Integrand size = 62, antiderivative size = 23 \[ \int \frac {e^{\frac {1}{225} \left (x^2+6 x \log (x)+9 \log ^2(x)\right )} \left (4950+132 x+50 x^2+2 x^3+\left (396+150 x+6 x^2\right ) \log (x)\right )}{108900+9900 x+225 x^2} \, dx=\frac {e^{\frac {1}{25} \left (\frac {x}{3}+\log (x)\right )^2} x}{22+x} \]
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Leaf count is larger than twice the leaf count of optimal. \(69\) vs. \(2(23)=46\).
Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 3.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {27, 12, 2326} \[ \int \frac {e^{\frac {1}{225} \left (x^2+6 x \log (x)+9 \log ^2(x)\right )} \left (4950+132 x+50 x^2+2 x^3+\left (396+150 x+6 x^2\right ) \log (x)\right )}{108900+9900 x+225 x^2} \, dx=\frac {x^{2 x/75} e^{\frac {1}{225} \left (x^2+9 \log ^2(x)\right )} \left (x^3+25 x^2+3 \left (x^2+25 x+66\right ) \log (x)+66 x\right )}{(x+22)^2 \left (x+3 \log (x)+\frac {9 \log (x)}{x}+3\right )} \]
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Rule 12
Rule 27
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \int \frac {e^{\frac {1}{225} \left (x^2+6 x \log (x)+9 \log ^2(x)\right )} \left (4950+132 x+50 x^2+2 x^3+\left (396+150 x+6 x^2\right ) \log (x)\right )}{225 (22+x)^2} \, dx \\ & = \frac {1}{225} \int \frac {e^{\frac {1}{225} \left (x^2+6 x \log (x)+9 \log ^2(x)\right )} \left (4950+132 x+50 x^2+2 x^3+\left (396+150 x+6 x^2\right ) \log (x)\right )}{(22+x)^2} \, dx \\ & = \frac {e^{\frac {1}{225} \left (x^2+9 \log ^2(x)\right )} x^{2 x/75} \left (66 x+25 x^2+x^3+3 \left (66+25 x+x^2\right ) \log (x)\right )}{(22+x)^2 \left (3+x+3 \log (x)+\frac {9 \log (x)}{x}\right )} \\ \end{align*}
Time = 5.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.43 \[ \int \frac {e^{\frac {1}{225} \left (x^2+6 x \log (x)+9 \log ^2(x)\right )} \left (4950+132 x+50 x^2+2 x^3+\left (396+150 x+6 x^2\right ) \log (x)\right )}{108900+9900 x+225 x^2} \, dx=\frac {e^{\frac {x^2}{225}+\frac {\log ^2(x)}{25}} x^{1+\frac {2 x}{75}}}{22+x} \]
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Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13
method | result | size |
norman | \(\frac {x \,{\mathrm e}^{\frac {\ln \left (x \right )^{2}}{25}+\frac {2 x \ln \left (x \right )}{75}+\frac {x^{2}}{225}}}{22+x}\) | \(26\) |
risch | \(\frac {x \,x^{\frac {2 x}{75}} {\mathrm e}^{\frac {\ln \left (x \right )^{2}}{25}+\frac {x^{2}}{225}}}{22+x}\) | \(26\) |
parallelrisch | \(\frac {x \,{\mathrm e}^{\frac {\ln \left (x \right )^{2}}{25}+\frac {2 x \ln \left (x \right )}{75}+\frac {x^{2}}{225}}}{22+x}\) | \(26\) |
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Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {1}{225} \left (x^2+6 x \log (x)+9 \log ^2(x)\right )} \left (4950+132 x+50 x^2+2 x^3+\left (396+150 x+6 x^2\right ) \log (x)\right )}{108900+9900 x+225 x^2} \, dx=\frac {x e^{\left (\frac {1}{225} \, x^{2} + \frac {2}{75} \, x \log \left (x\right ) + \frac {1}{25} \, \log \left (x\right )^{2}\right )}}{x + 22} \]
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Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {e^{\frac {1}{225} \left (x^2+6 x \log (x)+9 \log ^2(x)\right )} \left (4950+132 x+50 x^2+2 x^3+\left (396+150 x+6 x^2\right ) \log (x)\right )}{108900+9900 x+225 x^2} \, dx=\frac {x e^{\frac {x^{2}}{225} + \frac {2 x \log {\left (x \right )}}{75} + \frac {\log {\left (x \right )}^{2}}{25}}}{x + 22} \]
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Time = 0.28 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {1}{225} \left (x^2+6 x \log (x)+9 \log ^2(x)\right )} \left (4950+132 x+50 x^2+2 x^3+\left (396+150 x+6 x^2\right ) \log (x)\right )}{108900+9900 x+225 x^2} \, dx=\frac {x e^{\left (\frac {1}{225} \, x^{2} + \frac {2}{75} \, x \log \left (x\right ) + \frac {1}{25} \, \log \left (x\right )^{2}\right )}}{x + 22} \]
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\[ \int \frac {e^{\frac {1}{225} \left (x^2+6 x \log (x)+9 \log ^2(x)\right )} \left (4950+132 x+50 x^2+2 x^3+\left (396+150 x+6 x^2\right ) \log (x)\right )}{108900+9900 x+225 x^2} \, dx=\int { \frac {2 \, {\left (x^{3} + 25 \, x^{2} + 3 \, {\left (x^{2} + 25 \, x + 66\right )} \log \left (x\right ) + 66 \, x + 2475\right )} e^{\left (\frac {1}{225} \, x^{2} + \frac {2}{75} \, x \log \left (x\right ) + \frac {1}{25} \, \log \left (x\right )^{2}\right )}}{225 \, {\left (x^{2} + 44 \, x + 484\right )}} \,d x } \]
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Time = 13.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {1}{225} \left (x^2+6 x \log (x)+9 \log ^2(x)\right )} \left (4950+132 x+50 x^2+2 x^3+\left (396+150 x+6 x^2\right ) \log (x)\right )}{108900+9900 x+225 x^2} \, dx=\frac {x\,x^{\frac {2\,x}{75}}\,{\mathrm {e}}^{\frac {x^2}{225}+\frac {{\ln \left (x\right )}^2}{25}}}{x+22} \]
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