3.100.46 \(\int \frac {1}{10} e^{-136-x} (1-2 \log (4)+(1-x) \log (x)) \, dx\)

Optimal. Leaf size=21 \[ \frac {1}{5} e^{-136-x} \left (\log (4)+\frac {1}{2} x \log (x)\right ) \]

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Rubi [B]  time = 0.10, antiderivative size = 60, normalized size of antiderivative = 2.86, number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {12, 6742, 2194, 2176, 2554} \begin {gather*} \frac {e^{-x-136}}{10}+\frac {1}{10} e^{-x-136} \log (x)-\frac {1}{10} e^{-x-136} (1-x) \log (x)-\frac {1}{10} e^{-x-136} (1-\log (16)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(-136 - x)*(1 - 2*Log[4] + (1 - x)*Log[x]))/10,x]

[Out]

E^(-136 - x)/10 - (E^(-136 - x)*(1 - Log[16]))/10 + (E^(-136 - x)*Log[x])/10 - (E^(-136 - x)*(1 - x)*Log[x])/1
0

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 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 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}{10} \int e^{-136-x} (1-2 \log (4)+(1-x) \log (x)) \, dx\\ &=\frac {1}{10} \int \left (e^{-136-x} (1-2 \log (4))-e^{-136-x} (-1+x) \log (x)\right ) \, dx\\ &=-\left (\frac {1}{10} \int e^{-136-x} (-1+x) \log (x) \, dx\right )+\frac {1}{10} (1-2 \log (4)) \int e^{-136-x} \, dx\\ &=-\frac {1}{10} e^{-136-x} (1-\log (16))+\frac {1}{10} e^{-136-x} \log (x)-\frac {1}{10} e^{-136-x} (1-x) \log (x)-\frac {1}{10} \int e^{-136-x} \, dx\\ &=\frac {e^{-136-x}}{10}-\frac {1}{10} e^{-136-x} (1-\log (16))+\frac {1}{10} e^{-136-x} \log (x)-\frac {1}{10} e^{-136-x} (1-x) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 18, normalized size = 0.86 \begin {gather*} \frac {1}{10} e^{-136-x} (\log (16)+x \log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-136 - x)*(1 - 2*Log[4] + (1 - x)*Log[x]))/10,x]

[Out]

(E^(-136 - x)*(Log[16] + x*Log[x]))/10

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fricas [A]  time = 1.64, size = 22, normalized size = 1.05 \begin {gather*} \frac {1}{10} \, x e^{\left (-x - 136\right )} \log \relax (x) + \frac {2}{5} \, e^{\left (-x - 136\right )} \log \relax (2) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((-x+1)*log(x)-4*log(2)+1)/exp(x+136),x, algorithm="fricas")

[Out]

1/10*x*e^(-x - 136)*log(x) + 2/5*e^(-x - 136)*log(2)

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giac [A]  time = 1.04, size = 21, normalized size = 1.00 \begin {gather*} \frac {1}{10} \, {\left (x e^{\left (-x\right )} \log \relax (x) + 4 \, e^{\left (-x\right )} \log \relax (2)\right )} e^{\left (-136\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((-x+1)*log(x)-4*log(2)+1)/exp(x+136),x, algorithm="giac")

[Out]

1/10*(x*e^(-x)*log(x) + 4*e^(-x)*log(2))*e^(-136)

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maple [A]  time = 0.05, size = 18, normalized size = 0.86




method result size



norman \(\left (\frac {x \ln \relax (x )}{10}+\frac {2 \ln \relax (2)}{5}\right ) {\mathrm e}^{-x -136}\) \(18\)
risch \(\frac {x \,{\mathrm e}^{-x -136} \ln \relax (x )}{10}+\frac {2 \,{\mathrm e}^{-x -136} \ln \relax (2)}{5}\) \(23\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/10*((1-x)*ln(x)-4*ln(2)+1)/exp(x+136),x,method=_RETURNVERBOSE)

[Out]

(1/10*x*ln(x)+2/5*ln(2))/exp(x+136)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{10} \, {\left (x + 1\right )} e^{\left (-x - 136\right )} \log \relax (x) + \frac {1}{10} \, {\rm Ei}\left (-x\right ) e^{\left (-136\right )} + \frac {2}{5} \, e^{\left (-x - 136\right )} \log \relax (2) - \frac {1}{10} \, e^{\left (-x - 136\right )} \log \relax (x) - \frac {1}{10} \, e^{\left (-x - 136\right )} - \frac {1}{10} \, \int \frac {{\left (x + 1\right )} e^{\left (-x - 136\right )}}{x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((-x+1)*log(x)-4*log(2)+1)/exp(x+136),x, algorithm="maxima")

[Out]

1/10*(x + 1)*e^(-x - 136)*log(x) + 1/10*Ei(-x)*e^(-136) + 2/5*e^(-x - 136)*log(2) - 1/10*e^(-x - 136)*log(x) -
 1/10*e^(-x - 136) - 1/10*integrate((x + 1)*e^(-x - 136)/x, x)

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mupad [B]  time = 7.41, size = 17, normalized size = 0.81 \begin {gather*} {\mathrm {e}}^{-x-136}\,\left (\frac {\ln \left (16\right )}{10}+\frac {x\,\ln \relax (x)}{10}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(- x - 136)*((2*log(2))/5 + (log(x)*(x - 1))/10 - 1/10),x)

[Out]

exp(- x - 136)*(log(16)/10 + (x*log(x))/10)

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sympy [A]  time = 0.34, size = 17, normalized size = 0.81 \begin {gather*} \frac {\left (x \log {\relax (x )} + 4 \log {\relax (2 )}\right ) e^{- x - 136}}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/10*((-x+1)*ln(x)-4*ln(2)+1)/exp(x+136),x)

[Out]

(x*log(x) + 4*log(2))*exp(-x - 136)/10

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