3.57.72 \(\int \frac {5-x-10 \log (x) \log (-\frac {x}{-5+x})}{(-5 x+x^2) \log (x)} \, dx\)

Optimal. Leaf size=19 \[ 4+\log ^2\left (\frac {x}{5-x}\right )-\log (\log (x)) \]

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Rubi [A]  time = 0.22, antiderivative size = 18, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {1593, 6742, 2302, 29, 2505} \begin {gather*} \log ^2\left (\frac {x}{5-x}\right )-\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - x - 10*Log[x]*Log[-(x/(-5 + x))])/((-5*x + x^2)*Log[x]),x]

[Out]

Log[x/(5 - x)]^2 - Log[Log[x]]

Rule 29

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

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 2505

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)*(u_), x_Symbol] :> Wi
th[{h = Simplify[u*(a + b*x)*(c + d*x)]}, Simp[(h*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s + 1))/(p*r*(s + 1)*(
b*c - a*d)), x] /; FreeQ[h, x]] /; FreeQ[{a, b, c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && EqQ[p + q,
0] && NeQ[s, -1]

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 {5-x-10 \log (x) \log \left (-\frac {x}{-5+x}\right )}{(-5+x) x \log (x)} \, dx\\ &=\int \left (-\frac {1}{x \log (x)}-\frac {10 \log \left (-\frac {x}{-5+x}\right )}{(-5+x) x}\right ) \, dx\\ &=-\left (10 \int \frac {\log \left (-\frac {x}{-5+x}\right )}{(-5+x) x} \, dx\right )-\int \frac {1}{x \log (x)} \, dx\\ &=\log ^2\left (\frac {x}{5-x}\right )-\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right )\\ &=\log ^2\left (\frac {x}{5-x}\right )-\log (\log (x))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 17, normalized size = 0.89 \begin {gather*} \log ^2\left (-\frac {x}{-5+x}\right )-\log (\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 - x - 10*Log[x]*Log[-(x/(-5 + x))])/((-5*x + x^2)*Log[x]),x]

[Out]

Log[-(x/(-5 + x))]^2 - Log[Log[x]]

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fricas [A]  time = 0.91, size = 17, normalized size = 0.89 \begin {gather*} \log \left (-\frac {x}{x - 5}\right )^{2} - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*log(-x/(x-5))*log(x)+5-x)/(x^2-5*x)/log(x),x, algorithm="fricas")

[Out]

log(-x/(x - 5))^2 - log(log(x))

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giac [C]  time = 0.19, size = 45, normalized size = 2.37 \begin {gather*} -2 i \, \pi \log \left (x - 5\right ) + 2 \, {\left (\log \left (x - 5\right ) - \log \relax (x)\right )} \log \left (x - 5\right ) - \log \left (x - 5\right )^{2} + 2 i \, \pi \log \relax (x) + \log \relax (x)^{2} - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*log(-x/(x-5))*log(x)+5-x)/(x^2-5*x)/log(x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.44, size = 18, normalized size = 0.95

method	result	size
norman	\(\ln \left (-\frac {x}{x -5}\right )^{2}-\ln \left (\ln \relax (x )\right )\)	\(18\)
default	\(-\ln \left (\ln \relax (x )\right )+\ln \left (-1-\frac {5}{x -5}\right )^{2}\)	\(19\)
risch	\(\ln \left (x -5\right )^{2}-2 \ln \relax (x ) \ln \left (x -5\right )+i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}+2 i \pi \ln \relax (x )-i \pi \ln \left (x -5\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{3}-i \pi \ln \left (x -5\right ) \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}-i \pi \ln \left (x -5\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}+i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{3}-i \pi \ln \relax (x ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )-2 i \pi \ln \left (x -5\right )+2 i \pi \ln \left (x -5\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}+i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}+i \pi \ln \left (x -5\right ) \mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (\frac {i}{x -5}\right ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )-2 i \pi \ln \relax (x ) \mathrm {csgn}\left (\frac {i x}{x -5}\right )^{2}+\ln \relax (x )^{2}-\ln \left (\ln \relax (x )\right )\)	\(281\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-10*ln(-x/(x-5))*ln(x)+5-x)/(x^2-5*x)/ln(x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.38, size = 28, normalized size = 1.47 \begin {gather*} \log \relax (x)^{2} - 2 \, \log \relax (x) \log \left (-x + 5\right ) + \log \left (-x + 5\right )^{2} - \log \left (\log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*log(-x/(x-5))*log(x)+5-x)/(x^2-5*x)/log(x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.86, size = 17, normalized size = 0.89 \begin {gather*} {\ln \left (-\frac {x}{x-5}\right )}^2-\ln \left (\ln \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 10*log(-x/(x - 5))*log(x) - 5)/(log(x)*(5*x - x^2)),x)

[Out]

log(-x/(x - 5))^2 - log(log(x))

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sympy [A]  time = 0.30, size = 14, normalized size = 0.74 \begin {gather*} \log {\left (- \frac {x}{x - 5} \right )}^{2} - \log {\left (\log {\relax (x )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-10*ln(-x/(x-5))*ln(x)+5-x)/(x**2-5*x)/ln(x),x)

[Out]

log(-x/(x - 5))**2 - log(log(x))

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