Integrand size = 40, antiderivative size = 22 \[ \int \frac {e^{33-3 e^5+3 e^x+96 x^2} \left (1+\left (3 e^x x+192 x^2\right ) \log (x)\right )}{x} \, dx=e^{3 \left (11-e^5+e^x+32 x^2\right )} \log (x) \]
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Leaf count is larger than twice the leaf count of optimal. \(48\) vs. \(2(22)=44\).
Time = 0.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.18, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {2326} \[ \int \frac {e^{33-3 e^5+3 e^x+96 x^2} \left (1+\left (3 e^x x+192 x^2\right ) \log (x)\right )}{x} \, dx=\frac {e^{96 x^2+3 e^x+3 \left (11-e^5\right )} \left (64 x^2+e^x x\right ) \log (x)}{x \left (64 x+e^x\right )} \]
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Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {e^{3 e^x+3 \left (11-e^5\right )+96 x^2} \left (e^x x+64 x^2\right ) \log (x)}{x \left (e^x+64 x\right )} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{33-3 e^5+3 e^x+96 x^2} \left (1+\left (3 e^x x+192 x^2\right ) \log (x)\right )}{x} \, dx=e^{33-3 e^5+3 e^x+96 x^2} \log (x) \]
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Time = 0.15 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91
method | result | size |
risch | \(\ln \left (x \right ) {\mathrm e}^{3 \,{\mathrm e}^{x}-3 \,{\mathrm e}^{5}+96 x^{2}+33}\) | \(20\) |
parallelrisch | \(\ln \left (x \right ) {\mathrm e}^{3 \,{\mathrm e}^{x}-3 \,{\mathrm e}^{5}+96 x^{2}+33}\) | \(20\) |
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Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{33-3 e^5+3 e^x+96 x^2} \left (1+\left (3 e^x x+192 x^2\right ) \log (x)\right )}{x} \, dx=e^{\left (96 \, x^{2} - 3 \, e^{5} + 3 \, e^{x} + 33\right )} \log \left (x\right ) \]
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Time = 2.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{33-3 e^5+3 e^x+96 x^2} \left (1+\left (3 e^x x+192 x^2\right ) \log (x)\right )}{x} \, dx=e^{96 x^{2} + 3 e^{x} - 3 e^{5} + 33} \log {\left (x \right )} \]
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Time = 0.38 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{33-3 e^5+3 e^x+96 x^2} \left (1+\left (3 e^x x+192 x^2\right ) \log (x)\right )}{x} \, dx=e^{\left (96 \, x^{2} - 3 \, e^{5} + 3 \, e^{x} + 33\right )} \log \left (x\right ) \]
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Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {e^{33-3 e^5+3 e^x+96 x^2} \left (1+\left (3 e^x x+192 x^2\right ) \log (x)\right )}{x} \, dx=e^{\left (96 \, x^{2} - 3 \, e^{5} + 3 \, e^{x} + 33\right )} \log \left (x\right ) \]
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Timed out. \[ \int \frac {e^{33-3 e^5+3 e^x+96 x^2} \left (1+\left (3 e^x x+192 x^2\right ) \log (x)\right )}{x} \, dx=\int \frac {{\mathrm {e}}^{3\,{\mathrm {e}}^x-3\,{\mathrm {e}}^5+96\,x^2+33}\,\left (\ln \left (x\right )\,\left (3\,x\,{\mathrm {e}}^x+192\,x^2\right )+1\right )}{x} \,d x \]
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