3.74.92 \(\int \frac {4 e^{-3-x} (-1-x) \log ^2(2) \log ^2(\log (2))}{x^2} \, dx\)

Optimal. Leaf size=21 \[ \frac {4 e^{-3-x} \log ^2(2) \log ^2(\log (2))}{x} \]

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Rubi [A]  time = 0.04, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 2197} \begin {gather*} \frac {4 e^{-x-3} \log ^2(2) \log ^2(\log (2))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*E^(-3 - x)*(-1 - x)*Log[2]^2*Log[Log[2]]^2)/x^2,x]

[Out]

(4*E^(-3 - x)*Log[2]^2*Log[Log[2]]^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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.00 \begin {gather*} \frac {4 e^{-3-x} \log ^2(2) \log ^2(\log (2))}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*E^(-3 - x)*(-1 - x)*Log[2]^2*Log[Log[2]]^2)/x^2,x]

[Out]

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

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fricas [A]  time = 0.64, size = 20, normalized size = 0.95 \begin {gather*} \frac {4 \, e^{\left (-x - 3\right )} \log \relax (2)^{2} \log \left (\log \relax (2)\right )^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*e^(-x - 3)*log(2)^2*log(log(2))^2/x

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giac [A]  time = 0.21, size = 20, normalized size = 0.95 \begin {gather*} \frac {4 \, e^{\left (-x - 3\right )} \log \relax (2)^{2} \log \left (\log \relax (2)\right )^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*e^(-x - 3)*log(2)^2*log(log(2))^2/x

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maple [A]  time = 0.05, size = 21, normalized size = 1.00




method result size



gosper \(\frac {4 \ln \relax (2)^{2} \ln \left (\ln \relax (2)\right )^{2} {\mathrm e}^{-3-x}}{x}\) \(21\)
derivativedivides \(\frac {4 \ln \relax (2)^{2} \ln \left (\ln \relax (2)\right )^{2} {\mathrm e}^{-3-x}}{x}\) \(21\)
default \(\frac {4 \ln \relax (2)^{2} \ln \left (\ln \relax (2)\right )^{2} {\mathrm e}^{-3-x}}{x}\) \(21\)
norman \(\frac {4 \ln \relax (2)^{2} \ln \left (\ln \relax (2)\right )^{2} {\mathrm e}^{-3-x}}{x}\) \(21\)
risch \(\frac {4 \ln \relax (2)^{2} \ln \left (\ln \relax (2)\right )^{2} {\mathrm e}^{-3-x}}{x}\) \(21\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*ln(ln(2))^2*ln(2)^2/x/exp(3+x)

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maxima [C]  time = 0.40, size = 26, normalized size = 1.24 \begin {gather*} -4 \, {\left ({\rm Ei}\left (-x\right ) e^{\left (-3\right )} - e^{\left (-3\right )} \Gamma \left (-1, x\right )\right )} \log \relax (2)^{2} \log \left (\log \relax (2)\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*(Ei(-x)*e^(-3) - e^(-3)*gamma(-1, x))*log(2)^2*log(log(2))^2

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mupad [B]  time = 0.09, size = 20, normalized size = 0.95 \begin {gather*} \frac {4\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-3}\,{\ln \relax (2)}^2\,{\ln \left (\ln \relax (2)\right )}^2}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*exp(-x)*exp(-3)*log(2)^2*log(log(2))^2)/x

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sympy [A]  time = 0.14, size = 20, normalized size = 0.95 \begin {gather*} \frac {4 e^{- x - 3} \log {\relax (2 )}^{2} \log {\left (\log {\relax (2 )} \right )}^{2}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*exp(-x - 3)*log(2)**2*log(log(2))**2/x

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