\(\int \frac {16+4 x-8 \log (x)}{(4 x+x^2) \log (x)} \, dx\) [6824]

Optimal result

Integrand size = 23, antiderivative size = 17 \[ \int \frac {16+4 x-8 \log (x)}{\left (4 x+x^2\right ) \log (x)} \, dx=2 \log \left (\frac {5 (4+x) \log ^2(x)}{4 x}\right ) \]

[Out]

2*ln(5/4*ln(x)^2/x*(4+x))

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {1607, 6873, 12, 6874, 36, 29, 31, 2339} \[ \int \frac {16+4 x-8 \log (x)}{\left (4 x+x^2\right ) \log (x)} \, dx=-2 \log (x)+2 \log (x+4)+4 \log (\log (x)) \]

[In]

Int[(16 + 4*x - 8*Log[x])/((4*x + x^2)*Log[x]),x]

[Out]

-2*Log[x] + 2*Log[4 + x] + 4*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 29

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

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 6873

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

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+4 x-8 \log (x)}{x (4+x) \log (x)} \, dx \\ & = \int \frac {4 (4+x-2 \log (x))}{x (4+x) \log (x)} \, dx \\ & = 4 \int \frac {4+x-2 \log (x)}{x (4+x) \log (x)} \, dx \\ & = 4 \int \left (-\frac {2}{x (4+x)}+\frac {1}{x \log (x)}\right ) \, dx \\ & = 4 \int \frac {1}{x \log (x)} \, dx-8 \int \frac {1}{x (4+x)} \, dx \\ & = -\left (2 \int \frac {1}{x} \, dx\right )+2 \int \frac {1}{4+x} \, dx+4 \text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (x)\right ) \\ & = -2 \log (x)+2 \log (4+x)+4 \log (\log (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.18 \[ \int \frac {16+4 x-8 \log (x)}{\left (4 x+x^2\right ) \log (x)} \, dx=4 \left (-\frac {\log (x)}{2}+\frac {1}{2} \log (4+x)+\log (\log (x))\right ) \]

[In]

Integrate[(16 + 4*x - 8*Log[x])/((4*x + x^2)*Log[x]),x]

[Out]

4*(-1/2*Log[x] + Log[4 + x]/2 + Log[Log[x]])

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00

method	result	size
default	\(4 \ln \left (\ln \left (x \right )\right )-2 \ln \left (x \right )+2 \ln \left (4+x \right )\)	\(17\)
norman	\(4 \ln \left (\ln \left (x \right )\right )-2 \ln \left (x \right )+2 \ln \left (4+x \right )\)	\(17\)
risch	\(4 \ln \left (\ln \left (x \right )\right )-2 \ln \left (x \right )+2 \ln \left (4+x \right )\)	\(17\)
parallelrisch	\(4 \ln \left (\ln \left (x \right )\right )-2 \ln \left (x \right )+2 \ln \left (4+x \right )\)	\(17\)
parts	\(4 \ln \left (\ln \left (x \right )\right )-2 \ln \left (x \right )+2 \ln \left (4+x \right )\)	\(17\)

[In]

int((-8*ln(x)+4*x+16)/(x^2+4*x)/ln(x),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {16+4 x-8 \log (x)}{\left (4 x+x^2\right ) \log (x)} \, dx=2 \, \log \left (x + 4\right ) - 2 \, \log \left (x\right ) + 4 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate((-8*log(x)+4*x+16)/(x^2+4*x)/log(x),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {16+4 x-8 \log (x)}{\left (4 x+x^2\right ) \log (x)} \, dx=- 2 \log {\left (x \right )} + 2 \log {\left (x + 4 \right )} + 4 \log {\left (\log {\left (x \right )} \right )} \]

[In]

integrate((-8*ln(x)+4*x+16)/(x**2+4*x)/ln(x),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {16+4 x-8 \log (x)}{\left (4 x+x^2\right ) \log (x)} \, dx=2 \, \log \left (x + 4\right ) - 2 \, \log \left (x\right ) + 4 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate((-8*log(x)+4*x+16)/(x^2+4*x)/log(x),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {16+4 x-8 \log (x)}{\left (4 x+x^2\right ) \log (x)} \, dx=2 \, \log \left (x + 4\right ) - 2 \, \log \left (x\right ) + 4 \, \log \left (\log \left (x\right )\right ) \]

[In]

integrate((-8*log(x)+4*x+16)/(x^2+4*x)/log(x),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 11.72 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94 \[ \int \frac {16+4 x-8 \log (x)}{\left (4 x+x^2\right ) \log (x)} \, dx=2\,\ln \left (x+4\right )+4\,\ln \left (\ln \left (x\right )\right )-2\,\ln \left (x\right ) \]

[In]

int((4*x - 8*log(x) + 16)/(log(x)*(4*x + x^2)),x)

[Out]

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