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

Optimal. Leaf size=18 \[ \frac {1+x-\log (x)}{2 \left (e+e^x\right )} \]

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Rubi [F]  time = 1.04, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[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 {gather*} \begin {aligned} \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 \operatorname {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 \operatorname {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} \operatorname {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 {\operatorname {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 {\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )}{2 e}+\frac {\operatorname {Subst}\left (\int \frac {1}{e+x} \, dx,x,e^x\right )}{2 e}+\frac {\operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 18, normalized size = 1.00 \begin {gather*} \frac {1+x-\log (x)}{2 \left (e+e^x\right )} \end {gather*}

Antiderivative was successfully verified.

[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))

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fricas [A]  time = 0.58, size = 24, normalized size = 1.33 \begin {gather*} \frac {{\left (x + 1\right )} e - e \log \relax (x)}{2 \, {\left (e^{2} + e^{\left (x + 1\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 16, normalized size = 0.89 \begin {gather*} \frac {x - \log \relax (x) + 1}{2 \, {\left (e + e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x - log(x) + 1)/(e + e^x)

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maple [A]  time = 0.24, size = 18, normalized size = 1.00

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.38, size = 16, normalized size = 0.89 \begin {gather*} \frac {x - \log \relax (x) + 1}{2 \, {\left (e + e^{x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(x - log(x) + 1)/(e + e^x)

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mupad [B]  time = 3.65, size = 19, normalized size = 1.06 \begin {gather*} \frac {x-\ln \relax (x)+1}{2\,\left (\mathrm {e}+{\mathrm {e}}^x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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sympy [A]  time = 0.23, size = 15, normalized size = 0.83 \begin {gather*} \frac {x - \log {\relax (x )} + 1}{2 e^{x} + 2 e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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