3.44.69 \(\int \frac {1-8 x \log (x)}{2 x \log (x)} \, dx\)

Optimal. Leaf size=13 \[ -4 x+\frac {1}{4} \log \left (\log ^2(x)\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 11, normalized size of antiderivative = 0.85, number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {12, 6688, 2302, 29} \begin {gather*} \frac {1}{2} \log (\log (x))-4 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 8*x*Log[x])/(2*x*Log[x]),x]

[Out]

-4*x + Log[Log[x]]/2

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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} &=\frac {1}{2} \int \frac {1-8 x \log (x)}{x \log (x)} \, dx\\ &=\frac {1}{2} \int \left (-8+\frac {1}{x \log (x)}\right ) \, dx\\ &=-4 x+\frac {1}{2} \int \frac {1}{x \log (x)} \, dx\\ &=-4 x+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=-4 x+\frac {1}{2} \log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 11, normalized size = 0.85 \begin {gather*} -4 x+\frac {1}{2} \log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 8*x*Log[x])/(2*x*Log[x]),x]

[Out]

-4*x + Log[Log[x]]/2

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fricas [A]  time = 0.52, size = 9, normalized size = 0.69 \begin {gather*} -4 \, x + \frac {1}{2} \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-8*x*log(x)+1)/x/log(x),x, algorithm="fricas")

[Out]

-4*x + 1/2*log(log(x))

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giac [A]  time = 0.13, size = 9, normalized size = 0.69 \begin {gather*} -4 \, x + \frac {1}{2} \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-8*x*log(x)+1)/x/log(x),x, algorithm="giac")

[Out]

-4*x + 1/2*log(log(x))

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maple [A]  time = 0.02, size = 10, normalized size = 0.77

method	result	size
default	\(-4 x +\frac {\ln \left (\ln \relax (x )\right )}{2}\)	\(10\)
norman	\(-4 x +\frac {\ln \left (\ln \relax (x )\right )}{2}\)	\(10\)
risch	\(-4 x +\frac {\ln \left (\ln \relax (x )\right )}{2}\)	\(10\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/2*(-8*x*ln(x)+1)/x/ln(x),x,method=_RETURNVERBOSE)

[Out]

-4*x+1/2*ln(ln(x))

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maxima [A]  time = 0.40, size = 9, normalized size = 0.69 \begin {gather*} -4 \, x + \frac {1}{2} \, \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-8*x*log(x)+1)/x/log(x),x, algorithm="maxima")

[Out]

-4*x + 1/2*log(log(x))

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mupad [B]  time = 3.27, size = 9, normalized size = 0.69 \begin {gather*} \frac {\ln \left (\ln \relax (x)\right )}{2}-4\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*x*log(x) - 1/2)/(x*log(x)),x)

[Out]

log(log(x))/2 - 4*x

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sympy [A]  time = 0.09, size = 8, normalized size = 0.62 \begin {gather*} - 4 x + \frac {\log {\left (\log {\relax (x )} \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/2*(-8*x*ln(x)+1)/x/ln(x),x)

[Out]

-4*x + log(log(x))/2

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