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

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

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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} $$\begin{array}{c}-\frac{1}{5(\frac{x}{\log (-\frac{1}{4(2-x){\log }^2(12(-x+\log (x)+2))})}+1)}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{2(2-3x+x^2)+(-2+x-\log (x))\log (12(2-x+\log (x)))(x+(-2+x)\log \left(\frac{1}{4(-2+x){\log }^2(12(2-x+\log (x)))}\right))}{5(2-x)(2-x+\log (x))\log (12(2-x+\log (x))){(x+\log \left(\frac{1}{4(-2+x){\log }^2(12(2-x+\log (x)))}\right))}^2}dx\\ & =\frac{1}{5}\int \frac{2(2-3x+x^2)+(-2+x-\log (x))\log (12(2-x+\log (x)))(x+(-2+x)\log \left(\frac{1}{4(-2+x){\log }^2(12(2-x+\log (x)))}\right))}{(2-x)(2-x+\log (x))\log (12(2-x+\log (x))){(x+\log \left(\frac{1}{4(-2+x){\log }^2(12(2-x+\log (x)))}\right))}^2}dx\\ & =\frac{1}{5}\mathrm{Subst}(\int \frac{1}{(1+x{)}^2}dx,x,\frac{x}{\log \left(\frac{1}{4(-2+x){\log }^2(12(2-x+\log (x)))}\right)})\\ & =-\frac{1}{5(1+\frac{x}{\log (-\frac{1}{4(2-x){\log }^2(12(2-x+\log (x)))})})}\end{array}\end{array}$$

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

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)]))

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

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)))

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

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

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

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

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)))

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

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

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

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)

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

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)))

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