Integrand size = 17, antiderivative size = 14 \[ \int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx=3-\frac {5 e^{-4+x}}{2 x^3} \]
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Time = 0.02 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {12, 2228} \[ \int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx=-\frac {5 e^{x-4}}{2 x^3} \]
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Rule 12
Rule 2228
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {e^{-4+x} (15-5 x)}{x^4} \, dx \\ & = -\frac {5 e^{-4+x}}{2 x^3} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx=-\frac {5 e^{-4+x}}{2 x^3} \]
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Time = 1.30 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.71
method | result | size |
risch | \(-\frac {5 \,{\mathrm e}^{x -4}}{2 x^{3}}\) | \(10\) |
gosper | \(-\frac {5 \,{\mathrm e}^{x -4}}{2 x^{3}}\) | \(14\) |
norman | \(-\frac {5 \,{\mathrm e}^{x -4}}{2 x^{3}}\) | \(14\) |
parallelrisch | \(-\frac {5 \,{\mathrm e}^{x -4}}{2 x^{3}}\) | \(14\) |
derivativedivides | \(-\frac {5 \,{\mathrm e}^{x -4} \left (\left (-x +4\right )^{2}+9 x -8\right )}{12 x^{3}}+\frac {5 \,{\mathrm e}^{x -4} \left (\left (-x +4\right )^{2}+9 x -14\right )}{12 x^{3}}\) | \(44\) |
default | \(-\frac {5 \,{\mathrm e}^{x -4} \left (\left (-x +4\right )^{2}+9 x -8\right )}{12 x^{3}}+\frac {5 \,{\mathrm e}^{x -4} \left (\left (-x +4\right )^{2}+9 x -14\right )}{12 x^{3}}\) | \(44\) |
meijerg | \(-\frac {15 \,{\mathrm e}^{x -16-{\mathrm e}^{-4} x} \left (\frac {{\mathrm e}^{12}}{3 x^{3}}+\frac {{\mathrm e}^{8}}{2 x^{2}}+\frac {{\mathrm e}^{4}}{2 x}+\frac {35}{36}-\frac {\ln \left (x \right )}{6}-\frac {i \pi }{6}-\frac {{\mathrm e}^{12} \left (22 x^{3} {\mathrm e}^{-12}+36 x^{2} {\mathrm e}^{-8}+36 \,{\mathrm e}^{-4} x +24\right )}{72 x^{3}}+\frac {{\mathrm e}^{12+{\mathrm e}^{-4} x} \left (4 x^{2} {\mathrm e}^{-8}+4 \,{\mathrm e}^{-4} x +8\right )}{24 x^{3}}+\frac {\ln \left (-{\mathrm e}^{-4} x \right )}{6}+\frac {\operatorname {Ei}_{1}\left (-{\mathrm e}^{-4} x \right )}{6}\right )}{2}-\frac {5 \,{\mathrm e}^{x -12-{\mathrm e}^{-4} x} \left (-\frac {{\mathrm e}^{8}}{2 x^{2}}-\frac {{\mathrm e}^{4}}{x}-\frac {11}{4}+\frac {\ln \left (x \right )}{2}+\frac {i \pi }{2}+\frac {{\mathrm e}^{8} \left (9 x^{2} {\mathrm e}^{-8}+12 \,{\mathrm e}^{-4} x +6\right )}{12 x^{2}}-\frac {{\mathrm e}^{8+{\mathrm e}^{-4} x} \left (3+3 \,{\mathrm e}^{-4} x \right )}{6 x^{2}}-\frac {\ln \left (-{\mathrm e}^{-4} x \right )}{2}-\frac {\operatorname {Ei}_{1}\left (-{\mathrm e}^{-4} x \right )}{2}\right )}{2}\) | \(207\) |
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Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64 \[ \int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx=-\frac {5 \, e^{\left (x - 4\right )}}{2 \, x^{3}} \]
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Time = 0.05 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.86 \[ \int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx=- \frac {5 e^{x - 4}}{2 x^{3}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.36 \[ \int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx=\frac {5}{2} \, e^{\left (-4\right )} \Gamma \left (-2, -x\right ) + \frac {15}{2} \, e^{\left (-4\right )} \Gamma \left (-3, -x\right ) \]
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Time = 0.25 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64 \[ \int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx=-\frac {5 \, e^{\left (x - 4\right )}}{2 \, x^{3}} \]
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Time = 0.06 (sec) , antiderivative size = 9, normalized size of antiderivative = 0.64 \[ \int \frac {e^{-4+x} (15-5 x)}{2 x^4} \, dx=-\frac {5\,{\mathrm {e}}^{-4}\,{\mathrm {e}}^x}{2\,x^3} \]
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