\(\int \frac {e (-1+x)+e^x (-1-x^2)+e^x x \log (x)}{2 e^2 x+2 e^{2 x} x+4 e^{1+x} x} \, dx\) [5061]

Optimal result

Integrand size = 50, antiderivative size = 18 \[ \int \frac {e (-1+x)+e^x \left (-1-x^2\right )+e^x x \log (x)}{2 e^2 x+2 e^{2 x} x+4 e^{1+x} x} \, dx=\frac {1+x-\log (x)}{2 \left (e+e^x\right )} \]

[Out]

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

Rubi [F]

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

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e (-1+x)-e^x \left (1+x^2\right )+e^x x \log (x)}{2 \left (e+e^x\right )^2 x} \, dx \\ & = \frac {1}{2} \int \frac {e (-1+x)-e^x \left (1+x^2\right )+e^x x \log (x)}{\left (e+e^x\right )^2 x} \, dx \\ & = \frac {1}{2} \int \left (\frac {e (1+x-\log (x))}{\left (e+e^x\right )^2}-\frac {1+x^2-x \log (x)}{\left (e+e^x\right ) x}\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {1+x^2-x \log (x)}{\left (e+e^x\right ) x} \, dx\right )+\frac {1}{2} e \int \frac {1+x-\log (x)}{\left (e+e^x\right )^2} \, dx \\ & = -\left (\frac {1}{2} \int \left (\frac {1}{\left (e+e^x\right ) x}+\frac {x}{e+e^x}-\frac {\log (x)}{e+e^x}\right ) \, dx\right )+\frac {1}{2} e \int \left (\frac {1}{\left (e+e^x\right )^2}+\frac {x}{\left (e+e^x\right )^2}-\frac {\log (x)}{\left (e+e^x\right )^2}\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {1}{\left (e+e^x\right ) x} \, dx\right )-\frac {1}{2} \int \frac {x}{e+e^x} \, dx+\frac {1}{2} \int \frac {\log (x)}{e+e^x} \, dx+\frac {1}{2} e \int \frac {1}{\left (e+e^x\right )^2} \, dx+\frac {1}{2} e \int \frac {x}{\left (e+e^x\right )^2} \, dx-\frac {1}{2} e \int \frac {\log (x)}{\left (e+e^x\right )^2} \, dx \\ & = -\frac {x^2}{4 e}-\frac {1}{2} \int \frac {1}{\left (e+e^x\right ) x} \, dx-\frac {1}{2} \int \frac {e^x x}{\left (e+e^x\right )^2} \, dx+\frac {1}{2} \int \frac {x}{e+e^x} \, dx+\frac {1}{2} \int \frac {\log (x)}{e+e^x} \, dx+\frac {\int \frac {e^x x}{e+e^x} \, dx}{2 e}-\frac {1}{2} e \int \frac {\log (x)}{\left (e+e^x\right )^2} \, dx+\frac {1}{2} e \text {Subst}\left (\int \frac {1}{x (e+x)^2} \, dx,x,e^x\right ) \\ & = \frac {x}{2 \left (e+e^x\right )}+\frac {x \log \left (1+e^{-1+x}\right )}{2 e}-\frac {1}{2} \int \frac {1}{e+e^x} \, dx-\frac {1}{2} \int \frac {1}{\left (e+e^x\right ) x} \, dx+\frac {1}{2} \int \frac {\log (x)}{e+e^x} \, dx-\frac {\int \frac {e^x x}{e+e^x} \, dx}{2 e}-\frac {\int \log \left (1+e^{-1+x}\right ) \, dx}{2 e}-\frac {1}{2} e \int \frac {\log (x)}{\left (e+e^x\right )^2} \, dx+\frac {1}{2} e \text {Subst}\left (\int \left (\frac {1}{e^2 x}-\frac {1}{e (e+x)^2}-\frac {1}{e^2 (e+x)}\right ) \, dx,x,e^x\right ) \\ & = \frac {1}{2 \left (e+e^x\right )}+\frac {x}{2 e}+\frac {x}{2 \left (e+e^x\right )}-\frac {\log \left (e+e^x\right )}{2 e}-\frac {1}{2} \int \frac {1}{\left (e+e^x\right ) x} \, dx+\frac {1}{2} \int \frac {\log (x)}{e+e^x} \, dx-\frac {1}{2} \text {Subst}\left (\int \frac {1}{x (e+x)} \, dx,x,e^x\right )+\frac {\int \log \left (1+e^{-1+x}\right ) \, dx}{2 e}-\frac {\text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{-1+x}\right )}{2 e}-\frac {1}{2} e \int \frac {\log (x)}{\left (e+e^x\right )^2} \, dx \\ & = \frac {1}{2 \left (e+e^x\right )}+\frac {x}{2 e}+\frac {x}{2 \left (e+e^x\right )}-\frac {\log \left (e+e^x\right )}{2 e}+\frac {\text {Li}_2\left (-e^{-1+x}\right )}{2 e}-\frac {1}{2} \int \frac {1}{\left (e+e^x\right ) x} \, dx+\frac {1}{2} \int \frac {\log (x)}{e+e^x} \, dx-\frac {\text {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )}{2 e}+\frac {\text {Subst}\left (\int \frac {1}{e+x} \, dx,x,e^x\right )}{2 e}+\frac {\text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{-1+x}\right )}{2 e}-\frac {1}{2} e \int \frac {\log (x)}{\left (e+e^x\right )^2} \, dx \\ & = \frac {1}{2 \left (e+e^x\right )}+\frac {x}{2 \left (e+e^x\right )}-\frac {1}{2} \int \frac {1}{\left (e+e^x\right ) x} \, dx+\frac {1}{2} \int \frac {\log (x)}{e+e^x} \, dx-\frac {1}{2} e \int \frac {\log (x)}{\left (e+e^x\right )^2} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {e (-1+x)+e^x \left (-1-x^2\right )+e^x x \log (x)}{2 e^2 x+2 e^{2 x} x+4 e^{1+x} x} \, dx=\frac {1+x-\log (x)}{2 \left (e+e^x\right )} \]

[In]

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

[Out]

(1 + x - Log[x])/(2*(E + E^x))

Maple [A] (verified)

Time = 1.71 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

method	result	size
parallelrisch	\(\frac {1-\ln \left (x \right )+x}{2 \,{\mathrm e}+2 \,{\mathrm e}^{x}}\)	\(17\)
norman	\(\frac {-\frac {\ln \left (x \right )}{2}+\frac {x}{2}+\frac {1}{2}}{{\mathrm e}+{\mathrm e}^{x}}\)	\(18\)
risch	\(-\frac {\ln \left (x \right )}{2 \left ({\mathrm e}+{\mathrm e}^{x}\right )}+\frac {1+x}{2 \,{\mathrm e}+2 \,{\mathrm e}^{x}}\)	\(25\)

[In]

int((x*exp(x)*ln(x)+(-x^2-1)*exp(x)+(-1+x)*exp(1))/(2*x*exp(x)^2+4*x*exp(1)*exp(x)+2*x*exp(1)^2),x,method=_RET
URNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int \frac {e (-1+x)+e^x \left (-1-x^2\right )+e^x x \log (x)}{2 e^2 x+2 e^{2 x} x+4 e^{1+x} x} \, dx=\frac {{\left (x + 1\right )} e - e \log \left (x\right )}{2 \, {\left (e^{2} + e^{\left (x + 1\right )}\right )}} \]

[In]

integrate((x*exp(x)*log(x)+(-x^2-1)*exp(x)+(-1+x)*exp(1))/(2*x*exp(x)^2+4*x*exp(1)*exp(x)+2*x*exp(1)^2),x, alg
orithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {e (-1+x)+e^x \left (-1-x^2\right )+e^x x \log (x)}{2 e^2 x+2 e^{2 x} x+4 e^{1+x} x} \, dx=\frac {x - \log {\left (x \right )} + 1}{2 e^{x} + 2 e} \]

[In]

integrate((x*exp(x)*ln(x)+(-x**2-1)*exp(x)+(-1+x)*exp(1))/(2*x*exp(x)**2+4*x*exp(1)*exp(x)+2*x*exp(1)**2),x)

[Out]

(x - log(x) + 1)/(2*exp(x) + 2*E)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {e (-1+x)+e^x \left (-1-x^2\right )+e^x x \log (x)}{2 e^2 x+2 e^{2 x} x+4 e^{1+x} x} \, dx=\frac {x - \log \left (x\right ) + 1}{2 \, {\left (e + e^{x}\right )}} \]

[In]

integrate((x*exp(x)*log(x)+(-x^2-1)*exp(x)+(-1+x)*exp(1))/(2*x*exp(x)^2+4*x*exp(1)*exp(x)+2*x*exp(1)^2),x, alg
orithm="maxima")

[Out]

1/2*(x - log(x) + 1)/(e + e^x)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {e (-1+x)+e^x \left (-1-x^2\right )+e^x x \log (x)}{2 e^2 x+2 e^{2 x} x+4 e^{1+x} x} \, dx=\frac {x - \log \left (x\right ) + 1}{2 \, {\left (e + e^{x}\right )}} \]

[In]

integrate((x*exp(x)*log(x)+(-x^2-1)*exp(x)+(-1+x)*exp(1))/(2*x*exp(x)^2+4*x*exp(1)*exp(x)+2*x*exp(1)^2),x, alg
orithm="giac")

[Out]

1/2*(x - log(x) + 1)/(e + e^x)

Mupad [B] (verification not implemented)

Time = 12.35 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06 \[ \int \frac {e (-1+x)+e^x \left (-1-x^2\right )+e^x x \log (x)}{2 e^2 x+2 e^{2 x} x+4 e^{1+x} x} \, dx=\frac {x-\ln \left (x\right )+1}{2\,\left (\mathrm {e}+{\mathrm {e}}^x\right )} \]

[In]

int((exp(1)*(x - 1) - exp(x)*(x^2 + 1) + x*exp(x)*log(x))/(2*x*exp(2*x) + 2*x*exp(2) + 4*x*exp(1)*exp(x)),x)

[Out]

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