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

Optimal. Leaf size=16 \[ \frac {1}{256} e^{-2 x} x^2 \log ^2(x) \]

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Rubi [A]  time = 0.07, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {12, 6741, 2288} \begin {gather*} \frac {1}{256} e^{-2 x} x^2 \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Log[x]^2)/(256*E^(2*x))

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 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{128} \int e^{-2 x} \left (x \log (x)+\left (x-x^2\right ) \log ^2(x)\right ) \, dx\\ &=\frac {1}{128} \int e^{-2 x} x \log (x) (1+\log (x)-x \log (x)) \, dx\\ &=\frac {1}{256} e^{-2 x} x^2 \log ^2(x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 16, normalized size = 1.00 \begin {gather*} \frac {1}{256} e^{-2 x} x^2 \log ^2(x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*Log[x]^2)/(256*E^(2*x))

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fricas [A]  time = 1.02, size = 13, normalized size = 0.81 \begin {gather*} \frac {1}{256} \, x^{2} e^{\left (-2 \, x\right )} \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*x^2*e^(-2*x)*log(x)^2

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giac [A]  time = 0.25, size = 13, normalized size = 0.81 \begin {gather*} \frac {1}{256} \, x^{2} e^{\left (-2 \, x\right )} \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*x^2*e^(-2*x)*log(x)^2

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maple [A]  time = 0.03, size = 14, normalized size = 0.88

method	result	size
risch	\(\frac {x^{2} \ln \relax (x )^{2} {\mathrm e}^{-2 x}}{256}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*x^2*ln(x)^2*exp(-2*x)

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maxima [A]  time = 0.46, size = 13, normalized size = 0.81 \begin {gather*} \frac {1}{256} \, x^{2} e^{\left (-2 \, x\right )} \log \relax (x)^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*x^2*e^(-2*x)*log(x)^2

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mupad [B]  time = 1.06, size = 13, normalized size = 0.81 \begin {gather*} \frac {x^2\,{\mathrm {e}}^{-2\,x}\,{\ln \relax (x)}^2}{256} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^2*exp(-2*x)*log(x)^2)/256

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sympy [A]  time = 0.31, size = 14, normalized size = 0.88 \begin {gather*} \frac {x^{2} e^{- 2 x} \log {\relax (x )}^{2}}{256} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**2*exp(-2*x)*log(x)**2/256

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