Integrand size = 80, antiderivative size = 28 \[ \int \frac {-e+e^{\log ^2\left (\frac {3}{2}\right )-2 \log \left (\frac {3}{2}\right ) \log (3)+\log ^2(3)}-x-x^2+\left (e-e^{\log ^2\left (\frac {3}{2}\right )-2 \log \left (\frac {3}{2}\right ) \log (3)+\log ^2(3)}+2 x\right ) \log (x)}{x^2-2 x \log (x)+\log ^2(x)} \, dx=\frac {x \left (e-e^{\left (-\log \left (\frac {3}{2}\right )+\log (3)\right )^2}+x\right )}{-x+\log (x)} \] Output:
x/(ln(x)-x)*(exp(1)-exp((ln(3)+ln(2/3))^2)+x)
Time = 0.30 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-e+e^{\log ^2\left (\frac {3}{2}\right )-2 \log \left (\frac {3}{2}\right ) \log (3)+\log ^2(3)}-x-x^2+\left (e-e^{\log ^2\left (\frac {3}{2}\right )-2 \log \left (\frac {3}{2}\right ) \log (3)+\log ^2(3)}+2 x\right ) \log (x)}{x^2-2 x \log (x)+\log ^2(x)} \, dx=\frac {x \left (e-e^{\log ^2(2)}+x\right )}{-x+\log (x)} \] Input:
Integrate[(-E + E^(Log[3/2]^2 - 2*Log[3/2]*Log[3] + Log[3]^2) - x - x^2 + (E - E^(Log[3/2]^2 - 2*Log[3/2]*Log[3] + Log[3]^2) + 2*x)*Log[x])/(x^2 - 2 *x*Log[x] + Log[x]^2),x]
Output:
(x*(E - E^Log[2]^2 + x))/(-x + Log[x])
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.
\(\displaystyle \int \frac {-x^2-x+\left (2 x+e-e^{\log ^2\left (\frac {3}{2}\right )+\log ^2(3)-2 \log \left (\frac {3}{2}\right ) \log (3)}\right ) \log (x)-e+e^{\log ^2\left (\frac {3}{2}\right )+\log ^2(3)-2 \log \left (\frac {3}{2}\right ) \log (3)}}{x^2+\log ^2(x)-2 x \log (x)} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {-x (x+1)+\left (2 x+e-e^{\log ^2(2)}\right ) \log (x)-e \left (1-e^{\log ^2(2)-1}\right )}{(x-\log (x))^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-2 x-e+e^{\log ^2(2)}}{x-\log (x)}-\frac {(x-1) \left (-x-e+e^{\log ^2(2)}\right )}{(x-\log (x))^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {x^2}{(x-\log (x))^2}dx-\left (\left (e-e^{\log ^2(2)}\right ) \int \frac {1}{(x-\log (x))^2}dx\right )-\left (1-e+e^{\log ^2(2)}\right ) \int \frac {x}{(x-\log (x))^2}dx-\left (e-e^{\log ^2(2)}\right ) \int \frac {1}{x-\log (x)}dx-2 \int \frac {x}{x-\log (x)}dx\) |
Input:
Int[(-E + E^(Log[3/2]^2 - 2*Log[3/2]*Log[3] + Log[3]^2) - x - x^2 + (E - E ^(Log[3/2]^2 - 2*Log[3/2]*Log[3] + Log[3]^2) + 2*x)*Log[x])/(x^2 - 2*x*Log [x] + Log[x]^2),x]
Output:
$Aborted
Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\frac {\left (-x +{\mathrm e}^{\ln \left (2\right )^{2}}-{\mathrm e}\right ) x}{x -\ln \left (x \right )}\) | \(24\) |
default | \(-\frac {\left ({\mathrm e}^{\ln \left (2\right )^{2}}-{\mathrm e}\right ) x -x^{2}}{\ln \left (x \right )-x}\) | \(29\) |
norman | \(\frac {\left ({\mathrm e}^{\ln \left (2\right )^{2}}-{\mathrm e}\right ) \ln \left (x \right )-x^{2}}{x -\ln \left (x \right )}\) | \(29\) |
parallelrisch | \(\frac {{\mathrm e}^{\ln \left (3\right )^{2}+2 \ln \left (\frac {2}{3}\right ) \ln \left (3\right )+\ln \left (\frac {2}{3}\right )^{2}} x -x \,{\mathrm e}-x^{2}}{x -\ln \left (x \right )}\) | \(39\) |
Input:
int(((-exp(ln(3)^2+2*ln(2/3)*ln(3)+ln(2/3)^2)+exp(1)+2*x)*ln(x)+exp(ln(3)^ 2+2*ln(2/3)*ln(3)+ln(2/3)^2)-exp(1)-x^2-x)/(ln(x)^2-2*x*ln(x)+x^2),x,metho d=_RETURNVERBOSE)
Output:
(-x+exp(ln(2)^2)-exp(1))*x/(x-ln(x))
Time = 0.10 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.32 \[ \int \frac {-e+e^{\log ^2\left (\frac {3}{2}\right )-2 \log \left (\frac {3}{2}\right ) \log (3)+\log ^2(3)}-x-x^2+\left (e-e^{\log ^2\left (\frac {3}{2}\right )-2 \log \left (\frac {3}{2}\right ) \log (3)+\log ^2(3)}+2 x\right ) \log (x)}{x^2-2 x \log (x)+\log ^2(x)} \, dx=-\frac {x^{2} + x e - x e^{\left (\log \left (3\right )^{2} + 2 \, \log \left (3\right ) \log \left (\frac {2}{3}\right ) + \log \left (\frac {2}{3}\right )^{2}\right )}}{x - \log \left (x\right )} \] Input:
integrate(((-exp(log(3)^2+2*log(2/3)*log(3)+log(2/3)^2)+exp(1)+2*x)*log(x) +exp(log(3)^2+2*log(2/3)*log(3)+log(2/3)^2)-exp(1)-x^2-x)/(log(x)^2-2*x*lo g(x)+x^2),x, algorithm="fricas")
Output:
-(x^2 + x*e - x*e^(log(3)^2 + 2*log(3)*log(2/3) + log(2/3)^2))/(x - log(x) )
Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {-e+e^{\log ^2\left (\frac {3}{2}\right )-2 \log \left (\frac {3}{2}\right ) \log (3)+\log ^2(3)}-x-x^2+\left (e-e^{\log ^2\left (\frac {3}{2}\right )-2 \log \left (\frac {3}{2}\right ) \log (3)+\log ^2(3)}+2 x\right ) \log (x)}{x^2-2 x \log (x)+\log ^2(x)} \, dx=\frac {x^{2} - x e^{\log {\left (2 \right )}^{2}} + e x}{- x + \log {\left (x \right )}} \] Input:
integrate(((-exp(ln(3)**2+2*ln(2/3)*ln(3)+ln(2/3)**2)+exp(1)+2*x)*ln(x)+ex p(ln(3)**2+2*ln(2/3)*ln(3)+ln(2/3)**2)-exp(1)-x**2-x)/(ln(x)**2-2*x*ln(x)+ x**2),x)
Output:
(x**2 - x*exp(log(2)**2) + E*x)/(-x + log(x))
Time = 0.14 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {-e+e^{\log ^2\left (\frac {3}{2}\right )-2 \log \left (\frac {3}{2}\right ) \log (3)+\log ^2(3)}-x-x^2+\left (e-e^{\log ^2\left (\frac {3}{2}\right )-2 \log \left (\frac {3}{2}\right ) \log (3)+\log ^2(3)}+2 x\right ) \log (x)}{x^2-2 x \log (x)+\log ^2(x)} \, dx=-\frac {x^{2} + x {\left (e - e^{\left (\log \left (2\right )^{2}\right )}\right )}}{x - \log \left (x\right )} \] Input:
integrate(((-exp(log(3)^2+2*log(2/3)*log(3)+log(2/3)^2)+exp(1)+2*x)*log(x) +exp(log(3)^2+2*log(2/3)*log(3)+log(2/3)^2)-exp(1)-x^2-x)/(log(x)^2-2*x*lo g(x)+x^2),x, algorithm="maxima")
Output:
-(x^2 + x*(e - e^(log(2)^2)))/(x - log(x))
Time = 0.12 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {-e+e^{\log ^2\left (\frac {3}{2}\right )-2 \log \left (\frac {3}{2}\right ) \log (3)+\log ^2(3)}-x-x^2+\left (e-e^{\log ^2\left (\frac {3}{2}\right )-2 \log \left (\frac {3}{2}\right ) \log (3)+\log ^2(3)}+2 x\right ) \log (x)}{x^2-2 x \log (x)+\log ^2(x)} \, dx=-\frac {x^{2} + x e - x e^{\left (\log \left (2\right )^{2}\right )}}{x - \log \left (x\right )} \] Input:
integrate(((-exp(log(3)^2+2*log(2/3)*log(3)+log(2/3)^2)+exp(1)+2*x)*log(x) +exp(log(3)^2+2*log(2/3)*log(3)+log(2/3)^2)-exp(1)-x^2-x)/(log(x)^2-2*x*lo g(x)+x^2),x, algorithm="giac")
Output:
-(x^2 + x*e - x*e^(log(2)^2))/(x - log(x))
Time = 4.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79 \[ \int \frac {-e+e^{\log ^2\left (\frac {3}{2}\right )-2 \log \left (\frac {3}{2}\right ) \log (3)+\log ^2(3)}-x-x^2+\left (e-e^{\log ^2\left (\frac {3}{2}\right )-2 \log \left (\frac {3}{2}\right ) \log (3)+\log ^2(3)}+2 x\right ) \log (x)}{x^2-2 x \log (x)+\log ^2(x)} \, dx=-\frac {x\,\left (x-{\mathrm {e}}^{{\ln \left (2\right )}^2}+\mathrm {e}\right )}{x-\ln \left (x\right )} \] Input:
int(-(x + exp(1) - exp(2*log(3)*log(2/3) + log(3)^2 + log(2/3)^2) - log(x) *(2*x + exp(1) - exp(2*log(3)*log(2/3) + log(3)^2 + log(2/3)^2)) + x^2)/(l og(x)^2 - 2*x*log(x) + x^2),x)
Output:
-(x*(x - exp(log(2)^2) + exp(1)))/(x - log(x))
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.11 \[ \int \frac {-e+e^{\log ^2\left (\frac {3}{2}\right )-2 \log \left (\frac {3}{2}\right ) \log (3)+\log ^2(3)}-x-x^2+\left (e-e^{\log ^2\left (\frac {3}{2}\right )-2 \log \left (\frac {3}{2}\right ) \log (3)+\log ^2(3)}+2 x\right ) \log (x)}{x^2-2 x \log (x)+\log ^2(x)} \, dx=\frac {-e^{\mathrm {log}\left (\frac {2}{3}\right )^{2}+\mathrm {log}\left (3\right )^{2}} 2^{2 \,\mathrm {log}\left (3\right )} \mathrm {log}\left (x \right )+3^{2 \,\mathrm {log}\left (3\right )} \mathrm {log}\left (x \right ) e +3^{2 \,\mathrm {log}\left (3\right )} x^{2}}{3^{2 \,\mathrm {log}\left (3\right )} \left (\mathrm {log}\left (x \right )-x \right )} \] Input:
int(((-exp(log(3)^2+2*log(2/3)*log(3)+log(2/3)^2)+exp(1)+2*x)*log(x)+exp(l og(3)^2+2*log(2/3)*log(3)+log(2/3)^2)-exp(1)-x^2-x)/(log(x)^2-2*x*log(x)+x ^2),x)
Output:
( - e**(log(2/3)**2 + log(3)**2)*2**(2*log(3))*log(x) + 3**(2*log(3))*log( x)*e + 3**(2*log(3))*x**2)/(3**(2*log(3))*(log(x) - x))