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

Optimal. Leaf size=22 \[ \frac {e^{\frac {2}{2 x+\log (2 x)}} \log (\log (2))}{\log (3)} \]

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Rubi [A]  time = 0.57, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {12, 6688, 6706} \begin {gather*} \frac {\log (\log (2)) e^{\frac {2}{2 x+\log (2 x)}}}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

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

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Mathematica [A]  time = 0.36, size = 22, normalized size = 1.00 \begin {gather*} \frac {e^{\frac {2}{2 x+\log (2 x)}} \log (\log (2))}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.65, size = 21, normalized size = 0.95 \begin {gather*} \frac {e^{\left (\frac {2}{2 \, x + \log \left (2 \, x\right )}\right )} \log \left (\log \relax (2)\right )}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 21, normalized size = 0.95 \begin {gather*} \frac {e^{\left (\frac {2}{2 \, x + \log \left (2 \, x\right )}\right )} \log \left (\log \relax (2)\right )}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 22, normalized size = 1.00

method	result	size
risch	\(\frac {{\mathrm e}^{\frac {2}{\ln \left (2 x \right )+2 x}} \ln \left (\ln \relax (2)\right )}{\ln \relax (3)}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.46, size = 55, normalized size = 2.50 \begin {gather*} {\left (\frac {2 \, x e^{\left (\frac {2}{2 \, x + \log \relax (2) + \log \relax (x)}\right )}}{2 \, x \log \relax (3) + \log \relax (3)} + \frac {e^{\left (\frac {2}{2 \, x + \log \relax (2) + \log \relax (x)}\right )}}{2 \, x \log \relax (3) + \log \relax (3)}\right )} \log \left (\log \relax (2)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.78, size = 22, normalized size = 1.00 \begin {gather*} \ln \left ({\ln \relax (2)}^{\frac {1}{\ln \relax (3)}}\right )\,{\mathrm {e}}^{\frac {2}{2\,x+\ln \left (2\,x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(2)^(1/log(3)))*exp(2/(2*x + log(2*x)))

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sympy [A]  time = 0.47, size = 19, normalized size = 0.86 \begin {gather*} \frac {e^{\frac {2}{2 x + \log {\left (2 x \right )}}} \log {\left (\log {\relax (2 )} \right )}}{\log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

exp(2/(2*x + log(2*x)))*log(log(2))/log(3)

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