3.25.1 \(\int \frac {e^{5+e^{\frac {2-x \log (3)}{x}}-3 x+3 x \log (x)} (-2 e^{\frac {2-x \log (3)}{x}}+3 x^2 \log (x))}{x^2} \, dx\)

Optimal. Leaf size=24 \[ e^{5+\frac {e^{2/x}}{3}-3 x (1-\log (x))} \]

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Rubi [A]  time = 0.49, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {6706} \begin {gather*} e^{-3 x+\frac {e^{2/x}}{3}+5} x^{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(5 + E^((2 - x*Log[3])/x) - 3*x + 3*x*Log[x])*(-2*E^((2 - x*Log[3])/x) + 3*x^2*Log[x]))/x^2,x]

[Out]

E^(5 + E^(2/x)/3 - 3*x)*x^(3*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^{5+\frac {e^{2/x}}{3}-3 x} x^{3 x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.27, size = 24, normalized size = 1.00 \begin {gather*} e^{5+\frac {e^{2/x}}{3}-3 x} x^{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5 + E^((2 - x*Log[3])/x) - 3*x + 3*x*Log[x])*(-2*E^((2 - x*Log[3])/x) + 3*x^2*Log[x]))/x^2,x]

[Out]

E^(5 + E^(2/x)/3 - 3*x)*x^(3*x)

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fricas [A]  time = 0.76, size = 23, normalized size = 0.96 \begin {gather*} e^{\left (3 \, x \log \relax (x) - 3 \, x + e^{\left (-\frac {x \log \relax (3) - 2}{x}\right )} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2*log(x)-2*exp((-x*log(3)+2)/x))*exp(3*x*log(x)+exp((-x*log(3)+2)/x)-3*x+5)/x^2,x, algorithm="f
ricas")

[Out]

e^(3*x*log(x) - 3*x + e^(-(x*log(3) - 2)/x) + 5)

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giac [A]  time = 0.20, size = 22, normalized size = 0.92 \begin {gather*} e^{\left (3 \, x \log \relax (x) - 3 \, x + e^{\left (\frac {2}{x} - \log \relax (3)\right )} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2*log(x)-2*exp((-x*log(3)+2)/x))*exp(3*x*log(x)+exp((-x*log(3)+2)/x)-3*x+5)/x^2,x, algorithm="g
iac")

[Out]

e^(3*x*log(x) - 3*x + e^(2/x - log(3)) + 5)

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maple [A]  time = 0.04, size = 21, normalized size = 0.88

method	result	size
risch	\(x^{3 x} {\mathrm e}^{5+\frac {{\mathrm e}^{\frac {2}{x}}}{3}-3 x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x^2*ln(x)-2*exp((-x*ln(3)+2)/x))*exp(3*x*ln(x)+exp((-x*ln(3)+2)/x)-3*x+5)/x^2,x,method=_RETURNVERBOSE)

[Out]

x^(3*x)*exp(5+1/3*exp(2/x)-3*x)

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maxima [A]  time = 0.63, size = 19, normalized size = 0.79 \begin {gather*} e^{\left (3 \, x \log \relax (x) - 3 \, x + \frac {1}{3} \, e^{\frac {2}{x}} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x^2*log(x)-2*exp((-x*log(3)+2)/x))*exp(3*x*log(x)+exp((-x*log(3)+2)/x)-3*x+5)/x^2,x, algorithm="m
axima")

[Out]

e^(3*x*log(x) - 3*x + 1/3*e^(2/x) + 5)

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mupad [B]  time = 1.40, size = 21, normalized size = 0.88 \begin {gather*} x^{3\,x}\,{\mathrm {e}}^{-3\,x}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{2/x}}{3}}\,{\mathrm {e}}^5 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(-(x*log(3) - 2)/x) - 3*x + 3*x*log(x) + 5)*(2*exp(-(x*log(3) - 2)/x) - 3*x^2*log(x)))/x^2,x)

[Out]

x^(3*x)*exp(-3*x)*exp(exp(2/x)/3)*exp(5)

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sympy [A]  time = 0.44, size = 22, normalized size = 0.92 \begin {gather*} e^{3 x \log {\relax (x )} - 3 x + e^{\frac {- x \log {\relax (3 )} + 2}{x}} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*x**2*ln(x)-2*exp((-x*ln(3)+2)/x))*exp(3*x*ln(x)+exp((-x*ln(3)+2)/x)-3*x+5)/x**2,x)

[Out]

exp(3*x*log(x) - 3*x + exp((-x*log(3) + 2)/x) + 5)

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