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

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

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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} $$\begin{array}{c}\frac{\log (\log (2))e^{\frac{2}{2x+\log (2x)}}}{\log (3)}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\log (\log (2))\int \frac{e^{\frac{2}{2x+\log (2x)}}(-2-4x)}{4x^3\log (3)+4x^2\log (3)\log (2x)+x\log (3){\log }^2(2x)}dx\\ & =\log (\log (2))\int \frac{2e^{\frac{2}{2x+\log (2x)}}(-1-2x)}{x\log (3)(2x+\log (2x){)}^2}dx\\ & =\frac{(2\log (\log (2)))\int \frac{e^{\frac{2}{2x+\log (2x)}}(-1-2x)}{x(2x+\log (2x){)}^2}dx}{\log (3)}\\ & =\frac{e^{\frac{2}{2x+\log (2x)}}\log (\log (2))}{\log (3)}\end{array}\end{array}$$

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

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 $$\begin{array}{c}\frac{e^{\left(\frac{2}{2x+\log \left(2x\right)}\right)}\log (\log (2))}{\log (3)}\end{array}$$

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 $$\begin{array}{c}\frac{e^{\left(\frac{2}{2x+\log \left(2x\right)}\right)}\log (\log (2))}{\log (3)}\end{array}$$

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	$\frac{{\mathrm{e}}^{\frac{2}{\ln \left(2x\right)+2x}}\ln (\ln (2))}{\ln (3)}$	$22$

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 $$\begin{array}{c}(\frac{2x{e}^{\left(\frac{2}{2x+\log (2)+\log (x)}\right)}}{2x\log (3)+\log (3)}+\frac{e^{\left(\frac{2}{2x+\log (2)+\log (x)}\right)}}{2x\log (3)+\log (3)})\log (\log (2))\end{array}$$

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 $$\begin{array}{c}\ln \left({\ln (2)}^{\frac{1}{\ln (3)}}\right){\mathrm{e}}^{\frac{2}{2x+\ln \left(2x\right)}}\end{array}$$

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 $$\begin{array}{c}\frac{e^{\frac{2}{2x+\log \left(2x\right)}}\log (\log (2))}{\log (3)}\end{array}$$

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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