\(\int \frac {e^{-x} (-x^2+(8 x^2-x^3+(-1-x) \log (2)) \log (x)+(-x^2+x^3) \log (x) \log (\log (x)))}{x^2 \log (x)} \, dx\) [6029]

Optimal result

Integrand size = 56, antiderivative size = 31 \[ \int \frac {e^{-x} \left (-x^2+\left (8 x^2-x^3+(-1-x) \log (2)\right ) \log (x)+\left (-x^2+x^3\right ) \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx=e^{-x} \left (-3+2 x+\frac {x-x (5+x)+\log (2)}{x}-x \log (\log (x))\right ) \]

[Out]

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

Rubi [F]

\[ \int \frac {e^{-x} \left (-x^2+\left (8 x^2-x^3+(-1-x) \log (2)\right ) \log (x)+\left (-x^2+x^3\right ) \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx=\int \frac {e^{-x} \left (-x^2+\left (8 x^2-x^3+(-1-x) \log (2)\right ) \log (x)+\left (-x^2+x^3\right ) \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx \]

[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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.43 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int \frac {e^{-x} \left (-x^2+\left (8 x^2-x^3+(-1-x) \log (2)\right ) \log (x)+\left (-x^2+x^3\right ) \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx=e^{-x} \left (-7+x+\frac {\log (2)}{x}\right )-e^{-x} x \log (\log (x)) \]

[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

Maple [A] (verified)

Time = 3.09 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{-x} \left (-x^{2} \ln \left (\ln \left (x \right )\right )+x^{2}+\ln \left (2\right )-7 x \right )}{x}\)	\(26\)
risch	\(-x \,{\mathrm e}^{-x} \ln \left (\ln \left (x \right )\right )+\frac {\left (x^{2}+\ln \left (2\right )-7 x \right ) {\mathrm e}^{-x}}{x}\)	\(29\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-x} \left (-x^2+\left (8 x^2-x^3+(-1-x) \log (2)\right ) \log (x)+\left (-x^2+x^3\right ) \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx=-\frac {x^{2} e^{\left (-x\right )} \log \left (\log \left (x\right )\right ) - {\left (x^{2} - 7 \, x + \log \left (2\right )\right )} e^{\left (-x\right )}}{x} \]

[In]

integrate(((x^3-x^2)*log(x)*log(log(x))+((-1-x)*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

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.71 \[ \int \frac {e^{-x} \left (-x^2+\left (8 x^2-x^3+(-1-x) \log (2)\right ) \log (x)+\left (-x^2+x^3\right ) \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx=\frac {\left (- x^{2} \log {\left (\log {\left (x \right )} \right )} + x^{2} - 7 x + \log {\left (2 \right )}\right ) e^{- x}}{x} \]

[In]

integrate(((x**3-x**2)*ln(x)*ln(ln(x))+((-1-x)*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

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.24 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int \frac {e^{-x} \left (-x^2+\left (8 x^2-x^3+(-1-x) \log (2)\right ) \log (x)+\left (-x^2+x^3\right ) \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx=-x e^{\left (-x\right )} \log \left (\log \left (x\right )\right ) + {\left (x + 1\right )} e^{\left (-x\right )} - {\rm Ei}\left (-x\right ) \log \left (2\right ) + \Gamma \left (-1, x\right ) \log \left (2\right ) - 8 \, e^{\left (-x\right )} \]

[In]

integrate(((x^3-x^2)*log(x)*log(log(x))+((-1-x)*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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.32 \[ \int \frac {e^{-x} \left (-x^2+\left (8 x^2-x^3+(-1-x) \log (2)\right ) \log (x)+\left (-x^2+x^3\right ) \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx=-\frac {x^{2} e^{\left (-x\right )} \log \left (\log \left (x\right )\right ) - x^{2} e^{\left (-x\right )} + 7 \, x e^{\left (-x\right )} - e^{\left (-x\right )} \log \left (2\right )}{x} \]

[In]

integrate(((x^3-x^2)*log(x)*log(log(x))+((-1-x)*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

Mupad [B] (verification not implemented)

Time = 12.14 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {e^{-x} \left (-x^2+\left (8 x^2-x^3+(-1-x) \log (2)\right ) \log (x)+\left (-x^2+x^3\right ) \log (x) \log (\log (x))\right )}{x^2 \log (x)} \, dx=\frac {{\mathrm {e}}^{-x}\,\left (x^2-7\,x+\ln \left (2\right )\right )}{x}-x\,\ln \left (\ln \left (x\right )\right )\,{\mathrm {e}}^{-x} \]

[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)