\(\int \frac {e^{-e^x} (2 x^3-e^x x^4+e^{5+e^x+x^2+x \log ^2(x)} (4-8 x^2-8 x \log (x)-4 x \log ^2(x)))}{4 x^2} \, dx\) [3204]

Optimal result

Integrand size = 64, antiderivative size = 32 \[ \int \frac {e^{-e^x} \left (2 x^3-e^x x^4+e^{5+e^x+x^2+x \log ^2(x)} \left (4-8 x^2-8 x \log (x)-4 x \log ^2(x)\right )\right )}{4 x^2} \, dx=-\frac {e^{5+x \left (x+\log ^2(x)\right )}}{x}+\frac {1}{4} e^{-e^x} x^2 \]

[Out]

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

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {12, 6874, 2326} \[ \int \frac {e^{-e^x} \left (2 x^3-e^x x^4+e^{5+e^x+x^2+x \log ^2(x)} \left (4-8 x^2-8 x \log (x)-4 x \log ^2(x)\right )\right )}{4 x^2} \, dx=\frac {1}{4} e^{-e^x} x^2-\frac {e^{x^2+x \log ^2(x)+5} \left (2 x^2+x \log ^2(x)+2 x \log (x)\right )}{x^2 \left (2 x+\log ^2(x)+2 \log (x)\right )} \]

[In]

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

[Out]

x^2/(4*E^E^x) - (E^(5 + x^2 + x*Log[x]^2)*(2*x^2 + 2*x*Log[x] + x*Log[x]^2))/(x^2*(2*x + 2*Log[x] + Log[x]^2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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

Mathematica [A] (verified)

Time = 0.46 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-e^x} \left (2 x^3-e^x x^4+e^{5+e^x+x^2+x \log ^2(x)} \left (4-8 x^2-8 x \log (x)-4 x \log ^2(x)\right )\right )}{4 x^2} \, dx=\frac {1}{4} \left (-\frac {4 e^{5+x^2+x \log ^2(x)}}{x}+e^{-e^x} x^2\right ) \]

[In]

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

[Out]

((-4*E^(5 + x^2 + x*Log[x]^2))/x + x^2/E^E^x)/4

Maple [A] (verified)

Time = 1.49 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91

method	result	size
risch	\(\frac {{\mathrm e}^{-{\mathrm e}^{x}} x^{2}}{4}-\frac {{\mathrm e}^{x \ln \left (x \right )^{2}+x^{2}+5}}{x}\)	\(29\)
parallelrisch	\(\frac {\left (x^{3}-4 \,{\mathrm e}^{{\mathrm e}^{x}} {\mathrm e}^{x \ln \left (x \right )^{2}+x^{2}+5}\right ) {\mathrm e}^{-{\mathrm e}^{x}}}{4 x}\)	\(32\)

[In]

int(1/4*((-4*x*ln(x)^2-8*x*ln(x)-8*x^2+4)*exp(exp(x))*exp(x*ln(x)^2+x^2+5)-exp(x)*x^4+2*x^3)/exp(exp(x))/x^2,x
,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-e^x} \left (2 x^3-e^x x^4+e^{5+e^x+x^2+x \log ^2(x)} \left (4-8 x^2-8 x \log (x)-4 x \log ^2(x)\right )\right )}{4 x^2} \, dx=\frac {x^{3} e^{\left (-e^{x}\right )} - 4 \, e^{\left (x \log \left (x\right )^{2} + x^{2} + 5\right )}}{4 \, x} \]

[In]

integrate(1/4*((-4*x*log(x)^2-8*x*log(x)-8*x^2+4)*exp(exp(x))*exp(x*log(x)^2+x^2+5)-exp(x)*x^4+2*x^3)/exp(exp(
x))/x^2,x, algorithm="fricas")

[Out]

1/4*(x^3*e^(-e^x) - 4*e^(x*log(x)^2 + x^2 + 5))/x

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.75 \[ \int \frac {e^{-e^x} \left (2 x^3-e^x x^4+e^{5+e^x+x^2+x \log ^2(x)} \left (4-8 x^2-8 x \log (x)-4 x \log ^2(x)\right )\right )}{4 x^2} \, dx=\frac {x^{2} e^{- e^{x}}}{4} - \frac {e^{x^{2} + x \log {\left (x \right )}^{2} + 5}}{x} \]

[In]

integrate(1/4*((-4*x*ln(x)**2-8*x*ln(x)-8*x**2+4)*exp(exp(x))*exp(x*ln(x)**2+x**2+5)-exp(x)*x**4+2*x**3)/exp(e
xp(x))/x**2,x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^{-e^x} \left (2 x^3-e^x x^4+e^{5+e^x+x^2+x \log ^2(x)} \left (4-8 x^2-8 x \log (x)-4 x \log ^2(x)\right )\right )}{4 x^2} \, dx=\frac {x^{3} e^{\left (-e^{x}\right )} - 4 \, e^{\left (x \log \left (x\right )^{2} + x^{2} + 5\right )}}{4 \, x} \]

[In]

integrate(1/4*((-4*x*log(x)^2-8*x*log(x)-8*x^2+4)*exp(exp(x))*exp(x*log(x)^2+x^2+5)-exp(x)*x^4+2*x^3)/exp(exp(
x))/x^2,x, algorithm="maxima")

[Out]

1/4*(x^3*e^(-e^x) - 4*e^(x*log(x)^2 + x^2 + 5))/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-e^x} \left (2 x^3-e^x x^4+e^{5+e^x+x^2+x \log ^2(x)} \left (4-8 x^2-8 x \log (x)-4 x \log ^2(x)\right )\right )}{4 x^2} \, dx=\frac {{\left (x^{3} e^{\left (x - e^{x}\right )} - 4 \, e^{\left (x \log \left (x\right )^{2} + x^{2} + x + 5\right )}\right )} e^{\left (-x\right )}}{4 \, x} \]

[In]

integrate(1/4*((-4*x*log(x)^2-8*x*log(x)-8*x^2+4)*exp(exp(x))*exp(x*log(x)^2+x^2+5)-exp(x)*x^4+2*x^3)/exp(exp(
x))/x^2,x, algorithm="giac")

[Out]

1/4*(x^3*e^(x - e^x) - 4*e^(x*log(x)^2 + x^2 + x + 5))*e^(-x)/x

Mupad [F(-1)]

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

[In]

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

[Out]

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