Integrand size = 47, antiderivative size = 31 \[ \int \frac {e^{-3-x} \left (e^{2 x} \left (x^2+x^3\right )+e^{2+x} \left (3 e+2 e^{2 x} x^2 \log (5)\right )\right )}{x^2} \, dx=6-\frac {3-e^{-1+2 x} x \left (e^{-2-x} x+\log (5)\right )}{x} \]
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Time = 0.15 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {6820, 2207, 2225} \[ \int \frac {e^{-3-x} \left (e^{2 x} \left (x^2+x^3\right )+e^{2+x} \left (3 e+2 e^{2 x} x^2 \log (5)\right )\right )}{x^2} \, dx=e^{x-3} (x+1)-e^{x-3}-\frac {3}{x}+\frac {1}{2} e^{2 x-1} \log (25) \]
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Rule 2207
Rule 2225
Rule 6820
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {3}{x^2}+e^{-3+x} (1+x)+e^{-1+2 x} \log (25)\right ) \, dx \\ & = -\frac {3}{x}+\log (25) \int e^{-1+2 x} \, dx+\int e^{-3+x} (1+x) \, dx \\ & = -\frac {3}{x}+e^{-3+x} (1+x)+\frac {1}{2} e^{-1+2 x} \log (25)-\int e^{-3+x} \, dx \\ & = -e^{-3+x}-\frac {3}{x}+e^{-3+x} (1+x)+\frac {1}{2} e^{-1+2 x} \log (25) \\ \end{align*}
Time = 0.07 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-3-x} \left (e^{2 x} \left (x^2+x^3\right )+e^{2+x} \left (3 e+2 e^{2 x} x^2 \log (5)\right )\right )}{x^2} \, dx=-\frac {3}{x}+e^{-3+x} x+\frac {1}{2} e^{-1+2 x} \log (25) \]
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Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71
method | result | size |
risch | \(-\frac {3}{x}+\ln \left (5\right ) {\mathrm e}^{-1+2 x}+x \,{\mathrm e}^{-3+x}\) | \(22\) |
parts | \({\mathrm e}^{x} {\mathrm e}^{-1} {\mathrm e}^{-2} x +{\mathrm e}^{-1} \ln \left (5\right ) {\mathrm e}^{2 x}-\frac {3}{x}\) | \(30\) |
default | \({\mathrm e}^{-1} \left ({\mathrm e}^{-2} {\mathrm e}^{x}+{\mathrm e}^{-2} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )-\frac {3 \,{\mathrm e}}{x}+\ln \left (5\right ) {\mathrm e}^{2 x}\right )\) | \(42\) |
norman | \(\frac {\left ({\mathrm e}^{-1} {\mathrm e}^{-2} x^{2} {\mathrm e}^{2 x}+{\mathrm e}^{-1} \ln \left (5\right ) x \,{\mathrm e}^{3 x}-3 \,{\mathrm e}^{x}\right ) {\mathrm e}^{-x}}{x}\) | \(42\) |
parallelrisch | \(\frac {{\mathrm e}^{-1} \left (\ln \left (5\right ) {\mathrm e}^{2+x} {\mathrm e}^{2 x} x +{\mathrm e}^{2 x} x^{2}-3 \,{\mathrm e} \,{\mathrm e}^{2+x}\right ) {\mathrm e}^{-2-x}}{x}\) | \(44\) |
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Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {e^{-3-x} \left (e^{2 x} \left (x^2+x^3\right )+e^{2+x} \left (3 e+2 e^{2 x} x^2 \log (5)\right )\right )}{x^2} \, dx=\frac {{\left (x^{2} e^{\left (x + 2\right )} + x e^{\left (2 \, x + 4\right )} \log \left (5\right ) - 3 \, e^{5}\right )} e^{\left (-5\right )}}{x} \]
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Time = 0.09 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-3-x} \left (e^{2 x} \left (x^2+x^3\right )+e^{2+x} \left (3 e+2 e^{2 x} x^2 \log (5)\right )\right )}{x^2} \, dx=\frac {e x \sqrt {e^{2 x}} + e^{3} e^{2 x} \log {\left (5 \right )}}{e^{4}} - \frac {3}{x} \]
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Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-3-x} \left (e^{2 x} \left (x^2+x^3\right )+e^{2+x} \left (3 e+2 e^{2 x} x^2 \log (5)\right )\right )}{x^2} \, dx={\left (x - 1\right )} e^{\left (x - 3\right )} + e^{\left (2 \, x - 1\right )} \log \left (5\right ) - \frac {3}{x} + e^{\left (x - 3\right )} \]
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Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (27) = 54\).
Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.13 \[ \int \frac {e^{-3-x} \left (e^{2 x} \left (x^2+x^3\right )+e^{2+x} \left (3 e+2 e^{2 x} x^2 \log (5)\right )\right )}{x^2} \, dx=\frac {{\left (x + 2\right )}^{2} e^{\left (x + 2\right )} + {\left (x + 2\right )} e^{\left (2 \, x + 4\right )} \log \left (5\right ) - 4 \, {\left (x + 2\right )} e^{\left (x + 2\right )} - 2 \, e^{\left (2 \, x + 4\right )} \log \left (5\right ) - 3 \, e^{5} + 4 \, e^{\left (x + 2\right )}}{{\left (x + 2\right )} e^{5} - 2 \, e^{5}} \]
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Time = 0.10 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.68 \[ \int \frac {e^{-3-x} \left (e^{2 x} \left (x^2+x^3\right )+e^{2+x} \left (3 e+2 e^{2 x} x^2 \log (5)\right )\right )}{x^2} \, dx=x\,{\mathrm {e}}^{-3}\,{\mathrm {e}}^x-\frac {3}{x}+{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-1}\,\ln \left (5\right ) \]
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