3.15.54 \(\int \frac {1}{128} e^{-x} (e^{2 e^2} x+e^{2 e^2} (2 x-x^2) \log (x)) \, dx\)

Optimal. Leaf size=20 \[ \frac {1}{128} e^{2 e^2-x} x^2 \log (x) \]

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Rubi [A]  time = 0.31, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.184, Rules used = {12, 6688, 6742, 2176, 2194, 2196, 2554} \begin {gather*} \frac {1}{128} e^{2 e^2-x} x^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*E^2)*x + E^(2*E^2)*(2*x - x^2)*Log[x])/(128*E^x),x]

[Out]

(E^(2*E^2 - x)*x^2*Log[x])/128

Rule 12

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, 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}{128} \int e^{-x} \left (e^{2 e^2} x+e^{2 e^2} \left (2 x-x^2\right ) \log (x)\right ) \, dx\\ &=\frac {1}{128} \int e^{2 e^2-x} x (1-(-2+x) \log (x)) \, dx\\ &=\frac {1}{128} \int \left (e^{2 e^2-x} x-e^{2 e^2-x} (-2+x) x \log (x)\right ) \, dx\\ &=\frac {1}{128} \int e^{2 e^2-x} x \, dx-\frac {1}{128} \int e^{2 e^2-x} (-2+x) x \log (x) \, dx\\ &=-\frac {1}{128} e^{2 e^2-x} x+\frac {1}{128} e^{2 e^2-x} x^2 \log (x)+\frac {1}{128} \int e^{2 e^2-x} \, dx-\frac {1}{128} \int e^{2 e^2-x} x \, dx\\ &=-\frac {1}{128} e^{2 e^2-x}+\frac {1}{128} e^{2 e^2-x} x^2 \log (x)-\frac {1}{128} \int e^{2 e^2-x} \, dx\\ &=\frac {1}{128} e^{2 e^2-x} x^2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 20, normalized size = 1.00 \begin {gather*} \frac {1}{128} e^{2 e^2-x} x^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*E^2)*x + E^(2*E^2)*(2*x - x^2)*Log[x])/(128*E^x),x]

[Out]

(E^(2*E^2 - x)*x^2*Log[x])/128

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fricas [A]  time = 1.02, size = 16, normalized size = 0.80 \begin {gather*} \frac {1}{128} \, x^{2} e^{\left (-x + 2 \, e^{2}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/128*((-x^2+2*x)*exp(exp(2))^2*log(x)+x*exp(exp(2))^2)/exp(x),x, algorithm="fricas")

[Out]

1/128*x^2*e^(-x + 2*e^2)*log(x)

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giac [A]  time = 0.25, size = 16, normalized size = 0.80 \begin {gather*} \frac {1}{128} \, x^{2} e^{\left (-x + 2 \, e^{2}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/128*((-x^2+2*x)*exp(exp(2))^2*log(x)+x*exp(exp(2))^2)/exp(x),x, algorithm="giac")

[Out]

1/128*x^2*e^(-x + 2*e^2)*log(x)

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maple [A]  time = 0.05, size = 17, normalized size = 0.85

method	result	size
norman	\(\frac {{\mathrm e}^{2 \,{\mathrm e}^{2}} {\mathrm e}^{-x} x^{2} \ln \relax (x )}{128}\)	\(17\)
risch	\(\frac {x^{2} \ln \relax (x ) {\mathrm e}^{2 \,{\mathrm e}^{2}-x}}{128}\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/128*((-x^2+2*x)*exp(exp(2))^2*ln(x)+x*exp(exp(2))^2)/exp(x),x,method=_RETURNVERBOSE)

[Out]

1/128*exp(exp(2))^2/exp(x)*x^2*ln(x)

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maxima [B]  time = 0.56, size = 50, normalized size = 2.50 \begin {gather*} \frac {1}{128} \, {\left (x^{2} e^{\left (2 \, e^{2}\right )} \log \relax (x) + x e^{\left (2 \, e^{2}\right )} + e^{\left (2 \, e^{2}\right )}\right )} e^{\left (-x\right )} - \frac {1}{128} \, {\left (x e^{\left (2 \, e^{2}\right )} + e^{\left (2 \, e^{2}\right )}\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/128*((-x^2+2*x)*exp(exp(2))^2*log(x)+x*exp(exp(2))^2)/exp(x),x, algorithm="maxima")

[Out]

1/128*(x^2*e^(2*e^2)*log(x) + x*e^(2*e^2) + e^(2*e^2))*e^(-x) - 1/128*(x*e^(2*e^2) + e^(2*e^2))*e^(-x)

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mupad [B]  time = 1.14, size = 16, normalized size = 0.80 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{2\,{\mathrm {e}}^2}\,{\mathrm {e}}^{-x}\,\ln \relax (x)}{128} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-x)*((x*exp(2*exp(2)))/128 + (exp(2*exp(2))*log(x)*(2*x - x^2))/128),x)

[Out]

(x^2*exp(2*exp(2))*exp(-x)*log(x))/128

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sympy [A]  time = 0.31, size = 17, normalized size = 0.85 \begin {gather*} \frac {x^{2} e^{- x} e^{2 e^{2}} \log {\relax (x )}}{128} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/128*((-x**2+2*x)*exp(exp(2))**2*ln(x)+x*exp(exp(2))**2)/exp(x),x)

[Out]

x**2*exp(-x)*exp(2*exp(2))*log(x)/128

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