Integrand size = 53, antiderivative size = 20 \[ \int \frac {e^{1+x+x^2} \left (-1-x-x^2-2 x^3\right )+e^{1+x+x^2} \left (-x-2 x^2\right ) \log (x)}{x \log (5)} \, dx=6-\frac {e^{1+x+x^2} (x+\log (x))}{\log (5)} \]
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Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60, number of steps used = 20, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {12, 14, 2266, 2235, 2272, 2273, 2634} \[ \int \frac {e^{1+x+x^2} \left (-1-x-x^2-2 x^3\right )+e^{1+x+x^2} \left (-x-2 x^2\right ) \log (x)}{x \log (5)} \, dx=-\frac {e^{x^2+x+1} x}{\log (5)}-\frac {e^{x^2+x+1} \log (x)}{\log (5)} \]
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Rule 12
Rule 14
Rule 2235
Rule 2266
Rule 2272
Rule 2273
Rule 2634
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^{1+x+x^2} \left (-1-x-x^2-2 x^3\right )+e^{1+x+x^2} \left (-x-2 x^2\right ) \log (x)}{x} \, dx}{\log (5)} \\ & = \frac {\int \left (-e^{1+x+x^2}-\frac {e^{1+x+x^2}}{x}-e^{1+x+x^2} x-2 e^{1+x+x^2} x^2-e^{1+x+x^2} \log (x)-2 e^{1+x+x^2} x \log (x)\right ) \, dx}{\log (5)} \\ & = -\frac {\int e^{1+x+x^2} \, dx}{\log (5)}-\frac {\int \frac {e^{1+x+x^2}}{x} \, dx}{\log (5)}-\frac {\int e^{1+x+x^2} x \, dx}{\log (5)}-\frac {\int e^{1+x+x^2} \log (x) \, dx}{\log (5)}-\frac {2 \int e^{1+x+x^2} x^2 \, dx}{\log (5)}-\frac {2 \int e^{1+x+x^2} x \log (x) \, dx}{\log (5)} \\ & = -\frac {e^{1+x+x^2}}{2 \log (5)}-\frac {e^{1+x+x^2} x}{\log (5)}-\frac {e^{1+x+x^2} \log (x)}{\log (5)}+\frac {\int e^{1+x+x^2} \, dx}{2 \log (5)}+\frac {\int e^{1+x+x^2} \, dx}{\log (5)}-\frac {\int \frac {e^{1+x+x^2}}{x} \, dx}{\log (5)}+\frac {\int e^{1+x+x^2} x \, dx}{\log (5)}+\frac {\int \frac {e^{3/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2}+x\right )}{2 x} \, dx}{\log (5)}+\frac {2 \int \frac {2 e^{1+x+x^2}-e^{3/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2}+x\right )}{4 x} \, dx}{\log (5)}-\frac {e^{3/4} \int e^{\frac {1}{4} (1+2 x)^2} \, dx}{\log (5)} \\ & = -\frac {e^{1+x+x^2} x}{\log (5)}-\frac {e^{3/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (1+2 x)\right )}{2 \log (5)}-\frac {e^{1+x+x^2} \log (x)}{\log (5)}-\frac {\int e^{1+x+x^2} \, dx}{2 \log (5)}+\frac {\int \frac {2 e^{1+x+x^2}-e^{3/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2}+x\right )}{x} \, dx}{2 \log (5)}-\frac {\int \frac {e^{1+x+x^2}}{x} \, dx}{\log (5)}+\frac {e^{3/4} \int e^{\frac {1}{4} (1+2 x)^2} \, dx}{2 \log (5)}+\frac {e^{3/4} \int e^{\frac {1}{4} (1+2 x)^2} \, dx}{\log (5)}+\frac {\left (e^{3/4} \sqrt {\pi }\right ) \int \frac {\text {erfi}\left (\frac {1}{2}+x\right )}{x} \, dx}{2 \log (5)} \\ & = -\frac {e^{1+x+x^2} x}{\log (5)}+\frac {e^{3/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} (1+2 x)\right )}{4 \log (5)}-\frac {e^{1+x+x^2} \log (x)}{\log (5)}+\frac {\int \left (\frac {2 e^{1+x+x^2}}{x}-\frac {e^{3/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2}+x\right )}{x}\right ) \, dx}{2 \log (5)}-\frac {\int \frac {e^{1+x+x^2}}{x} \, dx}{\log (5)}-\frac {e^{3/4} \int e^{\frac {1}{4} (1+2 x)^2} \, dx}{2 \log (5)}+\frac {\left (e^{3/4} \sqrt {\pi }\right ) \int \frac {\text {erfi}\left (\frac {1}{2}+x\right )}{x} \, dx}{2 \log (5)} \\ & = -\frac {e^{1+x+x^2} x}{\log (5)}-\frac {e^{1+x+x^2} \log (x)}{\log (5)} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{1+x+x^2} \left (-1-x-x^2-2 x^3\right )+e^{1+x+x^2} \left (-x-2 x^2\right ) \log (x)}{x \log (5)} \, dx=-\frac {e^{1+x+x^2} (x+\log (x))}{\log (5)} \]
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Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.40
method | result | size |
parallelrisch | \(\frac {-{\mathrm e}^{x^{2}+x +1} x -{\mathrm e}^{x^{2}+x +1} \ln \left (x \right )}{\ln \left (5\right )}\) | \(28\) |
norman | \(-\frac {x \,{\mathrm e}^{x^{2}+x +1}}{\ln \left (5\right )}-\frac {{\mathrm e}^{x^{2}+x +1} \ln \left (x \right )}{\ln \left (5\right )}\) | \(31\) |
risch | \(-\frac {x \,{\mathrm e}^{x^{2}+x +1}}{\ln \left (5\right )}-\frac {{\mathrm e}^{x^{2}+x +1} \ln \left (x \right )}{\ln \left (5\right )}\) | \(31\) |
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Time = 0.25 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {e^{1+x+x^2} \left (-1-x-x^2-2 x^3\right )+e^{1+x+x^2} \left (-x-2 x^2\right ) \log (x)}{x \log (5)} \, dx=-\frac {x e^{\left (x^{2} + x + 1\right )} + e^{\left (x^{2} + x + 1\right )} \log \left (x\right )}{\log \left (5\right )} \]
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Time = 0.14 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {e^{1+x+x^2} \left (-1-x-x^2-2 x^3\right )+e^{1+x+x^2} \left (-x-2 x^2\right ) \log (x)}{x \log (5)} \, dx=\frac {\left (- x - \log {\left (x \right )}\right ) e^{x^{2} + x + 1}}{\log {\left (5 \right )}} \]
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\[ \int \frac {e^{1+x+x^2} \left (-1-x-x^2-2 x^3\right )+e^{1+x+x^2} \left (-x-2 x^2\right ) \log (x)}{x \log (5)} \, dx=\int { -\frac {{\left (2 \, x^{2} + x\right )} e^{\left (x^{2} + x + 1\right )} \log \left (x\right ) + {\left (2 \, x^{3} + x^{2} + x + 1\right )} e^{\left (x^{2} + x + 1\right )}}{x \log \left (5\right )} \,d x } \]
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Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {e^{1+x+x^2} \left (-1-x-x^2-2 x^3\right )+e^{1+x+x^2} \left (-x-2 x^2\right ) \log (x)}{x \log (5)} \, dx=-\frac {x e^{\left (x^{2} + x + 1\right )} + e^{\left (x^{2} + x + 1\right )} \log \left (x\right )}{\log \left (5\right )} \]
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Time = 9.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{1+x+x^2} \left (-1-x-x^2-2 x^3\right )+e^{1+x+x^2} \left (-x-2 x^2\right ) \log (x)}{x \log (5)} \, dx=-\frac {{\mathrm {e}}^{x^2}\,\mathrm {e}\,{\mathrm {e}}^x\,\left (x+\ln \left (x\right )\right )}{\ln \left (5\right )} \]
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