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

Optimal. Leaf size=21 \[ 2+e^3-(-1-x)^2-x+\log (\log (\log (x))) \]

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Rubi [A]  time = 0.17, antiderivative size = 13, normalized size of antiderivative = 0.62, number of steps used = 5, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6688, 2302, 29} \begin {gather*} -x^2-3 x+\log (\log (\log (x))) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-3*x - x^2 + Log[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]]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 13, normalized size = 0.62 \begin {gather*} -3 x-x^2+\log (\log (\log (x))) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-3*x - x^2 + Log[Log[Log[x]]]

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fricas [A]  time = 0.52, size = 13, normalized size = 0.62 \begin {gather*} -x^{2} - 3 \, x + \log \left (\log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2 - 3*x + log(log(log(x)))

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giac [A]  time = 0.17, size = 13, normalized size = 0.62 \begin {gather*} -x^{2} - 3 \, x + \log \left (\log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2 - 3*x + log(log(log(x)))

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maple [A]  time = 0.02, size = 14, normalized size = 0.67




method result size



default \(-3 x +\ln \left (\ln \left (\ln \relax (x )\right )\right )-x^{2}\) \(14\)
norman \(-3 x +\ln \left (\ln \left (\ln \relax (x )\right )\right )-x^{2}\) \(14\)
risch \(-3 x +\ln \left (\ln \left (\ln \relax (x )\right )\right )-x^{2}\) \(14\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*x+ln(ln(ln(x)))-x^2

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maxima [A]  time = 0.36, size = 13, normalized size = 0.62 \begin {gather*} -x^{2} - 3 \, x + \log \left (\log \left (\log \relax (x)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x^2 - 3*x + log(log(log(x)))

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mupad [B]  time = 7.60, size = 13, normalized size = 0.62 \begin {gather*} \ln \left (\ln \left (\ln \relax (x)\right )\right )-3\,x-x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(log(log(x))) - 3*x - x^2

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sympy [A]  time = 0.25, size = 12, normalized size = 0.57 \begin {gather*} - x^{2} - 3 x + \log {\left (\log {\left (\log {\relax (x )} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2 - 3*x + log(log(log(x)))

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