\(\int \frac {e^{\frac {e}{\log (e^{3/x}+x)}} (-3 e^{1+\frac {3}{x}}+e x^2)}{(e^{3/x} x^2+x^3) \log ^2(e^{3/x}+x)} \, dx\) [6636]

Optimal result

Integrand size = 63, antiderivative size = 20 \[ \int \frac {e^{\frac {e}{\log \left (e^{3/x}+x\right )}} \left (-3 e^{1+\frac {3}{x}}+e x^2\right )}{\left (e^{3/x} x^2+x^3\right ) \log ^2\left (e^{3/x}+x\right )} \, dx=-7-e^{\frac {e}{\log \left (e^{3/x}+x\right )}} \]

[Out]

-exp(exp(1)/ln(exp(3/x)+x))-7

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {6838} \[ \int \frac {e^{\frac {e}{\log \left (e^{3/x}+x\right )}} \left (-3 e^{1+\frac {3}{x}}+e x^2\right )}{\left (e^{3/x} x^2+x^3\right ) \log ^2\left (e^{3/x}+x\right )} \, dx=-e^{\frac {e}{\log \left (x+e^{3/x}\right )}} \]

[In]

Int[(E^(E/Log[E^(3/x) + x])*(-3*E^(1 + 3/x) + E*x^2))/((E^(3/x)*x^2 + x^3)*Log[E^(3/x) + x]^2),x]

[Out]

-E^(E/Log[E^(3/x) + x])

Rule 6838

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{align*} \text {integral}& = -e^{\frac {e}{\log \left (e^{3/x}+x\right )}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {e}{\log \left (e^{3/x}+x\right )}} \left (-3 e^{1+\frac {3}{x}}+e x^2\right )}{\left (e^{3/x} x^2+x^3\right ) \log ^2\left (e^{3/x}+x\right )} \, dx=-e^{\frac {e}{\log \left (e^{3/x}+x\right )}} \]

[In]

Integrate[(E^(E/Log[E^(3/x) + x])*(-3*E^(1 + 3/x) + E*x^2))/((E^(3/x)*x^2 + x^3)*Log[E^(3/x) + x]^2),x]

[Out]

-E^(E/Log[E^(3/x) + x])

Maple [A] (verified)

Time = 38.36 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90

method	result	size
risch	\(-{\mathrm e}^{\frac {{\mathrm e}}{\ln \left ({\mathrm e}^{\frac {3}{x}}+x \right )}}\)	\(18\)
parallelrisch	\(-{\mathrm e}^{\frac {{\mathrm e}}{\ln \left ({\mathrm e}^{\frac {3}{x}}+x \right )}}\)	\(18\)

[In]

int((-3*exp(1)*exp(3/x)+x^2*exp(1))*exp(exp(1)/ln(exp(3/x)+x))/(x^2*exp(3/x)+x^3)/ln(exp(3/x)+x)^2,x,method=_R
ETURNVERBOSE)

[Out]

-exp(exp(1)/ln(exp(3/x)+x))

Fricas [A] (verification not implemented)

none

Time = 0.47 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.25 \[ \int \frac {e^{\frac {e}{\log \left (e^{3/x}+x\right )}} \left (-3 e^{1+\frac {3}{x}}+e x^2\right )}{\left (e^{3/x} x^2+x^3\right ) \log ^2\left (e^{3/x}+x\right )} \, dx=-e^{\left (\frac {e}{\log \left ({\left (x e + e^{\left (\frac {x + 3}{x}\right )}\right )} e^{\left (-1\right )}\right )}\right )} \]

[In]

integrate((-3*exp(1)*exp(3/x)+x^2*exp(1))*exp(exp(1)/log(exp(3/x)+x))/(x^2*exp(3/x)+x^3)/log(exp(3/x)+x)^2,x,
algorithm="fricas")

[Out]

-e^(e/log((x*e + e^((x + 3)/x))*e^(-1)))

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{\frac {e}{\log \left (e^{3/x}+x\right )}} \left (-3 e^{1+\frac {3}{x}}+e x^2\right )}{\left (e^{3/x} x^2+x^3\right ) \log ^2\left (e^{3/x}+x\right )} \, dx=\text {Timed out} \]

[In]

integrate((-3*exp(1)*exp(3/x)+x**2*exp(1))*exp(exp(1)/ln(exp(3/x)+x))/(x**2*exp(3/x)+x**3)/ln(exp(3/x)+x)**2,x
)

[Out]

Timed out

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 72 vs. \(2 (19) = 38\).

Time = 0.20 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.60 \[ \int \frac {e^{\frac {e}{\log \left (e^{3/x}+x\right )}} \left (-3 e^{1+\frac {3}{x}}+e x^2\right )}{\left (e^{3/x} x^2+x^3\right ) \log ^2\left (e^{3/x}+x\right )} \, dx=-\frac {x^{2} e^{\left (\frac {e}{\log \left (x + e^{\frac {3}{x}}\right )}\right )}}{x^{2} - 3 \, e^{\frac {3}{x}}} + \frac {3 \, e^{\left (\frac {e}{\log \left (x + e^{\frac {3}{x}}\right )} + \frac {3}{x}\right )}}{x^{2} - 3 \, e^{\frac {3}{x}}} \]

[In]

integrate((-3*exp(1)*exp(3/x)+x^2*exp(1))*exp(exp(1)/log(exp(3/x)+x))/(x^2*exp(3/x)+x^3)/log(exp(3/x)+x)^2,x,
algorithm="maxima")

[Out]

-x^2*e^(e/log(x + e^(3/x)))/(x^2 - 3*e^(3/x)) + 3*e^(e/log(x + e^(3/x)) + 3/x)/(x^2 - 3*e^(3/x))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {e}{\log \left (e^{3/x}+x\right )}} \left (-3 e^{1+\frac {3}{x}}+e x^2\right )}{\left (e^{3/x} x^2+x^3\right ) \log ^2\left (e^{3/x}+x\right )} \, dx=-e^{\left (\frac {e}{\log \left (x + e^{\frac {3}{x}}\right )}\right )} \]

[In]

integrate((-3*exp(1)*exp(3/x)+x^2*exp(1))*exp(exp(1)/log(exp(3/x)+x))/(x^2*exp(3/x)+x^3)/log(exp(3/x)+x)^2,x,
algorithm="giac")

[Out]

-e^(e/log(x + e^(3/x)))

Mupad [B] (verification not implemented)

Time = 11.61 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {e}{\log \left (e^{3/x}+x\right )}} \left (-3 e^{1+\frac {3}{x}}+e x^2\right )}{\left (e^{3/x} x^2+x^3\right ) \log ^2\left (e^{3/x}+x\right )} \, dx=-{\mathrm {e}}^{\frac {\mathrm {e}}{\ln \left (x+{\mathrm {e}}^{3/x}\right )}} \]

[In]

int(-(exp(exp(1)/log(x + exp(3/x)))*(3*exp(1)*exp(3/x) - x^2*exp(1)))/(log(x + exp(3/x))^2*(x^2*exp(3/x) + x^3
)),x)

[Out]

-exp(exp(1)/log(x + exp(3/x)))