Integrand size = 33, antiderivative size = 20 \[ \int \frac {2}{-10+5 x+(-2+x) \log \left (\frac {\left (100-100 x+25 x^2\right ) \log (5)}{16 e^5}\right )} \, dx=\log \left (2 \left (5+\log \left (\frac {25 (-2+x)^2 \log (5)}{16 e^5}\right )\right )\right ) \]
[Out]
Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {12, 6873, 2437, 2339, 29} \[ \int \frac {2}{-10+5 x+(-2+x) \log \left (\frac {\left (100-100 x+25 x^2\right ) \log (5)}{16 e^5}\right )} \, dx=\log \left (\log \left (\frac {25 (2-x)^2 \log (5)}{16 e^5}\right )+5\right ) \]
[In]
[Out]
Rule 12
Rule 29
Rule 2339
Rule 2437
Rule 6873
Rubi steps \begin{align*} \text {integral}& = 2 \int \frac {1}{-10+5 x+(-2+x) \log \left (\frac {\left (100-100 x+25 x^2\right ) \log (5)}{16 e^5}\right )} \, dx \\ & = 2 \int \frac {1}{(-2+x) \left (5+\log \left (\frac {25 (-2+x)^2 \log (5)}{16 e^5}\right )\right )} \, dx \\ & = 2 \text {Subst}\left (\int \frac {1}{x \left (5+\log \left (\frac {25 x^2 \log (5)}{16 e^5}\right )\right )} \, dx,x,-2+x\right ) \\ & = \text {Subst}\left (\int \frac {1}{x} \, dx,x,5+\log \left (\frac {25 (-2+x)^2 \log (5)}{16 e^5}\right )\right ) \\ & = \log \left (5+\log \left (\frac {25 (-2+x)^2 \log (5)}{16 e^5}\right )\right ) \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {2}{-10+5 x+(-2+x) \log \left (\frac {\left (100-100 x+25 x^2\right ) \log (5)}{16 e^5}\right )} \, dx=\log \left (5+\log \left (\frac {25 (-2+x)^2 \log (5)}{16 e^5}\right )\right ) \]
[In]
[Out]
Time = 0.50 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05
method | result | size |
risch | \(\ln \left (\ln \left (\frac {\left (25 x^{2}-100 x +100\right ) \ln \left (5\right ) {\mathrm e}^{-5}}{16}\right )+5\right )\) | \(21\) |
parallelrisch | \(\ln \left (\ln \left (\frac {25 \left (x^{2}-4 x +4\right ) \ln \left (5\right ) {\mathrm e}^{-5}}{16}\right )+5\right )\) | \(21\) |
default | \(\ln \left (2 \ln \left (5\right )-4 \ln \left (2\right )+\ln \left (\ln \left (5\right )\right )+\ln \left (x^{2}-4 x +4\right )\right )\) | \(23\) |
norman | \(\ln \left (\ln \left (\frac {\left (25 x^{2}-100 x +100\right ) \ln \left (5\right ) {\mathrm e}^{-5}}{16}\right )+5\right )\) | \(23\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {2}{-10+5 x+(-2+x) \log \left (\frac {\left (100-100 x+25 x^2\right ) \log (5)}{16 e^5}\right )} \, dx=\log \left (\log \left (\frac {25}{16} \, {\left (x^{2} - 4 \, x + 4\right )} e^{\left (-5\right )} \log \left (5\right )\right ) + 5\right ) \]
[In]
[Out]
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.30 \[ \int \frac {2}{-10+5 x+(-2+x) \log \left (\frac {\left (100-100 x+25 x^2\right ) \log (5)}{16 e^5}\right )} \, dx=\log {\left (\log {\left (\frac {\left (\frac {25 x^{2}}{16} - \frac {25 x}{4} + \frac {25}{4}\right ) \log {\left (5 \right )}}{e^{5}} \right )} + 5 \right )} \]
[In]
[Out]
none
Time = 0.29 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {2}{-10+5 x+(-2+x) \log \left (\frac {\left (100-100 x+25 x^2\right ) \log (5)}{16 e^5}\right )} \, dx=\log \left (\log \left (5\right ) - 2 \, \log \left (2\right ) + \log \left (x - 2\right ) + \frac {1}{2} \, \log \left (\log \left (5\right )\right )\right ) \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.35 \[ \int \frac {2}{-10+5 x+(-2+x) \log \left (\frac {\left (100-100 x+25 x^2\right ) \log (5)}{16 e^5}\right )} \, dx=\log \left (2 \, \log \left (5\right ) - 4 \, \log \left (2\right ) + \log \left (x^{2} \log \left (5\right ) - 4 \, x \log \left (5\right ) + 4 \, \log \left (5\right )\right )\right ) \]
[In]
[Out]
Time = 1.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {2}{-10+5 x+(-2+x) \log \left (\frac {\left (100-100 x+25 x^2\right ) \log (5)}{16 e^5}\right )} \, dx=\ln \left (\ln \left (\frac {{\mathrm {e}}^{-5}\,\ln \left (5\right )\,\left (25\,x^2-100\,x+100\right )}{16}\right )+5\right ) \]
[In]
[Out]