Integrand size = 41, antiderivative size = 29 \[ \int \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x-5 \log (3)\right )} \left (-e^x x \log (4)+(-1+5 x) \log (4)\right )}{x^2} \, dx=\frac {e^{-e^x+5 \left (1+x+\frac {1}{4} (-1-\log (3))\right )} \log (4)}{x} \]
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Time = 0.40 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {2306, 12, 6873, 2326} \[ \int \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x-5 \log (3)\right )} \left (-e^x x \log (4)+(-1+5 x) \log (4)\right )}{x^2} \, dx=\frac {e^{\frac {1}{4} \left (20 x-4 e^x+15\right )} \left (5 x-e^x x\right ) \log (4)}{3 \sqrt [4]{3} \left (5-e^x\right ) x^2} \]
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Rule 12
Rule 2306
Rule 2326
Rule 6873
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x\right )} \left (-e^x x \log (4)+(-1+5 x) \log (4)\right )}{3 \sqrt [4]{3} x^2} \, dx \\ & = \frac {\int \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x\right )} \left (-e^x x \log (4)+(-1+5 x) \log (4)\right )}{x^2} \, dx}{3 \sqrt [4]{3}} \\ & = \frac {\int \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x\right )} \left (-1+5 x-e^x x\right ) \log (4)}{x^2} \, dx}{3 \sqrt [4]{3}} \\ & = \frac {\log (4) \int \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x\right )} \left (-1+5 x-e^x x\right )}{x^2} \, dx}{3 \sqrt [4]{3}} \\ & = \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x\right )} \left (5 x-e^x x\right ) \log (4)}{3 \sqrt [4]{3} \left (5-e^x\right ) x^2} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x-5 \log (3)\right )} \left (-e^x x \log (4)+(-1+5 x) \log (4)\right )}{x^2} \, dx=\frac {e^{\frac {15}{4}-e^x+5 x} \log (4)}{3 \sqrt [4]{3} x} \]
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Time = 0.13 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72
method | result | size |
risch | \(\frac {2 \ln \left (2\right ) 3^{\frac {3}{4}} {\mathrm e}^{-{\mathrm e}^{x}+\frac {15}{4}+5 x}}{9 x}\) | \(21\) |
norman | \(\frac {2 \ln \left (2\right ) {\mathrm e}^{-{\mathrm e}^{x}-\frac {5 \ln \left (3\right )}{4}+5 x +\frac {15}{4}}}{x}\) | \(22\) |
parallelrisch | \(\frac {2 \ln \left (2\right ) {\mathrm e}^{-{\mathrm e}^{x}-\frac {5 \ln \left (3\right )}{4}+5 x +\frac {15}{4}}}{x}\) | \(22\) |
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Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x-5 \log (3)\right )} \left (-e^x x \log (4)+(-1+5 x) \log (4)\right )}{x^2} \, dx=\frac {2 \, e^{\left (5 \, x - e^{x} - \frac {5}{4} \, \log \left (3\right ) + \frac {15}{4}\right )} \log \left (2\right )}{x} \]
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Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83 \[ \int \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x-5 \log (3)\right )} \left (-e^x x \log (4)+(-1+5 x) \log (4)\right )}{x^2} \, dx=\frac {2 \cdot 3^{\frac {3}{4}} e^{5 x - e^{x} + \frac {15}{4}} \log {\left (2 \right )}}{9 x} \]
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Time = 0.38 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x-5 \log (3)\right )} \left (-e^x x \log (4)+(-1+5 x) \log (4)\right )}{x^2} \, dx=\frac {2 \cdot 3^{\frac {3}{4}} e^{\left (5 \, x - e^{x} + \frac {15}{4}\right )} \log \left (2\right )}{9 \, x} \]
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Exception generated. \[ \int \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x-5 \log (3)\right )} \left (-e^x x \log (4)+(-1+5 x) \log (4)\right )}{x^2} \, dx=\text {Exception raised: TypeError} \]
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Time = 0.13 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {e^{\frac {1}{4} \left (15-4 e^x+20 x-5 \log (3)\right )} \left (-e^x x \log (4)+(-1+5 x) \log (4)\right )}{x^2} \, dx=\frac {2\,3^{3/4}\,{\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{15/4}\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,\ln \left (2\right )}{9\,x} \]
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