3.85.72 \(\int \frac {e^{16 e^{\frac {2 (-5 x+\log (x))}{x}}+\frac {2 (-5 x+\log (x))}{x}} (32-32 \log (x))}{x^2} \, dx\)

Optimal. Leaf size=15 \[ e^{16 e^{-10+\frac {2 \log (x)}{x}}} \]

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Rubi [F]  time = 1.09, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\exp \left (16 e^{\frac {2 (-5 x+\log (x))}{x}}+\frac {2 (-5 x+\log (x))}{x}\right ) (32-32 \log (x))}{x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(16*E^((2*(-5*x + Log[x]))/x) + (2*(-5*x + Log[x]))/x)*(32 - 32*Log[x]))/x^2,x]

[Out]

32*Defer[Int][E^(16*E^((2*(-5*x + Log[x]))/x) + (2*(-5*x + Log[x]))/x)/x^2, x] - 32*Defer[Int][(E^(16*E^((2*(-
5*x + Log[x]))/x) + (2*(-5*x + Log[x]))/x)*Log[x])/x^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32 \exp \left (16 e^{\frac {2 (-5 x+\log (x))}{x}}+\frac {2 (-5 x+\log (x))}{x}\right ) (1-\log (x))}{x^2} \, dx\\ &=32 \int \frac {\exp \left (16 e^{\frac {2 (-5 x+\log (x))}{x}}+\frac {2 (-5 x+\log (x))}{x}\right ) (1-\log (x))}{x^2} \, dx\\ &=32 \int \left (\frac {\exp \left (16 e^{\frac {2 (-5 x+\log (x))}{x}}+\frac {2 (-5 x+\log (x))}{x}\right )}{x^2}-\frac {\exp \left (16 e^{\frac {2 (-5 x+\log (x))}{x}}+\frac {2 (-5 x+\log (x))}{x}\right ) \log (x)}{x^2}\right ) \, dx\\ &=32 \int \frac {\exp \left (16 e^{\frac {2 (-5 x+\log (x))}{x}}+\frac {2 (-5 x+\log (x))}{x}\right )}{x^2} \, dx-32 \int \frac {\exp \left (16 e^{\frac {2 (-5 x+\log (x))}{x}}+\frac {2 (-5 x+\log (x))}{x}\right ) \log (x)}{x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 14, normalized size = 0.93 \begin {gather*} e^{\frac {16 x^{2/x}}{e^{10}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(16*E^((2*(-5*x + Log[x]))/x) + (2*(-5*x + Log[x]))/x)*(32 - 32*Log[x]))/x^2,x]

[Out]

E^((16*x^(2/x))/E^10)

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fricas [B]  time = 0.57, size = 43, normalized size = 2.87 \begin {gather*} e^{\left (\frac {2 \, {\left (8 \, x e^{\left (-\frac {2 \, {\left (5 \, x - \log \relax (x)\right )}}{x}\right )} - 5 \, x + \log \relax (x)\right )}}{x} + \frac {2 \, {\left (5 \, x - \log \relax (x)\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(x)+32)*exp((log(x)-5*x)/x)^2*exp(16*exp((log(x)-5*x)/x)^2)/x^2,x, algorithm="fricas")

[Out]

e^(2*(8*x*e^(-2*(5*x - log(x))/x) - 5*x + log(x))/x + 2*(5*x - log(x))/x)

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giac [B]  time = 0.24, size = 43, normalized size = 2.87 \begin {gather*} e^{\left (\frac {2 \, {\left (8 \, x e^{\left (-\frac {2 \, {\left (5 \, x - \log \relax (x)\right )}}{x}\right )} - 5 \, x + \log \relax (x)\right )}}{x} + \frac {2 \, {\left (5 \, x - \log \relax (x)\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(x)+32)*exp((log(x)-5*x)/x)^2*exp(16*exp((log(x)-5*x)/x)^2)/x^2,x, algorithm="giac")

[Out]

e^(2*(8*x*e^(-2*(5*x - log(x))/x) - 5*x + log(x))/x + 2*(5*x - log(x))/x)

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maple [A]  time = 0.02, size = 13, normalized size = 0.87

method	result	size
risch	\({\mathrm e}^{16 x^{\frac {2}{x}} {\mathrm e}^{-10}}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-32*ln(x)+32)*exp((ln(x)-5*x)/x)^2*exp(16*exp((ln(x)-5*x)/x)^2)/x^2,x,method=_RETURNVERBOSE)

[Out]

exp(16*(x^(1/x))^2*exp(-10))

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maxima [A]  time = 0.70, size = 13, normalized size = 0.87 \begin {gather*} e^{\left (16 \, e^{\left (\frac {2 \, \log \relax (x)}{x} - 10\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*log(x)+32)*exp((log(x)-5*x)/x)^2*exp(16*exp((log(x)-5*x)/x)^2)/x^2,x, algorithm="maxima")

[Out]

e^(16*e^(2*log(x)/x - 10))

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mupad [B]  time = 5.41, size = 12, normalized size = 0.80 \begin {gather*} {\mathrm {e}}^{16\,x^{2/x}\,{\mathrm {e}}^{-10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-(2*(5*x - log(x)))/x)*exp(16*exp(-(2*(5*x - log(x)))/x))*(32*log(x) - 32))/x^2,x)

[Out]

exp(16*x^(2/x)*exp(-10))

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sympy [A]  time = 0.53, size = 14, normalized size = 0.93 \begin {gather*} e^{16 e^{\frac {2 \left (- 5 x + \log {\relax (x )}\right )}{x}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*ln(x)+32)*exp((ln(x)-5*x)/x)**2*exp(16*exp((ln(x)-5*x)/x)**2)/x**2,x)

[Out]

exp(16*exp(2*(-5*x + log(x))/x))

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