\(\int e^{-2 e^x x} ((4 x-4 x^3) \log (x)+(4 x-8 x^3+e^x (-4 x^2-4 x^3+4 x^4+4 x^5)) \log ^2(x)) \, dx\) [5605]

Optimal result

Integrand size = 61, antiderivative size = 23 \[ \int e^{-2 e^x x} \left (\left (4 x-4 x^3\right ) \log (x)+\left (4 x-8 x^3+e^x \left (-4 x^2-4 x^3+4 x^4+4 x^5\right )\right ) \log ^2(x)\right ) \, dx=2 e^{-2 e^x x} \left (x^2-x^4\right ) \log ^2(x) \]

[Out]

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

Rubi [F]

\[ \int e^{-2 e^x x} \left (\left (4 x-4 x^3\right ) \log (x)+\left (4 x-8 x^3+e^x \left (-4 x^2-4 x^3+4 x^4+4 x^5\right )\right ) \log ^2(x)\right ) \, dx=\int e^{-2 e^x x} \left (\left (4 x-4 x^3\right ) \log (x)+\left (4 x-8 x^3+e^x \left (-4 x^2-4 x^3+4 x^4+4 x^5\right )\right ) \log ^2(x)\right ) \, dx \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 3.76 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int e^{-2 e^x x} \left (\left (4 x-4 x^3\right ) \log (x)+\left (4 x-8 x^3+e^x \left (-4 x^2-4 x^3+4 x^4+4 x^5\right )\right ) \log ^2(x)\right ) \, dx=-2 e^{-2 e^x x} (-1+x) x^2 (1+x) \log ^2(x) \]

[In]

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

[Out]

(-2*(-1 + x)*x^2*(1 + x)*Log[x]^2)/E^(2*E^x*x)

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91

method	result	size
risch	\(-2 \ln \left (x \right )^{2} \left (x^{2}-1\right ) x^{2} {\mathrm e}^{-2 \,{\mathrm e}^{x} x}\)	\(21\)
parallelrisch	\(\left (-2 x^{4} \ln \left (x \right )^{2}+2 x^{2} \ln \left (x \right )^{2}\right ) {\mathrm e}^{-2 \,{\mathrm e}^{x} x}\)	\(28\)

[In]

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

[Out]

-2*ln(x)^2*(x^2-1)*x^2*exp(-2*exp(x)*x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int e^{-2 e^x x} \left (\left (4 x-4 x^3\right ) \log (x)+\left (4 x-8 x^3+e^x \left (-4 x^2-4 x^3+4 x^4+4 x^5\right )\right ) \log ^2(x)\right ) \, dx=-2 \, {\left (x^{4} - x^{2}\right )} e^{\left (-2 \, x e^{x}\right )} \log \left (x\right )^{2} \]

[In]

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

[Out]

-2*(x^4 - x^2)*e^(-2*x*e^x)*log(x)^2

Sympy [A] (verification not implemented)

Time = 3.68 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int e^{-2 e^x x} \left (\left (4 x-4 x^3\right ) \log (x)+\left (4 x-8 x^3+e^x \left (-4 x^2-4 x^3+4 x^4+4 x^5\right )\right ) \log ^2(x)\right ) \, dx=\left (- 2 x^{4} \log {\left (x \right )}^{2} + 2 x^{2} \log {\left (x \right )}^{2}\right ) e^{- 2 x e^{x}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int e^{-2 e^x x} \left (\left (4 x-4 x^3\right ) \log (x)+\left (4 x-8 x^3+e^x \left (-4 x^2-4 x^3+4 x^4+4 x^5\right )\right ) \log ^2(x)\right ) \, dx=-2 \, {\left (x^{4} - x^{2}\right )} e^{\left (-2 \, x e^{x}\right )} \log \left (x\right )^{2} \]

[In]

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

[Out]

-2*(x^4 - x^2)*e^(-2*x*e^x)*log(x)^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.74 \[ \int e^{-2 e^x x} \left (\left (4 x-4 x^3\right ) \log (x)+\left (4 x-8 x^3+e^x \left (-4 x^2-4 x^3+4 x^4+4 x^5\right )\right ) \log ^2(x)\right ) \, dx=-2 \, {\left (x^{4} e^{\left (-2 \, x e^{x} + x\right )} \log \left (x\right )^{2} - x^{2} e^{\left (-2 \, x e^{x} + x\right )} \log \left (x\right )^{2}\right )} e^{\left (-x\right )} \]

[In]

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

[Out]

-2*(x^4*e^(-2*x*e^x + x)*log(x)^2 - x^2*e^(-2*x*e^x + x)*log(x)^2)*e^(-x)

Mupad [F(-1)]

Timed out. \[ \int e^{-2 e^x x} \left (\left (4 x-4 x^3\right ) \log (x)+\left (4 x-8 x^3+e^x \left (-4 x^2-4 x^3+4 x^4+4 x^5\right )\right ) \log ^2(x)\right ) \, dx=\int -{\mathrm {e}}^{-2\,x\,{\mathrm {e}}^x}\,\left ({\ln \left (x\right )}^2\,\left ({\mathrm {e}}^x\,\left (-4\,x^5-4\,x^4+4\,x^3+4\,x^2\right )-4\,x+8\,x^3\right )-\ln \left (x\right )\,\left (4\,x-4\,x^3\right )\right ) \,d x \]

[In]

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

[Out]

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