\(\int \frac {-16+x+(-16+2 x) \log (x)}{(-16 x+x^2) \log (x)} \, dx\) [4300]

Optimal result

Integrand size = 25, antiderivative size = 21 \[ \int \frac {-16+x+(-16+2 x) \log (x)}{\left (-16 x+x^2\right ) \log (x)} \, dx=\log \left (\frac {\left (-x+\frac {x^2}{16}\right ) \log (x)}{24 e^4}\right ) \]

[Out]

ln(1/24*ln(x)*(1/16*x^2-x)/exp(4))

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.57, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1607, 6874, 78, 2339, 29} \[ \int \frac {-16+x+(-16+2 x) \log (x)}{\left (-16 x+x^2\right ) \log (x)} \, dx=\log (16-x)+\log (x)+\log (\log (x)) \]

[In]

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

[Out]

Log[16 - x] + Log[x] + Log[Log[x]]

Rule 29

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 1607

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 2339

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 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-16+x+(-16+2 x) \log (x)}{(-16+x) x \log (x)} \, dx \\ & = \int \left (\frac {2 (-8+x)}{(-16+x) x}+\frac {1}{x \log (x)}\right ) \, dx \\ & = 2 \int \frac {-8+x}{(-16+x) x} \, dx+\int \frac {1}{x \log (x)} \, dx \\ & = 2 \int \left (\frac {1}{2 (-16+x)}+\frac {1}{2 x}\right ) \, dx+\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = \log (16-x)+\log (x)+\log (\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.57 \[ \int \frac {-16+x+(-16+2 x) \log (x)}{\left (-16 x+x^2\right ) \log (x)} \, dx=\log (16-x)+\log (x)+\log (\log (x)) \]

[In]

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

[Out]

Log[16 - x] + Log[x] + Log[Log[x]]

Maple [A] (verified)

Time = 1.10 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.52

method	result	size
default	\(\ln \left (\ln \left (x \right )\right )+\ln \left (x \left (x -16\right )\right )\)	\(11\)
norman	\(\ln \left (x \right )+\ln \left (\ln \left (x \right )\right )+\ln \left (x -16\right )\)	\(11\)
parallelrisch	\(\ln \left (x \right )+\ln \left (\ln \left (x \right )\right )+\ln \left (x -16\right )\)	\(11\)
parts	\(\ln \left (\ln \left (x \right )\right )+\ln \left (x \left (x -16\right )\right )\)	\(11\)
risch	\(\ln \left (x^{2}-16 x \right )+\ln \left (\ln \left (x \right )\right )\)	\(13\)

[In]

int(((2*x-16)*ln(x)+x-16)/(x^2-16*x)/ln(x),x,method=_RETURNVERBOSE)

[Out]

ln(ln(x))+ln(x*(x-16))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.57 \[ \int \frac {-16+x+(-16+2 x) \log (x)}{\left (-16 x+x^2\right ) \log (x)} \, dx=\log \left (x^{2} - 16 \, x\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

log(x^2 - 16*x) + log(log(x))

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.57 \[ \int \frac {-16+x+(-16+2 x) \log (x)}{\left (-16 x+x^2\right ) \log (x)} \, dx=\log {\left (x^{2} - 16 x \right )} + \log {\left (\log {\left (x \right )} \right )} \]

[In]

integrate(((2*x-16)*ln(x)+x-16)/(x**2-16*x)/ln(x),x)

[Out]

log(x**2 - 16*x) + log(log(x))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.48 \[ \int \frac {-16+x+(-16+2 x) \log (x)}{\left (-16 x+x^2\right ) \log (x)} \, dx=\log \left (x - 16\right ) + \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

log(x - 16) + log(x) + log(log(x))

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.48 \[ \int \frac {-16+x+(-16+2 x) \log (x)}{\left (-16 x+x^2\right ) \log (x)} \, dx=\log \left (x - 16\right ) + \log \left (x\right ) + \log \left (\log \left (x\right )\right ) \]

[In]

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

[Out]

log(x - 16) + log(x) + log(log(x))

Mupad [B] (verification not implemented)

Time = 10.59 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.48 \[ \int \frac {-16+x+(-16+2 x) \log (x)}{\left (-16 x+x^2\right ) \log (x)} \, dx=\ln \left (x-16\right )+\ln \left (\ln \left (x\right )\right )+\ln \left (x\right ) \]

[In]

int(-(x + log(x)*(2*x - 16) - 16)/(log(x)*(16*x - x^2)),x)

[Out]

log(x - 16) + log(log(x)) + log(x)