\(\int \frac {2-e^{-1+x}-e^x+(e^{-1+x} x+e^x x) \log (x)-x \log ^2(x)}{(2 x-e^{-1+x} x-e^x x) \log (x)+(4 x+x^2) \log ^2(x)} \, dx\) [9477]

Optimal result

Integrand size = 74, antiderivative size = 27 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=\log \left (-\frac {2}{4+x+\frac {2-e^{-1+x}-e^x}{\log (x)}}\right ) \]

[Out]

ln(-2/((-exp(x)-exp(-1+x)+2)/ln(x)+4+x))

Rubi [F]

\[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=\int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx \]

[In]

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]

[Out]

-x + Log[Log[x]] + 4*E*Defer[Int][1/(x*(-2*E + E^x*(1 + E) - 4*E*Log[x] - E*x*Log[x])), x] + E*Defer[Int][(2*E
- E^x*(1 + E) + 4*E*Log[x] + E*x*Log[x])^(-1), x] + 3*E*Defer[Int][Log[x]/(2*E - E^x*(1 + E) + 4*E*Log[x] + E
*x*Log[x]), x] + E*Defer[Int][(x*Log[x])/(2*E - E^x*(1 + E) + 4*E*Log[x] + E*x*Log[x]), x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {e \left (2-e^{-1+x} (1+e)+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)\right )}{x \log (x) \left (2 e-e^x (1+e)+4 e \log (x)+e x \log (x)\right )} \, dx \\ & = e \int \frac {2-e^{-1+x} (1+e)+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{x \log (x) \left (2 e-e^x (1+e)+4 e \log (x)+e x \log (x)\right )} \, dx \\ & = e \int \left (\frac {1-x \log (x)}{e x \log (x)}+\frac {-4+x+3 x \log (x)+x^2 \log (x)}{x \left (2 e-e^x (1+e)+4 e \log (x)+e x \log (x)\right )}\right ) \, dx \\ & = e \int \frac {-4+x+3 x \log (x)+x^2 \log (x)}{x \left (2 e-e^x (1+e)+4 e \log (x)+e x \log (x)\right )} \, dx+\int \frac {1-x \log (x)}{x \log (x)} \, dx \\ & = e \int \left (\frac {4}{x \left (-2 e+e^x (1+e)-4 e \log (x)-e x \log (x)\right )}+\frac {1}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)}+\frac {3 \log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)}+\frac {x \log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)}\right ) \, dx+\int \left (-1+\frac {1}{x \log (x)}\right ) \, dx \\ & = -x+e \int \frac {1}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+e \int \frac {x \log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(3 e) \int \frac {\log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(4 e) \int \frac {1}{x \left (-2 e+e^x (1+e)-4 e \log (x)-e x \log (x)\right )} \, dx+\int \frac {1}{x \log (x)} \, dx \\ & = -x+e \int \frac {1}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+e \int \frac {x \log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(3 e) \int \frac {\log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(4 e) \int \frac {1}{x \left (-2 e+e^x (1+e)-4 e \log (x)-e x \log (x)\right )} \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = -x+\log (\log (x))+e \int \frac {1}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+e \int \frac {x \log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(3 e) \int \frac {\log (x)}{2 e-e^x (1+e)+4 e \log (x)+e x \log (x)} \, dx+(4 e) \int \frac {1}{x \left (-2 e+e^x (1+e)-4 e \log (x)-e x \log (x)\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 3.82 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=\log (\log (x))-\log \left (-2 e+e^x+e^{1+x}-4 e \log (x)-e x \log (x)\right ) \]

[In]

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)*Lo
g[x] + (4*x + x^2)*Log[x]^2),x]

[Out]

Log[Log[x]] - Log[-2*E + E^x + E^(1 + x) - 4*E*Log[x] - E*x*Log[x]]

Maple [A] (verified)

Time = 1.01 (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	\(-\ln \left (x \,{\mathrm e} \ln \left (x \right )-{\mathrm e} \,{\mathrm e}^{x}+4 \,{\mathrm e} \ln \left (x \right )+2 \,{\mathrm e}-{\mathrm e}^{x}\right )+\ln \left (\ln \left (x \right )\right )\)	\(35\)

[In]

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)

[Out]

ln(ln(x))-ln(x*ln(x)+4*ln(x)-exp(x)-exp(-1+x)+2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.48 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=-\log \left (x + 4\right ) - \log \left (\frac {{\left (x + 4\right )} e \log \left (x\right ) - {\left (e + 1\right )} e^{x} + 2 \, e}{x + 4}\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

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="fricas")

[Out]

-log(x + 4) - log(((x + 4)*e*log(x) - (e + 1)*e^x + 2*e)/(x + 4)) + log(log(x))

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=- \log {\left (\frac {- e x \log {\left (x \right )} - 4 e \log {\left (x \right )} - 2 e}{1 + e} + e^{x} \right )} + \log {\left (\log {\left (x \right )} \right )} \]

[In]

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)

[Out]

-log((-E*x*log(x) - 4*E*log(x) - 2*E)/(1 + E) + exp(x)) + log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.44 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=-\log \left (\frac {{\left (e + 1\right )} e^{x} - {\left (x e + 4 \, e\right )} \log \left (x\right ) - 2 \, e}{e + 1}\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

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="maxima")

[Out]

-log(((e + 1)*e^x - (x*e + 4*e)*log(x) - 2*e)/(e + 1)) + log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=-\log \left (x e \log \left (x\right ) + 4 \, e \log \left (x\right ) + 2 \, e - e^{\left (x + 1\right )} - e^{x}\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

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")

[Out]

-log(x*e*log(x) + 4*e*log(x) + 2*e - e^(x + 1) - e^x) + log(log(x))

Mupad [F(-1)]

Timed out. \[ \int \frac {2-e^{-1+x}-e^x+\left (e^{-1+x} x+e^x x\right ) \log (x)-x \log ^2(x)}{\left (2 x-e^{-1+x} x-e^x x\right ) \log (x)+\left (4 x+x^2\right ) \log ^2(x)} \, dx=-\int -\frac {x\,{\ln \left (x\right )}^2+\left (-x\,{\mathrm {e}}^{x-1}-x\,{\mathrm {e}}^x\right )\,\ln \left (x\right )+{\mathrm {e}}^{x-1}+{\mathrm {e}}^x-2}{\ln \left (x\right )\,\left (x\,{\mathrm {e}}^{x-1}-2\,x+x\,{\mathrm {e}}^x\right )-{\ln \left (x\right )}^2\,\left (x^2+4\,x\right )} \,d x \]

[In]

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)

[Out]

-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)