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

Optimal. Leaf size=20 \[ -7-e^{\frac {e}{\log \left (e^{3/x}+x\right )}} \]

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Rubi [A]  time = 0.45, antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 1, number of rules used = 1, integrand size = 63, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.016, Rules used = {6706} \begin {gather*} -e^{\frac {e}{\log \left (x+e^{3/x}\right )}} \end {gather*}

Antiderivative was successfully verified.

[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 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}{\log \left (e^{3/x}+x\right )}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.36, size = 18, normalized size = 0.90 \begin {gather*} -e^{\frac {e}{\log \left (e^{3/x}+x\right )}} \end {gather*}

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.50, size = 25, normalized size = 1.25 \begin {gather*} -e^{\left (\frac {e}{\log \left ({\left (x e + e^{\left (\frac {x + 3}{x}\right )}\right )} e^{\left (-1\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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giac [A]  time = 0.15, size = 17, normalized size = 0.85 \begin {gather*} -e^{\left (\frac {e}{\log \left (x + e^{\frac {3}{x}}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maple [A]  time = 0.06, size = 18, normalized size = 0.90

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [B]  time = 0.39, size = 72, normalized size = 3.60 \begin {gather*} -\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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 4.40, size = 17, normalized size = 0.85 \begin {gather*} -{\mathrm {e}}^{\frac {\mathrm {e}}{\ln \left (x+{\mathrm {e}}^{3/x}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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