\(\int \frac {2+(2+4 x) \log (x)}{(81 x-36 x^2+4 x^3+(108 x-24 x^2) \log ^2(4)+(54 x-4 x^2) \log ^4(4)+12 x \log ^6(4)+x \log ^8(4)) \log (x)+(-18 x+4 x^2-12 x \log ^2(4)-2 x \log ^4(4)) \log (x) \log (x \log (x))+x \log (x) \log ^2(x \log (x))} \, dx\) [2386]

Optimal result

Integrand size = 114, antiderivative size = 23 \[ \int \frac {2+(2+4 x) \log (x)}{\left (81 x-36 x^2+4 x^3+\left (108 x-24 x^2\right ) \log ^2(4)+\left (54 x-4 x^2\right ) \log ^4(4)+12 x \log ^6(4)+x \log ^8(4)\right ) \log (x)+\left (-18 x+4 x^2-12 x \log ^2(4)-2 x \log ^4(4)\right ) \log (x) \log (x \log (x))+x \log (x) \log ^2(x \log (x))} \, dx=\frac {2}{-2 x+\left (3+\log ^2(4)\right )^2-\log (x \log (x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {6820, 12, 6818} \[ \int \frac {2+(2+4 x) \log (x)}{\left (81 x-36 x^2+4 x^3+\left (108 x-24 x^2\right ) \log ^2(4)+\left (54 x-4 x^2\right ) \log ^4(4)+12 x \log ^6(4)+x \log ^8(4)\right ) \log (x)+\left (-18 x+4 x^2-12 x \log ^2(4)-2 x \log ^4(4)\right ) \log (x) \log (x \log (x))+x \log (x) \log ^2(x \log (x))} \, dx=-\frac {2}{2 x+\log (x \log (x))-\left (3+\log ^2(4)\right )^2} \]

[In]

Int[(2 + (2 + 4*x)*Log[x])/((81*x - 36*x^2 + 4*x^3 + (108*x - 24*x^2)*Log[4]^2 + (54*x - 4*x^2)*Log[4]^4 + 12*
x*Log[4]^6 + x*Log[4]^8)*Log[x] + (-18*x + 4*x^2 - 12*x*Log[4]^2 - 2*x*Log[4]^4)*Log[x]*Log[x*Log[x]] + x*Log[
x]*Log[x*Log[x]]^2),x]

[Out]

-2/(2*x - (3 + Log[4]^2)^2 + Log[x*Log[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 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

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 {2 (1+\log (x)+2 x \log (x))}{x \log (x) \left (2 x-\left (3+\log ^2(4)\right )^2+\log (x \log (x))\right )^2} \, dx \\ & = 2 \int \frac {1+\log (x)+2 x \log (x)}{x \log (x) \left (2 x-\left (3+\log ^2(4)\right )^2+\log (x \log (x))\right )^2} \, dx \\ & = -\frac {2}{2 x-\left (3+\log ^2(4)\right )^2+\log (x \log (x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {2+(2+4 x) \log (x)}{\left (81 x-36 x^2+4 x^3+\left (108 x-24 x^2\right ) \log ^2(4)+\left (54 x-4 x^2\right ) \log ^4(4)+12 x \log ^6(4)+x \log ^8(4)\right ) \log (x)+\left (-18 x+4 x^2-12 x \log ^2(4)-2 x \log ^4(4)\right ) \log (x) \log (x \log (x))+x \log (x) \log ^2(x \log (x))} \, dx=-\frac {2}{2 x-\left (3+\log ^2(4)\right )^2+\log (x \log (x))} \]

[In]

Integrate[(2 + (2 + 4*x)*Log[x])/((81*x - 36*x^2 + 4*x^3 + (108*x - 24*x^2)*Log[4]^2 + (54*x - 4*x^2)*Log[4]^4
+ 12*x*Log[4]^6 + x*Log[4]^8)*Log[x] + (-18*x + 4*x^2 - 12*x*Log[4]^2 - 2*x*Log[4]^4)*Log[x]*Log[x*Log[x]] +
x*Log[x]*Log[x*Log[x]]^2),x]

[Out]

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

Maple [A] (verified)

Time = 2.88 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26

method	result	size
parallelrisch	\(\frac {2}{16 \ln \left (2\right )^{4}+24 \ln \left (2\right )^{2}-\ln \left (x \ln \left (x \right )\right )-2 x +9}\)	\(29\)
risch	\(\frac {4 i}{\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2}-\pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )-\pi \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{3}+\pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2}+32 i \ln \left (2\right )^{4}+48 i \ln \left (2\right )^{2}-4 i x -2 i \ln \left (x \right )-2 i \ln \left (\ln \left (x \right )\right )+18 i}\)	\(104\)
default	\(-\frac {4 i}{\left (2 x \ln \left (x \right )+\ln \left (x \right )+1\right ) \left (\pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )-\pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2}-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2}+\pi \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{3}-32 i \ln \left (2\right )^{4}-48 i \ln \left (2\right )^{2}+4 i x +2 i \ln \left (\ln \left (x \right )\right )+2 i \ln \left (x \right )-18 i\right )}-\frac {4 \ln \left (x \right ) \left (1+2 x \right )}{\left (2 x \ln \left (x \right )+\ln \left (x \right )+1\right ) \left (-i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )+i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2}+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2}-i \pi \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{3}-32 \ln \left (2\right )^{4}-48 \ln \left (2\right )^{2}+4 x +2 \ln \left (\ln \left (x \right )\right )+2 \ln \left (x \right )-18\right )}\)	\(236\)
parts	\(-\frac {4 i}{\left (2 x \ln \left (x \right )+\ln \left (x \right )+1\right ) \left (\pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )-\pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2}-\pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2}+\pi \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{3}-32 i \ln \left (2\right )^{4}-48 i \ln \left (2\right )^{2}+4 i x +2 i \ln \left (\ln \left (x \right )\right )+2 i \ln \left (x \right )-18 i\right )}-\frac {4 \ln \left (x \right ) \left (1+2 x \right )}{\left (2 x \ln \left (x \right )+\ln \left (x \right )+1\right ) \left (-i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )+i \pi \,\operatorname {csgn}\left (i \ln \left (x \right )\right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2}+i \pi \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{2}-i \pi \operatorname {csgn}\left (i x \ln \left (x \right )\right )^{3}-32 \ln \left (2\right )^{4}-48 \ln \left (2\right )^{2}+4 x +2 \ln \left (\ln \left (x \right )\right )+2 \ln \left (x \right )-18\right )}\)	\(236\)

[In]

int(((4*x+2)*ln(x)+2)/(x*ln(x)*ln(x*ln(x))^2+(-32*x*ln(2)^4-48*x*ln(2)^2+4*x^2-18*x)*ln(x)*ln(x*ln(x))+(256*x*
ln(2)^8+768*x*ln(2)^6+16*(-4*x^2+54*x)*ln(2)^4+4*(-24*x^2+108*x)*ln(2)^2+4*x^3-36*x^2+81*x)*ln(x)),x,method=_R
ETURNVERBOSE)

[Out]

2/(16*ln(2)^4+24*ln(2)^2-ln(x*ln(x))-2*x+9)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {2+(2+4 x) \log (x)}{\left (81 x-36 x^2+4 x^3+\left (108 x-24 x^2\right ) \log ^2(4)+\left (54 x-4 x^2\right ) \log ^4(4)+12 x \log ^6(4)+x \log ^8(4)\right ) \log (x)+\left (-18 x+4 x^2-12 x \log ^2(4)-2 x \log ^4(4)\right ) \log (x) \log (x \log (x))+x \log (x) \log ^2(x \log (x))} \, dx=\frac {2}{16 \, \log \left (2\right )^{4} + 24 \, \log \left (2\right )^{2} - 2 \, x - \log \left (x \log \left (x\right )\right ) + 9} \]

[In]

integrate(((4*x+2)*log(x)+2)/(x*log(x)*log(x*log(x))^2+(-32*x*log(2)^4-48*x*log(2)^2+4*x^2-18*x)*log(x)*log(x*
log(x))+(256*x*log(2)^8+768*x*log(2)^6+16*(-4*x^2+54*x)*log(2)^4+4*(-24*x^2+108*x)*log(2)^2+4*x^3-36*x^2+81*x)
*log(x)),x, algorithm="fricas")

[Out]

2/(16*log(2)^4 + 24*log(2)^2 - 2*x - log(x*log(x)) + 9)

Sympy [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {2+(2+4 x) \log (x)}{\left (81 x-36 x^2+4 x^3+\left (108 x-24 x^2\right ) \log ^2(4)+\left (54 x-4 x^2\right ) \log ^4(4)+12 x \log ^6(4)+x \log ^8(4)\right ) \log (x)+\left (-18 x+4 x^2-12 x \log ^2(4)-2 x \log ^4(4)\right ) \log (x) \log (x \log (x))+x \log (x) \log ^2(x \log (x))} \, dx=- \frac {2}{2 x + \log {\left (x \log {\left (x \right )} \right )} - 24 \log {\left (2 \right )}^{2} - 9 - 16 \log {\left (2 \right )}^{4}} \]

[In]

integrate(((4*x+2)*ln(x)+2)/(x*ln(x)*ln(x*ln(x))**2+(-32*x*ln(2)**4-48*x*ln(2)**2+4*x**2-18*x)*ln(x)*ln(x*ln(x
))+(256*x*ln(2)**8+768*x*ln(2)**6+16*(-4*x**2+54*x)*ln(2)**4+4*(-24*x**2+108*x)*ln(2)**2+4*x**3-36*x**2+81*x)*
ln(x)),x)

[Out]

-2/(2*x + log(x*log(x)) - 24*log(2)**2 - 9 - 16*log(2)**4)

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {2+(2+4 x) \log (x)}{\left (81 x-36 x^2+4 x^3+\left (108 x-24 x^2\right ) \log ^2(4)+\left (54 x-4 x^2\right ) \log ^4(4)+12 x \log ^6(4)+x \log ^8(4)\right ) \log (x)+\left (-18 x+4 x^2-12 x \log ^2(4)-2 x \log ^4(4)\right ) \log (x) \log (x \log (x))+x \log (x) \log ^2(x \log (x))} \, dx=\frac {2}{16 \, \log \left (2\right )^{4} + 24 \, \log \left (2\right )^{2} - 2 \, x - \log \left (x\right ) - \log \left (\log \left (x\right )\right ) + 9} \]

[In]

integrate(((4*x+2)*log(x)+2)/(x*log(x)*log(x*log(x))^2+(-32*x*log(2)^4-48*x*log(2)^2+4*x^2-18*x)*log(x)*log(x*
log(x))+(256*x*log(2)^8+768*x*log(2)^6+16*(-4*x^2+54*x)*log(2)^4+4*(-24*x^2+108*x)*log(2)^2+4*x^3-36*x^2+81*x)
*log(x)),x, algorithm="maxima")

[Out]

2/(16*log(2)^4 + 24*log(2)^2 - 2*x - log(x) - log(log(x)) + 9)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.30 \[ \int \frac {2+(2+4 x) \log (x)}{\left (81 x-36 x^2+4 x^3+\left (108 x-24 x^2\right ) \log ^2(4)+\left (54 x-4 x^2\right ) \log ^4(4)+12 x \log ^6(4)+x \log ^8(4)\right ) \log (x)+\left (-18 x+4 x^2-12 x \log ^2(4)-2 x \log ^4(4)\right ) \log (x) \log (x \log (x))+x \log (x) \log ^2(x \log (x))} \, dx=\frac {2}{16 \, \log \left (2\right )^{4} + 24 \, \log \left (2\right )^{2} - 2 \, x - \log \left (x\right ) - \log \left (\log \left (x\right )\right ) + 9} \]

[In]

integrate(((4*x+2)*log(x)+2)/(x*log(x)*log(x*log(x))^2+(-32*x*log(2)^4-48*x*log(2)^2+4*x^2-18*x)*log(x)*log(x*
log(x))+(256*x*log(2)^8+768*x*log(2)^6+16*(-4*x^2+54*x)*log(2)^4+4*(-24*x^2+108*x)*log(2)^2+4*x^3-36*x^2+81*x)
*log(x)),x, algorithm="giac")

[Out]

2/(16*log(2)^4 + 24*log(2)^2 - 2*x - log(x) - log(log(x)) + 9)

Mupad [B] (verification not implemented)

Time = 9.98 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {2+(2+4 x) \log (x)}{\left (81 x-36 x^2+4 x^3+\left (108 x-24 x^2\right ) \log ^2(4)+\left (54 x-4 x^2\right ) \log ^4(4)+12 x \log ^6(4)+x \log ^8(4)\right ) \log (x)+\left (-18 x+4 x^2-12 x \log ^2(4)-2 x \log ^4(4)\right ) \log (x) \log (x \log (x))+x \log (x) \log ^2(x \log (x))} \, dx=\frac {2}{24\,{\ln \left (2\right )}^2-\ln \left (x\,\ln \left (x\right )\right )-2\,x+16\,{\ln \left (2\right )}^4+9} \]

[In]

int((log(x)*(4*x + 2) + 2)/(log(x)*(81*x + 16*log(2)^4*(54*x - 4*x^2) + 4*log(2)^2*(108*x - 24*x^2) + 768*x*lo
g(2)^6 + 256*x*log(2)^8 - 36*x^2 + 4*x^3) - log(x*log(x))*log(x)*(18*x + 48*x*log(2)^2 + 32*x*log(2)^4 - 4*x^2
) + x*log(x*log(x))^2*log(x)),x)

[Out]

2/(24*log(2)^2 - log(x*log(x)) - 2*x + 16*log(2)^4 + 9)