3.29.70 \(\int \frac {(-4+4 \log (x)) \log (\frac {5 x}{4})-\log (x) \log (\frac {\log ^2(x)}{x^2})}{x \log (x) \log (\frac {5 x}{4}) \log (\frac {\log ^2(x)}{x^2})} \, dx\)

Optimal. Leaf size=24 \[ \log \left (\frac {1}{3 \log \left (\frac {5 x}{4}\right ) \log ^2\left (\frac {\log ^2(x)}{x^2}\right )}\right ) \]

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Rubi [A]  time = 0.31, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {6688, 14, 2302, 29, 6684} \begin {gather*} -2 \log \left (\log \left (\frac {\log ^2(x)}{x^2}\right )\right )-\log \left (\log \left (\frac {5 x}{4}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-4 + 4*Log[x])*Log[(5*x)/4] - Log[x]*Log[Log[x]^2/x^2])/(x*Log[x]*Log[(5*x)/4]*Log[Log[x]^2/x^2]),x]

[Out]

-Log[Log[(5*x)/4]] - 2*Log[Log[Log[x]^2/x^2]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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 \frac {-\frac {1}{\log \left (\frac {5 x}{4}\right )}+\frac {4 (-1+\log (x))}{\log (x) \log \left (\frac {\log ^2(x)}{x^2}\right )}}{x} \, dx\\ &=\int \left (-\frac {1}{x \log \left (\frac {5 x}{4}\right )}+\frac {4 (-1+\log (x))}{x \log (x) \log \left (\frac {\log ^2(x)}{x^2}\right )}\right ) \, dx\\ &=4 \int \frac {-1+\log (x)}{x \log (x) \log \left (\frac {\log ^2(x)}{x^2}\right )} \, dx-\int \frac {1}{x \log \left (\frac {5 x}{4}\right )} \, dx\\ &=-2 \log \left (\log \left (\frac {\log ^2(x)}{x^2}\right )\right )-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\frac {5 x}{4}\right )\right )\\ &=-\log \left (\log \left (\frac {5 x}{4}\right )\right )-2 \log \left (\log \left (\frac {\log ^2(x)}{x^2}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 22, normalized size = 0.92 \begin {gather*} -\log \left (\log \left (\frac {5 x}{4}\right )\right )-2 \log \left (\log \left (\frac {\log ^2(x)}{x^2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-4 + 4*Log[x])*Log[(5*x)/4] - Log[x]*Log[Log[x]^2/x^2])/(x*Log[x]*Log[(5*x)/4]*Log[Log[x]^2/x^2]),
x]

[Out]

-Log[Log[(5*x)/4]] - 2*Log[Log[Log[x]^2/x^2]]

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fricas [A]  time = 0.84, size = 21, normalized size = 0.88 \begin {gather*} -\log \left (\log \left (\frac {5}{4}\right ) + \log \relax (x)\right ) - 2 \, \log \left (\log \left (\frac {\log \relax (x)^{2}}{x^{2}}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)*log(log(x)^2/x^2)+(4*log(x)-4)*log(5/4*x))/x/log(x)/log(5/4*x)/log(log(x)^2/x^2),x, algorit
hm="fricas")

[Out]

-log(log(5/4) + log(x)) - 2*log(log(log(x)^2/x^2))

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giac [A]  time = 0.25, size = 32, normalized size = 1.33 \begin {gather*} -\log \left (-\log \relax (5) + 2 \, \log \relax (2) - \log \relax (x)\right ) - 2 \, \log \left (-\log \left (\log \relax (x)^{2}\right ) + 2 \, \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)*log(log(x)^2/x^2)+(4*log(x)-4)*log(5/4*x))/x/log(x)/log(5/4*x)/log(log(x)^2/x^2),x, algorit
hm="giac")

[Out]

-log(-log(5) + 2*log(2) - log(x)) - 2*log(-log(log(x)^2) + 2*log(x))

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maple [A]  time = 0.50, size = 26, normalized size = 1.08

method	result	size
default	\(-\ln \left (\ln \relax (5)-2 \ln \relax (2)+\ln \relax (x )\right )-2 \ln \left (\ln \left (\frac {\ln \relax (x )^{2}}{x^{2}}\right )\right )\)	\(26\)
risch	\(-\ln \left (\ln \relax (x )-\frac {i \left (2 i \ln \relax (5)-4 i \ln \relax (2)\right )}{2}\right )-2 \ln \left (\ln \left (\ln \relax (x )\right )+\frac {i \left (\pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2}\right ) \mathrm {csgn}\left (\frac {i \ln \relax (x )^{2}}{x^{2}}\right )+\pi \,\mathrm {csgn}\left (\frac {i}{x^{2}}\right ) \mathrm {csgn}\left (\frac {i \ln \relax (x )^{2}}{x^{2}}\right )^{2}+\pi \mathrm {csgn}\left (i x^{2}\right )^{3}-\pi \mathrm {csgn}\left (i \ln \relax (x )\right )^{2} \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )+2 \pi \,\mathrm {csgn}\left (i \ln \relax (x )\right ) \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{2}-\pi \mathrm {csgn}\left (i \ln \relax (x )^{2}\right )^{3}+\pi \,\mathrm {csgn}\left (i \ln \relax (x )^{2}\right ) \mathrm {csgn}\left (\frac {i \ln \relax (x )^{2}}{x^{2}}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i \ln \relax (x )^{2}}{x^{2}}\right )^{3}+4 i \ln \relax (x )\right )}{4}\right )\)	\(223\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-ln(x)*ln(ln(x)^2/x^2)+(4*ln(x)-4)*ln(5/4*x))/x/ln(x)/ln(5/4*x)/ln(ln(x)^2/x^2),x,method=_RETURNVERBOSE)

[Out]

-ln(ln(5)-2*ln(2)+ln(x))-2*ln(ln(ln(x)^2/x^2))

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maxima [A]  time = 0.42, size = 19, normalized size = 0.79 \begin {gather*} -2 \, \log \left (-\log \relax (x) + \log \left (\log \relax (x)\right )\right ) - \log \left (\log \left (\frac {5}{4} \, x\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-log(x)*log(log(x)^2/x^2)+(4*log(x)-4)*log(5/4*x))/x/log(x)/log(5/4*x)/log(log(x)^2/x^2),x, algorit
hm="maxima")

[Out]

-2*log(-log(x) + log(log(x))) - log(log(5/4*x))

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mupad [B]  time = 1.92, size = 20, normalized size = 0.83 \begin {gather*} -2\,\ln \left (\ln \left (\frac {{\ln \relax (x)}^2}{x^2}\right )\right )-\ln \left (\ln \left (\frac {5\,x}{4}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*log(log(x)^2/x^2) - log((5*x)/4)*(4*log(x) - 4))/(x*log((5*x)/4)*log(x)*log(log(x)^2/x^2)),x)

[Out]

- 2*log(log(log(x)^2/x^2)) - log(log((5*x)/4))

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sympy [A]  time = 0.36, size = 27, normalized size = 1.12 \begin {gather*} - \log {\left (\log {\relax (x )} - 2 \log {\relax (2 )} + \log {\relax (5 )} \right )} - 2 \log {\left (\log {\left (\frac {\log {\relax (x )}^{2}}{x^{2}} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-ln(x)*ln(ln(x)**2/x**2)+(4*ln(x)-4)*ln(5/4*x))/x/ln(x)/ln(5/4*x)/ln(ln(x)**2/x**2),x)

[Out]

-log(log(x) - 2*log(2) + log(5)) - 2*log(log(log(x)**2/x**2))

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