Integrand size = 47, antiderivative size = 26 \[ \int \frac {e^{x^2} \left (1+4 x-2 x^2\right )+\left (63-36 x+9 x^2\right ) \log (4)}{\left (36-36 x+9 x^2\right ) \log (4)} \, dx=5+x-\frac {1+x+\frac {e^{x^2}}{9 \log (4)}}{-2+x} \]
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Time = 0.12 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.106, Rules used = {12, 27, 6874, 2326, 697} \[ \int \frac {e^{x^2} \left (1+4 x-2 x^2\right )+\left (63-36 x+9 x^2\right ) \log (4)}{\left (36-36 x+9 x^2\right ) \log (4)} \, dx=\frac {e^{x^2} \left (2 x-x^2\right )}{9 (2-x)^2 x \log (4)}+x+\frac {3}{2-x} \]
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Rule 12
Rule 27
Rule 697
Rule 2326
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^{x^2} \left (1+4 x-2 x^2\right )+\left (63-36 x+9 x^2\right ) \log (4)}{36-36 x+9 x^2} \, dx}{\log (4)} \\ & = \frac {\int \frac {e^{x^2} \left (1+4 x-2 x^2\right )+\left (63-36 x+9 x^2\right ) \log (4)}{9 (-2+x)^2} \, dx}{\log (4)} \\ & = \frac {\int \frac {e^{x^2} \left (1+4 x-2 x^2\right )+\left (63-36 x+9 x^2\right ) \log (4)}{(-2+x)^2} \, dx}{9 \log (4)} \\ & = \frac {\int \left (-\frac {e^{x^2} \left (-1-4 x+2 x^2\right )}{(-2+x)^2}+\frac {9 \left (7-4 x+x^2\right ) \log (4)}{(-2+x)^2}\right ) \, dx}{9 \log (4)} \\ & = -\frac {\int \frac {e^{x^2} \left (-1-4 x+2 x^2\right )}{(-2+x)^2} \, dx}{9 \log (4)}+\int \frac {7-4 x+x^2}{(-2+x)^2} \, dx \\ & = \frac {e^{x^2} \left (2 x-x^2\right )}{9 (2-x)^2 x \log (4)}+\int \left (1+\frac {3}{(-2+x)^2}\right ) \, dx \\ & = \frac {3}{2-x}+x+\frac {e^{x^2} \left (2 x-x^2\right )}{9 (2-x)^2 x \log (4)} \\ \end{align*}
Time = 0.08 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.27 \[ \int \frac {e^{x^2} \left (1+4 x-2 x^2\right )+\left (63-36 x+9 x^2\right ) \log (4)}{\left (36-36 x+9 x^2\right ) \log (4)} \, dx=\frac {-e^{x^2}+9 \left (-3-2 x+x^2\right ) \log (4)}{9 (-2+x) \log (4)} \]
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Time = 0.43 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85
method | result | size |
norman | \(\frac {x^{2}-\frac {{\mathrm e}^{x^{2}}}{18 \ln \left (2\right )}-7}{-2+x}\) | \(22\) |
risch | \(x -\frac {3}{-2+x}-\frac {{\mathrm e}^{x^{2}}}{18 \ln \left (2\right ) \left (-2+x \right )}\) | \(25\) |
parts | \(x -\frac {3}{-2+x}-\frac {{\mathrm e}^{x^{2}}}{18 \ln \left (2\right ) \left (-2+x \right )}\) | \(25\) |
parallelrisch | \(\frac {18 x^{2} \ln \left (2\right )-{\mathrm e}^{x^{2}}-126 \ln \left (2\right )}{18 \ln \left (2\right ) \left (-2+x \right )}\) | \(30\) |
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Time = 0.28 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {e^{x^2} \left (1+4 x-2 x^2\right )+\left (63-36 x+9 x^2\right ) \log (4)}{\left (36-36 x+9 x^2\right ) \log (4)} \, dx=\frac {18 \, {\left (x^{2} - 2 \, x - 3\right )} \log \left (2\right ) - e^{\left (x^{2}\right )}}{18 \, {\left (x - 2\right )} \log \left (2\right )} \]
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Time = 0.09 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{x^2} \left (1+4 x-2 x^2\right )+\left (63-36 x+9 x^2\right ) \log (4)}{\left (36-36 x+9 x^2\right ) \log (4)} \, dx=x - \frac {e^{x^{2}}}{18 x \log {\left (2 \right )} - 36 \log {\left (2 \right )}} - \frac {3}{x - 2} \]
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Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (24) = 48\).
Time = 0.23 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.46 \[ \int \frac {e^{x^2} \left (1+4 x-2 x^2\right )+\left (63-36 x+9 x^2\right ) \log (4)}{\left (36-36 x+9 x^2\right ) \log (4)} \, dx=\frac {18 \, {\left (x - \frac {4}{x - 2} + 4 \, \log \left (x - 2\right )\right )} \log \left (2\right ) + 72 \, {\left (\frac {2}{x - 2} - \log \left (x - 2\right )\right )} \log \left (2\right ) - \frac {e^{\left (x^{2}\right )}}{x - 2} - \frac {126 \, \log \left (2\right )}{x - 2}}{18 \, \log \left (2\right )} \]
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Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \frac {e^{x^2} \left (1+4 x-2 x^2\right )+\left (63-36 x+9 x^2\right ) \log (4)}{\left (36-36 x+9 x^2\right ) \log (4)} \, dx=\frac {18 \, x^{2} \log \left (2\right ) - 36 \, x \log \left (2\right ) - e^{\left (x^{2}\right )} - 54 \, \log \left (2\right )}{18 \, {\left (x - 2\right )} \log \left (2\right )} \]
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Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{x^2} \left (1+4 x-2 x^2\right )+\left (63-36 x+9 x^2\right ) \log (4)}{\left (36-36 x+9 x^2\right ) \log (4)} \, dx=x-\frac {\frac {{\mathrm {e}}^{x^2}}{18}+\ln \left (8\right )}{\ln \left (2\right )\,\left (x-2\right )} \]
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