3.19.92 \(\int \frac {-8-7 x+(12+6 x) \log (x)}{-20 x-5 x^2+(12 x+3 x^2) \log (x)} \, dx\)

Optimal. Leaf size=17 \[ \log \left (20 x (4+x) \left (\frac {1}{3}-\frac {\log (x)}{5}\right )\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.38, antiderivative size = 14, normalized size of antiderivative = 0.82, number of steps used = 7, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {6741, 6742, 72, 2302, 29} \begin {gather*} \log (x)+\log (x+4)+\log (5-3 \log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8 - 7*x + (12 + 6*x)*Log[x])/(-20*x - 5*x^2 + (12*x + 3*x^2)*Log[x]),x]

[Out]

Log[x] + Log[4 + x] + Log[5 - 3*Log[x]]

Rule 29

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[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 6741

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

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 {8+7 x-(12+6 x) \log (x)}{x (4+x) (5-3 \log (x))} \, dx\\ &=\int \left (\frac {2 (2+x)}{x (4+x)}+\frac {3}{x (-5+3 \log (x))}\right ) \, dx\\ &=2 \int \frac {2+x}{x (4+x)} \, dx+3 \int \frac {1}{x (-5+3 \log (x))} \, dx\\ &=2 \int \left (\frac {1}{2 x}+\frac {1}{2 (4+x)}\right ) \, dx+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,-5+3 \log (x)\right )\\ &=\log (x)+\log (4+x)+\log (5-3 \log (x))\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 14, normalized size = 0.82 \begin {gather*} \log (x)+\log (4+x)+\log (5-3 \log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 - 7*x + (12 + 6*x)*Log[x])/(-20*x - 5*x^2 + (12*x + 3*x^2)*Log[x]),x]

[Out]

Log[x] + Log[4 + x] + Log[5 - 3*Log[x]]

________________________________________________________________________________________

fricas [A]  time = 0.92, size = 16, normalized size = 0.94 \begin {gather*} \log \left (x^{2} + 4 \, x\right ) + \log \left (3 \, \log \relax (x) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x+12)*log(x)-7*x-8)/((3*x^2+12*x)*log(x)-5*x^2-20*x),x, algorithm="fricas")

[Out]

log(x^2 + 4*x) + log(3*log(x) - 5)

________________________________________________________________________________________

giac [A]  time = 0.26, size = 14, normalized size = 0.82 \begin {gather*} \log \left (x + 4\right ) + \log \relax (x) + \log \left (3 \, \log \relax (x) - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x+12)*log(x)-7*x-8)/((3*x^2+12*x)*log(x)-5*x^2-20*x),x, algorithm="giac")

[Out]

log(x + 4) + log(x) + log(3*log(x) - 5)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 15, normalized size = 0.88

method	result	size
norman	\(\ln \relax (x )+\ln \left (4+x \right )+\ln \left (3 \ln \relax (x )-5\right )\)	\(15\)
risch	\(\ln \left (x^{2}+4 x \right )+\ln \left (\ln \relax (x )-\frac {5}{3}\right )\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x+12)*ln(x)-7*x-8)/((3*x^2+12*x)*ln(x)-5*x^2-20*x),x,method=_RETURNVERBOSE)

[Out]

ln(x)+ln(4+x)+ln(3*ln(x)-5)

________________________________________________________________________________________

maxima [A]  time = 0.86, size = 12, normalized size = 0.71 \begin {gather*} \log \left (x + 4\right ) + \log \relax (x) + \log \left (\log \relax (x) - \frac {5}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x+12)*log(x)-7*x-8)/((3*x^2+12*x)*log(x)-5*x^2-20*x),x, algorithm="maxima")

[Out]

log(x + 4) + log(x) + log(log(x) - 5/3)

________________________________________________________________________________________

mupad [B]  time = 1.25, size = 14, normalized size = 0.82 \begin {gather*} \ln \left (3\,\ln \relax (x)-5\right )+\ln \left (x+4\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((7*x - log(x)*(6*x + 12) + 8)/(20*x - log(x)*(12*x + 3*x^2) + 5*x^2),x)

[Out]

log(3*log(x) - 5) + log(x + 4) + log(x)

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 15, normalized size = 0.88 \begin {gather*} \log {\left (x^{2} + 4 x \right )} + \log {\left (\log {\relax (x )} - \frac {5}{3} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x+12)*ln(x)-7*x-8)/((3*x**2+12*x)*ln(x)-5*x**2-20*x),x)

[Out]

log(x**2 + 4*x) + log(log(x) - 5/3)

________________________________________________________________________________________