3.18.49 \(\int \frac {-16-5 x \log (x) \log ^2(\log (x))}{32 x \log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=15 \[ \frac {1}{32} \left (-5 x+\frac {16}{\log (\log (x))}\right ) \]

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Rubi [A]  time = 0.17, antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 6688, 2302, 30} \begin {gather*} \frac {1}{2 \log (\log (x))}-\frac {5 x}{32} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16 - 5*x*Log[x]*Log[Log[x]]^2)/(32*x*Log[x]*Log[Log[x]]^2),x]

[Out]

(-5*x)/32 + 1/(2*Log[Log[x]])

Rule 12

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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}{32} \int \frac {-16-5 x \log (x) \log ^2(\log (x))}{x \log (x) \log ^2(\log (x))} \, dx\\ &=\frac {1}{32} \int \left (-5-\frac {16}{x \log (x) \log ^2(\log (x))}\right ) \, dx\\ &=-\frac {5 x}{32}-\frac {1}{2} \int \frac {1}{x \log (x) \log ^2(\log (x))} \, dx\\ &=-\frac {5 x}{32}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,\log (x)\right )\\ &=-\frac {5 x}{32}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (\log (x))\right )\\ &=-\frac {5 x}{32}+\frac {1}{2 \log (\log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 15, normalized size = 1.00 \begin {gather*} -\frac {5 x}{32}+\frac {1}{2 \log (\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16 - 5*x*Log[x]*Log[Log[x]]^2)/(32*x*Log[x]*Log[Log[x]]^2),x]

[Out]

(-5*x)/32 + 1/(2*Log[Log[x]])

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fricas [A]  time = 0.88, size = 15, normalized size = 1.00 \begin {gather*} -\frac {5 \, x \log \left (\log \relax (x)\right ) - 16}{32 \, \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/32*(-5*x*log(x)*log(log(x))^2-16)/x/log(x)/log(log(x))^2,x, algorithm="fricas")

[Out]

-1/32*(5*x*log(log(x)) - 16)/log(log(x))

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giac [A]  time = 0.24, size = 11, normalized size = 0.73 \begin {gather*} -\frac {5}{32} \, x + \frac {1}{2 \, \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/32*(-5*x*log(x)*log(log(x))^2-16)/x/log(x)/log(log(x))^2,x, algorithm="giac")

[Out]

-5/32*x + 1/2/log(log(x))

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maple [A]  time = 0.02, size = 12, normalized size = 0.80

method	result	size
default	\(-\frac {5 x}{32}+\frac {1}{2 \ln \left (\ln \relax (x )\right )}\)	\(12\)
risch	\(-\frac {5 x}{32}+\frac {1}{2 \ln \left (\ln \relax (x )\right )}\)	\(12\)
norman	\(\frac {\frac {1}{2}-\frac {5 x \ln \left (\ln \relax (x )\right )}{32}}{\ln \left (\ln \relax (x )\right )}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/32*(-5*x*ln(x)*ln(ln(x))^2-16)/x/ln(x)/ln(ln(x))^2,x,method=_RETURNVERBOSE)

[Out]

-5/32*x+1/2/ln(ln(x))

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maxima [A]  time = 0.45, size = 11, normalized size = 0.73 \begin {gather*} -\frac {5}{32} \, x + \frac {1}{2 \, \log \left (\log \relax (x)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/32*(-5*x*log(x)*log(log(x))^2-16)/x/log(x)/log(log(x))^2,x, algorithm="maxima")

[Out]

-5/32*x + 1/2/log(log(x))

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mupad [B]  time = 1.25, size = 11, normalized size = 0.73 \begin {gather*} \frac {1}{2\,\ln \left (\ln \relax (x)\right )}-\frac {5\,x}{32} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((5*x*log(log(x))^2*log(x))/32 + 1/2)/(x*log(log(x))^2*log(x)),x)

[Out]

1/(2*log(log(x))) - (5*x)/32

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sympy [A]  time = 0.22, size = 12, normalized size = 0.80 \begin {gather*} - \frac {5 x}{32} + \frac {1}{2 \log {\left (\log {\relax (x )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/32*(-5*x*ln(x)*ln(ln(x))**2-16)/x/ln(x)/ln(ln(x))**2,x)

[Out]

-5*x/32 + 1/(2*log(log(x)))

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