3.75.89 \(\int \frac {5 x+e^{e^{\frac {-3 x+3 \log (x)}{x}}+\frac {-3 x+3 \log (x)}{x}} (6-6 \log (x))}{2 x^2} \, dx\)

Optimal. Leaf size=33 \[ e^{e^{\frac {3 (-x+\log (x))}{x}}}+\frac {5}{2} \log \left (-x+\frac {e^4 x}{5}\right ) \]

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Rubi [F]  time = 0.28, 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 {5 x+\exp \left (e^{\frac {-3 x+3 \log (x)}{x}}+\frac {-3 x+3 \log (x)}{x}\right ) (6-6 \log (x))}{2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5*x + E^(E^((-3*x + 3*Log[x])/x) + (-3*x + 3*Log[x])/x)*(6 - 6*Log[x]))/(2*x^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{2} \int \frac {5 x+\exp \left (e^{\frac {-3 x+3 \log (x)}{x}}+\frac {-3 x+3 \log (x)}{x}\right ) (6-6 \log (x))}{x^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {5}{x}-6 e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}} (-1+\log (x))\right ) \, dx\\ &=\frac {5 \log (x)}{2}-3 \int e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}} (-1+\log (x)) \, dx\\ &=\frac {5 \log (x)}{2}-3 \int \left (-e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}}+e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}} \log (x)\right ) \, dx\\ &=\frac {5 \log (x)}{2}+3 \int e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}} \, dx-3 \int e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}} \log (x) \, dx\\ &=\frac {5 \log (x)}{2}+3 \int e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}} \, dx+3 \int \frac {\int e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}} \, dx}{x} \, dx-(3 \log (x)) \int e^{-3+\frac {x^{3/x}}{e^3}} x^{-2+\frac {3}{x}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 24, normalized size = 0.73 \begin {gather*} \frac {1}{2} \left (2 e^{\frac {x^{3/x}}{e^3}}+5 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*x + E^(E^((-3*x + 3*Log[x])/x) + (-3*x + 3*Log[x])/x)*(6 - 6*Log[x]))/(2*x^2),x]

[Out]

(2*E^(x^(3/x)/E^3) + 5*Log[x])/2

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fricas [B]  time = 0.59, size = 60, normalized size = 1.82 \begin {gather*} \frac {1}{2} \, {\left (5 \, e^{\left (-\frac {3 \, {\left (x - \log \relax (x)\right )}}{x}\right )} \log \relax (x) + 2 \, e^{\left (\frac {x e^{\left (-\frac {3 \, {\left (x - \log \relax (x)\right )}}{x}\right )} - 3 \, x + 3 \, \log \relax (x)}{x}\right )}\right )} e^{\left (\frac {3 \, {\left (x - \log \relax (x)\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-6*log(x)+6)*exp((3*log(x)-3*x)/x)*exp(exp((3*log(x)-3*x)/x))+5*x)/x^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.21, size = 60, normalized size = 1.82 \begin {gather*} \frac {1}{2} \, {\left (5 \, e^{\left (-\frac {3 \, {\left (x - \log \relax (x)\right )}}{x}\right )} \log \relax (x) + 2 \, e^{\left (\frac {x e^{\left (-\frac {3 \, {\left (x - \log \relax (x)\right )}}{x}\right )} - 3 \, x + 3 \, \log \relax (x)}{x}\right )}\right )} e^{\left (\frac {3 \, {\left (x - \log \relax (x)\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-6*log(x)+6)*exp((3*log(x)-3*x)/x)*exp(exp((3*log(x)-3*x)/x))+5*x)/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.15, size = 17, normalized size = 0.52

method	result	size
risch	\(\frac {5 \ln \relax (x )}{2}+{\mathrm e}^{x^{\frac {3}{x}} {\mathrm e}^{-3}}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*((-6*ln(x)+6)*exp((3*ln(x)-3*x)/x)*exp(exp((3*ln(x)-3*x)/x))+5*x)/x^2,x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.51, size = 16, normalized size = 0.48 \begin {gather*} e^{\left (e^{\left (\frac {3 \, \log \relax (x)}{x} - 3\right )}\right )} + \frac {5}{2} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-6*log(x)+6)*exp((3*log(x)-3*x)/x)*exp(exp((3*log(x)-3*x)/x))+5*x)/x^2,x, algorithm="maxima")

[Out]

e^(e^(3*log(x)/x - 3)) + 5/2*log(x)

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mupad [B]  time = 5.88, size = 16, normalized size = 0.48 \begin {gather*} {\mathrm {e}}^{x^{3/x}\,{\mathrm {e}}^{-3}}+\frac {5\,\ln \relax (x)}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*x)/2 - (exp(-(3*x - 3*log(x))/x)*exp(exp(-(3*x - 3*log(x))/x))*(6*log(x) - 6))/2)/x^2,x)

[Out]

exp(x^(3/x)*exp(-3)) + (5*log(x))/2

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sympy [A]  time = 0.55, size = 19, normalized size = 0.58 \begin {gather*} e^{e^{\frac {- 3 x + 3 \log {\relax (x )}}{x}}} + \frac {5 \log {\relax (x )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*((-6*ln(x)+6)*exp((3*ln(x)-3*x)/x)*exp(exp((3*ln(x)-3*x)/x))+5*x)/x**2,x)

[Out]

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

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