Integrand size = 39, antiderivative size = 15 \[ \int \frac {225-225 \log (x)-\log ^2(x)}{50625 e x^2+450 e x^2 \log (x)+e x^2 \log ^2(x)} \, dx=\frac {1}{e \left (x+\frac {225 x}{\log (x)}\right )} \]
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Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {6820, 12, 6874, 2343, 2346, 2209} \[ \int \frac {225-225 \log (x)-\log ^2(x)}{50625 e x^2+450 e x^2 \log (x)+e x^2 \log ^2(x)} \, dx=\frac {1}{e x}-\frac {225}{e x (\log (x)+225)} \]
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Rule 12
Rule 2209
Rule 2343
Rule 2346
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {225-225 \log (x)-\log ^2(x)}{e x^2 (225+\log (x))^2} \, dx \\ & = \frac {\int \frac {225-225 \log (x)-\log ^2(x)}{x^2 (225+\log (x))^2} \, dx}{e} \\ & = \frac {\int \left (-\frac {1}{x^2}+\frac {225}{x^2 (225+\log (x))^2}+\frac {225}{x^2 (225+\log (x))}\right ) \, dx}{e} \\ & = \frac {1}{e x}+\frac {225 \int \frac {1}{x^2 (225+\log (x))^2} \, dx}{e}+\frac {225 \int \frac {1}{x^2 (225+\log (x))} \, dx}{e} \\ & = \frac {1}{e x}-\frac {225}{e x (225+\log (x))}-\frac {225 \int \frac {1}{x^2 (225+\log (x))} \, dx}{e}+\frac {225 \text {Subst}\left (\int \frac {e^{-x}}{225+x} \, dx,x,\log (x)\right )}{e} \\ & = \frac {1}{e x}+225 e^{224} \text {Ei}(-225-\log (x))-\frac {225}{e x (225+\log (x))}-\frac {225 \text {Subst}\left (\int \frac {e^{-x}}{225+x} \, dx,x,\log (x)\right )}{e} \\ & = \frac {1}{e x}-\frac {225}{e x (225+\log (x))} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {225-225 \log (x)-\log ^2(x)}{50625 e x^2+450 e x^2 \log (x)+e x^2 \log ^2(x)} \, dx=\frac {\log (x)}{e (225 x+x \log (x))} \]
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Time = 0.55 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13
method | result | size |
default | \(\frac {{\mathrm e}^{-1} \ln \left (x \right )}{x \left (\ln \left (x \right )+225\right )}\) | \(17\) |
norman | \(\frac {{\mathrm e}^{-1} \ln \left (x \right )}{x \left (\ln \left (x \right )+225\right )}\) | \(17\) |
parallelrisch | \(\frac {{\mathrm e}^{-1} \ln \left (x \right )}{x \left (\ln \left (x \right )+225\right )}\) | \(17\) |
risch | \(\frac {{\mathrm e}^{-1}}{x}-\frac {225 \,{\mathrm e}^{-1}}{x \left (\ln \left (x \right )+225\right )}\) | \(21\) |
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Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13 \[ \int \frac {225-225 \log (x)-\log ^2(x)}{50625 e x^2+450 e x^2 \log (x)+e x^2 \log ^2(x)} \, dx=\frac {\log \left (x\right )}{x e \log \left (x\right ) + 225 \, x e} \]
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Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.47 \[ \int \frac {225-225 \log (x)-\log ^2(x)}{50625 e x^2+450 e x^2 \log (x)+e x^2 \log ^2(x)} \, dx=- \frac {225}{e x \log {\left (x \right )} + 225 e x} + \frac {1}{e x} \]
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Time = 0.22 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13 \[ \int \frac {225-225 \log (x)-\log ^2(x)}{50625 e x^2+450 e x^2 \log (x)+e x^2 \log ^2(x)} \, dx=\frac {\log \left (x\right )}{x e \log \left (x\right ) + 225 \, x e} \]
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Time = 0.26 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.13 \[ \int \frac {225-225 \log (x)-\log ^2(x)}{50625 e x^2+450 e x^2 \log (x)+e x^2 \log ^2(x)} \, dx=\frac {\log \left (x\right )}{x e \log \left (x\right ) + 225 \, x e} \]
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Time = 12.43 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {225-225 \log (x)-\log ^2(x)}{50625 e x^2+450 e x^2 \log (x)+e x^2 \log ^2(x)} \, dx=\frac {{\mathrm {e}}^{-1}\,\ln \left (x\right )}{x\,\left (\ln \left (x\right )+225\right )} \]
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