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

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

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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 {gather*} \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 )} \end {gather*}

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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

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 {gather*} \frac {x^{3} e^{\left (-e^{x}\right )} - 4 \, e^{\left (x \log \relax (x)^{2} + x^{2} + 5\right )}}{4 \, x} \end {gather*}

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 {gather*} \frac {{\left (x^{3} e^{\left (x - e^{x}\right )} - 4 \, e^{\left (x \log \relax (x)^{2} + x^{2} + x + 5\right )}\right )} e^{\left (-x\right )}}{4 \, x} \end {gather*}

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	\(\frac {x^{2} {\mathrm e}^{-{\mathrm e}^{x}}}{4}-\frac {{\mathrm e}^{x \ln \relax (x )^{2}+x^{2}+5}}{x}\)	\(29\)

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 {gather*} \frac {x^{3} e^{\left (-e^{x}\right )} - 4 \, e^{\left (x \log \relax (x)^{2} + x^{2} + 5\right )}}{4 \, x} \end {gather*}

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 {gather*} \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 \relax (x)}^2+5}\,\left (8\,x^2+4\,x\,{\ln \relax (x)}^2+8\,x\,\ln \relax (x)-4\right )}{4}\right )}{x^2} \,d x \end {gather*}

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 {gather*} \frac {x^{2} e^{- e^{x}}}{4} - \frac {e^{x^{2} + x \log {\relax (x )}^{2} + 5}}{x} \end {gather*}

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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