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

Optimal. Leaf size=20 \[ 3+\frac {e^{-e^x-2 \log ^2(x)}}{x} \]

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Rubi [B]  time = 0.15, antiderivative size = 41, normalized size of antiderivative = 2.05, number of steps used = 1, number of rules used = 1, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {2288} \begin {gather*} \frac {e^{-e^x-2 \log ^2(x)} \left (e^x x+4 \log (x)\right )}{x^2 \left (e^x+\frac {4 \log (x)}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {e^{-e^x-2 \log ^2(x)} \left (e^x x+4 \log (x)\right )}{x^2 \left (e^x+\frac {4 \log (x)}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 18, normalized size = 0.90 \begin {gather*} \frac {e^{-e^x-2 \log ^2(x)}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-E^x - 2*Log[x]^2)/x

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fricas [A]  time = 0.80, size = 16, normalized size = 0.80 \begin {gather*} \frac {e^{\left (-2 \, \log \relax (x)^{2} - e^{x}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-2*log(x)^2 - e^x)/x

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giac [A]  time = 0.14, size = 16, normalized size = 0.80 \begin {gather*} \frac {e^{\left (-2 \, \log \relax (x)^{2} - e^{x}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-2*log(x)^2 - e^x)/x

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maple [A]  time = 0.05, size = 17, normalized size = 0.85

method	result	size
risch	\(\frac {{\mathrm e}^{-{\mathrm e}^{x}-2 \ln \relax (x )^{2}}}{x}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 16, normalized size = 0.80 \begin {gather*} \frac {e^{\left (-2 \, \log \relax (x)^{2} - e^{x}\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(-2*log(x)^2 - e^x)/x

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mupad [B]  time = 5.81, size = 16, normalized size = 0.80 \begin {gather*} \frac {{\mathrm {e}}^{-2\,{\ln \relax (x)}^2}\,{\mathrm {e}}^{-{\mathrm {e}}^x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(exp(-2*log(x)^2)*exp(-exp(x)))/x

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sympy [A]  time = 0.40, size = 15, normalized size = 0.75 \begin {gather*} \frac {e^{- e^{x}} e^{- 2 \log {\relax (x )}^{2}}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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