Integrand size = 33, antiderivative size = 18 \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=e^{-6+\frac {10}{x}} x (5+10 x \log (3)) \]
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Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.39 (sec) , antiderivative size = 90, normalized size of antiderivative = 5.00, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {6873, 6874, 2241, 2237, 2245} \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=\frac {50 \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right )}{e^6}-\frac {1000 \log (3) \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right )}{e^6}-\frac {50 (1-20 \log (3)) \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right )}{e^6}+10 e^{\frac {10}{x}-6} x^2 \log (3)+100 e^{\frac {10}{x}-6} x \log (3)+5 e^{\frac {10}{x}-6} x (1-20 \log (3)) \]
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Rule 2237
Rule 2241
Rule 2245
Rule 6873
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{-6+\frac {10}{x}} \left (-50+5 x (1-20 \log (3))+20 x^2 \log (3)\right )}{x} \, dx \\ & = \int \left (-\frac {50 e^{-6+\frac {10}{x}}}{x}+5 e^{-6+\frac {10}{x}} (1-20 \log (3))+20 e^{-6+\frac {10}{x}} x \log (3)\right ) \, dx \\ & = -\left (50 \int \frac {e^{-6+\frac {10}{x}}}{x} \, dx\right )+(5 (1-20 \log (3))) \int e^{-6+\frac {10}{x}} \, dx+(20 \log (3)) \int e^{-6+\frac {10}{x}} x \, dx \\ & = \frac {50 \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right )}{e^6}+5 e^{-6+\frac {10}{x}} x (1-20 \log (3))+10 e^{-6+\frac {10}{x}} x^2 \log (3)+(50 (1-20 \log (3))) \int \frac {e^{-6+\frac {10}{x}}}{x} \, dx+(100 \log (3)) \int e^{-6+\frac {10}{x}} \, dx \\ & = \frac {50 \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right )}{e^6}+5 e^{-6+\frac {10}{x}} x (1-20 \log (3))-\frac {50 \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right ) (1-20 \log (3))}{e^6}+100 e^{-6+\frac {10}{x}} x \log (3)+10 e^{-6+\frac {10}{x}} x^2 \log (3)+(1000 \log (3)) \int \frac {e^{-6+\frac {10}{x}}}{x} \, dx \\ & = \frac {50 \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right )}{e^6}+5 e^{-6+\frac {10}{x}} x (1-20 \log (3))-\frac {50 \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right ) (1-20 \log (3))}{e^6}+100 e^{-6+\frac {10}{x}} x \log (3)+10 e^{-6+\frac {10}{x}} x^2 \log (3)-\frac {1000 \operatorname {ExpIntegralEi}\left (\frac {10}{x}\right ) \log (3)}{e^6} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=5 e^{-6+\frac {10}{x}} x (1+x \log (9)) \]
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Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.28
method | result | size |
gosper | \(5 \left (2 x \ln \left (3\right )+1\right ) x \,{\mathrm e}^{-\frac {2 \left (3 x -5\right )}{x}}\) | \(23\) |
risch | \(\left (5 x +10 x^{2} \ln \left (3\right )\right ) {\mathrm e}^{-\frac {2 \left (3 x -5\right )}{x}}\) | \(24\) |
norman | \(\left (5 x +10 x^{2} \ln \left (3\right )\right ) {\mathrm e}^{-\frac {2 \left (3 x -5\right )}{x}}\) | \(25\) |
parallelrisch | \(\frac {\left (10 x^{3} \ln \left (3\right )+5 x^{2}\right ) {\mathrm e}^{-\frac {2 \left (3 x -5\right )}{x}}}{x}\) | \(30\) |
derivativedivides | \(5 \,{\mathrm e}^{-6+\frac {10}{x}} x +500 \ln \left (3\right ) \left (-\frac {{\mathrm e}^{-6+\frac {10}{x}} \left (-1-\frac {10}{x}\right ) x^{2}}{50}+2 \,{\mathrm e}^{-6} \operatorname {Ei}_{1}\left (-\frac {10}{x}\right )\right )-500 \ln \left (3\right ) \left (\frac {{\mathrm e}^{-6+\frac {10}{x}} x}{5}+2 \,{\mathrm e}^{-6} \operatorname {Ei}_{1}\left (-\frac {10}{x}\right )\right )\) | \(76\) |
default | \(5 \,{\mathrm e}^{-6+\frac {10}{x}} x +500 \ln \left (3\right ) \left (-\frac {{\mathrm e}^{-6+\frac {10}{x}} \left (-1-\frac {10}{x}\right ) x^{2}}{50}+2 \,{\mathrm e}^{-6} \operatorname {Ei}_{1}\left (-\frac {10}{x}\right )\right )-500 \ln \left (3\right ) \left (\frac {{\mathrm e}^{-6+\frac {10}{x}} x}{5}+2 \,{\mathrm e}^{-6} \operatorname {Ei}_{1}\left (-\frac {10}{x}\right )\right )\) | \(76\) |
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Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=5 \, {\left (2 \, x^{2} \log \left (3\right ) + x\right )} e^{\left (-\frac {2 \, {\left (3 \, x - 5\right )}}{x}\right )} \]
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Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=\left (10 x^{2} \log {\left (3 \right )} + 5 x\right ) e^{- \frac {2 \cdot \left (3 x - 5\right )}{x}} \]
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Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.22 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.67 \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=1000 \, e^{\left (-6\right )} \Gamma \left (-1, -\frac {10}{x}\right ) \log \left (3\right ) + 2000 \, e^{\left (-6\right )} \Gamma \left (-2, -\frac {10}{x}\right ) \log \left (3\right ) + 50 \, {\rm Ei}\left (\frac {10}{x}\right ) e^{\left (-6\right )} - 50 \, e^{\left (-6\right )} \Gamma \left (-1, -\frac {10}{x}\right ) \]
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Leaf count of result is larger than twice the leaf count of optimal. 77 vs. \(2 (18) = 36\).
Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 4.28 \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=\frac {25 \, {\left (10 \, e^{\left (-\frac {2 \, {\left (3 \, x - 5\right )}}{x}\right )} \log \left (3\right ) - \frac {{\left (3 \, x - 5\right )} e^{\left (-\frac {2 \, {\left (3 \, x - 5\right )}}{x}\right )}}{x} + 3 \, e^{\left (-\frac {2 \, {\left (3 \, x - 5\right )}}{x}\right )}\right )}}{\frac {{\left (3 \, x - 5\right )}^{2}}{x^{2}} - \frac {6 \, {\left (3 \, x - 5\right )}}{x} + 9} \]
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Time = 7.61 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-\frac {2 (-5+3 x)}{x}} \left (-50+5 x+\left (-100 x+20 x^2\right ) \log (3)\right )}{x} \, dx=5\,x\,{\mathrm {e}}^{\frac {10}{x}-6}\,\left (2\,x\,\ln \left (3\right )+1\right ) \]
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