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

Optimal. Leaf size=31 \[ e^{-x} \left (-3+2 x+\frac {x-x (5+x)+\log (2)}{x}-x \log (\log (x))\right ) \]

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Rubi [F]  time = 1.01, 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^{-x} \left (-x^2+\left (8 x^2-x^3+(-1-x) \log (2)\right ) \log (x)+\left (-x^2+x^3\right ) \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-7/E^x + x/E^x + Log[2]/(E^x*x) - Defer[Int][1/(E^x*Log[x]), x] - Defer[Int][Log[Log[x]]/E^x, x] + Defer[Int][
(x*Log[Log[x]])/E^x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {e^{-x} \left (-x^2+8 x^2 \log (x)-x^3 \log (x)-\log (2) \log (x)-x \log (2) \log (x)\right )}{x^2 \log (x)}+e^{-x} (-1+x) \log (\log (x))\right ) \, dx\\ &=\int \frac {e^{-x} \left (-x^2+8 x^2 \log (x)-x^3 \log (x)-\log (2) \log (x)-x \log (2) \log (x)\right )}{x^2 \log (x)} \, dx+\int e^{-x} (-1+x) \log (\log (x)) \, dx\\ &=\int \left (\frac {e^{-x} \left (8 x^2-x^3-\log (2)-x \log (2)\right )}{x^2}-\frac {e^{-x}}{\log (x)}\right ) \, dx+\int \left (-e^{-x} \log (\log (x))+e^{-x} x \log (\log (x))\right ) \, dx\\ &=\int \frac {e^{-x} \left (8 x^2-x^3-\log (2)-x \log (2)\right )}{x^2} \, dx-\int \frac {e^{-x}}{\log (x)} \, dx-\int e^{-x} \log (\log (x)) \, dx+\int e^{-x} x \log (\log (x)) \, dx\\ &=\int \left (8 e^{-x}-e^{-x} x-\frac {e^{-x} \log (2)}{x^2}-\frac {e^{-x} \log (2)}{x}\right ) \, dx-\int \frac {e^{-x}}{\log (x)} \, dx-\int e^{-x} \log (\log (x)) \, dx+\int e^{-x} x \log (\log (x)) \, dx\\ &=8 \int e^{-x} \, dx-\log (2) \int \frac {e^{-x}}{x^2} \, dx-\log (2) \int \frac {e^{-x}}{x} \, dx-\int e^{-x} x \, dx-\int \frac {e^{-x}}{\log (x)} \, dx-\int e^{-x} \log (\log (x)) \, dx+\int e^{-x} x \log (\log (x)) \, dx\\ &=-8 e^{-x}+e^{-x} x+\frac {e^{-x} \log (2)}{x}-\text {Ei}(-x) \log (2)+\log (2) \int \frac {e^{-x}}{x} \, dx-\int e^{-x} \, dx-\int \frac {e^{-x}}{\log (x)} \, dx-\int e^{-x} \log (\log (x)) \, dx+\int e^{-x} x \log (\log (x)) \, dx\\ &=-7 e^{-x}+e^{-x} x+\frac {e^{-x} \log (2)}{x}-\int \frac {e^{-x}}{\log (x)} \, dx-\int e^{-x} \log (\log (x)) \, dx+\int e^{-x} x \log (\log (x)) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 27, normalized size = 0.87 \begin {gather*} e^{-x} \left (-7+x+\frac {\log (2)}{x}\right )-e^{-x} x \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-7 + x + Log[2]/x)/E^x - (x*Log[Log[x]])/E^x

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fricas [A]  time = 0.63, size = 32, normalized size = 1.03 \begin {gather*} -\frac {x^{2} e^{\left (-x\right )} \log \left (\log \relax (x)\right ) - {\left (x^{2} - 7 \, x + \log \relax (2)\right )} e^{\left (-x\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-x^2)*log(x)*log(log(x))+((-x-1)*log(2)-x^3+8*x^2)*log(x)-x^2)/x^2/exp(x)/log(x),x, algorithm="
fricas")

[Out]

-(x^2*e^(-x)*log(log(x)) - (x^2 - 7*x + log(2))*e^(-x))/x

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giac [A]  time = 0.16, size = 41, normalized size = 1.32 \begin {gather*} -\frac {x^{2} e^{\left (-x\right )} \log \left (\log \relax (x)\right ) - x^{2} e^{\left (-x\right )} + 7 \, x e^{\left (-x\right )} - e^{\left (-x\right )} \log \relax (2)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-x^2)*log(x)*log(log(x))+((-x-1)*log(2)-x^3+8*x^2)*log(x)-x^2)/x^2/exp(x)/log(x),x, algorithm="
giac")

[Out]

-(x^2*e^(-x)*log(log(x)) - x^2*e^(-x) + 7*x*e^(-x) - e^(-x)*log(2))/x

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maple [A]  time = 0.03, size = 29, normalized size = 0.94

method	result	size
risch	\(-x \,{\mathrm e}^{-x} \ln \left (\ln \relax (x )\right )+\frac {\left (x^{2}+\ln \relax (2)-7 x \right ) {\mathrm e}^{-x}}{x}\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3-x^2)*ln(x)*ln(ln(x))+((-x-1)*ln(2)-x^3+8*x^2)*ln(x)-x^2)/x^2/exp(x)/ln(x),x,method=_RETURNVERBOSE)

[Out]

-x*exp(-x)*ln(ln(x))+(x^2+ln(2)-7*x)/x*exp(-x)

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maxima [C]  time = 0.40, size = 39, normalized size = 1.26 \begin {gather*} -x e^{\left (-x\right )} \log \left (\log \relax (x)\right ) + {\left (x + 1\right )} e^{\left (-x\right )} - {\rm Ei}\left (-x\right ) \log \relax (2) + \Gamma \left (-1, x\right ) \log \relax (2) - 8 \, e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-x^2)*log(x)*log(log(x))+((-x-1)*log(2)-x^3+8*x^2)*log(x)-x^2)/x^2/exp(x)/log(x),x, algorithm="
maxima")

[Out]

-x*e^(-x)*log(log(x)) + (x + 1)*e^(-x) - Ei(-x)*log(2) + gamma(-1, x)*log(2) - 8*e^(-x)

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mupad [B]  time = 4.49, size = 28, normalized size = 0.90 \begin {gather*} \frac {{\mathrm {e}}^{-x}\,\left (x^2-7\,x+\ln \relax (2)\right )}{x}-x\,\ln \left (\ln \relax (x)\right )\,{\mathrm {e}}^{-x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(-x)*(log(x)*(log(2)*(x + 1) - 8*x^2 + x^3) + x^2 + log(log(x))*log(x)*(x^2 - x^3)))/(x^2*log(x)),x)

[Out]

(exp(-x)*(log(2) - 7*x + x^2))/x - x*log(log(x))*exp(-x)

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sympy [A]  time = 0.37, size = 22, normalized size = 0.71 \begin {gather*} \frac {\left (- x^{2} \log {\left (\log {\relax (x )} \right )} + x^{2} - 7 x + \log {\relax (2 )}\right ) e^{- x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3-x**2)*ln(x)*ln(ln(x))+((-x-1)*ln(2)-x**3+8*x**2)*ln(x)-x**2)/x**2/exp(x)/ln(x),x)

[Out]

(-x**2*log(log(x)) + x**2 - 7*x + log(2))*exp(-x)/x

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