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

Optimal. Leaf size=29 \[ e^{\frac {e^{e^x}-e^x-2 x}{x}-x+x \log (\log (x))} \]

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Rubi [A]  time = 2.50, antiderivative size = 30, normalized size of antiderivative = 1.03, number of steps used = 1, number of rules used = 1, integrand size = 86, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {6706} \begin {gather*} e^{\frac {-x^2-2 x+e^{e^x}-e^x}{x}} \log ^x(x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]  time = 3.70, size = 29, normalized size = 1.00 \begin {gather*} e^{-2+\frac {e^{e^x}}{x}-\frac {e^x}{x}-x} \log ^x(x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-2 + E^E^x/x - E^x/x - x)*Log[x]^x

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fricas [A]  time = 0.69, size = 28, normalized size = 0.97 \begin {gather*} e^{\left (\frac {x^{2} \log \left (\log \relax (x)\right ) - x^{2} - 2 \, x - e^{x} + e^{\left (e^{x}\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (x^{2} \log \relax (x) \log \left (\log \relax (x)\right ) + {\left (x e^{x} - 1\right )} e^{\left (e^{x}\right )} \log \relax (x) + x^{2} - {\left (x^{2} + {\left (x - 1\right )} e^{x}\right )} \log \relax (x)\right )} e^{\left (\frac {x^{2} \log \left (\log \relax (x)\right ) - x^{2} - 2 \, x - e^{x} + e^{\left (e^{x}\right )}}{x}\right )}}{x^{2} \log \relax (x)}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 26, normalized size = 0.90

method	result	size
risch	\(\ln \relax (x )^{x} {\mathrm e}^{-\frac {x^{2}+{\mathrm e}^{x}-{\mathrm e}^{{\mathrm e}^{x}}+2 x}{x}}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.65, size = 25, normalized size = 0.86 \begin {gather*} e^{\left (x \log \left (\log \relax (x)\right ) - x - \frac {e^{x}}{x} + \frac {e^{\left (e^{x}\right )}}{x} - 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.41, size = 27, normalized size = 0.93 \begin {gather*} {\mathrm {e}}^{-x}\,{\mathrm {e}}^{-2}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^x}{x}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{{\mathrm {e}}^x}}{x}}\,{\ln \relax (x)}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 10.64, size = 26, normalized size = 0.90 \begin {gather*} e^{\frac {x^{2} \log {\left (\log {\relax (x )} \right )} - x^{2} - 2 x - e^{x} + e^{e^{x}}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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