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

Optimal. Leaf size=21 \[ 1-\frac {1}{e \log \left (e^{-x} x^{\frac {1}{x}}\right )} \]

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Rubi [F]  time = 0.25, 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 {1-x^2-\log (x)}{e x^2 \log ^2\left (e^{-x} x^{\frac {1}{x}}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {1-x^2-\log (x)}{x^2 \log ^2\left (e^{-x} x^{\frac {1}{x}}\right )} \, dx}{e}\\ &=\frac {\int \left (-\frac {1}{\log ^2\left (e^{-x} x^{\frac {1}{x}}\right )}+\frac {1}{x^2 \log ^2\left (e^{-x} x^{\frac {1}{x}}\right )}-\frac {\log (x)}{x^2 \log ^2\left (e^{-x} x^{\frac {1}{x}}\right )}\right ) \, dx}{e}\\ &=-\frac {\int \frac {1}{\log ^2\left (e^{-x} x^{\frac {1}{x}}\right )} \, dx}{e}+\frac {\int \frac {1}{x^2 \log ^2\left (e^{-x} x^{\frac {1}{x}}\right )} \, dx}{e}-\frac {\int \frac {\log (x)}{x^2 \log ^2\left (e^{-x} x^{\frac {1}{x}}\right )} \, dx}{e}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 19, normalized size = 0.90 \begin {gather*} -\frac {1}{e \log \left (e^{-x} x^{\frac {1}{x}}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(1/(E*Log[x^x^(-1)/E^x]))

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fricas [A]  time = 0.90, size = 17, normalized size = 0.81 \begin {gather*} \frac {x}{x^{2} e - e \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)-x^2+1)/x^2/exp(1)/log(exp(log(x)/x)/exp(x))^2,x, algorithm="fricas")

[Out]

x/(x^2*e - e*log(x))

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giac [A]  time = 0.20, size = 14, normalized size = 0.67 \begin {gather*} \frac {x e^{\left (-1\right )}}{x^{2} - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)-x^2+1)/x^2/exp(1)/log(exp(log(x)/x)/exp(x))^2,x, algorithm="giac")

[Out]

x*e^(-1)/(x^2 - log(x))

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maple [B]  time = 0.32, size = 53, normalized size = 2.52

method	result	size
default	\(\frac {{\mathrm e}^{-1} x}{x^{2}+\left (\ln \left ({\mathrm e}^{x}\right )-x \right ) x -\left (\ln \left ({\mathrm e}^{\frac {\ln \relax (x )}{x}} {\mathrm e}^{-x}\right )-\frac {\ln \relax (x )}{x}+\ln \left ({\mathrm e}^{x}\right )\right ) x -\ln \relax (x )}\)	\(53\)
risch	\(\frac {2 i {\mathrm e}^{-1}}{\pi \,\mathrm {csgn}\left (i x^{\frac {1}{x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} x^{\frac {1}{x}}\right )^{2}-\pi \,\mathrm {csgn}\left (i x^{\frac {1}{x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x} x^{\frac {1}{x}}\right ) \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )-\pi \mathrm {csgn}\left (i {\mathrm e}^{-x} x^{\frac {1}{x}}\right )^{3}+\pi \mathrm {csgn}\left (i {\mathrm e}^{-x} x^{\frac {1}{x}}\right )^{2} \mathrm {csgn}\left (i {\mathrm e}^{-x}\right )+2 i \ln \left ({\mathrm e}^{x}\right )-2 i \ln \left (x^{\frac {1}{x}}\right )}\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(x)-x^2+1)/x^2/exp(1)/ln(exp(ln(x)/x)/exp(x))^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.45, size = 15, normalized size = 0.71 \begin {gather*} \frac {e^{\left (-1\right )}}{x - \log \left (x^{\left (\frac {1}{x}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)-x^2+1)/x^2/exp(1)/log(exp(log(x)/x)/exp(x))^2,x, algorithm="maxima")

[Out]

e^(-1)/(x - log(x^(1/x)))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {{\mathrm {e}}^{-1}\,\left (\ln \relax (x)+x^2-1\right )}{x^2\,{\ln \left ({\mathrm {e}}^{-x}\,{\mathrm {e}}^{\frac {\ln \relax (x)}{x}}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.38, size = 15, normalized size = 0.71 \begin {gather*} - \frac {x}{- e x^{2} + e \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x/(-E*x**2 + E*log(x))

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