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

Optimal. Leaf size=23 \[ \frac {2}{5} e^{-3-x} \left (-e^{-3+e}+\log ^2(x)\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 29, normalized size of antiderivative = 1.26, number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12, 2288} \begin {gather*} -\frac {2 e^{-x-3} \left (e^{e-3} x-x \log ^2(x)\right )}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*E^(-3 - x)*(E^(-3 + E)*x - x*Log[x]^2))/(5*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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 28, normalized size = 1.22 \begin {gather*} \frac {1}{5} \left (-2 e^{-6+e-x}+2 e^{-3-x} \log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*E^(-6 + E - x) + 2*E^(-3 - x)*Log[x]^2)/5

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fricas [A]  time = 0.90, size = 23, normalized size = 1.00 \begin {gather*} \frac {2}{5} \, e^{\left (-x - 3\right )} \log \relax (x)^{2} - \frac {2}{5} \, e^{\left (-x + e - 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*e^(-x - 3)*log(x)^2 - 2/5*e^(-x + e - 6)

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giac [A]  time = 0.13, size = 24, normalized size = 1.04 \begin {gather*} \frac {2}{5} \, {\left (e^{\left (-x\right )} \log \relax (x)^{2} - e^{\left (-x + e - 3\right )}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 24, normalized size = 1.04

method	result	size
risch	\(\frac {2 \ln \relax (x )^{2} {\mathrm e}^{-3-x}}{5}-\frac {2 \,{\mathrm e}^{{\mathrm e}-6-x}}{5}\)	\(24\)
norman	\(\left (\frac {2 \,{\mathrm e}^{-3} \ln \relax (x )^{2}}{5}-\frac {2 \left ({\mathrm e}^{-3}\right )^{2} {\mathrm e}^{{\mathrm e}}}{5}\right ) {\mathrm e}^{-x}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*ln(x)^2*exp(-3-x)-2/5*exp(exp(1)-6-x)

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maxima [A]  time = 0.45, size = 23, normalized size = 1.00 \begin {gather*} \frac {2}{5} \, e^{\left (-x - 3\right )} \log \relax (x)^{2} - \frac {2}{5} \, e^{\left (-x + e - 6\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*e^(-x - 3)*log(x)^2 - 2/5*e^(-x + e - 6)

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mupad [B]  time = 5.82, size = 22, normalized size = 0.96 \begin {gather*} -{\mathrm {e}}^{-x-6}\,\left (\frac {2\,{\mathrm {e}}^{\mathrm {e}}}{5}-\frac {2\,{\mathrm {e}}^3\,{\ln \relax (x)}^2}{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-x)*exp(-3)*((4*log(x))/5 + (2*x*exp(exp(1) - 3))/5 - (2*x*log(x)^2)/5))/x,x)

[Out]

-exp(- x - 6)*((2*exp(exp(1)))/5 - (2*exp(3)*log(x)^2)/5)

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sympy [A]  time = 0.38, size = 24, normalized size = 1.04 \begin {gather*} \frac {\left (2 e^{3} \log {\relax (x )}^{2} - 2 e^{e}\right ) e^{- x}}{5 e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*exp(3)*log(x)**2 - 2*exp(E))*exp(-6)*exp(-x)/5

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