^{2 x}+3 x+\log (4)+e^x (1+\log (4))\right )^2 \log (x)} \, dx+\frac {1}{3} (1-\log (4)) \int \frac {1}{\left (e^{2 x}+3 x+\log (4)+e^x (1+\log (4))\right ) \log (x)} \, dx+(-1+\log (4)) \int \frac {x}{\left (e^{2 x}+3 x+\log (4)+e^x (1+\log (4))\right )^2 \log (x)} \, dx+\frac {1}{3} \log (4) \int \frac {1}{x \left (-e^{2 x}-3 x-\log (4)-e^x (1+\log (4))\right ) \log ^2(x)} \, dx+\frac {1}{3} \left (-\log ^2(4)+\log (4) \log (16)-\log (256)\right ) \int \frac {1}{\left (e^{2 x}+3 x+\log (4)+e^x (1+\log (4))\right )^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}
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Mathematica [A]
time = 0.22, size = 36, normalized size = 1.38 \begin {gather*} \frac {e^x+\log (4)}{3 \left (e^x+e^{2 x}+3 x+\log (4)+e^x \log (4)\right ) \log (x)} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.10, size = 36, normalized size = 1.38
method | result | size |
risch | \(\frac {{\mathrm e}^{x}+2 \ln \left (2\right )}{3 \left (2 \,{\mathrm e}^{x} \ln \left (2\right )+{\mathrm e}^{2 x}+2 \ln \left (2\right )+{\mathrm e}^{x}+3 x \right ) \ln \left (x \right )}\) | \(36\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.57, size = 41, normalized size = 1.58 \begin {gather*} \frac {e^{x} + 2 \, \log \left (2\right )}{3 \, {\left ({\left (2 \, \log \left (2\right ) + 1\right )} e^{x} \log \left (x\right ) + {\left (3 \, x + 2 \, \log \left (2\right )\right )} \log \left (x\right ) + e^{\left (2 \, x\right )} \log \left (x\right )\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.34, size = 36, normalized size = 1.38 \begin {gather*} \frac {e^{x} + 2 \, \log \left (2\right )}{3 \, {\left ({\left (2 \, \log \left (2\right ) + 1\right )} e^{x} + 3 \, x + e^{\left (2 \, x\right )} + 2 \, \log \left (2\right )\right )} \log \left (x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 49 vs.
\(2 (22) = 44\).
time = 0.17, size = 49, normalized size = 1.88 \begin {gather*} \frac {e^{x} + 2 \log {\left (2 \right )}}{9 x \log {\left (x \right )} + \left (3 \log {\left (x \right )} + 6 \log {\left (2 \right )} \log {\left (x \right )}\right ) e^{x} + 3 e^{2 x} \log {\left (x \right )} + 6 \log {\left (2 \right )} \log {\left (x \right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-1)] Timed out
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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {\frac {{\mathrm {e}}^{3\,x}}{3}+2\,x\,\ln \left (2\right )+\frac {{\mathrm {e}}^x\,\left (3\,x+4\,\ln \left (2\right )+4\,{\ln \left (2\right )}^2\right )}{3}+\frac {{\mathrm {e}}^{2\,x}\,\left (4\,\ln \left (2\right )+1\right )}{3}+\frac {\ln \left (x\right )\,\left (x\,{\mathrm {e}}^{3\,x}+6\,x\,\ln \left (2\right )+{\mathrm {e}}^x\,\left (3\,x+4\,x\,{\ln \left (2\right )}^2-3\,x^2\right )+4\,x\,{\mathrm {e}}^{2\,x}\,\ln \left (2\right )\right )}{3}+\frac {4\,{\ln \left (2\right )}^2}{3}}{{\ln \left (x\right )}^2\,\left (x\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{3\,x}\,\left (2\,x+4\,x\,\ln \left (2\right )\right )+{\mathrm {e}}^{2\,x}\,\left (x+8\,x\,\ln \left (2\right )+4\,x\,{\ln \left (2\right )}^2+6\,x^2\right )+4\,x\,{\ln \left (2\right )}^2+12\,x^2\,\ln \left (2\right )+{\mathrm {e}}^x\,\left (2\,\ln \left (2\right )\,\left (6\,x^2+2\,x\right )+8\,x\,{\ln \left (2\right )}^2+6\,x^2\right )+9\,x^3\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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