Integrand size = 57, antiderivative size = 26 \[ \int \frac {x^2+e^{\frac {-25-9 x-e^2 x-e^x x+3 x^2+x \log (x)}{x}} \left (25+x+3 x^2-e^x x^2\right )}{x^2} \, dx=x+e^{-9-e^2-e^x-\frac {25}{x}+3 x} x \]
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Leaf count is larger than twice the leaf count of optimal. \(57\) vs. \(2(26)=52\).
Time = 0.16 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.19, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {14, 2326} \[ \int \frac {x^2+e^{\frac {-25-9 x-e^2 x-e^x x+3 x^2+x \log (x)}{x}} \left (25+x+3 x^2-e^x x^2\right )}{x^2} \, dx=\frac {e^{3 x-e^x-\frac {25}{x}-e^2-9} \left (-e^x x^2+3 x^2+25\right )}{\left (\frac {25}{x^2}-e^x+3\right ) x}+x \]
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Rule 14
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \int \left (1+\frac {e^{-e^x-9 \left (1+\frac {e^2}{9}\right )-\frac {25}{x}+3 x} \left (25+x+3 x^2-e^x x^2\right )}{x}\right ) \, dx \\ & = x+\int \frac {e^{-e^x-9 \left (1+\frac {e^2}{9}\right )-\frac {25}{x}+3 x} \left (25+x+3 x^2-e^x x^2\right )}{x} \, dx \\ & = x+\frac {e^{-9-e^2-e^x-\frac {25}{x}+3 x} \left (25+3 x^2-e^x x^2\right )}{\left (3-e^x+\frac {25}{x^2}\right ) x} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {x^2+e^{\frac {-25-9 x-e^2 x-e^x x+3 x^2+x \log (x)}{x}} \left (25+x+3 x^2-e^x x^2\right )}{x^2} \, dx=\left (1+e^{-9-e^2-e^x-\frac {25}{x}+3 x}\right ) x \]
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Time = 1.86 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12
method | result | size |
risch | \(x +x \,{\mathrm e}^{-\frac {{\mathrm e}^{2} x +{\mathrm e}^{x} x -3 x^{2}+9 x +25}{x}}\) | \(29\) |
parallelrisch | \(x +{\mathrm e}^{-\frac {{\mathrm e}^{2} x -x \ln \left (x \right )+{\mathrm e}^{x} x -3 x^{2}+9 x +25}{x}}\) | \(32\) |
norman | \(\frac {x^{2}+x \,{\mathrm e}^{\frac {x \ln \left (x \right )-{\mathrm e}^{x} x -{\mathrm e}^{2} x +3 x^{2}-9 x -25}{x}}}{x}\) | \(40\) |
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Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {x^2+e^{\frac {-25-9 x-e^2 x-e^x x+3 x^2+x \log (x)}{x}} \left (25+x+3 x^2-e^x x^2\right )}{x^2} \, dx=x + e^{\left (\frac {3 \, x^{2} - x e^{2} - x e^{x} + x \log \left (x\right ) - 9 \, x - 25}{x}\right )} \]
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Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {x^2+e^{\frac {-25-9 x-e^2 x-e^x x+3 x^2+x \log (x)}{x}} \left (25+x+3 x^2-e^x x^2\right )}{x^2} \, dx=x + e^{\frac {3 x^{2} - x e^{x} + x \log {\left (x \right )} - 9 x - x e^{2} - 25}{x}} \]
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Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {x^2+e^{\frac {-25-9 x-e^2 x-e^x x+3 x^2+x \log (x)}{x}} \left (25+x+3 x^2-e^x x^2\right )}{x^2} \, dx=x e^{\left (3 \, x - \frac {25}{x} - e^{2} - e^{x} - 9\right )} + x \]
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\[ \int \frac {x^2+e^{\frac {-25-9 x-e^2 x-e^x x+3 x^2+x \log (x)}{x}} \left (25+x+3 x^2-e^x x^2\right )}{x^2} \, dx=\int { \frac {x^{2} - {\left (x^{2} e^{x} - 3 \, x^{2} - x - 25\right )} e^{\left (\frac {3 \, x^{2} - x e^{2} - x e^{x} + x \log \left (x\right ) - 9 \, x - 25}{x}\right )}}{x^{2}} \,d x } \]
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Time = 15.53 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {x^2+e^{\frac {-25-9 x-e^2 x-e^x x+3 x^2+x \log (x)}{x}} \left (25+x+3 x^2-e^x x^2\right )}{x^2} \, dx=x+x\,{\mathrm {e}}^{-{\mathrm {e}}^2}\,{\mathrm {e}}^{3\,x}\,{\mathrm {e}}^{-9}\,{\mathrm {e}}^{-\frac {25}{x}}\,{\mathrm {e}}^{-{\mathrm {e}}^x} \]
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