Integrand size = 39, antiderivative size = 20 \[ \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx=\frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2} \]
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Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {12, 14, 2208, 2209} \[ \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx=\frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2} \]
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Rule 12
Rule 14
Rule 2208
Rule 2209
Rubi steps \begin{align*} \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} \operatorname {ExpIntegralEi}\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{align*}
Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx=\frac {e^{x/4} \left (16+\log \left (\frac {\log (2)}{4}\right )\right )}{x^2} \]
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Time = 1.02 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.80
method | result | size |
gosper | \(\frac {\left (16+\ln \left (\frac {\ln \left (2\right )}{4}\right )\right ) {\mathrm e}^{\frac {x}{4}}}{x^{2}}\) | \(16\) |
norman | \(\frac {\left (-2 \ln \left (2\right )+\ln \left (\ln \left (2\right )\right )+16\right ) {\mathrm e}^{\frac {x}{4}}}{x^{2}}\) | \(18\) |
risch | \(-\frac {\left (2 \ln \left (2\right )-\ln \left (\ln \left (2\right )\right )-16\right ) {\mathrm e}^{\frac {x}{4}}}{x^{2}}\) | \(21\) |
parallelrisch | \(\frac {4 \,{\mathrm e}^{\frac {x}{4}} \ln \left (\frac {\ln \left (2\right )}{4}\right )+64 \,{\mathrm e}^{\frac {x}{4}}}{4 x^{2}}\) | \(24\) |
derivativedivides | \(\frac {16 \,{\mathrm e}^{\frac {x}{4}}}{x^{2}}+\frac {\ln \left (2\right ) \left (-\frac {8 \,{\mathrm e}^{\frac {x}{4}}}{x^{2}}-\frac {2 \,{\mathrm e}^{\frac {x}{4}}}{x}-\frac {\operatorname {Ei}_{1}\left (-\frac {x}{4}\right )}{2}\right )}{4}-\frac {\ln \left (\ln \left (2\right )\right ) \left (-\frac {8 \,{\mathrm e}^{\frac {x}{4}}}{x^{2}}-\frac {2 \,{\mathrm e}^{\frac {x}{4}}}{x}-\frac {\operatorname {Ei}_{1}\left (-\frac {x}{4}\right )}{2}\right )}{8}-\frac {\ln \left (2\right ) \left (-\frac {4 \,{\mathrm e}^{\frac {x}{4}}}{x}-\operatorname {Ei}_{1}\left (-\frac {x}{4}\right )\right )}{8}+\frac {\ln \left (\ln \left (2\right )\right ) \left (-\frac {4 \,{\mathrm e}^{\frac {x}{4}}}{x}-\operatorname {Ei}_{1}\left (-\frac {x}{4}\right )\right )}{16}\) | \(115\) |
default | \(\frac {16 \,{\mathrm e}^{\frac {x}{4}}}{x^{2}}+\frac {\ln \left (2\right ) \left (-\frac {8 \,{\mathrm e}^{\frac {x}{4}}}{x^{2}}-\frac {2 \,{\mathrm e}^{\frac {x}{4}}}{x}-\frac {\operatorname {Ei}_{1}\left (-\frac {x}{4}\right )}{2}\right )}{4}-\frac {\ln \left (\ln \left (2\right )\right ) \left (-\frac {8 \,{\mathrm e}^{\frac {x}{4}}}{x^{2}}-\frac {2 \,{\mathrm e}^{\frac {x}{4}}}{x}-\frac {\operatorname {Ei}_{1}\left (-\frac {x}{4}\right )}{2}\right )}{8}-\frac {\ln \left (2\right ) \left (-\frac {4 \,{\mathrm e}^{\frac {x}{4}}}{x}-\operatorname {Ei}_{1}\left (-\frac {x}{4}\right )\right )}{8}+\frac {\ln \left (\ln \left (2\right )\right ) \left (-\frac {4 \,{\mathrm e}^{\frac {x}{4}}}{x}-\operatorname {Ei}_{1}\left (-\frac {x}{4}\right )\right )}{16}\) | \(115\) |
meijerg | \(-4 \left (\frac {1}{4}+\frac {\ln \left (\frac {\ln \left (2\right )}{4}\right )}{64}\right ) \left (\frac {4}{x}+1-\ln \left (x \right )+2 \ln \left (2\right )-i \pi -\frac {2 \left (2+\frac {x}{2}\right )}{x}+\frac {4 \,{\mathrm e}^{\frac {x}{4}}}{x}+\ln \left (-\frac {x}{4}\right )+\operatorname {Ei}_{1}\left (-\frac {x}{4}\right )\right )-4 \left (\frac {\ln \left (\frac {\ln \left (2\right )}{4}\right )}{32}+\frac {1}{2}\right ) \left (-\frac {8}{x^{2}}-\frac {4}{x}-\frac {3}{4}+\frac {\ln \left (x \right )}{2}-\ln \left (2\right )+\frac {i \pi }{2}+\frac {\frac {3}{4} x^{2}+4 x +8}{x^{2}}-\frac {8 \left (3+\frac {3 x}{4}\right ) {\mathrm e}^{\frac {x}{4}}}{3 x^{2}}-\frac {\ln \left (-\frac {x}{4}\right )}{2}-\frac {\operatorname {Ei}_{1}\left (-\frac {x}{4}\right )}{2}\right )\) | \(137\) |
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Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx=\frac {e^{\left (\frac {1}{4} \, x\right )} \log \left (\frac {1}{4} \, \log \left (2\right )\right ) + 16 \, e^{\left (\frac {1}{4} \, x\right )}}{x^{2}} \]
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Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95 \[ \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx=\frac {\left (- 2 \log {\left (2 \right )} + \log {\left (\log {\left (2 \right )} \right )} + 16\right ) e^{\frac {x}{4}}}{x^{2}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.85 \[ \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx=\frac {1}{16} \, \Gamma \left (-1, -\frac {1}{4} \, x\right ) \log \left (\frac {1}{4} \, \log \left (2\right )\right ) + \frac {1}{8} \, \Gamma \left (-2, -\frac {1}{4} \, x\right ) \log \left (\frac {1}{4} \, \log \left (2\right )\right ) + \Gamma \left (-1, -\frac {1}{4} \, x\right ) + 2 \, \Gamma \left (-2, -\frac {1}{4} \, x\right ) \]
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Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.45 \[ \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx=-\frac {2 \, e^{\left (\frac {1}{4} \, x\right )} \log \left (2\right ) - e^{\left (\frac {1}{4} \, x\right )} \log \left (\log \left (2\right )\right ) - 16 \, e^{\left (\frac {1}{4} \, x\right )}}{x^{2}} \]
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Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {e^{x/4} (-128+16 x)+e^{x/4} (-8+x) \log \left (\frac {\log (2)}{4}\right )}{4 x^3} \, dx=\frac {{\mathrm {e}}^{x/4}\,\left (\ln \left (\frac {\ln \left (2\right )}{4}\right )+16\right )}{x^2} \]
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