\(\int \frac {4 e^{-3-x} (-1-x) \log ^2(2) \log ^2(\log (2))}{x^2} \, dx\) [7392]

Optimal result

Integrand size = 26, antiderivative size = 21 \[ \int \frac {4 e^{-3-x} (-1-x) \log ^2(2) \log ^2(\log (2))}{x^2} \, dx=\frac {4 e^{-3-x} \log ^2(2) \log ^2(\log (2))}{x} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 2228} \[ \int \frac {4 e^{-3-x} (-1-x) \log ^2(2) \log ^2(\log (2))}{x^2} \, dx=\frac {4 e^{-x-3} \log ^2(2) \log ^2(\log (2))}{x} \]

[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 2228

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{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {4 e^{-3-x} (-1-x) \log ^2(2) \log ^2(\log (2))}{x^2} \, dx=\frac {4 e^{-3-x} \log ^2(2) \log ^2(\log (2))}{x} \]

[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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00

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

[In]

int(4*(-1-x)*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)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {4 e^{-3-x} (-1-x) \log ^2(2) \log ^2(\log (2))}{x^2} \, dx=\frac {4 \, e^{\left (-x - 3\right )} \log \left (2\right )^{2} \log \left (\log \left (2\right )\right )^{2}}{x} \]

[In]

integrate(4*(-1-x)*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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {4 e^{-3-x} (-1-x) \log ^2(2) \log ^2(\log (2))}{x^2} \, dx=\frac {4 e^{- x - 3} \log {\left (2 \right )}^{2} \log {\left (\log {\left (2 \right )} \right )}^{2}}{x} \]

[In]

integrate(4*(-1-x)*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

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.24 \[ \int \frac {4 e^{-3-x} (-1-x) \log ^2(2) \log ^2(\log (2))}{x^2} \, dx=-4 \, {\left ({\rm Ei}\left (-x\right ) e^{\left (-3\right )} - e^{\left (-3\right )} \Gamma \left (-1, x\right )\right )} \log \left (2\right )^{2} \log \left (\log \left (2\right )\right )^{2} \]

[In]

integrate(4*(-1-x)*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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {4 e^{-3-x} (-1-x) \log ^2(2) \log ^2(\log (2))}{x^2} \, dx=\frac {4 \, e^{\left (-x - 3\right )} \log \left (2\right )^{2} \log \left (\log \left (2\right )\right )^{2}}{x} \]

[In]

integrate(4*(-1-x)*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

Mupad [B] (verification not implemented)

Time = 11.06 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {4 e^{-3-x} (-1-x) \log ^2(2) \log ^2(\log (2))}{x^2} \, dx=\frac {4\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-3}\,{\ln \left (2\right )}^2\,{\ln \left (\ln \left (2\right )\right )}^2}{x} \]

[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