Integrand size = 32, antiderivative size = 23 \[ \int \frac {20+\left (-7+4 x-2 x^2\right ) \log ^2(2)}{\left (1-2 x+x^2\right ) \log ^2(2)} \, dx=-2 x+\frac {5 \left (-x+\frac {4}{\log ^2(2)}\right )}{1-x} \]
[Out]
Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {12, 27, 1864} \[ \int \frac {20+\left (-7+4 x-2 x^2\right ) \log ^2(2)}{\left (1-2 x+x^2\right ) \log ^2(2)} \, dx=-2 x-\frac {5 \left (1-\frac {4}{\log ^2(2)}\right )}{1-x} \]
[In]
[Out]
Rule 12
Rule 27
Rule 1864
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {20+\left (-7+4 x-2 x^2\right ) \log ^2(2)}{1-2 x+x^2} \, dx}{\log ^2(2)} \\ & = \frac {\int \frac {20+\left (-7+4 x-2 x^2\right ) \log ^2(2)}{(-1+x)^2} \, dx}{\log ^2(2)} \\ & = \frac {\int \left (-2 \log ^2(2)-\frac {5 \left (-4+\log ^2(2)\right )}{(-1+x)^2}\right ) \, dx}{\log ^2(2)} \\ & = -2 x-\frac {5 \left (1-\frac {4}{\log ^2(2)}\right )}{1-x} \\ \end{align*}
Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {20+\left (-7+4 x-2 x^2\right ) \log ^2(2)}{\left (1-2 x+x^2\right ) \log ^2(2)} \, dx=\frac {-2 (-1+x) \log ^2(2)+\frac {5 \left (-4+\log ^2(2)\right )}{-1+x}}{\log ^2(2)} \]
[In]
[Out]
Time = 0.77 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00
method | result | size |
risch | \(-2 x +\frac {5}{-1+x}-\frac {20}{\ln \left (2\right )^{2} \left (-1+x \right )}\) | \(23\) |
gosper | \(-\frac {2 x^{2} \ln \left (2\right )^{2}+20-7 \ln \left (2\right )^{2}}{\ln \left (2\right )^{2} \left (-1+x \right )}\) | \(29\) |
default | \(\frac {-2 x \ln \left (2\right )^{2}-\frac {-5 \ln \left (2\right )^{2}+20}{-1+x}}{\ln \left (2\right )^{2}}\) | \(29\) |
parallelrisch | \(-\frac {2 x^{2} \ln \left (2\right )^{2}+20-7 \ln \left (2\right )^{2}}{\ln \left (2\right )^{2} \left (-1+x \right )}\) | \(29\) |
norman | \(\frac {-2 x^{2} \ln \left (2\right )+\frac {7 \ln \left (2\right )^{2}-20}{\ln \left (2\right )}}{\left (-1+x \right ) \ln \left (2\right )}\) | \(32\) |
meijerg | \(\frac {20 x}{\ln \left (2\right )^{2} \left (1-x \right )}-\frac {2 x \left (-3 x +6\right )}{3 \left (1-x \right )}-\frac {3 x}{1-x}\) | \(41\) |
[In]
[Out]
none
Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {20+\left (-7+4 x-2 x^2\right ) \log ^2(2)}{\left (1-2 x+x^2\right ) \log ^2(2)} \, dx=-\frac {{\left (2 \, x^{2} - 2 \, x - 5\right )} \log \left (2\right )^{2} + 20}{{\left (x - 1\right )} \log \left (2\right )^{2}} \]
[In]
[Out]
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {20+\left (-7+4 x-2 x^2\right ) \log ^2(2)}{\left (1-2 x+x^2\right ) \log ^2(2)} \, dx=- 2 x - \frac {20 - 5 \log {\left (2 \right )}^{2}}{x \log {\left (2 \right )}^{2} - \log {\left (2 \right )}^{2}} \]
[In]
[Out]
none
Time = 0.20 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {20+\left (-7+4 x-2 x^2\right ) \log ^2(2)}{\left (1-2 x+x^2\right ) \log ^2(2)} \, dx=-\frac {2 \, x \log \left (2\right )^{2} - \frac {5 \, {\left (\log \left (2\right )^{2} - 4\right )}}{x - 1}}{\log \left (2\right )^{2}} \]
[In]
[Out]
none
Time = 0.27 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {20+\left (-7+4 x-2 x^2\right ) \log ^2(2)}{\left (1-2 x+x^2\right ) \log ^2(2)} \, dx=-\frac {2 \, x \log \left (2\right )^{2} - \frac {5 \, {\left (\log \left (2\right )^{2} - 4\right )}}{x - 1}}{\log \left (2\right )^{2}} \]
[In]
[Out]
Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {20+\left (-7+4 x-2 x^2\right ) \log ^2(2)}{\left (1-2 x+x^2\right ) \log ^2(2)} \, dx=\frac {5\,{\ln \left (2\right )}^2-20}{{\ln \left (2\right )}^2\,\left (x-1\right )}-2\,x \]
[In]
[Out]