3.39.5 \(\int \frac {4+2 x^3+8 \log (\frac {5}{x})+e^{x^{\frac {1}{x}}} x^{\frac {1}{x}} (x-x \log (x))}{x^3} \, dx\)

Optimal. Leaf size=25 \[ \frac {3}{4}+e^{x^{\frac {1}{x}}}+2 x-\frac {4 \log \left (\frac {5}{x}\right )}{x^2} \]

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Rubi [F]  time = 0.22, 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 {4+2 x^3+8 \log \left (\frac {5}{x}\right )+e^{x^{\frac {1}{x}}} x^{\frac {1}{x}} (x-x \log (x))}{x^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(4 + 2*x^3 + 8*Log[5/x] + E^x^x^(-1)*x^x^(-1)*(x - x*Log[x]))/x^3,x]

[Out]

2*x - (4*Log[5/x])/x^2 + Defer[Int][E^x^x^(-1)*x^(-2 + x^(-1)), x] - Log[x]*Defer[Int][E^x^x^(-1)*x^(-2 + x^(-
1)), x] + Defer[Int][Defer[Int][E^x^x^(-1)*x^(-2 + x^(-1)), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {2 \left (2+x^3+4 \log \left (\frac {5}{x}\right )\right )}{x^3}-e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} (-1+\log (x))\right ) \, dx\\ &=2 \int \frac {2+x^3+4 \log \left (\frac {5}{x}\right )}{x^3} \, dx-\int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} (-1+\log (x)) \, dx\\ &=2 \int \left (\frac {2+x^3}{x^3}+\frac {4 \log \left (\frac {5}{x}\right )}{x^3}\right ) \, dx-\int \left (-e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}}+e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \log (x)\right ) \, dx\\ &=2 \int \frac {2+x^3}{x^3} \, dx+8 \int \frac {\log \left (\frac {5}{x}\right )}{x^3} \, dx+\int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \, dx-\int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \log (x) \, dx\\ &=\frac {2}{x^2}-\frac {4 \log \left (\frac {5}{x}\right )}{x^2}+2 \int \left (1+\frac {2}{x^3}\right ) \, dx-\log (x) \int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \, dx+\int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \, dx+\int \frac {\int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \, dx}{x} \, dx\\ &=2 x-\frac {4 \log \left (\frac {5}{x}\right )}{x^2}-\log (x) \int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \, dx+\int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \, dx+\int \frac {\int e^{x^{\frac {1}{x}}} x^{-2+\frac {1}{x}} \, dx}{x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 22, normalized size = 0.88 \begin {gather*} e^{x^{\frac {1}{x}}}+2 x-\frac {4 \log \left (\frac {5}{x}\right )}{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 + 2*x^3 + 8*Log[5/x] + E^x^x^(-1)*x^x^(-1)*(x - x*Log[x]))/x^3,x]

[Out]

E^x^x^(-1) + 2*x - (4*Log[5/x])/x^2

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fricas [A]  time = 0.59, size = 39, normalized size = 1.56 \begin {gather*} \frac {2 \, x^{3} + x^{2} e^{\left (e^{\left (\frac {\log \relax (5) - \log \left (\frac {5}{x}\right )}{x}\right )}\right )} - 4 \, \log \left (\frac {5}{x}\right )}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-x*log(x))*exp(log(x)/x)*exp(exp(log(x)/x))+8*log(5/x)+2*x^3+4)/x^3,x, algorithm="fricas")

[Out]

(2*x^3 + x^2*e^(e^((log(5) - log(5/x))/x)) - 4*log(5/x))/x^2

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giac [A]  time = 0.26, size = 24, normalized size = 0.96 \begin {gather*} 2 \, x - \frac {4 \, \log \relax (5)}{x^{2}} + \frac {4 \, \log \relax (x)}{x^{2}} + e^{\left (x^{\left (\frac {1}{x}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-x*log(x))*exp(log(x)/x)*exp(exp(log(x)/x))+8*log(5/x)+2*x^3+4)/x^3,x, algorithm="giac")

[Out]

2*x - 4*log(5)/x^2 + 4*log(x)/x^2 + e^(x^(1/x))

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maple [A]  time = 0.05, size = 24, normalized size = 0.96




method result size



default \({\mathrm e}^{{\mathrm e}^{\frac {\ln \relax (x )}{x}}}-\frac {4 \ln \left (\frac {5}{x}\right )}{x^{2}}+2 x\) \(24\)
risch \(\frac {4 \ln \relax (x )}{x^{2}}-\frac {2 \left (-x^{3}+2 \ln \relax (5)\right )}{x^{2}}+{\mathrm e}^{x^{\frac {1}{x}}}\) \(30\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x-x*ln(x))*exp(ln(x)/x)*exp(exp(ln(x)/x))+8*ln(5/x)+2*x^3+4)/x^3,x,method=_RETURNVERBOSE)

[Out]

exp(exp(ln(x)/x))-4*ln(5/x)/x^2+2*x

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maxima [A]  time = 0.48, size = 21, normalized size = 0.84 \begin {gather*} 2 \, x - \frac {4 \, \log \left (\frac {5}{x}\right )}{x^{2}} + e^{\left (x^{\left (\frac {1}{x}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-x*log(x))*exp(log(x)/x)*exp(exp(log(x)/x))+8*log(5/x)+2*x^3+4)/x^3,x, algorithm="maxima")

[Out]

2*x - 4*log(5/x)/x^2 + e^(x^(1/x))

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mupad [B]  time = 2.55, size = 26, normalized size = 1.04 \begin {gather*} 2\,x+{\mathrm {e}}^{x^{1/x}}-\frac {4\,\ln \left (\frac {1}{x}\right )}{x^2}-\frac {4\,\ln \relax (5)}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*log(5/x) + 2*x^3 + exp(log(x)/x)*exp(exp(log(x)/x))*(x - x*log(x)) + 4)/x^3,x)

[Out]

2*x + exp(x^(1/x)) - (4*log(1/x))/x^2 - (4*log(5))/x^2

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sympy [A]  time = 0.58, size = 27, normalized size = 1.08 \begin {gather*} 2 x + e^{e^{\frac {\log {\relax (x )}}{x}}} + \frac {4 \log {\relax (x )}}{x^{2}} - \frac {4 \log {\relax (5 )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x-x*ln(x))*exp(ln(x)/x)*exp(exp(ln(x)/x))+8*ln(5/x)+2*x**3+4)/x**3,x)

[Out]

2*x + exp(exp(log(x)/x)) + 4*log(x)/x**2 - 4*log(5)/x**2

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