Optimal. Leaf size=20 \[ \frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2} \]
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Rubi [A] time = 0.16, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 14, 2177, 2178} \begin {gather*} \frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 14
Rule 2177
Rule 2178
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{x^3} \, dx\\ &=\frac {1}{4} \int \left (-\frac {128 e^{x/4} \left (1+\frac {1}{16} \log \left (\frac {\log (2)}{4}\right )\right )}{x^3}+\frac {16 e^{x/4} \left (1+\frac {1}{16} \log \left (\frac {\log (2)}{4}\right )\right )}{x^2}\right ) \, dx\\ &=\frac {1}{4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right ) \int \frac {e^{x/4}}{x^2} \, dx-\left (2 \left (16+\log \left (\frac {\log (2)}{4}\right )\right )\right ) \int \frac {e^{x/4}}{x^3} \, dx\\ &=\frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2}-\frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{4 x}+\frac {1}{16} \left (16+\log \left (\frac {\log (2)}{4}\right )\right ) \int \frac {e^{x/4}}{x} \, dx-\frac {1}{4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right ) \int \frac {e^{x/4}}{x^2} \, dx\\ &=\frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2}+\frac {1}{16} \text {Ei}\left (\frac {x}{4}\right ) \left (16+\log \left (\frac {\log (2)}{4}\right )\right )-\frac {1}{16} \left (16+\log \left (\frac {\log (2)}{4}\right )\right ) \int \frac {e^{x/4}}{x} \, dx\\ &=\frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.01, size = 20, normalized size = 1.00 \begin {gather*} \frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 0.56, size = 21, normalized size = 1.05 \begin {gather*} \frac {e^{\left (\frac {1}{4} \, x\right )} \log \left (\frac {1}{4} \, \log \relax (2)\right ) + 16 \, e^{\left (\frac {1}{4} \, x\right )}}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.15, size = 29, normalized size = 1.45 \begin {gather*} -\frac {2 \, e^{\left (\frac {1}{4} \, x\right )} \log \relax (2) - e^{\left (\frac {1}{4} \, x\right )} \log \left (\log \relax (2)\right ) - 16 \, e^{\left (\frac {1}{4} \, x\right )}}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.11, size = 16, normalized size = 0.80
method | result | size |
gosper | \(\frac {\left (16+\ln \left (\frac {\ln \relax (2)}{4}\right )\right ) {\mathrm e}^{\frac {x}{4}}}{x^{2}}\) | \(16\) |
norman | \(\frac {\left (-2 \ln \relax (2)+\ln \left (\ln \relax (2)\right )+16\right ) {\mathrm e}^{\frac {x}{4}}}{x^{2}}\) | \(18\) |
risch | \(-\frac {\left (2 \ln \relax (2)-\ln \left (\ln \relax (2)\right )-16\right ) {\mathrm e}^{\frac {x}{4}}}{x^{2}}\) | \(21\) |
derivativedivides | \(\frac {16 \,{\mathrm e}^{\frac {x}{4}}}{x^{2}}-\frac {2 \,{\mathrm e}^{\frac {x}{4}} \ln \relax (2)}{x^{2}}+\frac {{\mathrm e}^{\frac {x}{4}} \ln \left (\ln \relax (2)\right )}{x^{2}}\) | \(33\) |
default | \(\frac {16 \,{\mathrm e}^{\frac {x}{4}}}{x^{2}}-\frac {2 \,{\mathrm e}^{\frac {x}{4}} \ln \relax (2)}{x^{2}}+\frac {{\mathrm e}^{\frac {x}{4}} \ln \left (\ln \relax (2)\right )}{x^{2}}\) | \(33\) |
meijerg | \(-4 \left (\frac {1}{4}+\frac {\ln \left (\frac {\ln \relax (2)}{4}\right )}{64}\right ) \left (-\frac {2 \left (2+\frac {x}{2}\right )}{x}+\frac {4 \,{\mathrm e}^{\frac {x}{4}}}{x}+\ln \left (-\frac {x}{4}\right )+\expIntegralEi \left (1, -\frac {x}{4}\right )+1-\ln \relax (x )+2 \ln \relax (2)-i \pi +\frac {4}{x}\right )-4 \left (\frac {\ln \left (\frac {\ln \relax (2)}{4}\right )}{32}+\frac {1}{2}\right ) \left (\frac {\frac {3}{4} x^{2}+4 x +8}{x^{2}}-\frac {8 \left (3+\frac {3 x}{4}\right ) {\mathrm e}^{\frac {x}{4}}}{3 x^{2}}-\frac {\ln \left (-\frac {x}{4}\right )}{2}-\frac {\expIntegralEi \left (1, -\frac {x}{4}\right )}{2}-\frac {3}{4}+\frac {\ln \relax (x )}{2}-\ln \relax (2)+\frac {i \pi }{2}-\frac {8}{x^{2}}-\frac {4}{x}\right )\) | \(137\) |
Verification of antiderivative is not currently implemented for this CAS.
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maxima [C] time = 0.37, size = 37, normalized size = 1.85 \begin {gather*} \frac {1}{16} \, \Gamma \left (-1, -\frac {1}{4} \, x\right ) \log \left (\frac {1}{4} \, \log \relax (2)\right ) + \frac {1}{8} \, \Gamma \left (-2, -\frac {1}{4} \, x\right ) \log \left (\frac {1}{4} \, \log \relax (2)\right ) + \Gamma \left (-1, -\frac {1}{4} \, x\right ) + 2 \, \Gamma \left (-2, -\frac {1}{4} \, x\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 0.07, size = 15, normalized size = 0.75 \begin {gather*} \frac {{\mathrm {e}}^{x/4}\,\left (\ln \left (\frac {\ln \relax (2)}{4}\right )+16\right )}{x^2} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 0.14, size = 19, normalized size = 0.95 \begin {gather*} \frac {\left (- 2 \log {\relax (2 )} + \log {\left (\log {\relax (2 )} \right )} + 16\right ) e^{\frac {x}{4}}}{x^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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