Integrand size = 50, antiderivative size = 28 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\frac {5+\frac {1}{2} \left (x+\log (x)+(1+x)^2 \log \left (\frac {\log (5)}{25}\right )\right )}{\log (x)} \]
[Out]
\[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx \]
[In]
[Out]
Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{x \log ^2(x)} \, dx \\ & = \frac {1}{2} \int \left (\frac {1+2 \log \left (\frac {\log (5)}{25}\right )+2 x \log \left (\frac {\log (5)}{25}\right )}{\log (x)}+\frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)}\right ) \, dx \\ & = \frac {1}{2} \int \frac {1+2 \log \left (\frac {\log (5)}{25}\right )+2 x \log \left (\frac {\log (5)}{25}\right )}{\log (x)} \, dx+\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx \\ & = \frac {1}{2} \int \left (\frac {2 x \log \left (\frac {\log (5)}{25}\right )}{\log (x)}+\frac {1+2 \log \left (\frac {\log (5)}{25}\right )}{\log (x)}\right ) \, dx+\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx \\ & = \frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx+\log \left (\frac {\log (5)}{25}\right ) \int \frac {x}{\log (x)} \, dx+\frac {1}{2} \left (1+2 \log \left (\frac {\log (5)}{25}\right )\right ) \int \frac {1}{\log (x)} \, dx \\ & = \frac {1}{2} \left (1+2 \log \left (\frac {\log (5)}{25}\right )\right ) \operatorname {LogIntegral}(x)+\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx+\log \left (\frac {\log (5)}{25}\right ) \text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (x)\right ) \\ & = \operatorname {ExpIntegralEi}(2 \log (x)) \log \left (\frac {\log (5)}{25}\right )+\frac {1}{2} \left (1+2 \log \left (\frac {\log (5)}{25}\right )\right ) \operatorname {LogIntegral}(x)+\frac {1}{2} \int \frac {-10+\log \left (\frac {25}{\log (5)}\right )-x^2 \log \left (\frac {\log (5)}{25}\right )-x (1-\log (625)+2 \log (\log (5)))}{x \log ^2(x)} \, dx \\ \end{align*}
Time = 0.19 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.39 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\frac {10+x+\log \left (\frac {\log (5)}{25}\right )+2 x \log \left (\frac {\log (5)}{25}\right )+x^2 \log \left (\frac {\log (5)}{25}\right )}{2 \log (x)} \]
[In]
[Out]
Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.14
method | result | size |
parallelrisch | \(\frac {\ln \left (\frac {\ln \left (5\right )}{25}\right ) x^{2}+2 \ln \left (\frac {\ln \left (5\right )}{25}\right ) x +10+\ln \left (\frac {\ln \left (5\right )}{25}\right )+x}{2 \ln \left (x \right )}\) | \(32\) |
norman | \(\frac {\left (-\ln \left (5\right )+\frac {\ln \left (\ln \left (5\right )\right )}{2}\right ) x^{2}+\left (-2 \ln \left (5\right )+\ln \left (\ln \left (5\right )\right )+\frac {1}{2}\right ) x +5-\ln \left (5\right )+\frac {\ln \left (\ln \left (5\right )\right )}{2}}{\ln \left (x \right )}\) | \(42\) |
risch | \(-\frac {2 x^{2} \ln \left (5\right )-\ln \left (\ln \left (5\right )\right ) x^{2}+4 x \ln \left (5\right )-2 x \ln \left (\ln \left (5\right )\right )+2 \ln \left (5\right )-\ln \left (\ln \left (5\right )\right )-x -10}{2 \ln \left (x \right )}\) | \(47\) |
default | \(2 \ln \left (5\right ) \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )-\ln \left (\ln \left (5\right )\right ) \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )+2 \ln \left (5\right ) \operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )-\ln \left (\ln \left (5\right )\right ) \operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )+\ln \left (5\right ) \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )-\frac {\ln \left (\ln \left (5\right )\right ) \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )}{2}+2 \ln \left (5\right ) \left (-\frac {x}{\ln \left (x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )\right )-\ln \left (\ln \left (5\right )\right ) \left (-\frac {x}{\ln \left (x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )\right )-\frac {\ln \left (5\right )}{\ln \left (x \right )}+\frac {\ln \left (\ln \left (5\right )\right )}{2 \ln \left (x \right )}+\frac {x}{2 \ln \left (x \right )}+\frac {5}{\ln \left (x \right )}\) | \(159\) |
parts | \(2 \ln \left (5\right ) \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )-\ln \left (\ln \left (5\right )\right ) \operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )+2 \ln \left (5\right ) \operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )-\ln \left (\ln \left (5\right )\right ) \operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )+\ln \left (5\right ) \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )-\frac {\ln \left (\ln \left (5\right )\right ) \left (-\frac {x^{2}}{\ln \left (x \right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (x \right )\right )\right )}{2}+2 \ln \left (5\right ) \left (-\frac {x}{\ln \left (x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )\right )-\ln \left (\ln \left (5\right )\right ) \left (-\frac {x}{\ln \left (x \right )}-\operatorname {Ei}_{1}\left (-\ln \left (x \right )\right )\right )-\frac {\ln \left (5\right )}{\ln \left (x \right )}+\frac {\ln \left (\ln \left (5\right )\right )}{2 \ln \left (x \right )}+\frac {x}{2 \ln \left (x \right )}+\frac {5}{\ln \left (x \right )}\) | \(159\) |
[In]
[Out]
none
Time = 0.25 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.82 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\frac {{\left (x^{2} + 2 \, x + 1\right )} \log \left (\frac {1}{25} \, \log \left (5\right )\right ) + x + 10}{2 \, \log \left (x\right )} \]
[In]
[Out]
Time = 0.05 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.75 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\frac {- 2 x^{2} \log {\left (5 \right )} + x^{2} \log {\left (\log {\left (5 \right )} \right )} - 4 x \log {\left (5 \right )} + 2 x \log {\left (\log {\left (5 \right )} \right )} + x - 2 \log {\left (5 \right )} + \log {\left (\log {\left (5 \right )} \right )} + 10}{2 \log {\left (x \right )}} \]
[In]
[Out]
Result contains higher order function than in optimal. Order 4 vs. order 3.
Time = 0.23 (sec) , antiderivative size = 77, normalized size of antiderivative = 2.75 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx={\rm Ei}\left (2 \, \log \left (x\right )\right ) \log \left (\frac {1}{25} \, \log \left (5\right )\right ) + {\rm Ei}\left (\log \left (x\right )\right ) \log \left (\frac {1}{25} \, \log \left (5\right )\right ) - \Gamma \left (-1, -\log \left (x\right )\right ) \log \left (\frac {1}{25} \, \log \left (5\right )\right ) - \Gamma \left (-1, -2 \, \log \left (x\right )\right ) \log \left (\frac {1}{25} \, \log \left (5\right )\right ) + \frac {\log \left (\frac {1}{25} \, \log \left (5\right )\right )}{2 \, \log \left (x\right )} + \frac {5}{\log \left (x\right )} + \frac {1}{2} \, {\rm Ei}\left (\log \left (x\right )\right ) - \frac {1}{2} \, \Gamma \left (-1, -\log \left (x\right )\right ) \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (22) = 44\).
Time = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.64 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=-\frac {2 \, x^{2} \log \left (5\right ) - x^{2} \log \left (\log \left (5\right )\right ) + 4 \, x \log \left (5\right ) - 2 \, x \log \left (\log \left (5\right )\right ) - x + 2 \, \log \left (5\right ) - \log \left (\log \left (5\right )\right ) - 10}{2 \, \log \left (x\right )} \]
[In]
[Out]
Time = 8.16 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.54 \[ \int \frac {-10-x+x \log (x)+\left (-1-2 x-x^2+\left (2 x+2 x^2\right ) \log (x)\right ) \log \left (\frac {\log (5)}{25}\right )}{2 x \log ^2(x)} \, dx=\frac {\ln \left (\frac {\ln \left (5\right )}{25}\right )\,x^4+\left (2\,\ln \left (\frac {\ln \left (5\right )}{25}\right )+1\right )\,x^3+\left (\ln \left (\frac {\ln \left (5\right )}{25}\right )+10\right )\,x^2}{2\,x^2\,\ln \left (x\right )} \]
[In]
[Out]