Optimal. Leaf size=27 \[ \frac {\left (3-e^x+6 x\right ) \log (\log (4))}{2 x (-4+2 x)} \]
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Rubi [B] time = 0.49, antiderivative size = 55, normalized size of antiderivative = 2.04, number of steps used = 16, number of rules used = 7, integrand size = 46, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {12, 1594, 27, 6742, 2177, 2178, 893} \begin {gather*} \frac {e^x \log (\log (4))}{8 x}-\frac {3 \log (\log (4))}{8 x}+\frac {e^x \log (\log (4))}{8 (2-x)}-\frac {15 \log (\log (4))}{8 (2-x)} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 27
Rule 893
Rule 1594
Rule 2177
Rule 2178
Rule 6742
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\log (\log (4)) \int \frac {6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )}{16 x^2-16 x^3+4 x^4} \, dx\\ &=\log (\log (4)) \int \frac {6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )}{x^2 \left (16-16 x+4 x^2\right )} \, dx\\ &=\log (\log (4)) \int \frac {6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )}{4 (-2+x)^2 x^2} \, dx\\ &=\frac {1}{4} \log (\log (4)) \int \frac {6-6 x-6 x^2+e^x \left (-2+4 x-x^2\right )}{(-2+x)^2 x^2} \, dx\\ &=\frac {1}{4} \log (\log (4)) \int \left (-\frac {e^x \left (2-4 x+x^2\right )}{(-2+x)^2 x^2}-\frac {6 \left (-1+x+x^2\right )}{(-2+x)^2 x^2}\right ) \, dx\\ &=-\left (\frac {1}{4} \log (\log (4)) \int \frac {e^x \left (2-4 x+x^2\right )}{(-2+x)^2 x^2} \, dx\right )-\frac {1}{2} (3 \log (\log (4))) \int \frac {-1+x+x^2}{(-2+x)^2 x^2} \, dx\\ &=-\left (\frac {1}{4} \log (\log (4)) \int \left (-\frac {e^x}{2 (-2+x)^2}+\frac {e^x}{2 (-2+x)}+\frac {e^x}{2 x^2}-\frac {e^x}{2 x}\right ) \, dx\right )-\frac {1}{2} (3 \log (\log (4))) \int \left (\frac {5}{4 (-2+x)^2}-\frac {1}{4 x^2}\right ) \, dx\\ &=-\frac {15 \log (\log (4))}{8 (2-x)}-\frac {3 \log (\log (4))}{8 x}+\frac {1}{8} \log (\log (4)) \int \frac {e^x}{(-2+x)^2} \, dx-\frac {1}{8} \log (\log (4)) \int \frac {e^x}{-2+x} \, dx-\frac {1}{8} \log (\log (4)) \int \frac {e^x}{x^2} \, dx+\frac {1}{8} \log (\log (4)) \int \frac {e^x}{x} \, dx\\ &=-\frac {15 \log (\log (4))}{8 (2-x)}+\frac {e^x \log (\log (4))}{8 (2-x)}-\frac {3 \log (\log (4))}{8 x}+\frac {e^x \log (\log (4))}{8 x}-\frac {1}{8} e^2 \text {Ei}(-2+x) \log (\log (4))+\frac {1}{8} \text {Ei}(x) \log (\log (4))+\frac {1}{8} \log (\log (4)) \int \frac {e^x}{-2+x} \, dx-\frac {1}{8} \log (\log (4)) \int \frac {e^x}{x} \, dx\\ &=-\frac {15 \log (\log (4))}{8 (2-x)}+\frac {e^x \log (\log (4))}{8 (2-x)}-\frac {3 \log (\log (4))}{8 x}+\frac {e^x \log (\log (4))}{8 x}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.10, size = 23, normalized size = 0.85 \begin {gather*} -\frac {\left (-3+e^x-6 x\right ) \log (\log (4))}{4 (-2+x) x} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.59, size = 25, normalized size = 0.93 \begin {gather*} \frac {{\left (6 \, x - e^{x} + 3\right )} \log \left (2 \, \log \relax (2)\right )}{4 \, {\left (x^{2} - 2 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.26, size = 25, normalized size = 0.93 \begin {gather*} \frac {{\left (6 \, x - e^{x} + 3\right )} \log \left (2 \, \log \relax (2)\right )}{4 \, {\left (x^{2} - 2 \, x\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.07, size = 37, normalized size = 1.37
method | result | size |
default | \(\frac {\ln \left (2 \ln \relax (2)\right ) \left (\frac {15}{2 \left (x -2\right )}-\frac {3}{2 x}+\frac {{\mathrm e}^{x}}{2 x}-\frac {{\mathrm e}^{x}}{2 \left (x -2\right )}\right )}{4}\) | \(37\) |
risch | \(\frac {\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) \left (\frac {3 x}{2}+\frac {3}{4}\right )}{\left (x -2\right ) x}-\frac {\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) {\mathrm e}^{x}}{4 \left (x -2\right ) x}\) | \(40\) |
norman | \(\frac {\left (-\frac {\ln \relax (2)}{4}-\frac {\ln \left (\ln \relax (2)\right )}{4}\right ) {\mathrm e}^{x}+\left (\frac {3 \ln \relax (2)}{2}+\frac {3 \ln \left (\ln \relax (2)\right )}{2}\right ) x +\frac {3 \ln \left (\ln \relax (2)\right )}{4}+\frac {3 \ln \relax (2)}{4}}{\left (x -2\right ) x}\) | \(45\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.48, size = 41, normalized size = 1.52 \begin {gather*} -\frac {1}{4} \, {\left (\frac {3 \, {\left (x - 1\right )}}{x^{2} - 2 \, x} + \frac {e^{x}}{x^{2} - 2 \, x} - \frac {9}{x - 2}\right )} \log \left (2 \, \log \relax (2)\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.56, size = 34, normalized size = 1.26 \begin {gather*} \frac {3\,\ln \left (\ln \relax (4)\right )+x^2\,\ln \left ({\ln \relax (4)}^3\right )-\ln \left (2\,\ln \relax (2)\right )\,{\mathrm {e}}^x}{4\,x\,\left (x-2\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [B] time = 0.46, size = 56, normalized size = 2.07 \begin {gather*} - \frac {x \left (- 6 \log {\relax (2 )} - 6 \log {\left (\log {\relax (2 )} \right )}\right ) - 3 \log {\relax (2 )} - 3 \log {\left (\log {\relax (2 )} \right )}}{4 x^{2} - 8 x} + \frac {\left (- \log {\relax (2 )} - \log {\left (\log {\relax (2 )} \right )}\right ) e^{x}}{4 x^{2} - 8 x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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