\(\int \frac {(x-2 x \log (4)+x \log ^2(4)) \log (\frac {x}{2})+(2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)) \log (\frac {x}{2}) \log (x)+(2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)) \log (\frac {x}{2}) \log ^2(x)+100 \log ^3(\log (\frac {x}{2}))}{(x-2 x \log (4)+x \log ^2(4)) \log (\frac {x}{2})} \, dx\) [10229]

Optimal result

Integrand size = 120, antiderivative size = 27 \[ \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx=x+x^2 \log ^2(x)+\frac {25 \log ^4\left (\log \left (\frac {x}{2}\right )\right )}{(-1+\log (4))^2} \]

[Out]

625*ln(ln(1/2*x))^4/(10*ln(2)-5)^2+x+x^2*ln(x)^2

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {6, 12, 6820, 2341, 2342, 2339, 30} \[ \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx=x^2 \log ^2(x)+x+\frac {25 \log ^4\left (\log \left (\frac {x}{2}\right )\right )}{(1-\log (4))^2} \]

[In]

Int[((x - 2*x*Log[4] + x*Log[4]^2)*Log[x/2] + (2*x^2 - 4*x^2*Log[4] + 2*x^2*Log[4]^2)*Log[x/2]*Log[x] + (2*x^2
- 4*x^2*Log[4] + 2*x^2*Log[4]^2)*Log[x/2]*Log[x]^2 + 100*Log[Log[x/2]]^3)/((x - 2*x*Log[4] + x*Log[4]^2)*Log[
x/2]),x]

[Out]

x + x^2*Log[x]^2 + (25*Log[Log[x/2]]^4)/(1 - Log[4])^2

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2341

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^
n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^(m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2342

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Lo
g[c*x^n])^p/(d*(m + 1))), x] - Dist[b*n*(p/(m + 1)), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rule 6820

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x (1-2 \log (4))+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx \\ & = \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{x \left (1-2 \log (4)+\log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx \\ & = \frac {\int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{x \log \left (\frac {x}{2}\right )} \, dx}{1-2 \log (4)+\log ^2(4)} \\ & = \frac {\int \left ((-1+\log (4))^2 \left (1+2 x \log (x)+2 x \log ^2(x)\right )+\frac {100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{x \log \left (\frac {x}{2}\right )}\right ) \, dx}{1-2 \log (4)+\log ^2(4)} \\ & = \frac {100 \int \frac {\log ^3\left (\log \left (\frac {x}{2}\right )\right )}{x \log \left (\frac {x}{2}\right )} \, dx}{(1-\log (4))^2}+\int \left (1+2 x \log (x)+2 x \log ^2(x)\right ) \, dx \\ & = x+2 \int x \log (x) \, dx+2 \int x \log ^2(x) \, dx+\frac {100 \text {Subst}\left (\int \frac {\log ^3(x)}{x} \, dx,x,\log \left (\frac {x}{2}\right )\right )}{(1-\log (4))^2} \\ & = x-\frac {x^2}{2}+x^2 \log (x)+x^2 \log ^2(x)-2 \int x \log (x) \, dx+\frac {100 \text {Subst}\left (\int x^3 \, dx,x,\log \left (\log \left (\frac {x}{2}\right )\right )\right )}{(1-\log (4))^2} \\ & = x+x^2 \log ^2(x)+\frac {25 \log ^4\left (\log \left (\frac {x}{2}\right )\right )}{(1-\log (4))^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx=x+x^2 \log ^2(x)+\frac {25 \log ^4\left (\log \left (\frac {x}{2}\right )\right )}{(-1+\log (4))^2} \]

[In]

Integrate[((x - 2*x*Log[4] + x*Log[4]^2)*Log[x/2] + (2*x^2 - 4*x^2*Log[4] + 2*x^2*Log[4]^2)*Log[x/2]*Log[x] +
(2*x^2 - 4*x^2*Log[4] + 2*x^2*Log[4]^2)*Log[x/2]*Log[x]^2 + 100*Log[Log[x/2]]^3)/((x - 2*x*Log[4] + x*Log[4]^2
)*Log[x/2]),x]

[Out]

x + x^2*Log[x]^2 + (25*Log[Log[x/2]]^4)/(-1 + Log[4])^2

Maple [A] (verified)

Time = 2.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26

method	result	size
parts	\(x +x^{2} \ln \left (x \right )^{2}+\frac {25 \ln \left (\ln \left (\frac {x}{2}\right )\right )^{4}}{4 \ln \left (2\right )^{2}-4 \ln \left (2\right )+1}\)	\(34\)
default	\(x +x^{2} \ln \left (x \right )^{2}+\frac {25 \ln \left (\ln \left (x \right )-\ln \left (2\right )\right )^{4}}{4 \ln \left (2\right )^{2}-4 \ln \left (2\right )+1}\)	\(37\)
risch	\(x +x^{2} \ln \left (x \right )^{2}+\frac {25 \ln \left (\ln \left (x \right )-\ln \left (2\right )\right )^{4}}{4 \ln \left (2\right )^{2}-4 \ln \left (2\right )+1}\)	\(37\)
parallelrisch	\(\frac {x +4 x \ln \left (2\right )^{2}-4 x \ln \left (2\right )+25 \ln \left (\ln \left (\frac {x}{2}\right )\right )^{4}+4 \ln \left (x \right )^{2} \ln \left (2\right )^{2} x^{2}-4 x^{2} \ln \left (2\right ) \ln \left (x \right )^{2}+x^{2} \ln \left (x \right )^{2}}{4 \ln \left (2\right )^{2}-4 \ln \left (2\right )+1}\)	\(71\)

[In]

int((100*ln(ln(1/2*x))^3+(8*x^2*ln(2)^2-8*x^2*ln(2)+2*x^2)*ln(1/2*x)*ln(x)^2+(8*x^2*ln(2)^2-8*x^2*ln(2)+2*x^2)
*ln(1/2*x)*ln(x)+(4*x*ln(2)^2-4*x*ln(2)+x)*ln(1/2*x))/(4*x*ln(2)^2-4*x*ln(2)+x)/ln(1/2*x),x,method=_RETURNVERB
OSE)

[Out]

x+x^2*ln(x)^2+25/(4*ln(2)^2-4*ln(2)+1)*ln(ln(1/2*x))^4

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 119 vs. \(2 (27) = 54\).

Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 4.41 \[ \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx=\frac {4 \, x^{2} \log \left (2\right )^{4} - 4 \, x^{2} \log \left (2\right )^{3} + 25 \, \log \left (\log \left (\frac {1}{2} \, x\right )\right )^{4} + {\left (x^{2} + 4 \, x\right )} \log \left (2\right )^{2} + {\left (4 \, x^{2} \log \left (2\right )^{2} - 4 \, x^{2} \log \left (2\right ) + x^{2}\right )} \log \left (\frac {1}{2} \, x\right )^{2} - 4 \, x \log \left (2\right ) + 2 \, {\left (4 \, x^{2} \log \left (2\right )^{3} - 4 \, x^{2} \log \left (2\right )^{2} + x^{2} \log \left (2\right )\right )} \log \left (\frac {1}{2} \, x\right ) + x}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} \]

[In]

integrate((100*log(log(1/2*x))^3+(8*x^2*log(2)^2-8*x^2*log(2)+2*x^2)*log(1/2*x)*log(x)^2+(8*x^2*log(2)^2-8*x^2
*log(2)+2*x^2)*log(1/2*x)*log(x)+(4*x*log(2)^2-4*x*log(2)+x)*log(1/2*x))/(4*x*log(2)^2-4*x*log(2)+x)/log(1/2*x
),x, algorithm="fricas")

[Out]

(4*x^2*log(2)^4 - 4*x^2*log(2)^3 + 25*log(log(1/2*x))^4 + (x^2 + 4*x)*log(2)^2 + (4*x^2*log(2)^2 - 4*x^2*log(2
) + x^2)*log(1/2*x)^2 - 4*x*log(2) + 2*(4*x^2*log(2)^3 - 4*x^2*log(2)^2 + x^2*log(2))*log(1/2*x) + x)/(4*log(2
)^2 - 4*log(2) + 1)

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx=x^{2} \log {\left (x \right )}^{2} + x + \frac {25 \log {\left (\log {\left (x \right )} - \log {\left (2 \right )} \right )}^{4}}{- 4 \log {\left (2 \right )} + 1 + 4 \log {\left (2 \right )}^{2}} \]

[In]

integrate((100*ln(ln(1/2*x))**3+(8*x**2*ln(2)**2-8*x**2*ln(2)+2*x**2)*ln(1/2*x)*ln(x)**2+(8*x**2*ln(2)**2-8*x*
*2*ln(2)+2*x**2)*ln(1/2*x)*ln(x)+(4*x*ln(2)**2-4*x*ln(2)+x)*ln(1/2*x))/(4*x*ln(2)**2-4*x*ln(2)+x)/ln(1/2*x),x)

[Out]

x**2*log(x)**2 + x + 25*log(log(x) - log(2))**4/(-4*log(2) + 1 + 4*log(2)**2)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 354 vs. \(2 (27) = 54\).

Time = 0.41 (sec) , antiderivative size = 354, normalized size of antiderivative = 13.11 \[ \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx=\frac {75 \, \log \left (-\log \left (2\right ) + \log \left (x\right )\right )^{4}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} - \frac {150 \, \log \left (-\log \left (2\right ) + \log \left (x\right )\right )^{2} \log \left (\log \left (\frac {1}{2} \, x\right )\right )^{2}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} + \frac {100 \, \log \left (-\log \left (2\right ) + \log \left (x\right )\right ) \log \left (\log \left (\frac {1}{2} \, x\right )\right )^{3}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} + \frac {2 \, {\left (2 \, x^{2} \log \left (x\right )^{2} - 2 \, x^{2} \log \left (x\right ) + x^{2}\right )} \log \left (2\right )^{2}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} + \frac {2 \, {\left (2 \, x^{2} \log \left (x\right ) - x^{2}\right )} \log \left (2\right )^{2}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} + \frac {4 \, x \log \left (2\right )^{2}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} - \frac {2 \, {\left (2 \, x^{2} \log \left (x\right )^{2} - 2 \, x^{2} \log \left (x\right ) + x^{2}\right )} \log \left (2\right )}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} - \frac {2 \, {\left (2 \, x^{2} \log \left (x\right ) - x^{2}\right )} \log \left (2\right )}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} - \frac {4 \, x \log \left (2\right )}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} + \frac {2 \, x^{2} \log \left (x\right )^{2} - 2 \, x^{2} \log \left (x\right ) + x^{2}}{2 \, {\left (4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1\right )}} + \frac {2 \, x^{2} \log \left (x\right ) - x^{2}}{2 \, {\left (4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1\right )}} + \frac {x}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} \]

[In]

integrate((100*log(log(1/2*x))^3+(8*x^2*log(2)^2-8*x^2*log(2)+2*x^2)*log(1/2*x)*log(x)^2+(8*x^2*log(2)^2-8*x^2
*log(2)+2*x^2)*log(1/2*x)*log(x)+(4*x*log(2)^2-4*x*log(2)+x)*log(1/2*x))/(4*x*log(2)^2-4*x*log(2)+x)/log(1/2*x
),x, algorithm="maxima")

[Out]

75*log(-log(2) + log(x))^4/(4*log(2)^2 - 4*log(2) + 1) - 150*log(-log(2) + log(x))^2*log(log(1/2*x))^2/(4*log(
2)^2 - 4*log(2) + 1) + 100*log(-log(2) + log(x))*log(log(1/2*x))^3/(4*log(2)^2 - 4*log(2) + 1) + 2*(2*x^2*log(
x)^2 - 2*x^2*log(x) + x^2)*log(2)^2/(4*log(2)^2 - 4*log(2) + 1) + 2*(2*x^2*log(x) - x^2)*log(2)^2/(4*log(2)^2
- 4*log(2) + 1) + 4*x*log(2)^2/(4*log(2)^2 - 4*log(2) + 1) - 2*(2*x^2*log(x)^2 - 2*x^2*log(x) + x^2)*log(2)/(4
*log(2)^2 - 4*log(2) + 1) - 2*(2*x^2*log(x) - x^2)*log(2)/(4*log(2)^2 - 4*log(2) + 1) - 4*x*log(2)/(4*log(2)^2
- 4*log(2) + 1) + 1/2*(2*x^2*log(x)^2 - 2*x^2*log(x) + x^2)/(4*log(2)^2 - 4*log(2) + 1) + 1/2*(2*x^2*log(x) -
x^2)/(4*log(2)^2 - 4*log(2) + 1) + x/(4*log(2)^2 - 4*log(2) + 1)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.33 \[ \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx=x^{2} \log \left (x\right )^{2} + \frac {25 \, \log \left (-\log \left (2\right ) + \log \left (x\right )\right )^{4}}{4 \, \log \left (2\right )^{2} - 4 \, \log \left (2\right ) + 1} + x \]

[In]

integrate((100*log(log(1/2*x))^3+(8*x^2*log(2)^2-8*x^2*log(2)+2*x^2)*log(1/2*x)*log(x)^2+(8*x^2*log(2)^2-8*x^2
*log(2)+2*x^2)*log(1/2*x)*log(x)+(4*x*log(2)^2-4*x*log(2)+x)*log(1/2*x))/(4*x*log(2)^2-4*x*log(2)+x)/log(1/2*x
),x, algorithm="giac")

[Out]

x^2*log(x)^2 + 25*log(-log(2) + log(x))^4/(4*log(2)^2 - 4*log(2) + 1) + x

Mupad [B] (verification not implemented)

Time = 15.89 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.15 \[ \int \frac {\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log (x)+\left (2 x^2-4 x^2 \log (4)+2 x^2 \log ^2(4)\right ) \log \left (\frac {x}{2}\right ) \log ^2(x)+100 \log ^3\left (\log \left (\frac {x}{2}\right )\right )}{\left (x-2 x \log (4)+x \log ^2(4)\right ) \log \left (\frac {x}{2}\right )} \, dx=\frac {\left (8\,{\ln \left (2\right )}^2-\ln \left (256\right )+2\right )\,x^2\,{\ln \left (x\right )}^2}{2\,\left (4\,{\ln \left (2\right )}^2-\ln \left (16\right )+1\right )}+x+\frac {{\ln \left (\ln \left (\frac {x}{2}\right )\right )}^4}{4\,\left (\frac {{\ln \left (2\right )}^2}{25}-\frac {\ln \left (16\right )}{100}+\frac {1}{100}\right )} \]

[In]

int((100*log(log(x/2))^3 + log(x/2)*(x - 4*x*log(2) + 4*x*log(2)^2) + log(x/2)*log(x)^2*(8*x^2*log(2)^2 - 8*x^
2*log(2) + 2*x^2) + log(x/2)*log(x)*(8*x^2*log(2)^2 - 8*x^2*log(2) + 2*x^2))/(log(x/2)*(x - 4*x*log(2) + 4*x*l
og(2)^2)),x)

[Out]

x + log(log(x/2))^4/(4*(log(2)^2/25 - log(16)/100 + 1/100)) + (x^2*log(x)^2*(8*log(2)^2 - log(256) + 2))/(2*(4
*log(2)^2 - log(16) + 1))