3.37.49 \(\int \frac {(16-8 x+x^2) \log ^2(3)+(-16-60 x+32 x^2-4 x^3) \log (x)-4 x \log ^2(x)+((16+60 x-32 x^2+4 x^3) \log (x)+4 x \log ^2(x)) \log (\log (x))+((-4-15 x+8 x^2-x^3) \log (x)-x \log ^2(x)) \log ^2(\log (x))}{(64 x-32 x^2+4 x^3) \log (x)+(-64 x+32 x^2-4 x^3) \log (x) \log (\log (x))+(16 x-8 x^2+x^3) \log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=27 \[ 16-x+\frac {\log (x)}{-4+x}+\frac {\log ^2(3)}{2-\log (\log (x))} \]

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Rubi [A]  time = 0.20, antiderivative size = 38, normalized size of antiderivative = 1.41, number of steps used = 9, number of rules used = 6, integrand size = 162, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6688, 893, 2314, 31, 2302, 30} \begin {gather*} -x+\frac {\log ^2(3)}{2-\log (\log (x))}-\frac {x \log (x)}{4 (4-x)}-\frac {\log (x)}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((16 - 8*x + x^2)*Log[3]^2 + (-16 - 60*x + 32*x^2 - 4*x^3)*Log[x] - 4*x*Log[x]^2 + ((16 + 60*x - 32*x^2 +
4*x^3)*Log[x] + 4*x*Log[x]^2)*Log[Log[x]] + ((-4 - 15*x + 8*x^2 - x^3)*Log[x] - x*Log[x]^2)*Log[Log[x]]^2)/((6
4*x - 32*x^2 + 4*x^3)*Log[x] + (-64*x + 32*x^2 - 4*x^3)*Log[x]*Log[Log[x]] + (16*x - 8*x^2 + x^3)*Log[x]*Log[L
og[x]]^2),x]

[Out]

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

Rule 30

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

Rule 31

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

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

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 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 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 \left (\frac {1+4 x-x^2}{(-4+x) x}-\frac {\log (x)}{(-4+x)^2}+\frac {\log ^2(3)}{x \log (x) (-2+\log (\log (x)))^2}\right ) \, dx\\ &=\log ^2(3) \int \frac {1}{x \log (x) (-2+\log (\log (x)))^2} \, dx+\int \frac {1+4 x-x^2}{(-4+x) x} \, dx-\int \frac {\log (x)}{(-4+x)^2} \, dx\\ &=-\frac {x \log (x)}{4 (4-x)}-\frac {1}{4} \int \frac {1}{-4+x} \, dx+\log ^2(3) \operatorname {Subst}\left (\int \frac {1}{x (-2+\log (x))^2} \, dx,x,\log (x)\right )+\int \left (-1+\frac {1}{4 (-4+x)}-\frac {1}{4 x}\right ) \, dx\\ &=-x-\frac {\log (x)}{4}-\frac {x \log (x)}{4 (4-x)}+\log ^2(3) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,-2+\log (\log (x))\right )\\ &=-x-\frac {\log (x)}{4}-\frac {x \log (x)}{4 (4-x)}+\frac {\log ^2(3)}{2-\log (\log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 25, normalized size = 0.93 \begin {gather*} -x+\frac {\log (x)}{-4+x}-\frac {\log ^2(3)}{-2+\log (\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((16 - 8*x + x^2)*Log[3]^2 + (-16 - 60*x + 32*x^2 - 4*x^3)*Log[x] - 4*x*Log[x]^2 + ((16 + 60*x - 32*
x^2 + 4*x^3)*Log[x] + 4*x*Log[x]^2)*Log[Log[x]] + ((-4 - 15*x + 8*x^2 - x^3)*Log[x] - x*Log[x]^2)*Log[Log[x]]^
2)/((64*x - 32*x^2 + 4*x^3)*Log[x] + (-64*x + 32*x^2 - 4*x^3)*Log[x]*Log[Log[x]] + (16*x - 8*x^2 + x^3)*Log[x]
*Log[Log[x]]^2),x]

[Out]

-x + Log[x]/(-4 + x) - Log[3]^2/(-2 + Log[Log[x]])

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fricas [A]  time = 0.58, size = 52, normalized size = 1.93 \begin {gather*} -\frac {{\left (x - 4\right )} \log \relax (3)^{2} - 2 \, x^{2} + {\left (x^{2} - 4 \, x - \log \relax (x)\right )} \log \left (\log \relax (x)\right ) + 8 \, x + 2 \, \log \relax (x)}{{\left (x - 4\right )} \log \left (\log \relax (x)\right ) - 2 \, x + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(x)^2+(-x^3+8*x^2-15*x-4)*log(x))*log(log(x))^2+(4*x*log(x)^2+(4*x^3-32*x^2+60*x+16)*log(x))
*log(log(x))-4*x*log(x)^2+(-4*x^3+32*x^2-60*x-16)*log(x)+(x^2-8*x+16)*log(3)^2)/((x^3-8*x^2+16*x)*log(x)*log(l
og(x))^2+(-4*x^3+32*x^2-64*x)*log(x)*log(log(x))+(4*x^3-32*x^2+64*x)*log(x)),x, algorithm="fricas")

[Out]

-((x - 4)*log(3)^2 - 2*x^2 + (x^2 - 4*x - log(x))*log(log(x)) + 8*x + 2*log(x))/((x - 4)*log(log(x)) - 2*x + 8
)

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giac [A]  time = 2.99, size = 25, normalized size = 0.93 \begin {gather*} -x - \frac {\log \relax (3)^{2}}{\log \left (\log \relax (x)\right ) - 2} + \frac {\log \relax (x)}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(x)^2+(-x^3+8*x^2-15*x-4)*log(x))*log(log(x))^2+(4*x*log(x)^2+(4*x^3-32*x^2+60*x+16)*log(x))
*log(log(x))-4*x*log(x)^2+(-4*x^3+32*x^2-60*x-16)*log(x)+(x^2-8*x+16)*log(3)^2)/((x^3-8*x^2+16*x)*log(x)*log(l
og(x))^2+(-4*x^3+32*x^2-64*x)*log(x)*log(log(x))+(4*x^3-32*x^2+64*x)*log(x)),x, algorithm="giac")

[Out]

-x - log(3)^2/(log(log(x)) - 2) + log(x)/(x - 4)

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maple [A]  time = 0.05, size = 32, normalized size = 1.19

method	result	size
default	\(-\frac {\ln \relax (3)^{2}}{\ln \left (\ln \relax (x )\right )-2}+\frac {x \ln \relax (x )}{4 x -16}-x -\frac {\ln \relax (x )}{4}\)	\(32\)
risch	\(-\frac {x^{2}-4 x -\ln \relax (x )}{x -4}-\frac {\ln \relax (3)^{2}}{\ln \left (\ln \relax (x )\right )-2}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*ln(x)^2+(-x^3+8*x^2-15*x-4)*ln(x))*ln(ln(x))^2+(4*x*ln(x)^2+(4*x^3-32*x^2+60*x+16)*ln(x))*ln(ln(x))-4
*x*ln(x)^2+(-4*x^3+32*x^2-60*x-16)*ln(x)+(x^2-8*x+16)*ln(3)^2)/((x^3-8*x^2+16*x)*ln(x)*ln(ln(x))^2+(-4*x^3+32*
x^2-64*x)*ln(x)*ln(ln(x))+(4*x^3-32*x^2+64*x)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 1.40, size = 55, normalized size = 2.04 \begin {gather*} -\frac {{\left (\log \relax (3)^{2} + 8\right )} x - 2 \, x^{2} - 4 \, \log \relax (3)^{2} + {\left (x^{2} - 4 \, x - \log \relax (x)\right )} \log \left (\log \relax (x)\right ) + 2 \, \log \relax (x)}{{\left (x - 4\right )} \log \left (\log \relax (x)\right ) - 2 \, x + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*log(x)^2+(-x^3+8*x^2-15*x-4)*log(x))*log(log(x))^2+(4*x*log(x)^2+(4*x^3-32*x^2+60*x+16)*log(x))
*log(log(x))-4*x*log(x)^2+(-4*x^3+32*x^2-60*x-16)*log(x)+(x^2-8*x+16)*log(3)^2)/((x^3-8*x^2+16*x)*log(x)*log(l
og(x))^2+(-4*x^3+32*x^2-64*x)*log(x)*log(log(x))+(4*x^3-32*x^2+64*x)*log(x)),x, algorithm="maxima")

[Out]

-((log(3)^2 + 8)*x - 2*x^2 - 4*log(3)^2 + (x^2 - 4*x - log(x))*log(log(x)) + 2*log(x))/((x - 4)*log(log(x)) -
2*x + 8)

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mupad [B]  time = 2.25, size = 25, normalized size = 0.93 \begin {gather*} \frac {\ln \relax (x)}{x-4}-\frac {{\ln \relax (3)}^2}{\ln \left (\ln \relax (x)\right )-2}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(log(x))^2*(x*log(x)^2 + log(x)*(15*x - 8*x^2 + x^3 + 4)) - log(log(x))*(4*x*log(x)^2 + log(x)*(60*x
- 32*x^2 + 4*x^3 + 16)) + 4*x*log(x)^2 - log(3)^2*(x^2 - 8*x + 16) + log(x)*(60*x - 32*x^2 + 4*x^3 + 16))/(log
(x)*(64*x - 32*x^2 + 4*x^3) - log(log(x))*log(x)*(64*x - 32*x^2 + 4*x^3) + log(log(x))^2*log(x)*(16*x - 8*x^2
+ x^3)),x)

[Out]

log(x)/(x - 4) - log(3)^2/(log(log(x)) - 2) - x

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sympy [A]  time = 0.37, size = 19, normalized size = 0.70 \begin {gather*} - x - \frac {\log {\relax (3 )}^{2}}{\log {\left (\log {\relax (x )} \right )} - 2} + \frac {\log {\relax (x )}}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*ln(x)**2+(-x**3+8*x**2-15*x-4)*ln(x))*ln(ln(x))**2+(4*x*ln(x)**2+(4*x**3-32*x**2+60*x+16)*ln(x)
)*ln(ln(x))-4*x*ln(x)**2+(-4*x**3+32*x**2-60*x-16)*ln(x)+(x**2-8*x+16)*ln(3)**2)/((x**3-8*x**2+16*x)*ln(x)*ln(
ln(x))**2+(-4*x**3+32*x**2-64*x)*ln(x)*ln(ln(x))+(4*x**3-32*x**2+64*x)*ln(x)),x)

[Out]

-x - log(3)**2/(log(log(x)) - 2) + log(x)/(x - 4)

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