3.96.44 \(\int \frac {e^{\frac {2 e^{\frac {5 e^5}{x}}-x^2+2 \log (2 x)}{x}} (e^{\frac {5 e^5}{x}} (-10 e^5-2 x)+2 x-x^3-2 x \log (2 x))}{x^3} \, dx\)

Optimal. Leaf size=26 \[ e^{x \left (-1+\frac {2 \left (e^{\frac {5 e^5}{x}}+\log (2 x)\right )}{x^2}\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 1.11, antiderivative size = 39, normalized size of antiderivative = 1.50, number of steps used = 1, number of rules used = 1, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {6706} \begin {gather*} 2^{2/x} e^{\frac {2 e^{\frac {5 e^5}{x}}-x^2}{x}} x^{2/x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((2*E^((5*E^5)/x) - x^2 + 2*Log[2*x])/x)*(E^((5*E^5)/x)*(-10*E^5 - 2*x) + 2*x - x^3 - 2*x*Log[2*x]))/x^
3,x]

[Out]

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

________________________________________________________________________________________

Mathematica [F]  time = 2.80, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {2 e^{\frac {5 e^5}{x}}-x^2+2 \log (2 x)}{x}} \left (e^{\frac {5 e^5}{x}} \left (-10 e^5-2 x\right )+2 x-x^3-2 x \log (2 x)\right )}{x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((2*E^((5*E^5)/x) - x^2 + 2*Log[2*x])/x)*(E^((5*E^5)/x)*(-10*E^5 - 2*x) + 2*x - x^3 - 2*x*Log[2*x
]))/x^3,x]

[Out]

Integrate[(E^((2*E^((5*E^5)/x) - x^2 + 2*Log[2*x])/x)*(E^((5*E^5)/x)*(-10*E^5 - 2*x) + 2*x - x^3 - 2*x*Log[2*x
]))/x^3, x]

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 26, normalized size = 1.00 \begin {gather*} e^{\left (-\frac {x^{2} - 2 \, e^{\left (\frac {5 \, e^{5}}{x}\right )} - 2 \, \log \left (2 \, x\right )}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(2*x)+(-10*exp(5)-2*x)*exp(5*exp(5)/x)-x^3+2*x)*exp((2*log(2*x)+2*exp(5*exp(5)/x)-x^2)/x)/x
^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.26, size = 27, normalized size = 1.04 \begin {gather*} e^{\left (-x + \frac {2 \, e^{\left (\frac {5 \, e^{5}}{x}\right )}}{x} + \frac {2 \, \log \left (2 \, x\right )}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(2*x)+(-10*exp(5)-2*x)*exp(5*exp(5)/x)-x^3+2*x)*exp((2*log(2*x)+2*exp(5*exp(5)/x)-x^2)/x)/x
^3,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.03, size = 28, normalized size = 1.08

method	result	size
risch	\({\mathrm e}^{\frac {2 \ln \left (2 x \right )+2 \,{\mathrm e}^{\frac {5 \,{\mathrm e}^{5}}{x}}-x^{2}}{x}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x*ln(2*x)+(-10*exp(5)-2*x)*exp(5*exp(5)/x)-x^3+2*x)*exp((2*ln(2*x)+2*exp(5*exp(5)/x)-x^2)/x)/x^3,x,met
hod=_RETURNVERBOSE)

[Out]

exp((2*ln(2*x)+2*exp(5*exp(5)/x)-x^2)/x)

________________________________________________________________________________________

maxima [A]  time = 0.55, size = 32, normalized size = 1.23 \begin {gather*} e^{\left (-x + \frac {2 \, e^{\left (\frac {5 \, e^{5}}{x}\right )}}{x} + \frac {2 \, \log \relax (2)}{x} + \frac {2 \, \log \relax (x)}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*log(2*x)+(-10*exp(5)-2*x)*exp(5*exp(5)/x)-x^3+2*x)*exp((2*log(2*x)+2*exp(5*exp(5)/x)-x^2)/x)/x
^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 9.76, size = 33, normalized size = 1.27 \begin {gather*} 2^{2/x}\,x^{2/x}\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{\frac {5\,{\mathrm {e}}^5}{x}}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((2*exp((5*exp(5))/x) + 2*log(2*x) - x^2)/x)*(2*x*log(2*x) - 2*x + exp((5*exp(5))/x)*(2*x + 10*exp(5)
) + x^3))/x^3,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.55, size = 22, normalized size = 0.85 \begin {gather*} e^{\frac {- x^{2} + 2 e^{\frac {5 e^{5}}{x}} + 2 \log {\left (2 x \right )}}{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x*ln(2*x)+(-10*exp(5)-2*x)*exp(5*exp(5)/x)-x**3+2*x)*exp((2*ln(2*x)+2*exp(5*exp(5)/x)-x**2)/x)/x
**3,x)

[Out]

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

________________________________________________________________________________________