3.24.84 \(\int \frac {3840-8960 x+2560 x^2+(3840 x-1280 x^2) \log (-3+x)+(-7680 x+1280 x^2+(3840 x-1280 x^2) \log (-3+x)) \log (x)+(-480+640 x-160 x^2+(480 x-160 x^2) \log (x)) \log (\log (3))+(15-5 x) \log ^2(\log (3))}{-768 x+256 x^2} \, dx\)

Optimal. Leaf size=29 \[ 5 \log (x) \left (x (x-\log (-3+x))-\left (-1+x+\frac {1}{16} \log (\log (3))\right )^2\right ) \]

________________________________________________________________________________________

Rubi [B]  time = 0.46, antiderivative size = 110, normalized size of antiderivative = 3.79, number of steps used = 18, number of rules used = 13, integrand size = 99, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.131, Rules used = {1593, 6688, 2389, 2295, 2370, 2411, 43, 2351, 2317, 2391, 2357, 2316, 2315} \begin {gather*} -10 x+5 x (\log (x)+1)+\frac {5}{8} x (16-\log (\log (3)))+\frac {5}{8} x (8-\log (\log (3))) \log (x)-\frac {5}{8} x (8-\log (\log (3)))-5 (3-x) \log (x-3)-15 \log (3) \log (x-3)-15 \log (x-3) \log \left (\frac {x}{3}\right )+5 (3-x) \log (x-3) (\log (x)+1)-\frac {5}{256} (16-\log (\log (3)))^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3840 - 8960*x + 2560*x^2 + (3840*x - 1280*x^2)*Log[-3 + x] + (-7680*x + 1280*x^2 + (3840*x - 1280*x^2)*Lo
g[-3 + x])*Log[x] + (-480 + 640*x - 160*x^2 + (480*x - 160*x^2)*Log[x])*Log[Log[3]] + (15 - 5*x)*Log[Log[3]]^2
)/(-768*x + 256*x^2),x]

[Out]

-10*x - 5*(3 - x)*Log[-3 + x] - 15*Log[3]*Log[-3 + x] - 15*Log[-3 + x]*Log[x/3] + 5*x*(1 + Log[x]) + 5*(3 - x)
*Log[-3 + x]*(1 + Log[x]) - (5*x*(8 - Log[Log[3]]))/8 + (5*x*Log[x]*(8 - Log[Log[3]]))/8 + (5*x*(16 - Log[Log[
3]]))/8 - (5*Log[x]*(16 - Log[Log[3]])^2)/256

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2316

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[((a + b*Log[-((c*d)/e)])*Log[d + e*
x])/e, x] + Dist[b, Int[Log[-((e*x)/d)]/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[-((c*d)/e), 0]

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2370

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> With[
{u = IntHide[Log[d*(e + f*x^m)^r], x]}, Dist[(a + b*Log[c*x^n])^p, u, x] - Dist[b*n*p, Int[Dist[(a + b*Log[c*x
^n])^(p - 1)/x, u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, r, m, n}, x] && IGtQ[p, 0] && RationalQ[m] && (EqQ[
p, 1] || (FractionQ[m] && IntegerQ[1/m]) || (EqQ[r, 1] && EqQ[m, 1] && EqQ[d*e, 1]))

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

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 {3840-8960 x+2560 x^2+\left (3840 x-1280 x^2\right ) \log (-3+x)+\left (-7680 x+1280 x^2+\left (3840 x-1280 x^2\right ) \log (-3+x)\right ) \log (x)+\left (-480+640 x-160 x^2+\left (480 x-160 x^2\right ) \log (x)\right ) \log (\log (3))+(15-5 x) \log ^2(\log (3))}{x (-768+256 x)} \, dx\\ &=\int \left (-5 \log (-3+x) (1+\log (x))-\frac {5 \log (x) (48+x (-8+\log (\log (3)))-3 \log (\log (3)))}{8 (-3+x)}-\frac {5 (-16+\log (\log (3))) (-16+32 x+\log (\log (3)))}{256 x}\right ) \, dx\\ &=-\left (\frac {5}{8} \int \frac {\log (x) (48+x (-8+\log (\log (3)))-3 \log (\log (3)))}{-3+x} \, dx\right )-5 \int \log (-3+x) (1+\log (x)) \, dx+\frac {1}{256} (5 (16-\log (\log (3)))) \int \frac {-16+32 x+\log (\log (3))}{x} \, dx\\ &=5 x (1+\log (x))+5 (3-x) \log (-3+x) (1+\log (x))-\frac {5}{8} \int \left (\frac {24 \log (x)}{-3+x}+\log (x) (-8+\log (\log (3)))\right ) \, dx+5 \int \left (-1-\frac {(3-x) \log (-3+x)}{x}\right ) \, dx+\frac {1}{256} (5 (16-\log (\log (3)))) \int \left (32+\frac {-16+\log (\log (3))}{x}\right ) \, dx\\ &=-5 x+5 x (1+\log (x))+5 (3-x) \log (-3+x) (1+\log (x))+\frac {5}{8} x (16-\log (\log (3)))-\frac {5}{256} \log (x) (16-\log (\log (3)))^2-5 \int \frac {(3-x) \log (-3+x)}{x} \, dx-15 \int \frac {\log (x)}{-3+x} \, dx+\frac {1}{8} (5 (8-\log (\log (3)))) \int \log (x) \, dx\\ &=-5 x-15 \log (3) \log (-3+x)+5 x (1+\log (x))+5 (3-x) \log (-3+x) (1+\log (x))-\frac {5}{8} x (8-\log (\log (3)))+\frac {5}{8} x \log (x) (8-\log (\log (3)))+\frac {5}{8} x (16-\log (\log (3)))-\frac {5}{256} \log (x) (16-\log (\log (3)))^2+5 \operatorname {Subst}\left (\int \frac {x \log (x)}{3+x} \, dx,x,-3+x\right )-15 \int \frac {\log \left (\frac {x}{3}\right )}{-3+x} \, dx\\ &=-5 x-15 \log (3) \log (-3+x)+5 x (1+\log (x))+5 (3-x) \log (-3+x) (1+\log (x))-\frac {5}{8} x (8-\log (\log (3)))+\frac {5}{8} x \log (x) (8-\log (\log (3)))+\frac {5}{8} x (16-\log (\log (3)))-\frac {5}{256} \log (x) (16-\log (\log (3)))^2+15 \text {Li}_2\left (1-\frac {x}{3}\right )+5 \operatorname {Subst}\left (\int \left (\log (x)-\frac {3 \log (x)}{3+x}\right ) \, dx,x,-3+x\right )\\ &=-5 x-15 \log (3) \log (-3+x)+5 x (1+\log (x))+5 (3-x) \log (-3+x) (1+\log (x))-\frac {5}{8} x (8-\log (\log (3)))+\frac {5}{8} x \log (x) (8-\log (\log (3)))+\frac {5}{8} x (16-\log (\log (3)))-\frac {5}{256} \log (x) (16-\log (\log (3)))^2+15 \text {Li}_2\left (1-\frac {x}{3}\right )+5 \operatorname {Subst}(\int \log (x) \, dx,x,-3+x)-15 \operatorname {Subst}\left (\int \frac {\log (x)}{3+x} \, dx,x,-3+x\right )\\ &=-10 x-5 (3-x) \log (-3+x)-15 \log (3) \log (-3+x)-15 \log (-3+x) \log \left (\frac {x}{3}\right )+5 x (1+\log (x))+5 (3-x) \log (-3+x) (1+\log (x))-\frac {5}{8} x (8-\log (\log (3)))+\frac {5}{8} x \log (x) (8-\log (\log (3)))+\frac {5}{8} x (16-\log (\log (3)))-\frac {5}{256} \log (x) (16-\log (\log (3)))^2+15 \text {Li}_2\left (1-\frac {x}{3}\right )+15 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{3}\right )}{x} \, dx,x,-3+x\right )\\ &=-10 x-5 (3-x) \log (-3+x)-15 \log (3) \log (-3+x)-15 \log (-3+x) \log \left (\frac {x}{3}\right )+5 x (1+\log (x))+5 (3-x) \log (-3+x) (1+\log (x))-\frac {5}{8} x (8-\log (\log (3)))+\frac {5}{8} x \log (x) (8-\log (\log (3)))+\frac {5}{8} x (16-\log (\log (3)))-\frac {5}{256} \log (x) (16-\log (\log (3)))^2\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [C]  time = 0.08, size = 89, normalized size = 3.07 \begin {gather*} -15 \log (3) \log (-3+x)-5 \log (x)+10 x \log (x)+15 \log \left (1-\frac {x}{3}\right ) \log (x)-5 x \log (-3+x) \log (x)+\frac {5}{8} \log (x) \log (\log (3))-\frac {5}{8} x \log (x) \log (\log (3))-\frac {5}{256} \log (x) \log ^2(\log (3))+15 \text {Li}_2\left (1-\frac {x}{3}\right )+15 \text {Li}_2\left (\frac {x}{3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3840 - 8960*x + 2560*x^2 + (3840*x - 1280*x^2)*Log[-3 + x] + (-7680*x + 1280*x^2 + (3840*x - 1280*x
^2)*Log[-3 + x])*Log[x] + (-480 + 640*x - 160*x^2 + (480*x - 160*x^2)*Log[x])*Log[Log[3]] + (15 - 5*x)*Log[Log
[3]]^2)/(-768*x + 256*x^2),x]

[Out]

-15*Log[3]*Log[-3 + x] - 5*Log[x] + 10*x*Log[x] + 15*Log[1 - x/3]*Log[x] - 5*x*Log[-3 + x]*Log[x] + (5*Log[x]*
Log[Log[3]])/8 - (5*x*Log[x]*Log[Log[3]])/8 - (5*Log[x]*Log[Log[3]]^2)/256 + 15*PolyLog[2, 1 - x/3] + 15*PolyL
og[2, x/3]

________________________________________________________________________________________

fricas [A]  time = 0.95, size = 35, normalized size = 1.21 \begin {gather*} -\frac {5}{8} \, {\left (x - 1\right )} \log \relax (x) \log \left (\log \relax (3)\right ) - \frac {5}{256} \, \log \relax (x) \log \left (\log \relax (3)\right )^{2} - 5 \, {\left (x \log \left (x - 3\right ) - 2 \, x + 1\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15-5*x)*log(log(3))^2+((-160*x^2+480*x)*log(x)-160*x^2+640*x-480)*log(log(3))+((-1280*x^2+3840*x)*
log(x-3)+1280*x^2-7680*x)*log(x)+(-1280*x^2+3840*x)*log(x-3)+2560*x^2-8960*x+3840)/(256*x^2-768*x),x, algorith
m="fricas")

[Out]

-5/8*(x - 1)*log(x)*log(log(3)) - 5/256*log(x)*log(log(3))^2 - 5*(x*log(x - 3) - 2*x + 1)*log(x)

________________________________________________________________________________________

giac [A]  time = 0.19, size = 36, normalized size = 1.24 \begin {gather*} -\frac {5}{8} \, x {\left (\log \left (\log \relax (3)\right ) - 16\right )} \log \relax (x) - 5 \, x \log \left (x - 3\right ) \log \relax (x) - \frac {5}{256} \, {\left (\log \left (\log \relax (3)\right )^{2} - 32 \, \log \left (\log \relax (3)\right ) + 256\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15-5*x)*log(log(3))^2+((-160*x^2+480*x)*log(x)-160*x^2+640*x-480)*log(log(3))+((-1280*x^2+3840*x)*
log(x-3)+1280*x^2-7680*x)*log(x)+(-1280*x^2+3840*x)*log(x-3)+2560*x^2-8960*x+3840)/(256*x^2-768*x),x, algorith
m="giac")

[Out]

-5/8*x*(log(log(3)) - 16)*log(x) - 5*x*log(x - 3)*log(x) - 5/256*(log(log(3))^2 - 32*log(log(3)) + 256)*log(x)

________________________________________________________________________________________

maple [A]  time = 0.51, size = 39, normalized size = 1.34

method	result	size
norman	\(\left (-5+\frac {5 \ln \left (\ln \relax (3)\right )}{8}-\frac {5 \ln \left (\ln \relax (3)\right )^{2}}{256}\right ) \ln \relax (x )+\left (10-\frac {5 \ln \left (\ln \relax (3)\right )}{8}\right ) x \ln \relax (x )-5 \ln \relax (x ) \ln \left (x -3\right ) x\)	\(39\)
risch	\(-5 \ln \relax (x ) \ln \left (x -3\right ) x -\frac {5 \ln \relax (x ) \ln \left (\ln \relax (3)\right ) x}{8}+10 x \ln \relax (x )-\frac {5 \ln \relax (x ) \ln \left (\ln \relax (3)\right )^{2}}{256}+\frac {5 \ln \left (\ln \relax (3)\right ) \ln \relax (x )}{8}-5 \ln \relax (x )\)	\(44\)
default	\(-15 \ln \left (x -3\right )+\frac {\left (2560-160 \ln \left (\ln \relax (3)\right )\right ) x \ln \relax (x )}{256}+5 \ln \left (x -3\right ) x -5 \ln \relax (x ) \ln \left (x -3\right ) x -\frac {5 \ln \relax (x ) \ln \left (\ln \relax (3)\right )^{2}}{256}-5 \ln \relax (x )+\frac {5 \ln \left (\ln \relax (3)\right ) \ln \relax (x )}{8}-5 \left (x -3\right ) \ln \left (x -3\right )-15\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((15-5*x)*ln(ln(3))^2+((-160*x^2+480*x)*ln(x)-160*x^2+640*x-480)*ln(ln(3))+((-1280*x^2+3840*x)*ln(x-3)+128
0*x^2-7680*x)*ln(x)+(-1280*x^2+3840*x)*ln(x-3)+2560*x^2-8960*x+3840)/(256*x^2-768*x),x,method=_RETURNVERBOSE)

[Out]

(-5+5/8*ln(ln(3))-5/256*ln(ln(3))^2)*ln(x)+(10-5/8*ln(ln(3)))*x*ln(x)-5*ln(x)*ln(x-3)*x

________________________________________________________________________________________

maxima [B]  time = 0.72, size = 132, normalized size = 4.55 \begin {gather*} -\frac {5}{8} \, x {\left (\log \left (\log \relax (3)\right ) - 16\right )} \log \relax (x) + \frac {5}{256} \, {\left (\log \left (x - 3\right ) - \log \relax (x)\right )} \log \left (\log \relax (3)\right )^{2} - \frac {5}{256} \, \log \left (x - 3\right ) \log \left (\log \relax (3)\right )^{2} + \frac {5}{8} \, x {\left (\log \left (\log \relax (3)\right ) - 24\right )} - 5 \, {\left (x \log \relax (x) - x + 3\right )} \log \left (x - 3\right ) - 5 \, {\left (x + 3 \, \log \left (x - 3\right )\right )} \log \left (x - 3\right ) + 15 \, \log \left (x - 3\right )^{2} - \frac {5}{8} \, {\left (x + 3 \, \log \left (x - 3\right )\right )} \log \left (\log \relax (3)\right ) - \frac {5}{8} \, {\left (\log \left (x - 3\right ) - \log \relax (x)\right )} \log \left (\log \relax (3)\right ) + \frac {5}{2} \, \log \left (x - 3\right ) \log \left (\log \relax (3)\right ) + 15 \, x + 15 \, \log \left (x - 3\right ) - 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15-5*x)*log(log(3))^2+((-160*x^2+480*x)*log(x)-160*x^2+640*x-480)*log(log(3))+((-1280*x^2+3840*x)*
log(x-3)+1280*x^2-7680*x)*log(x)+(-1280*x^2+3840*x)*log(x-3)+2560*x^2-8960*x+3840)/(256*x^2-768*x),x, algorith
m="maxima")

[Out]

-5/8*x*(log(log(3)) - 16)*log(x) + 5/256*(log(x - 3) - log(x))*log(log(3))^2 - 5/256*log(x - 3)*log(log(3))^2
+ 5/8*x*(log(log(3)) - 24) - 5*(x*log(x) - x + 3)*log(x - 3) - 5*(x + 3*log(x - 3))*log(x - 3) + 15*log(x - 3)
^2 - 5/8*(x + 3*log(x - 3))*log(log(3)) - 5/8*(log(x - 3) - log(x))*log(log(3)) + 5/2*log(x - 3)*log(log(3)) +
15*x + 15*log(x - 3) - 5*log(x)

________________________________________________________________________________________

mupad [B]  time = 1.83, size = 32, normalized size = 1.10 \begin {gather*} -\frac {5\,\ln \relax (x)\,\left (256\,x\,\ln \left (x-3\right )-32\,\ln \left (\ln \relax (3)\right )-512\,x+{\ln \left (\ln \relax (3)\right )}^2+32\,x\,\ln \left (\ln \relax (3)\right )+256\right )}{256} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x - 3)*(3840*x - 1280*x^2) - 8960*x + log(log(3))*(640*x + log(x)*(480*x - 160*x^2) - 160*x^2 - 480)
+ log(x)*(log(x - 3)*(3840*x - 1280*x^2) - 7680*x + 1280*x^2) - log(log(3))^2*(5*x - 15) + 2560*x^2 + 3840)/(
768*x - 256*x^2),x)

[Out]

-(5*log(x)*(256*x*log(x - 3) - 32*log(log(3)) - 512*x + log(log(3))^2 + 32*x*log(log(3)) + 256))/256

________________________________________________________________________________________

sympy [B]  time = 1.93, size = 97, normalized size = 3.34 \begin {gather*} \left (- 5 x \log {\relax (x )} - \frac {15}{4}\right ) \log {\left (x - 3 \right )} + \left (- \frac {5 x \log {\left (\log {\relax (3 )} \right )}}{8} + 10 x\right ) \log {\relax (x )} - \left (- \frac {5 \log {\left (\log {\relax (3 )} \right )}}{8} + \frac {5 \log {\left (\log {\relax (3 )} \right )}^{2}}{256} + 5\right ) \log {\relax (x )} + \frac {15 \log {\left (x + \frac {-6720 - 15 \log {\left (\log {\relax (3 )} \right )}^{2} + 480 \log {\left (\log {\relax (3 )} \right )}}{- 160 \log {\left (\log {\relax (3 )} \right )} + 5 \log {\left (\log {\relax (3 )} \right )}^{2} + 2240} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((15-5*x)*ln(ln(3))**2+((-160*x**2+480*x)*ln(x)-160*x**2+640*x-480)*ln(ln(3))+((-1280*x**2+3840*x)*l
n(x-3)+1280*x**2-7680*x)*ln(x)+(-1280*x**2+3840*x)*ln(x-3)+2560*x**2-8960*x+3840)/(256*x**2-768*x),x)

[Out]

(-5*x*log(x) - 15/4)*log(x - 3) + (-5*x*log(log(3))/8 + 10*x)*log(x) - (-5*log(log(3))/8 + 5*log(log(3))**2/25
6 + 5)*log(x) + 15*log(x + (-6720 - 15*log(log(3))**2 + 480*log(log(3)))/(-160*log(log(3)) + 5*log(log(3))**2
+ 2240))/4

________________________________________________________________________________________