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

Optimal. Leaf size=27 \[ x+x^2 \log ^2(x)+\frac {25 \log ^4\left (\log \left (\frac {x}{2}\right )\right )}{(-1+\log (4))^2} \]

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Rubi [A]  time = 0.25, antiderivative size = 29, normalized size of antiderivative = 1.07, number of steps used = 12, number of rules used = 7, integrand size = 120, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {6, 12, 6688, 2304, 2305, 2302, 30} \begin {gather*} x^2 \log ^2(x)+x+\frac {25 \log ^4\left (\log \left (\frac {x}{2}\right )\right )}{(1-\log (4))^2} \end {gather*}

Antiderivative was successfully verified.

[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 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 2304

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 2305

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 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \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 \operatorname {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 \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 27, normalized size = 1.00 \begin {gather*} x+x^2 \log ^2(x)+\frac {25 \log ^4\left (\log \left (\frac {x}{2}\right )\right )}{(-1+\log (4))^2} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [B]  time = 0.58, size = 119, normalized size = 4.41 \begin {gather*} \frac {4 \, x^{2} \log \relax (2)^{4} - 4 \, x^{2} \log \relax (2)^{3} + 25 \, \log \left (\log \left (\frac {1}{2} \, x\right )\right )^{4} + {\left (x^{2} + 4 \, x\right )} \log \relax (2)^{2} + {\left (4 \, x^{2} \log \relax (2)^{2} - 4 \, x^{2} \log \relax (2) + x^{2}\right )} \log \left (\frac {1}{2} \, x\right )^{2} - 4 \, x \log \relax (2) + 2 \, {\left (4 \, x^{2} \log \relax (2)^{3} - 4 \, x^{2} \log \relax (2)^{2} + x^{2} \log \relax (2)\right )} \log \left (\frac {1}{2} \, x\right ) + x}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.20, size = 36, normalized size = 1.33 \begin {gather*} x^{2} \log \relax (x)^{2} + \frac {25 \, \log \left (-\log \relax (2) + \log \relax (x)\right )^{4}}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) + 1} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.11, size = 37, normalized size = 1.37

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

Verification of antiderivative is not currently implemented for this CAS.

[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(x)-ln(2))^4

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maxima [B]  time = 0.59, size = 354, normalized size = 13.11 \begin {gather*} \frac {75 \, \log \left (-\log \relax (2) + \log \relax (x)\right )^{4}}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) + 1} - \frac {150 \, \log \left (-\log \relax (2) + \log \relax (x)\right )^{2} \log \left (\log \left (\frac {1}{2} \, x\right )\right )^{2}}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) + 1} + \frac {100 \, \log \left (-\log \relax (2) + \log \relax (x)\right ) \log \left (\log \left (\frac {1}{2} \, x\right )\right )^{3}}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) + 1} + \frac {2 \, {\left (2 \, x^{2} \log \relax (x)^{2} - 2 \, x^{2} \log \relax (x) + x^{2}\right )} \log \relax (2)^{2}}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) + 1} + \frac {2 \, {\left (2 \, x^{2} \log \relax (x) - x^{2}\right )} \log \relax (2)^{2}}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) + 1} + \frac {4 \, x \log \relax (2)^{2}}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) + 1} - \frac {2 \, {\left (2 \, x^{2} \log \relax (x)^{2} - 2 \, x^{2} \log \relax (x) + x^{2}\right )} \log \relax (2)}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) + 1} - \frac {2 \, {\left (2 \, x^{2} \log \relax (x) - x^{2}\right )} \log \relax (2)}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) + 1} - \frac {4 \, x \log \relax (2)}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) + 1} + \frac {2 \, x^{2} \log \relax (x)^{2} - 2 \, x^{2} \log \relax (x) + x^{2}}{2 \, {\left (4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) + 1\right )}} + \frac {2 \, x^{2} \log \relax (x) - x^{2}}{2 \, {\left (4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) + 1\right )}} + \frac {x}{4 \, \log \relax (2)^{2} - 4 \, \log \relax (2) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 6.66, size = 58, normalized size = 2.15 \begin {gather*} \frac {\left (8\,{\ln \relax (2)}^2-\ln \left (256\right )+2\right )\,x^2\,{\ln \relax (x)}^2}{2\,\left (4\,{\ln \relax (2)}^2-\ln \left (16\right )+1\right )}+x+\frac {{\ln \left (\ln \left (\frac {x}{2}\right )\right )}^4}{4\,\left (\frac {{\ln \relax (2)}^2}{25}-\frac {\ln \left (16\right )}{100}+\frac {1}{100}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.51, size = 34, normalized size = 1.26 \begin {gather*} x^{2} \log {\relax (x )}^{2} + x + \frac {25 \log {\left (\log {\relax (x )} - \log {\relax (2 )} \right )}^{4}}{- 4 \log {\relax (2 )} + 1 + 4 \log {\relax (2 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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