3.86.64 \(\int \frac {e^{\frac {x^2+\log (x)}{x}} ((120-240 x+120 x^2) \log (2)-120 \log (2) \log (x))}{x^4} \, dx\)

Optimal. Leaf size=17 \[ \frac {120 e^{x+\frac {\log (x)}{x}} \log (2)}{x^2} \]

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Rubi [F]  time = 0.48, 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 {e^{\frac {x^2+\log (x)}{x}} \left (\left (120-240 x+120 x^2\right ) \log (2)-120 \log (2) \log (x)\right )}{x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((x^2 + Log[x])/x)*((120 - 240*x + 120*x^2)*Log[2] - 120*Log[2]*Log[x]))/x^4,x]

[Out]

120*Log[2]*Defer[Int][E^x*x^(-4 + x^(-1)), x] - 120*Log[2]*Log[x]*Defer[Int][E^x*x^(-4 + x^(-1)), x] - 240*Log
[2]*Defer[Int][E^x*x^(-3 + x^(-1)), x] + 120*Log[2]*Defer[Int][E^x*x^(-2 + x^(-1)), x] + 120*Log[2]*Defer[Int]
[Defer[Int][E^x*x^(-4 + x^(-1)), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int 120 e^x x^{-4+\frac {1}{x}} \log (2) \left ((-1+x)^2-\log (x)\right ) \, dx\\ &=(120 \log (2)) \int e^x x^{-4+\frac {1}{x}} \left ((-1+x)^2-\log (x)\right ) \, dx\\ &=(120 \log (2)) \int \left (e^x (-1+x)^2 x^{-4+\frac {1}{x}}-e^x x^{-4+\frac {1}{x}} \log (x)\right ) \, dx\\ &=(120 \log (2)) \int e^x (-1+x)^2 x^{-4+\frac {1}{x}} \, dx-(120 \log (2)) \int e^x x^{-4+\frac {1}{x}} \log (x) \, dx\\ &=(120 \log (2)) \int \left (e^x x^{-4+\frac {1}{x}}-2 e^x x^{-3+\frac {1}{x}}+e^x x^{-2+\frac {1}{x}}\right ) \, dx+(120 \log (2)) \int \frac {\int e^x x^{-4+\frac {1}{x}} \, dx}{x} \, dx-(120 \log (2) \log (x)) \int e^x x^{-4+\frac {1}{x}} \, dx\\ &=(120 \log (2)) \int e^x x^{-4+\frac {1}{x}} \, dx+(120 \log (2)) \int e^x x^{-2+\frac {1}{x}} \, dx+(120 \log (2)) \int \frac {\int e^x x^{-4+\frac {1}{x}} \, dx}{x} \, dx-(240 \log (2)) \int e^x x^{-3+\frac {1}{x}} \, dx-(120 \log (2) \log (x)) \int e^x x^{-4+\frac {1}{x}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 14, normalized size = 0.82 \begin {gather*} 120 e^x x^{-2+\frac {1}{x}} \log (2) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((x^2 + Log[x])/x)*((120 - 240*x + 120*x^2)*Log[2] - 120*Log[2]*Log[x]))/x^4,x]

[Out]

120*E^x*x^(-2 + x^(-1))*Log[2]

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fricas [A]  time = 0.72, size = 18, normalized size = 1.06 \begin {gather*} \frac {120 \, e^{\left (\frac {x^{2} + \log \relax (x)}{x}\right )} \log \relax (2)}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-120*log(2)*log(x)+(120*x^2-240*x+120)*log(2))*exp((log(x)+x^2)/x)/x^4,x, algorithm="fricas")

[Out]

120*e^((x^2 + log(x))/x)*log(2)/x^2

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giac [A]  time = 0.65, size = 18, normalized size = 1.06 \begin {gather*} \frac {120 \, e^{\left (\frac {x^{2} + \log \relax (x)}{x}\right )} \log \relax (2)}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-120*log(2)*log(x)+(120*x^2-240*x+120)*log(2))*exp((log(x)+x^2)/x)/x^4,x, algorithm="giac")

[Out]

120*e^((x^2 + log(x))/x)*log(2)/x^2

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maple [A]  time = 0.06, size = 15, normalized size = 0.88

method	result	size
risch	\(\frac {120 \ln \relax (2) x^{\frac {1}{x}} {\mathrm e}^{x}}{x^{2}}\)	\(15\)
norman	\(\frac {120 \ln \relax (2) {\mathrm e}^{\frac {\ln \relax (x )+x^{2}}{x}}}{x^{2}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-120*ln(2)*ln(x)+(120*x^2-240*x+120)*ln(2))*exp((ln(x)+x^2)/x)/x^4,x,method=_RETURNVERBOSE)

[Out]

120/x^2*ln(2)*x^(1/x)*exp(x)

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maxima [A]  time = 0.51, size = 16, normalized size = 0.94 \begin {gather*} \frac {120 \, e^{\left (x + \frac {\log \relax (x)}{x}\right )} \log \relax (2)}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-120*log(2)*log(x)+(120*x^2-240*x+120)*log(2))*exp((log(x)+x^2)/x)/x^4,x, algorithm="maxima")

[Out]

120*e^(x + log(x)/x)*log(2)/x^2

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mupad [B]  time = 5.32, size = 13, normalized size = 0.76 \begin {gather*} 120\,x^{\frac {1}{x}-2}\,{\mathrm {e}}^x\,\ln \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((log(x) + x^2)/x)*(log(2)*(120*x^2 - 240*x + 120) - 120*log(2)*log(x)))/x^4,x)

[Out]

120*x^(1/x - 2)*exp(x)*log(2)

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sympy [A]  time = 0.26, size = 17, normalized size = 1.00 \begin {gather*} \frac {120 e^{\frac {x^{2} + \log {\relax (x )}}{x}} \log {\relax (2 )}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-120*ln(2)*ln(x)+(120*x**2-240*x+120)*ln(2))*exp((ln(x)+x**2)/x)/x**4,x)

[Out]

120*exp((x**2 + log(x))/x)*log(2)/x**2

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