Integrand size = 56, antiderivative size = 21 \[ \int \frac {-80 \log \left (\frac {1}{2 x}\right )+\left (80-80 \log \left (\frac {1}{2 x}\right )\right ) \log (x)+\left (20-20 \log \left (\frac {1}{2 x}\right )\right ) \log ^2(x)}{16+8 \log (x)+\log ^2(x)} \, dx=24-\frac {20 x \log \left (\frac {1}{2 x}\right ) \log (x)}{4+\log (x)} \] Output:
24-20/(ln(x)+4)*ln(x)*ln(1/2/x)*x
Time = 0.11 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {-80 \log \left (\frac {1}{2 x}\right )+\left (80-80 \log \left (\frac {1}{2 x}\right )\right ) \log (x)+\left (20-20 \log \left (\frac {1}{2 x}\right )\right ) \log ^2(x)}{16+8 \log (x)+\log ^2(x)} \, dx=-\frac {20 x \log \left (\frac {1}{2 x}\right ) \log (x)}{4+\log (x)} \] Input:
Integrate[(-80*Log[1/(2*x)] + (80 - 80*Log[1/(2*x)])*Log[x] + (20 - 20*Log [1/(2*x)])*Log[x]^2)/(16 + 8*Log[x] + Log[x]^2),x]
Output:
(-20*x*Log[1/(2*x)]*Log[x])/(4 + Log[x])
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.49 (sec) , antiderivative size = 79, normalized size of antiderivative = 3.76, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {7239, 7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (20-20 \log \left (\frac {1}{2 x}\right )\right ) \log ^2(x)+\left (80-80 \log \left (\frac {1}{2 x}\right )\right ) \log (x)-80 \log \left (\frac {1}{2 x}\right )}{\log ^2(x)+8 \log (x)+16} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {20 \log (x) (\log (x)+4)-20 \log \left (\frac {1}{2 x}\right ) (\log (x)+2)^2}{(\log (x)+4)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {80 \left (\log \left (\frac {1}{2 x}\right )-1\right )}{\log (x)+4}-20 \left (\log \left (\frac {1}{2 x}\right )-1\right )-\frac {80 \log \left (\frac {1}{2 x}\right )}{(\log (x)+4)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {80 \left (1-\log \left (\frac {1}{2 x}\right )\right ) \operatorname {ExpIntegralEi}(\log (x)+4)}{e^4}-\frac {80 \log \left (\frac {1}{2 x}\right ) \operatorname {ExpIntegralEi}(\log (x)+4)}{e^4}+\frac {80 \operatorname {ExpIntegralEi}(\log (x)+4)}{e^4}-20 x \log \left (\frac {1}{2 x}\right )+\frac {80 x \log \left (\frac {1}{2 x}\right )}{\log (x)+4}\) |
Input:
Int[(-80*Log[1/(2*x)] + (80 - 80*Log[1/(2*x)])*Log[x] + (20 - 20*Log[1/(2* x)])*Log[x]^2)/(16 + 8*Log[x] + Log[x]^2),x]
Output:
(80*ExpIntegralEi[4 + Log[x]])/E^4 - (80*ExpIntegralEi[4 + Log[x]]*(1 - Lo g[1/(2*x)]))/E^4 - 20*x*Log[1/(2*x)] - (80*ExpIntegralEi[4 + Log[x]]*Log[1 /(2*x)])/E^4 + (80*x*Log[1/(2*x)])/(4 + Log[x])
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.81 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86
method | result | size |
norman | \(-\frac {20 \ln \left (x \right ) \ln \left (\frac {1}{2 x}\right ) x}{\ln \left (x \right )+4}\) | \(18\) |
parallelrisch | \(-\frac {20 \ln \left (x \right ) \ln \left (\frac {1}{2 x}\right ) x}{\ln \left (x \right )+4}\) | \(18\) |
risch | \(20 x \ln \left (x \right )+20 x \ln \left (2\right )-80 x -\frac {40 x \left (2 \ln \left (2\right )-8\right )}{\ln \left (x \right )+4}\) | \(30\) |
default | \(100 x +80 \,{\mathrm e}^{-4} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )-4\right )+\frac {20 \ln \left (2\right ) \ln \left (x \right ) x}{\ln \left (x \right )+4}-20 \left (3 \ln \left (\frac {1}{x}\right )+2 \ln \left (x \right )+9\right ) x +\frac {20 x \left (-64-\left (\ln \left (\frac {1}{x}\right )+\ln \left (x \right )\right )^{3}-12 \left (\ln \left (\frac {1}{x}\right )+\ln \left (x \right )\right )^{2}-48 \ln \left (x \right )-48 \ln \left (\frac {1}{x}\right )\right )}{-\ln \left (x \right )-4}-20 \left (-16-\left (\ln \left (\frac {1}{x}\right )+\ln \left (x \right )\right )^{3}-9 \left (\ln \left (\frac {1}{x}\right )+\ln \left (x \right )\right )^{2}-24 \ln \left (x \right )-24 \ln \left (\frac {1}{x}\right )\right ) {\mathrm e}^{-\ln \left (x \right )-\ln \left (\frac {1}{x}\right )-4} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )-4\right )+20 \left (\ln \left (\frac {1}{x}\right )+\ln \left (x \right )\right )^{2} \left (\frac {x \left (-\ln \left (x \right )-\ln \left (\frac {1}{x}\right )-4\right )}{-\ln \left (x \right )-4}-\left (-\ln \left (x \right )-\ln \left (\frac {1}{x}\right )-3\right ) {\mathrm e}^{-\ln \left (x \right )-\ln \left (\frac {1}{x}\right )-4} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )-4\right )\right )+\frac {80 x \left (-\ln \left (x \right )-\ln \left (\frac {1}{x}\right )-4\right )}{-\ln \left (x \right )-4}-80 \left (-\ln \left (x \right )-\ln \left (\frac {1}{x}\right )-3\right ) {\mathrm e}^{-\ln \left (x \right )-\ln \left (\frac {1}{x}\right )-4} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )-4\right )+\frac {80 x \left (16+\left (\ln \left (\frac {1}{x}\right )+\ln \left (x \right )\right )^{2}+8 \ln \left (x \right )+8 \ln \left (\frac {1}{x}\right )\right )}{-\ln \left (x \right )-4}+80 \left (-8-\left (\ln \left (\frac {1}{x}\right )+\ln \left (x \right )\right )^{2}-6 \ln \left (x \right )-6 \ln \left (\frac {1}{x}\right )\right ) {\mathrm e}^{-\ln \left (x \right )-\ln \left (\frac {1}{x}\right )-4} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )-4\right )+80 \left (\ln \left (\frac {1}{x}\right )+\ln \left (x \right )\right ) \left (\frac {x \left (-\ln \left (x \right )-\ln \left (\frac {1}{x}\right )-4\right )}{-\ln \left (x \right )-4}-\left (-\ln \left (x \right )-\ln \left (\frac {1}{x}\right )-3\right ) {\mathrm e}^{-\ln \left (x \right )-\ln \left (\frac {1}{x}\right )-4} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )-4\right )\right )-40 \left (\ln \left (\frac {1}{x}\right )+\ln \left (x \right )\right ) \left (-x -\frac {x \left (16+\left (\ln \left (\frac {1}{x}\right )+\ln \left (x \right )\right )^{2}+8 \ln \left (x \right )+8 \ln \left (\frac {1}{x}\right )\right )}{-\ln \left (x \right )-4}-\left (-8-\left (\ln \left (\frac {1}{x}\right )+\ln \left (x \right )\right )^{2}-6 \ln \left (x \right )-6 \ln \left (\frac {1}{x}\right )\right ) {\mathrm e}^{-\ln \left (x \right )-\ln \left (\frac {1}{x}\right )-4} \operatorname {expIntegral}_{1}\left (-\ln \left (x \right )-4\right )\right )\) | \(510\) |
Input:
int(((-20*ln(1/2/x)+20)*ln(x)^2+(-80*ln(1/2/x)+80)*ln(x)-80*ln(1/2/x))/(ln (x)^2+8*ln(x)+16),x,method=_RETURNVERBOSE)
Output:
-20/(ln(x)+4)*ln(x)*ln(1/2/x)*x
Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.67 \[ \int \frac {-80 \log \left (\frac {1}{2 x}\right )+\left (80-80 \log \left (\frac {1}{2 x}\right )\right ) \log (x)+\left (20-20 \log \left (\frac {1}{2 x}\right )\right ) \log ^2(x)}{16+8 \log (x)+\log ^2(x)} \, dx=-\frac {20 \, {\left (x \log \left (2\right ) \log \left (\frac {1}{2 \, x}\right ) + x \log \left (\frac {1}{2 \, x}\right )^{2}\right )}}{\log \left (2\right ) + \log \left (\frac {1}{2 \, x}\right ) - 4} \] Input:
integrate(((-20*log(1/2/x)+20)*log(x)^2+(-80*log(1/2/x)+80)*log(x)-80*log( 1/2/x))/(log(x)^2+8*log(x)+16),x, algorithm="fricas")
Output:
-20*(x*log(2)*log(1/2/x) + x*log(1/2/x)^2)/(log(2) + log(1/2/x) - 4)
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.38 \[ \int \frac {-80 \log \left (\frac {1}{2 x}\right )+\left (80-80 \log \left (\frac {1}{2 x}\right )\right ) \log (x)+\left (20-20 \log \left (\frac {1}{2 x}\right )\right ) \log ^2(x)}{16+8 \log (x)+\log ^2(x)} \, dx=20 x \log {\left (x \right )} + x \left (-80 + 20 \log {\left (2 \right )}\right ) + \frac {- 80 x \log {\left (2 \right )} + 320 x}{\log {\left (x \right )} + 4} \] Input:
integrate(((-20*ln(1/2/x)+20)*ln(x)**2+(-80*ln(1/2/x)+80)*ln(x)-80*ln(1/2/ x))/(ln(x)**2+8*ln(x)+16),x)
Output:
20*x*log(x) + x*(-80 + 20*log(2)) + (-80*x*log(2) + 320*x)/(log(x) + 4)
Time = 0.15 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-80 \log \left (\frac {1}{2 x}\right )+\left (80-80 \log \left (\frac {1}{2 x}\right )\right ) \log (x)+\left (20-20 \log \left (\frac {1}{2 x}\right )\right ) \log ^2(x)}{16+8 \log (x)+\log ^2(x)} \, dx=\frac {20 \, {\left (x \log \left (2\right ) \log \left (x\right ) + x \log \left (x\right )^{2}\right )}}{\log \left (x\right ) + 4} \] Input:
integrate(((-20*log(1/2/x)+20)*log(x)^2+(-80*log(1/2/x)+80)*log(x)-80*log( 1/2/x))/(log(x)^2+8*log(x)+16),x, algorithm="maxima")
Output:
20*(x*log(2)*log(x) + x*log(x)^2)/(log(x) + 4)
Time = 0.11 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {-80 \log \left (\frac {1}{2 x}\right )+\left (80-80 \log \left (\frac {1}{2 x}\right )\right ) \log (x)+\left (20-20 \log \left (\frac {1}{2 x}\right )\right ) \log ^2(x)}{16+8 \log (x)+\log ^2(x)} \, dx=\frac {20 \, x \log \left (2\right ) \log \left (x\right )}{\log \left (x\right ) + 4} + \frac {20 \, x \log \left (x\right )^{2}}{\log \left (x\right ) + 4} \] Input:
integrate(((-20*log(1/2/x)+20)*log(x)^2+(-80*log(1/2/x)+80)*log(x)-80*log( 1/2/x))/(log(x)^2+8*log(x)+16),x, algorithm="giac")
Output:
20*x*log(2)*log(x)/(log(x) + 4) + 20*x*log(x)^2/(log(x) + 4)
Time = 2.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {-80 \log \left (\frac {1}{2 x}\right )+\left (80-80 \log \left (\frac {1}{2 x}\right )\right ) \log (x)+\left (20-20 \log \left (\frac {1}{2 x}\right )\right ) \log ^2(x)}{16+8 \log (x)+\log ^2(x)} \, dx=-\frac {20\,x\,\ln \left (\frac {1}{2\,x}\right )\,\ln \left (x\right )}{\ln \left (x\right )+4} \] Input:
int(-(80*log(1/(2*x)) + log(x)^2*(20*log(1/(2*x)) - 20) + log(x)*(80*log(1 /(2*x)) - 80))/(8*log(x) + log(x)^2 + 16),x)
Output:
-(20*x*log(1/(2*x))*log(x))/(log(x) + 4)
Time = 0.22 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {-80 \log \left (\frac {1}{2 x}\right )+\left (80-80 \log \left (\frac {1}{2 x}\right )\right ) \log (x)+\left (20-20 \log \left (\frac {1}{2 x}\right )\right ) \log ^2(x)}{16+8 \log (x)+\log ^2(x)} \, dx=\frac {20 \,\mathrm {log}\left (2 x \right ) \mathrm {log}\left (x \right ) x}{\mathrm {log}\left (x \right )+4} \] Input:
int(((-20*log(1/2/x)+20)*log(x)^2+(-80*log(1/2/x)+80)*log(x)-80*log(1/2/x) )/(log(x)^2+8*log(x)+16),x)
Output:
(20*log(2*x)*log(x)*x)/(log(x) + 4)