Integrand size = 74, antiderivative size = 27 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=\log \left (-\frac {2}{4+x+\frac {2-e^{-1+x}-e^x}{\log (x)}}\right ) \] Output:
ln(-2/((-exp(x)-exp(-1+x)+2)/ln(x)+4+x))
Time = 2.49 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=\log (\log (x))-\log \left (-2 e+e^x+e^{1+x}-4 e \log (x)-e x \log (x)\right ) \] Input:
Integrate[(2 - E^(-1 + x) - E^x + (E^(-1 + x)*x + E^x*x)*Log[x] - x*Log[x] ^2)/((2*x - E^(-1 + x)*x - E^x*x)*Log[x] + (4*x + x^2)*Log[x]^2),x]
Output:
Log[Log[x]] - Log[-2*E + E^x + E^(1 + x) - 4*E*Log[x] - E*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.
\(\displaystyle \int \frac {-e^{x-1}-e^x-x \log ^2(x)+\left (e^{x-1} x+e^x x\right ) \log (x)+2}{\left (x^2+4 x\right ) \log ^2(x)+\left (-e^{x-1} x-e^x x+2 x\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e \left (-(1+e) e^{x-1}-x \log ^2(x)+\left (e^{x-1} x+e^x x\right ) \log (x)+2\right )}{x \log (x) \left (-(1+e) e^x+e x \log (x)+4 e \log (x)+2 e\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e \int \frac {-x \log ^2(x)+\left (e^{x-1} x+e^x x\right ) \log (x)-e^{x-1} (1+e)+2}{x \log (x) \left (e x \log (x)+4 e \log (x)-e^x (1+e)+2 e\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle e \int \left (\frac {1-x \log (x)}{e x \log (x)}+\frac {\log (x) x^2+3 \log (x) x+x-4}{x \left (e x \log (x)+4 e \log (x)-e^x (1+e)+2 e\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle e \left (4 \int \frac {1}{x \left (-e x \log (x)-4 e \log (x)+e^x (1+e)-2 e\right )}dx+\int \frac {1}{e x \log (x)+4 e \log (x)-e^x (1+e)+2 e}dx+3 \int \frac {\log (x)}{e x \log (x)+4 e \log (x)-e^x (1+e)+2 e}dx+\int \frac {x \log (x)}{e x \log (x)+4 e \log (x)-e^x (1+e)+2 e}dx-\frac {x}{e}+\frac {\log (\log (x))}{e}\right )\) |
Input:
Int[(2 - E^(-1 + x) - E^x + (E^(-1 + x)*x + E^x*x)*Log[x] - x*Log[x]^2)/(( 2*x - E^(-1 + x)*x - E^x*x)*Log[x] + (4*x + x^2)*Log[x]^2),x]
Output:
$Aborted
Time = 0.42 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04
method | result | size |
parallelrisch | \(\ln \left (\ln \left (x \right )\right )-\ln \left (x \ln \left (x \right )+4 \ln \left (x \right )-{\mathrm e}^{x}-{\mathrm e}^{-1+x}+2\right )\) | \(28\) |
risch | \(-\ln \left (4+x \right )+\ln \left (\ln \left (x \right )\right )-\ln \left (\ln \left (x \right )-\frac {{\mathrm e}^{-1+x}+{\mathrm e}^{x}-2}{4+x}\right )\) | \(32\) |
norman | \(-\ln \left (x \,{\mathrm e} \ln \left (x \right )-{\mathrm e} \,{\mathrm e}^{x}+4 \,{\mathrm e} \ln \left (x \right )+2 \,{\mathrm e}-{\mathrm e}^{x}\right )+\ln \left (\ln \left (x \right )\right )\) | \(35\) |
Input:
int((-x*ln(x)^2+(exp(x)*x+x*exp(-1+x))*ln(x)-exp(x)-exp(-1+x)+2)/((x^2+4*x )*ln(x)^2+(-exp(x)*x-x*exp(-1+x)+2*x)*ln(x)),x,method=_RETURNVERBOSE)
Output:
ln(ln(x))-ln(x*ln(x)+4*ln(x)-exp(x)-exp(-1+x)+2)
Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=-\log \left (x + 4\right ) - \log \left (\frac {{\left (x + 4\right )} e \log \left (x\right ) - {\left (e + 1\right )} e^{x} + 2 \, e}{x + 4}\right ) + \log \left (\log \left (x\right )\right ) \] Input:
integrate((-x*log(x)^2+(exp(x)*x+x*exp(-1+x))*log(x)-exp(x)-exp(-1+x)+2)/( (x^2+4*x)*log(x)^2+(-exp(x)*x-x*exp(-1+x)+2*x)*log(x)),x, algorithm="frica s")
Output:
-log(x + 4) - log(((x + 4)*e*log(x) - (e + 1)*e^x + 2*e)/(x + 4)) + log(lo g(x))
Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=- \log {\left (\frac {- e x \log {\left (x \right )} - 4 e \log {\left (x \right )} - 2 e}{1 + e} + e^{x} \right )} + \log {\left (\log {\left (x \right )} \right )} \] Input:
integrate((-x*ln(x)**2+(exp(x)*x+x*exp(-1+x))*ln(x)-exp(x)-exp(-1+x)+2)/(( x**2+4*x)*ln(x)**2+(-exp(x)*x-x*exp(-1+x)+2*x)*ln(x)),x)
Output:
-log((-E*x*log(x) - 4*E*log(x) - 2*E)/(1 + E) + exp(x)) + log(log(x))
Time = 0.08 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=-\log \left (\frac {{\left (e + 1\right )} e^{x} - {\left (x e + 4 \, e\right )} \log \left (x\right ) - 2 \, e}{e + 1}\right ) + \log \left (\log \left (x\right )\right ) \] Input:
integrate((-x*log(x)^2+(exp(x)*x+x*exp(-1+x))*log(x)-exp(x)-exp(-1+x)+2)/( (x^2+4*x)*log(x)^2+(-exp(x)*x-x*exp(-1+x)+2*x)*log(x)),x, algorithm="maxim a")
Output:
-log(((e + 1)*e^x - (x*e + 4*e)*log(x) - 2*e)/(e + 1)) + log(log(x))
Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=-\log \left (x e \log \left (x\right ) + 4 \, e \log \left (x\right ) + 2 \, e - e^{\left (x + 1\right )} - e^{x}\right ) + \log \left (\log \left (x\right )\right ) \] Input:
integrate((-x*log(x)^2+(exp(x)*x+x*exp(-1+x))*log(x)-exp(x)-exp(-1+x)+2)/( (x^2+4*x)*log(x)^2+(-exp(x)*x-x*exp(-1+x)+2*x)*log(x)),x, algorithm="giac" )
Output:
-log(x*e*log(x) + 4*e*log(x) + 2*e - e^(x + 1) - e^x) + log(log(x))
Timed out. \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=-\int -\frac {x\,{\ln \left (x\right )}^2+\left (-x\,{\mathrm {e}}^{x-1}-x\,{\mathrm {e}}^x\right )\,\ln \left (x\right )+{\mathrm {e}}^{x-1}+{\mathrm {e}}^x-2}{\ln \left (x\right )\,\left (x\,{\mathrm {e}}^{x-1}-2\,x+x\,{\mathrm {e}}^x\right )-{\ln \left (x\right )}^2\,\left (x^2+4\,x\right )} \,d x \] Input:
int((exp(x - 1) + exp(x) + x*log(x)^2 - log(x)*(x*exp(x - 1) + x*exp(x)) - 2)/(log(x)*(x*exp(x - 1) - 2*x + x*exp(x)) - log(x)^2*(4*x + x^2)),x)
Output:
-int(-(exp(x - 1) + exp(x) + x*log(x)^2 - log(x)*(x*exp(x - 1) + x*exp(x)) - 2)/(log(x)*(x*exp(x - 1) - 2*x + x*exp(x)) - log(x)^2*(4*x + x^2)), x)
Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=\mathrm {log}\left (\mathrm {log}\left (x \right )\right )-\mathrm {log}\left (e^{x} e +e^{x}-\mathrm {log}\left (x \right ) e x -4 \,\mathrm {log}\left (x \right ) e -2 e \right ) \] Input:
int((-x*log(x)^2+(exp(x)*x+x*exp(-1+x))*log(x)-exp(x)-exp(-1+x)+2)/((x^2+4 *x)*log(x)^2+(-exp(x)*x-x*exp(-1+x)+2*x)*log(x)),x)
Output:
log(log(x)) - log(e**x*e + e**x - log(x)*e*x - 4*log(x)*e - 2*e)