3.33.4 $\int \frac{e^{-e^x}(2x^3-e^x{x}^4+e^{5+e^x+x^2+x{\log }^2(x)}(4-8x^2-8x\log (x)-4x{\log }^2(x)))}{4x^2}dx$

Optimal. Leaf size=32 $$-\frac{e^{5+x(x+{\log }^2(x))}}{x}+\frac{1}{4}{e}^{-e^x}{x}^2$$

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Rubi [A]  time = 1.07, antiderivative size = 64, normalized size of antiderivative = 2.00, number of steps used = 5, number of rules used = 3, integrand size = 64, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.047, Rules used = {12, 6742, 2288} $$\begin{array}{c}\frac{1}{4}{e}^{-e^x}{x}^2-\frac{e^{x^2+x{\log }^2(x)+5}(2x^2+x{\log }^2(x)+2x\log (x))}{x^2(2x+{\log }^2(x)+2\log (x))}\end{array}$$

Antiderivative was successfully verified.

[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 2288

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 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{4}\int \frac{e^{-e^x}(2x^3-e^x{x}^4+e^{5+e^x+x^2+x{\log }^2(x)}(4-8x^2-8x\log (x)-4x{\log }^2(x)))}{x^2}dx\\ & =\frac{1}{4}\int (-e^{-e^x}x(-2+e^{x}x)-\frac{4e^{5+x^2+x{\log }^2(x)}(-1+2x^2+2x\log (x)+x{\log }^2(x))}{x^2})dx\\ & =-(\frac{1}{4}\int e^{-e^x}x(-2+e^{x}x)dx)-\int \frac{e^{5+x^2+x{\log }^2(x)}(-1+2x^2+2x\log (x)+x{\log }^2(x))}{x^2}dx\\ & =\frac{1}{4}{e}^{-e^x}{x}^2-\frac{e^{5+x^2+x{\log }^2(x)}(2x^2+2x\log (x)+x{\log }^2(x))}{x^2(2x+2\log (x)+{\log }^2(x))}\end{array}\end{array}$$

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Mathematica [A]  time = 0.45, size = 34, normalized size = 1.06 $$\begin{array}{c}\frac{1}{4}(-\frac{4e^{5+x^2+x{\log }^2(x)}}{x}+e^{-e^x}{x}^2)\end{array}$$

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.51, size = 29, normalized size = 0.91 $$\begin{array}{c}\frac{x^3{e}^{(-e^x)}-4e^{(x\log (x{)}^2+x^2+5)}}{4x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.21, size = 36, normalized size = 1.12 $$\begin{array}{c}\frac{(x^3{e}^{(x-e^x)}-4e^{(x\log (x{)}^2+x^2+x+5)})e^{(-x)}}{4x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.06, size = 29, normalized size = 0.91

method	result	size
risch	$\frac{x^2{\mathrm{e}}^{-{\mathrm{e}}^x}}{4}-\frac{{\mathrm{e}}^{x\ln (x{)}^2+x^2+5}}{x}$	$29$

Verification of antiderivative is not currently implemented for this CAS.

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

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maxima [A]  time = 0.61, size = 29, normalized size = 0.91 $$\begin{array}{c}\frac{x^3{e}^{(-e^x)}-4e^{(x\log (x{)}^2+x^2+5)}}{4x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 $$\begin{array}{c}\int -\frac{{\mathrm{e}}^{-{\mathrm{e}}^x}(\frac{x^4{\mathrm{e}}^x}{4}-\frac{x^3}{2}+\frac{{\mathrm{e}}^{{\mathrm{e}}^x}{\mathrm{e}}^{x^2+x{\ln (x)}^2+5}(8x^2+4x{\ln (x)}^2+8x\ln (x)-4)}{4})}{x^2}dx\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.89, size = 24, normalized size = 0.75 $$\begin{array}{c}\frac{x^2{e}^{-e^x}}{4}-\frac{e^{x^2+x\log {(x)}^2+5}}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[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

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