3.33.51 \(\int \frac {-4+6 x-2 x^2+(2 x-x^2+x \log (x)) \log (24-12 x+12 \log (x))+(-4+4 x-x^2+(-2+x) \log (x)) \log (24-12 x+12 \log (x)) \log (\frac {1}{(-8+4 x) \log ^2(24-12 x+12 \log (x))})}{(-20 x^2+20 x^3-5 x^4+(-10 x^2+5 x^3) \log (x)) \log (24-12 x+12 \log (x))+(-40 x+40 x^2-10 x^3+(-20 x+10 x^2) \log (x)) \log (24-12 x+12 \log (x)) \log (\frac {1}{(-8+4 x) \log ^2(24-12 x+12 \log (x))})+(-20+20 x-5 x^2+(-10+5 x) \log (x)) \log (24-12 x+12 \log (x)) \log ^2(\frac {1}{(-8+4 x) \log ^2(24-12 x+12 \log (x))})} \, dx\)

Optimal. Leaf size=31 \[ \frac {x}{5 \left (x+\log \left (\frac {1}{4 (-2+x) \log ^2(12 (2-x+\log (x)))}\right )\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 2.22, antiderivative size = 36, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 4, integrand size = 237, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.017, Rules used = {6688, 12, 6711, 32} \begin {gather*} -\frac {1}{5 \left (\frac {x}{\log \left (-\frac {1}{4 (2-x) \log ^2(12 (-x+\log (x)+2))}\right )}+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 + 6*x - 2*x^2 + (2*x - x^2 + x*Log[x])*Log[24 - 12*x + 12*Log[x]] + (-4 + 4*x - x^2 + (-2 + x)*Log[x])
*Log[24 - 12*x + 12*Log[x]]*Log[1/((-8 + 4*x)*Log[24 - 12*x + 12*Log[x]]^2)])/((-20*x^2 + 20*x^3 - 5*x^4 + (-1
0*x^2 + 5*x^3)*Log[x])*Log[24 - 12*x + 12*Log[x]] + (-40*x + 40*x^2 - 10*x^3 + (-20*x + 10*x^2)*Log[x])*Log[24
- 12*x + 12*Log[x]]*Log[1/((-8 + 4*x)*Log[24 - 12*x + 12*Log[x]]^2)] + (-20 + 20*x - 5*x^2 + (-10 + 5*x)*Log[
x])*Log[24 - 12*x + 12*Log[x]]*Log[1/((-8 + 4*x)*Log[24 - 12*x + 12*Log[x]]^2)]^2),x]

[Out]

-1/5*1/(1 + x/Log[-1/4*1/((2 - x)*Log[12*(2 - x + Log[x])]^2)])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 {2 \left (2-3 x+x^2\right )+(-2+x-\log (x)) \log (12 (2-x+\log (x))) \left (x+(-2+x) \log \left (\frac {1}{4 (-2+x) \log ^2(12 (2-x+\log (x)))}\right )\right )}{5 (2-x) (2-x+\log (x)) \log (12 (2-x+\log (x))) \left (x+\log \left (\frac {1}{4 (-2+x) \log ^2(12 (2-x+\log (x)))}\right )\right )^2} \, dx\\ &=\frac {1}{5} \int \frac {2 \left (2-3 x+x^2\right )+(-2+x-\log (x)) \log (12 (2-x+\log (x))) \left (x+(-2+x) \log \left (\frac {1}{4 (-2+x) \log ^2(12 (2-x+\log (x)))}\right )\right )}{(2-x) (2-x+\log (x)) \log (12 (2-x+\log (x))) \left (x+\log \left (\frac {1}{4 (-2+x) \log ^2(12 (2-x+\log (x)))}\right )\right )^2} \, dx\\ &=\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,\frac {x}{\log \left (\frac {1}{4 (-2+x) \log ^2(12 (2-x+\log (x)))}\right )}\right )\\ &=-\frac {1}{5 \left (1+\frac {x}{\log \left (-\frac {1}{4 (2-x) \log ^2(12 (2-x+\log (x)))}\right )}\right )}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.18, size = 31, normalized size = 1.00 \begin {gather*} \frac {x}{5 \left (x+\log \left (\frac {1}{4 (-2+x) \log ^2(12 (2-x+\log (x)))}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 + 6*x - 2*x^2 + (2*x - x^2 + x*Log[x])*Log[24 - 12*x + 12*Log[x]] + (-4 + 4*x - x^2 + (-2 + x)*L
og[x])*Log[24 - 12*x + 12*Log[x]]*Log[1/((-8 + 4*x)*Log[24 - 12*x + 12*Log[x]]^2)])/((-20*x^2 + 20*x^3 - 5*x^4
+ (-10*x^2 + 5*x^3)*Log[x])*Log[24 - 12*x + 12*Log[x]] + (-40*x + 40*x^2 - 10*x^3 + (-20*x + 10*x^2)*Log[x])*
Log[24 - 12*x + 12*Log[x]]*Log[1/((-8 + 4*x)*Log[24 - 12*x + 12*Log[x]]^2)] + (-20 + 20*x - 5*x^2 + (-10 + 5*x
)*Log[x])*Log[24 - 12*x + 12*Log[x]]*Log[1/((-8 + 4*x)*Log[24 - 12*x + 12*Log[x]]^2)]^2),x]

[Out]

x/(5*(x + Log[1/(4*(-2 + x)*Log[12*(2 - x + Log[x])]^2)]))

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 27, normalized size = 0.87 \begin {gather*} \frac {x}{5 \, {\left (x + \log \left (\frac {1}{4 \, {\left (x - 2\right )} \log \left (-12 \, x + 12 \, \log \relax (x) + 24\right )^{2}}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x-2)*log(x)-x^2+4*x-4)*log(12*log(x)-12*x+24)*log(1/(4*x-8)/log(12*log(x)-12*x+24)^2)+(x*log(x)-x
^2+2*x)*log(12*log(x)-12*x+24)-2*x^2+6*x-4)/(((5*x-10)*log(x)-5*x^2+20*x-20)*log(12*log(x)-12*x+24)*log(1/(4*x
-8)/log(12*log(x)-12*x+24)^2)^2+((10*x^2-20*x)*log(x)-10*x^3+40*x^2-40*x)*log(12*log(x)-12*x+24)*log(1/(4*x-8)
/log(12*log(x)-12*x+24)^2)+((5*x^3-10*x^2)*log(x)-5*x^4+20*x^3-20*x^2)*log(12*log(x)-12*x+24)),x, algorithm="f
ricas")

[Out]

1/5*x/(x + log(1/4/((x - 2)*log(-12*x + 12*log(x) + 24)^2)))

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x-2)*log(x)-x^2+4*x-4)*log(12*log(x)-12*x+24)*log(1/(4*x-8)/log(12*log(x)-12*x+24)^2)+(x*log(x)-x
^2+2*x)*log(12*log(x)-12*x+24)-2*x^2+6*x-4)/(((5*x-10)*log(x)-5*x^2+20*x-20)*log(12*log(x)-12*x+24)*log(1/(4*x
-8)/log(12*log(x)-12*x+24)^2)^2+((10*x^2-20*x)*log(x)-10*x^3+40*x^2-40*x)*log(12*log(x)-12*x+24)*log(1/(4*x-8)
/log(12*log(x)-12*x+24)^2)+((5*x^3-10*x^2)*log(x)-5*x^4+20*x^3-20*x^2)*log(12*log(x)-12*x+24)),x, algorithm="g
iac")

[Out]

Timed out

________________________________________________________________________________________

maple [C]  time = 0.68, size = 283, normalized size = 9.13

method	result	size
risch	\(\frac {2 x}{5 \left (-i \pi \,\mathrm {csgn}\left (\frac {i}{x -2}\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (12 \ln \relax (x )-12 x +24\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (12 \ln \relax (x )-12 x +24\right )^{2} \left (x -2\right )}\right )+i \pi \,\mathrm {csgn}\left (\frac {i}{x -2}\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (12 \ln \relax (x )-12 x +24\right )^{2} \left (x -2\right )}\right )^{2}+i \pi \,\mathrm {csgn}\left (\frac {i}{\ln \left (12 \ln \relax (x )-12 x +24\right )^{2}}\right ) \mathrm {csgn}\left (\frac {i}{\ln \left (12 \ln \relax (x )-12 x +24\right )^{2} \left (x -2\right )}\right )^{2}+i \pi \mathrm {csgn}\left (i \ln \left (12 \ln \relax (x )-12 x +24\right )\right )^{2} \mathrm {csgn}\left (i \ln \left (12 \ln \relax (x )-12 x +24\right )^{2}\right )-2 i \pi \,\mathrm {csgn}\left (i \ln \left (12 \ln \relax (x )-12 x +24\right )\right ) \mathrm {csgn}\left (i \ln \left (12 \ln \relax (x )-12 x +24\right )^{2}\right )^{2}+i \pi \mathrm {csgn}\left (i \ln \left (12 \ln \relax (x )-12 x +24\right )^{2}\right )^{3}-i \pi \mathrm {csgn}\left (\frac {i}{\ln \left (12 \ln \relax (x )-12 x +24\right )^{2} \left (x -2\right )}\right )^{3}-4 \ln \relax (2)+2 x -2 \ln \left (x -2\right )-4 \ln \left (\ln \left (12 \ln \relax (x )-12 x +24\right )\right )\right )}\)	\(283\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x-2)*ln(x)-x^2+4*x-4)*ln(12*ln(x)-12*x+24)*ln(1/(4*x-8)/ln(12*ln(x)-12*x+24)^2)+(x*ln(x)-x^2+2*x)*ln(12
*ln(x)-12*x+24)-2*x^2+6*x-4)/(((5*x-10)*ln(x)-5*x^2+20*x-20)*ln(12*ln(x)-12*x+24)*ln(1/(4*x-8)/ln(12*ln(x)-12*
x+24)^2)^2+((10*x^2-20*x)*ln(x)-10*x^3+40*x^2-40*x)*ln(12*ln(x)-12*x+24)*ln(1/(4*x-8)/ln(12*ln(x)-12*x+24)^2)+
((5*x^3-10*x^2)*ln(x)-5*x^4+20*x^3-20*x^2)*ln(12*ln(x)-12*x+24)),x,method=_RETURNVERBOSE)

[Out]

2/5*x/(-I*Pi*csgn(I/(x-2))*csgn(I/ln(12*ln(x)-12*x+24)^2)*csgn(I/ln(12*ln(x)-12*x+24)^2/(x-2))+I*Pi*csgn(I/(x-
2))*csgn(I/ln(12*ln(x)-12*x+24)^2/(x-2))^2+I*Pi*csgn(I/ln(12*ln(x)-12*x+24)^2)*csgn(I/ln(12*ln(x)-12*x+24)^2/(
x-2))^2+I*Pi*csgn(I*ln(12*ln(x)-12*x+24))^2*csgn(I*ln(12*ln(x)-12*x+24)^2)-2*I*Pi*csgn(I*ln(12*ln(x)-12*x+24))
*csgn(I*ln(12*ln(x)-12*x+24)^2)^2+I*Pi*csgn(I*ln(12*ln(x)-12*x+24)^2)^3-I*Pi*csgn(I/ln(12*ln(x)-12*x+24)^2/(x-
2))^3-4*ln(2)+2*x-2*ln(x-2)-4*ln(ln(12*ln(x)-12*x+24)))

________________________________________________________________________________________

maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x-2)*log(x)-x^2+4*x-4)*log(12*log(x)-12*x+24)*log(1/(4*x-8)/log(12*log(x)-12*x+24)^2)+(x*log(x)-x
^2+2*x)*log(12*log(x)-12*x+24)-2*x^2+6*x-4)/(((5*x-10)*log(x)-5*x^2+20*x-20)*log(12*log(x)-12*x+24)*log(1/(4*x
-8)/log(12*log(x)-12*x+24)^2)^2+((10*x^2-20*x)*log(x)-10*x^3+40*x^2-40*x)*log(12*log(x)-12*x+24)*log(1/(4*x-8)
/log(12*log(x)-12*x+24)^2)+((5*x^3-10*x^2)*log(x)-5*x^4+20*x^3-20*x^2)*log(12*log(x)-12*x+24)),x, algorithm="m
axima")

[Out]

Timed out

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {6\,x+\ln \left (12\,\ln \relax (x)-12\,x+24\right )\,\left (2\,x+x\,\ln \relax (x)-x^2\right )-2\,x^2+\ln \left (\frac {1}{{\ln \left (12\,\ln \relax (x)-12\,x+24\right )}^2\,\left (4\,x-8\right )}\right )\,\ln \left (12\,\ln \relax (x)-12\,x+24\right )\,\left (4\,x+\ln \relax (x)\,\left (x-2\right )-x^2-4\right )-4}{-\ln \left (12\,\ln \relax (x)-12\,x+24\right )\,\left (20\,x+\ln \relax (x)\,\left (5\,x-10\right )-5\,x^2-20\right )\,{\ln \left (\frac {1}{{\ln \left (12\,\ln \relax (x)-12\,x+24\right )}^2\,\left (4\,x-8\right )}\right )}^2+\ln \left (12\,\ln \relax (x)-12\,x+24\right )\,\left (40\,x+\ln \relax (x)\,\left (20\,x-10\,x^2\right )-40\,x^2+10\,x^3\right )\,\ln \left (\frac {1}{{\ln \left (12\,\ln \relax (x)-12\,x+24\right )}^2\,\left (4\,x-8\right )}\right )+\ln \left (12\,\ln \relax (x)-12\,x+24\right )\,\left (\ln \relax (x)\,\left (10\,x^2-5\,x^3\right )+20\,x^2-20\,x^3+5\,x^4\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(6*x + log(12*log(x) - 12*x + 24)*(2*x + x*log(x) - x^2) - 2*x^2 + log(1/(log(12*log(x) - 12*x + 24)^2*(4
*x - 8)))*log(12*log(x) - 12*x + 24)*(4*x + log(x)*(x - 2) - x^2 - 4) - 4)/(log(12*log(x) - 12*x + 24)*(log(x)
*(10*x^2 - 5*x^3) + 20*x^2 - 20*x^3 + 5*x^4) - log(1/(log(12*log(x) - 12*x + 24)^2*(4*x - 8)))^2*log(12*log(x)
- 12*x + 24)*(20*x + log(x)*(5*x - 10) - 5*x^2 - 20) + log(1/(log(12*log(x) - 12*x + 24)^2*(4*x - 8)))*log(12
*log(x) - 12*x + 24)*(40*x + log(x)*(20*x - 10*x^2) - 40*x^2 + 10*x^3)),x)

[Out]

-int((6*x + log(12*log(x) - 12*x + 24)*(2*x + x*log(x) - x^2) - 2*x^2 + log(1/(log(12*log(x) - 12*x + 24)^2*(4
*x - 8)))*log(12*log(x) - 12*x + 24)*(4*x + log(x)*(x - 2) - x^2 - 4) - 4)/(log(12*log(x) - 12*x + 24)*(log(x)
*(10*x^2 - 5*x^3) + 20*x^2 - 20*x^3 + 5*x^4) - log(1/(log(12*log(x) - 12*x + 24)^2*(4*x - 8)))^2*log(12*log(x)
- 12*x + 24)*(20*x + log(x)*(5*x - 10) - 5*x^2 - 20) + log(1/(log(12*log(x) - 12*x + 24)^2*(4*x - 8)))*log(12
*log(x) - 12*x + 24)*(40*x + log(x)*(20*x - 10*x^2) - 40*x^2 + 10*x^3)), x)

________________________________________________________________________________________

sympy [A]  time = 3.55, size = 27, normalized size = 0.87 \begin {gather*} \frac {x}{5 x + 5 \log {\left (\frac {1}{\left (4 x - 8\right ) \log {\left (- 12 x + 12 \log {\relax (x )} + 24 \right )}^{2}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x-2)*ln(x)-x**2+4*x-4)*ln(12*ln(x)-12*x+24)*ln(1/(4*x-8)/ln(12*ln(x)-12*x+24)**2)+(x*ln(x)-x**2+2
*x)*ln(12*ln(x)-12*x+24)-2*x**2+6*x-4)/(((5*x-10)*ln(x)-5*x**2+20*x-20)*ln(12*ln(x)-12*x+24)*ln(1/(4*x-8)/ln(1
2*ln(x)-12*x+24)**2)**2+((10*x**2-20*x)*ln(x)-10*x**3+40*x**2-40*x)*ln(12*ln(x)-12*x+24)*ln(1/(4*x-8)/ln(12*ln
(x)-12*x+24)**2)+((5*x**3-10*x**2)*ln(x)-5*x**4+20*x**3-20*x**2)*ln(12*ln(x)-12*x+24)),x)

[Out]

x/(5*x + 5*log(1/((4*x - 8)*log(-12*x + 12*log(x) + 24)**2)))

________________________________________________________________________________________