Integrand size = 58, antiderivative size = 19 \[ \int \frac {10 x+4 x^2-6 \log (5)+\left (4 x^2-2 \log (5)\right ) \log (x)+x^2 \log ^2(x)}{4 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx=5+x+\frac {2 \left (-5+\frac {\log (5)}{x}\right )}{2+\log (x)} \]
[Out]
Time = 0.28 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {6820, 6874, 2395, 2339, 30, 2343, 2346, 2209} \[ \int \frac {10 x+4 x^2-6 \log (5)+\left (4 x^2-2 \log (5)\right ) \log (x)+x^2 \log ^2(x)}{4 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx=x-\frac {10}{\log (x)+2}+\frac {2 \log (5)}{x (\log (x)+2)} \]
[In]
[Out]
Rule 30
Rule 2209
Rule 2339
Rule 2343
Rule 2346
Rule 2395
Rule 6820
Rule 6874
Rubi steps \begin{align*} \text {integral}& = \int \frac {10 x+4 x^2-6 \log (5)+\left (4 x^2-2 \log (5)\right ) \log (x)+x^2 \log ^2(x)}{x^2 (2+\log (x))^2} \, dx \\ & = \int \left (1+\frac {2 (5 x-\log (5))}{x^2 (2+\log (x))^2}-\frac {2 \log (5)}{x^2 (2+\log (x))}\right ) \, dx \\ & = x+2 \int \frac {5 x-\log (5)}{x^2 (2+\log (x))^2} \, dx-(2 \log (5)) \int \frac {1}{x^2 (2+\log (x))} \, dx \\ & = x+2 \int \left (\frac {5}{x (2+\log (x))^2}-\frac {\log (5)}{x^2 (2+\log (x))^2}\right ) \, dx-(2 \log (5)) \text {Subst}\left (\int \frac {e^{-x}}{2+x} \, dx,x,\log (x)\right ) \\ & = x-2 e^2 \text {Ei}(-2-\log (x)) \log (5)+10 \int \frac {1}{x (2+\log (x))^2} \, dx-(2 \log (5)) \int \frac {1}{x^2 (2+\log (x))^2} \, dx \\ & = x-2 e^2 \text {Ei}(-2-\log (x)) \log (5)+\frac {2 \log (5)}{x (2+\log (x))}+10 \text {Subst}\left (\int \frac {1}{x^2} \, dx,x,2+\log (x)\right )+(2 \log (5)) \int \frac {1}{x^2 (2+\log (x))} \, dx \\ & = x-2 e^2 \text {Ei}(-2-\log (x)) \log (5)-\frac {10}{2+\log (x)}+\frac {2 \log (5)}{x (2+\log (x))}+(2 \log (5)) \text {Subst}\left (\int \frac {e^{-x}}{2+x} \, dx,x,\log (x)\right ) \\ & = x-\frac {10}{2+\log (x)}+\frac {2 \log (5)}{x (2+\log (x))} \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {10 x+4 x^2-6 \log (5)+\left (4 x^2-2 \log (5)\right ) \log (x)+x^2 \log ^2(x)}{4 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx=x+\frac {-10 x+\log (25)}{x (2+\log (x))} \]
[In]
[Out]
Time = 0.77 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05
method | result | size |
risch | \(x +\frac {2 \ln \left (5\right )-10 x}{\left (\ln \left (x \right )+2\right ) x}\) | \(20\) |
default | \(x -\frac {10}{\ln \left (x \right )+2}+\frac {2 \ln \left (5\right )}{\left (\ln \left (x \right )+2\right ) x}\) | \(24\) |
norman | \(\frac {x^{2} \ln \left (x \right )-10 x +2 x^{2}+2 \ln \left (5\right )}{\left (\ln \left (x \right )+2\right ) x}\) | \(30\) |
parallelrisch | \(\frac {x^{2} \ln \left (x \right )-10 x +2 x^{2}+2 \ln \left (5\right )}{\left (\ln \left (x \right )+2\right ) x}\) | \(30\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {10 x+4 x^2-6 \log (5)+\left (4 x^2-2 \log (5)\right ) \log (x)+x^2 \log ^2(x)}{4 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx=\frac {x^{2} \log \left (x\right ) + 2 \, x^{2} - 10 \, x + 2 \, \log \left (5\right )}{x \log \left (x\right ) + 2 \, x} \]
[In]
[Out]
Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89 \[ \int \frac {10 x+4 x^2-6 \log (5)+\left (4 x^2-2 \log (5)\right ) \log (x)+x^2 \log ^2(x)}{4 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx=x + \frac {- 10 x + 2 \log {\left (5 \right )}}{x \log {\left (x \right )} + 2 x} \]
[In]
[Out]
none
Time = 0.31 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.58 \[ \int \frac {10 x+4 x^2-6 \log (5)+\left (4 x^2-2 \log (5)\right ) \log (x)+x^2 \log ^2(x)}{4 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx=\frac {x^{2} \log \left (x\right ) + 2 \, x^{2} - 10 \, x + 2 \, \log \left (5\right )}{x \log \left (x\right ) + 2 \, x} \]
[In]
[Out]
none
Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {10 x+4 x^2-6 \log (5)+\left (4 x^2-2 \log (5)\right ) \log (x)+x^2 \log ^2(x)}{4 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx=x - \frac {2 \, {\left (5 \, x - \log \left (5\right )\right )}}{x \log \left (x\right ) + 2 \, x} \]
[In]
[Out]
Time = 9.01 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.63 \[ \int \frac {10 x+4 x^2-6 \log (5)+\left (4 x^2-2 \log (5)\right ) \log (x)+x^2 \log ^2(x)}{4 x^2+4 x^2 \log (x)+x^2 \log ^2(x)} \, dx=\frac {x^2+5\,x}{x}-\frac {10\,x-\ln \left (25\right )}{x\,\left (\ln \left (x\right )+2\right )} \]
[In]
[Out]