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

Optimal. Leaf size=23 \[ e^{\frac {1}{x+\log \left (\frac {3}{\log \left (\frac {2}{x}\right )}\right )}}+\log \left (\log ^2(x)\right ) \]

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Rubi [A]  time = 2.50, antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 147, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.034, Rules used = {6688, 6742, 2302, 29, 6706} \begin {gather*} 2 \log (\log (x))+e^{\frac {1}{x+\log \left (\frac {3}{\log \left (\frac {2}{x}\right )}\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(x + Log[3/Log[2/x]])^(-1) + 2*Log[Log[x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, 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]

Rule 6742

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

Rubi steps

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

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Mathematica [A]  time = 0.12, size = 23, normalized size = 1.00 \begin {gather*} e^{\frac {1}{x+\log \left (\frac {3}{\log \left (\frac {2}{x}\right )}\right )}}+2 \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(x + Log[3/Log[2/x]])^(-1) + 2*Log[Log[x]]

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fricas [A]  time = 0.85, size = 31, normalized size = 1.35 \begin {gather*} e^{\left (\frac {1}{x + \log \left (\frac {3}{\log \left (\frac {2}{x}\right )}\right )}\right )} + 2 \, \log \left (-\log \relax (2) + \log \left (\frac {2}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(1/(x + log(3/log(2/x)))) + 2*log(-log(2) + log(2/x))

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giac [A]  time = 0.43, size = 23, normalized size = 1.00 \begin {gather*} e^{\left (\frac {1}{x + \log \relax (3) - \log \left (\log \relax (2) - \log \relax (x)\right )}\right )} + 2 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e^(1/(x + log(3) - log(log(2) - log(x)))) + 2*log(log(x))

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maple [C]  time = 1.40, size = 150, normalized size = 6.52

method	result	size
risch	\(2 \ln \left (\ln \relax (x )\right )+{\mathrm e}^{\frac {2}{i \pi \mathrm {csgn}\left (\frac {1}{2 i \ln \relax (2)-2 i \ln \relax (x )}\right )^{3}+i \pi \mathrm {csgn}\left (\frac {1}{2 i \ln \relax (2)-2 i \ln \relax (x )}\right )^{2} \mathrm {csgn}\left (\frac {i}{2 i \ln \relax (2)-2 i \ln \relax (x )}\right )-i \pi \mathrm {csgn}\left (\frac {1}{2 i \ln \relax (2)-2 i \ln \relax (x )}\right )^{2}-i \pi \,\mathrm {csgn}\left (\frac {1}{2 i \ln \relax (2)-2 i \ln \relax (x )}\right ) \mathrm {csgn}\left (\frac {i}{2 i \ln \relax (2)-2 i \ln \relax (x )}\right )+i \pi -2 \ln \left (2 i \ln \relax (2)-2 i \ln \relax (x )\right )+2 \ln \relax (6)+2 x}}\)	\(150\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*ln(2/x)-1)*ln(x)*exp(1/(ln(3/ln(2/x))+x))+2*ln(2/x)*ln(3/ln(2/x))^2+4*x*ln(2/x)*ln(3/ln(2/x))+2*x^2*l
n(2/x))/(x*ln(2/x)*ln(x)*ln(3/ln(2/x))^2+2*x^2*ln(2/x)*ln(x)*ln(3/ln(2/x))+x^3*ln(2/x)*ln(x)),x,method=_RETURN
VERBOSE)

[Out]

2*ln(ln(x))+exp(2/(I*Pi*csgn(1/(2*I*ln(2)-2*I*ln(x)))^3+I*Pi*csgn(1/(2*I*ln(2)-2*I*ln(x)))^2*csgn(I/(2*I*ln(2)
-2*I*ln(x)))-I*Pi*csgn(1/(2*I*ln(2)-2*I*ln(x)))^2-I*Pi*csgn(1/(2*I*ln(2)-2*I*ln(x)))*csgn(I/(2*I*ln(2)-2*I*ln(
x)))+I*Pi-2*ln(2*I*ln(2)-2*I*ln(x))+2*ln(6)+2*x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left (x \log \relax (2) - x \log \relax (x) + 1\right )} e^{\left (\frac {1}{x + \log \relax (3) - \log \left (\log \relax (2) - \log \relax (x)\right )}\right )}}{x^{3} \log \relax (2) + 2 \, x^{2} \log \relax (3) \log \relax (2) + x \log \relax (3)^{2} \log \relax (2) + {\left (x \log \relax (2) - x \log \relax (x)\right )} \log \left (\log \relax (2) - \log \relax (x)\right )^{2} - {\left (x^{3} + 2 \, x^{2} \log \relax (3) + x \log \relax (3)^{2}\right )} \log \relax (x) - 2 \, {\left (x^{2} \log \relax (2) + x \log \relax (3) \log \relax (2) - {\left (x^{2} + x \log \relax (3)\right )} \log \relax (x)\right )} \log \left (\log \relax (2) - \log \relax (x)\right )}\,{d x} + 2 \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-integrate((x*log(2) - x*log(x) + 1)*e^(1/(x + log(3) - log(log(2) - log(x))))/(x^3*log(2) + 2*x^2*log(3)*log(
2) + x*log(3)^2*log(2) + (x*log(2) - x*log(x))*log(log(2) - log(x))^2 - (x^3 + 2*x^2*log(3) + x*log(3)^2)*log(
x) - 2*(x^2*log(2) + x*log(3)*log(2) - (x^2 + x*log(3))*log(x))*log(log(2) - log(x))), x) + 2*log(log(x))

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mupad [B]  time = 1.55, size = 22, normalized size = 0.96 \begin {gather*} 2\,\ln \left (\ln \relax (x)\right )+{\mathrm {e}}^{\frac {1}{x+\ln \left (\frac {3}{\ln \left (\frac {2}{x}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.91, size = 20, normalized size = 0.87 \begin {gather*} e^{\frac {1}{x + \log {\left (\frac {3}{- \log {\relax (x )} + \log {\relax (2 )}} \right )}}} + 2 \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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