3.75.6 \(\int \frac {-32+(32-2 x) \log (x)+(2 x \log (x)-32 \log (x) \log (\frac {x}{\log (x)})) \log (-x+16 \log (\frac {x}{\log (x)}))}{-x^3 \log (x)+16 x^2 \log (x) \log (\frac {x}{\log (x)})+(-2 x^2 \log (x)+32 x \log (x) \log (\frac {x}{\log (x)})) \log (-x+16 \log (\frac {x}{\log (x)}))+(-x \log (x)+16 \log (x) \log (\frac {x}{\log (x)})) \log ^2(-x+16 \log (\frac {x}{\log (x)}))} \, dx\)

Optimal. Leaf size=25 \[ -5+\log (5)-\frac {2 x}{x+\log \left (-x+16 \log \left (\frac {x}{\log (x)}\right )\right )} \]

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Rubi [A]  time = 0.40, antiderivative size = 24, normalized size of antiderivative = 0.96, number of steps used = 3, number of rules used = 3, integrand size = 136, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.022, Rules used = {6688, 6711, 32} \begin {gather*} \frac {2}{\frac {x}{\log \left (16 \log \left (\frac {x}{\log (x)}\right )-x\right )}+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-32 + (32 - 2*x)*Log[x] + (2*x*Log[x] - 32*Log[x]*Log[x/Log[x]])*Log[-x + 16*Log[x/Log[x]]])/(-(x^3*Log[x
]) + 16*x^2*Log[x]*Log[x/Log[x]] + (-2*x^2*Log[x] + 32*x*Log[x]*Log[x/Log[x]])*Log[-x + 16*Log[x/Log[x]]] + (-
(x*Log[x]) + 16*Log[x]*Log[x/Log[x]])*Log[-x + 16*Log[x/Log[x]]]^2),x]

[Out]

2/(1 + x/Log[-x + 16*Log[x/Log[x]]])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6688

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

Rule 6711

Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w
, x])]}, Dist[c*p, Subst[Int[(b + a*x^p)^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}
, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {32+2 \log (x) \left (-16+x-\left (x-16 \log \left (\frac {x}{\log (x)}\right )\right ) \log \left (-x+16 \log \left (\frac {x}{\log (x)}\right )\right )\right )}{\log (x) \left (x-16 \log \left (\frac {x}{\log (x)}\right )\right ) \left (x+\log \left (-x+16 \log \left (\frac {x}{\log (x)}\right )\right )\right )^2} \, dx\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,\frac {x}{\log \left (-x+16 \log \left (\frac {x}{\log (x)}\right )\right )}\right )\right )\\ &=\frac {2}{1+\frac {x}{\log \left (-x+16 \log \left (\frac {x}{\log (x)}\right )\right )}}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.64, size = 21, normalized size = 0.84 \begin {gather*} -\frac {2 x}{x+\log \left (-x+16 \log \left (\frac {x}{\log (x)}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-32 + (32 - 2*x)*Log[x] + (2*x*Log[x] - 32*Log[x]*Log[x/Log[x]])*Log[-x + 16*Log[x/Log[x]]])/(-(x^3
*Log[x]) + 16*x^2*Log[x]*Log[x/Log[x]] + (-2*x^2*Log[x] + 32*x*Log[x]*Log[x/Log[x]])*Log[-x + 16*Log[x/Log[x]]
] + (-(x*Log[x]) + 16*Log[x]*Log[x/Log[x]])*Log[-x + 16*Log[x/Log[x]]]^2),x]

[Out]

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

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fricas [A]  time = 0.67, size = 21, normalized size = 0.84 \begin {gather*} -\frac {2 \, x}{x + \log \left (-x + 16 \, \log \left (\frac {x}{\log \relax (x)}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*log(x)*log(x/log(x))+2*x*log(x))*log(16*log(x/log(x))-x)+(-2*x+32)*log(x)-32)/((16*log(x)*log(
x/log(x))-x*log(x))*log(16*log(x/log(x))-x)^2+(32*x*log(x)*log(x/log(x))-2*x^2*log(x))*log(16*log(x/log(x))-x)
+16*x^2*log(x)*log(x/log(x))-x^3*log(x)),x, algorithm="fricas")

[Out]

-2*x/(x + log(-x + 16*log(x/log(x))))

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giac [A]  time = 0.52, size = 21, normalized size = 0.84 \begin {gather*} -\frac {2 \, x}{x + \log \left (-x + 16 \, \log \relax (x) - 16 \, \log \left (\log \relax (x)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*log(x)*log(x/log(x))+2*x*log(x))*log(16*log(x/log(x))-x)+(-2*x+32)*log(x)-32)/((16*log(x)*log(
x/log(x))-x*log(x))*log(16*log(x/log(x))-x)^2+(32*x*log(x)*log(x/log(x))-2*x^2*log(x))*log(16*log(x/log(x))-x)
+16*x^2*log(x)*log(x/log(x))-x^3*log(x)),x, algorithm="giac")

[Out]

-2*x/(x + log(-x + 16*log(x) - 16*log(log(x))))

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maple [C]  time = 0.08, size = 72, normalized size = 2.88

method	result	size
risch	\(-\frac {2 x}{x +\ln \left (16 \ln \relax (x )-16 \ln \left (\ln \relax (x )\right )-8 i \pi \,\mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right ) \left (-\mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (\frac {i x}{\ln \relax (x )}\right )+\mathrm {csgn}\left (\frac {i}{\ln \relax (x )}\right )\right )-x \right )}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-32*ln(x)*ln(x/ln(x))+2*x*ln(x))*ln(16*ln(x/ln(x))-x)+(-2*x+32)*ln(x)-32)/((16*ln(x)*ln(x/ln(x))-x*ln(x)
)*ln(16*ln(x/ln(x))-x)^2+(32*x*ln(x)*ln(x/ln(x))-2*x^2*ln(x))*ln(16*ln(x/ln(x))-x)+16*x^2*ln(x)*ln(x/ln(x))-x^
3*ln(x)),x,method=_RETURNVERBOSE)

[Out]

-2*x/(x+ln(16*ln(x)-16*ln(ln(x))-8*I*Pi*csgn(I*x/ln(x))*(-csgn(I*x/ln(x))+csgn(I*x))*(-csgn(I*x/ln(x))+csgn(I/
ln(x)))-x))

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maxima [A]  time = 0.41, size = 21, normalized size = 0.84 \begin {gather*} -\frac {2 \, x}{x + \log \left (-x + 16 \, \log \relax (x) - 16 \, \log \left (\log \relax (x)\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*log(x)*log(x/log(x))+2*x*log(x))*log(16*log(x/log(x))-x)+(-2*x+32)*log(x)-32)/((16*log(x)*log(
x/log(x))-x*log(x))*log(16*log(x/log(x))-x)^2+(32*x*log(x)*log(x/log(x))-2*x^2*log(x))*log(16*log(x/log(x))-x)
+16*x^2*log(x)*log(x/log(x))-x^3*log(x)),x, algorithm="maxima")

[Out]

-2*x/(x + log(-x + 16*log(x) - 16*log(log(x))))

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mupad [B]  time = 8.09, size = 21, normalized size = 0.84 \begin {gather*} -\frac {2\,x}{x+\ln \left (16\,\ln \left (\frac {x}{\ln \relax (x)}\right )-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(2*x - 32) - log(16*log(x/log(x)) - x)*(2*x*log(x) - 32*log(x/log(x))*log(x)) + 32)/(x^3*log(x) +
log(16*log(x/log(x)) - x)^2*(x*log(x) - 16*log(x/log(x))*log(x)) + log(16*log(x/log(x)) - x)*(2*x^2*log(x) - 3
2*x*log(x/log(x))*log(x)) - 16*x^2*log(x/log(x))*log(x)),x)

[Out]

-(2*x)/(x + log(16*log(x/log(x)) - x))

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sympy [A]  time = 0.77, size = 17, normalized size = 0.68 \begin {gather*} - \frac {2 x}{x + \log {\left (- x + 16 \log {\left (\frac {x}{\log {\relax (x )}} \right )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*ln(x)*ln(x/ln(x))+2*x*ln(x))*ln(16*ln(x/ln(x))-x)+(-2*x+32)*ln(x)-32)/((16*ln(x)*ln(x/ln(x))-x
*ln(x))*ln(16*ln(x/ln(x))-x)**2+(32*x*ln(x)*ln(x/ln(x))-2*x**2*ln(x))*ln(16*ln(x/ln(x))-x)+16*x**2*ln(x)*ln(x/
ln(x))-x**3*ln(x)),x)

[Out]

-2*x/(x + log(-x + 16*log(x/log(x))))

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