\(\int \frac {(-6-60 x+24 x^2+e^x (-12 x-12 x^2)) \log ^2(x)+((2+20 x-8 x^2+e^x (4 x+4 x^2)) \log (x)+((60 x+12 e^x x-12 x^2) \log (x)+6 \log ^2(x)) \log (10 x+2 e^x x-2 x^2+\log (x))) \log (\log (10 x+2 e^x x-2 x^2+\log (x)))+(-20 x-4 e^x x+4 x^2-2 \log (x)) \log (10 x+2 e^x x-2 x^2+\log (x)) \log ^2(\log (10 x+2 e^x x-2 x^2+\log (x)))}{((10 x^2+2 e^x x^2-2 x^3) \log ^3(x)+x \log ^4(x)) \log (10 x+2 e^x x-2 x^2+\log (x))} \, dx\) [6684]

Optimal result

Integrand size = 233, antiderivative size = 31 \[ \int \frac {\left (-6-60 x+24 x^2+e^x \left (-12 x-12 x^2\right )\right ) \log ^2(x)+\left (\left (2+20 x-8 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (\left (60 x+12 e^x x-12 x^2\right ) \log (x)+6 \log ^2(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right ) \log \left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )+\left (-20 x-4 e^x x+4 x^2-2 \log (x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right ) \log ^2\left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )}{\left (\left (10 x^2+2 e^x x^2-2 x^3\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )} \, dx=1+\left (3-\frac {\log \left (\log \left (\left (4+2 \left (3+e^x-x\right )\right ) x+\log (x)\right )\right )}{\log (x)}\right )^2 \]

[Out]

1+(3-ln(ln((10+2*exp(x)-2*x)*x+ln(x)))/ln(x))^2

Rubi [A] (verified)

Time = 4.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {6820, 12, 6844} \[ \int \frac {\left (-6-60 x+24 x^2+e^x \left (-12 x-12 x^2\right )\right ) \log ^2(x)+\left (\left (2+20 x-8 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (\left (60 x+12 e^x x-12 x^2\right ) \log (x)+6 \log ^2(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right ) \log \left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )+\left (-20 x-4 e^x x+4 x^2-2 \log (x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right ) \log ^2\left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )}{\left (\left (10 x^2+2 e^x x^2-2 x^3\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )} \, dx=\frac {\log ^2\left (\log \left (2 \left (-x+e^x+5\right ) x+\log (x)\right )\right )}{\log ^2(x)}-\frac {6 \log \left (\log \left (2 \left (-x+e^x+5\right ) x+\log (x)\right )\right )}{\log (x)} \]

[In]

Int[((-6 - 60*x + 24*x^2 + E^x*(-12*x - 12*x^2))*Log[x]^2 + ((2 + 20*x - 8*x^2 + E^x*(4*x + 4*x^2))*Log[x] + (
(60*x + 12*E^x*x - 12*x^2)*Log[x] + 6*Log[x]^2)*Log[10*x + 2*E^x*x - 2*x^2 + Log[x]])*Log[Log[10*x + 2*E^x*x -
2*x^2 + Log[x]]] + (-20*x - 4*E^x*x + 4*x^2 - 2*Log[x])*Log[10*x + 2*E^x*x - 2*x^2 + Log[x]]*Log[Log[10*x + 2
*E^x*x - 2*x^2 + Log[x]]]^2)/(((10*x^2 + 2*E^x*x^2 - 2*x^3)*Log[x]^3 + x*Log[x]^4)*Log[10*x + 2*E^x*x - 2*x^2
+ Log[x]]),x]

[Out]

(-6*Log[Log[2*(5 + E^x - x)*x + Log[x]]])/Log[x] + Log[Log[2*(5 + E^x - x)*x + Log[x]]]^2/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 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6844

Int[(u_)*(v_)^(r_.)*((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)*q, Subst[Int[(a + b*x^q)^m, x], x, v^(m*p + r + 1)*w], x] /; FreeQ[c, x]] /; Fre
eQ[{a, b, m, p, q, r}, x] && EqQ[p + q*(m*p + r + 1), 0] && IntegerQ[q] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \int \frac {2 \left (3 \log (x)-\log \left (\log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right )\right )\right ) \left (2 x \left (-5-e^x+x\right ) \log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right ) \log \left (\log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right )\right )+\log (x) \left (1+2 \left (5+e^x\right ) x+2 \left (-2+e^x\right ) x^2-\log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right ) \log \left (\log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right )\right )\right )\right )}{x \left (2 x \left (-5-e^x+x\right )-\log (x)\right ) \log ^3(x) \log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right )} \, dx \\ & = 2 \int \frac {\left (3 \log (x)-\log \left (\log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right )\right )\right ) \left (2 x \left (-5-e^x+x\right ) \log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right ) \log \left (\log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right )\right )+\log (x) \left (1+2 \left (5+e^x\right ) x+2 \left (-2+e^x\right ) x^2-\log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right ) \log \left (\log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right )\right )\right )\right )}{x \left (2 x \left (-5-e^x+x\right )-\log (x)\right ) \log ^3(x) \log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right )} \, dx \\ & = -\left (2 \text {Subst}\left (\int (3-x) \, dx,x,\frac {\log \left (\log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right )\right )}{\log (x)}\right )\right ) \\ & = -\frac {6 \log \left (\log \left (2 \left (5+e^x-x\right ) x+\log (x)\right )\right )}{\log (x)}+\frac {\log ^2\left (\log \left (2 \left (5+e^x-x\right ) x+\log (x)\right )\right )}{\log ^2(x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \frac {\left (-6-60 x+24 x^2+e^x \left (-12 x-12 x^2\right )\right ) \log ^2(x)+\left (\left (2+20 x-8 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (\left (60 x+12 e^x x-12 x^2\right ) \log (x)+6 \log ^2(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right ) \log \left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )+\left (-20 x-4 e^x x+4 x^2-2 \log (x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right ) \log ^2\left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )}{\left (\left (10 x^2+2 e^x x^2-2 x^3\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )} \, dx=2 \left (-\frac {3 \log \left (\log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right )\right )}{\log (x)}+\frac {\log ^2\left (\log \left (-2 x \left (-5-e^x+x\right )+\log (x)\right )\right )}{2 \log ^2(x)}\right ) \]

[In]

Integrate[((-6 - 60*x + 24*x^2 + E^x*(-12*x - 12*x^2))*Log[x]^2 + ((2 + 20*x - 8*x^2 + E^x*(4*x + 4*x^2))*Log[
x] + ((60*x + 12*E^x*x - 12*x^2)*Log[x] + 6*Log[x]^2)*Log[10*x + 2*E^x*x - 2*x^2 + Log[x]])*Log[Log[10*x + 2*E
^x*x - 2*x^2 + Log[x]]] + (-20*x - 4*E^x*x + 4*x^2 - 2*Log[x])*Log[10*x + 2*E^x*x - 2*x^2 + Log[x]]*Log[Log[10
*x + 2*E^x*x - 2*x^2 + Log[x]]]^2)/(((10*x^2 + 2*E^x*x^2 - 2*x^3)*Log[x]^3 + x*Log[x]^4)*Log[10*x + 2*E^x*x -
2*x^2 + Log[x]]),x]

[Out]

2*((-3*Log[Log[-2*x*(-5 - E^x + x) + Log[x]]])/Log[x] + Log[Log[-2*x*(-5 - E^x + x) + Log[x]]]^2/(2*Log[x]^2))

Maple [A] (verified)

Time = 240.28 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65

method	result	size
risch	\(\frac {{\ln \left (\ln \left (\ln \left (x \right )+2 \,{\mathrm e}^{x} x -2 x^{2}+10 x \right )\right )}^{2}}{\ln \left (x \right )^{2}}-\frac {6 \ln \left (\ln \left (\ln \left (x \right )+2 \,{\mathrm e}^{x} x -2 x^{2}+10 x \right )\right )}{\ln \left (x \right )}\)	\(51\)
parallelrisch	\(-\frac {24 \ln \left (x \right ) \ln \left (\ln \left (\ln \left (x \right )+2 \,{\mathrm e}^{x} x -2 x^{2}+10 x \right )\right )-4 {\ln \left (\ln \left (\ln \left (x \right )+2 \,{\mathrm e}^{x} x -2 x^{2}+10 x \right )\right )}^{2}}{4 \ln \left (x \right )^{2}}\)	\(52\)

[In]

int(((-2*ln(x)-4*exp(x)*x+4*x^2-20*x)*ln(ln(x)+2*exp(x)*x-2*x^2+10*x)*ln(ln(ln(x)+2*exp(x)*x-2*x^2+10*x))^2+((
6*ln(x)^2+(12*exp(x)*x-12*x^2+60*x)*ln(x))*ln(ln(x)+2*exp(x)*x-2*x^2+10*x)+((4*x^2+4*x)*exp(x)-8*x^2+20*x+2)*l
n(x))*ln(ln(ln(x)+2*exp(x)*x-2*x^2+10*x))+((-12*x^2-12*x)*exp(x)+24*x^2-60*x-6)*ln(x)^2)/(x*ln(x)^4+(2*exp(x)*
x^2-2*x^3+10*x^2)*ln(x)^3)/ln(ln(x)+2*exp(x)*x-2*x^2+10*x),x,method=_RETURNVERBOSE)

[Out]

1/ln(x)^2*ln(ln(ln(x)+2*exp(x)*x-2*x^2+10*x))^2-6/ln(x)*ln(ln(ln(x)+2*exp(x)*x-2*x^2+10*x))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \frac {\left (-6-60 x+24 x^2+e^x \left (-12 x-12 x^2\right )\right ) \log ^2(x)+\left (\left (2+20 x-8 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (\left (60 x+12 e^x x-12 x^2\right ) \log (x)+6 \log ^2(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right ) \log \left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )+\left (-20 x-4 e^x x+4 x^2-2 \log (x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right ) \log ^2\left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )}{\left (\left (10 x^2+2 e^x x^2-2 x^3\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )} \, dx=-\frac {6 \, \log \left (x\right ) \log \left (\log \left (-2 \, x^{2} + 2 \, x e^{x} + 10 \, x + \log \left (x\right )\right )\right ) - \log \left (\log \left (-2 \, x^{2} + 2 \, x e^{x} + 10 \, x + \log \left (x\right )\right )\right )^{2}}{\log \left (x\right )^{2}} \]

[In]

integrate(((-2*log(x)-4*exp(x)*x+4*x^2-20*x)*log(log(x)+2*exp(x)*x-2*x^2+10*x)*log(log(log(x)+2*exp(x)*x-2*x^2
+10*x))^2+((6*log(x)^2+(12*exp(x)*x-12*x^2+60*x)*log(x))*log(log(x)+2*exp(x)*x-2*x^2+10*x)+((4*x^2+4*x)*exp(x)
-8*x^2+20*x+2)*log(x))*log(log(log(x)+2*exp(x)*x-2*x^2+10*x))+((-12*x^2-12*x)*exp(x)+24*x^2-60*x-6)*log(x)^2)/
(x*log(x)^4+(2*exp(x)*x^2-2*x^3+10*x^2)*log(x)^3)/log(log(x)+2*exp(x)*x-2*x^2+10*x),x, algorithm="fricas")

[Out]

-(6*log(x)*log(log(-2*x^2 + 2*x*e^x + 10*x + log(x))) - log(log(-2*x^2 + 2*x*e^x + 10*x + log(x)))^2)/log(x)^2

Sympy [F(-2)]

Exception generated. \[ \int \frac {\left (-6-60 x+24 x^2+e^x \left (-12 x-12 x^2\right )\right ) \log ^2(x)+\left (\left (2+20 x-8 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (\left (60 x+12 e^x x-12 x^2\right ) \log (x)+6 \log ^2(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right ) \log \left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )+\left (-20 x-4 e^x x+4 x^2-2 \log (x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right ) \log ^2\left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )}{\left (\left (10 x^2+2 e^x x^2-2 x^3\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(((-2*ln(x)-4*exp(x)*x+4*x**2-20*x)*ln(ln(x)+2*exp(x)*x-2*x**2+10*x)*ln(ln(ln(x)+2*exp(x)*x-2*x**2+10
*x))**2+((6*ln(x)**2+(12*exp(x)*x-12*x**2+60*x)*ln(x))*ln(ln(x)+2*exp(x)*x-2*x**2+10*x)+((4*x**2+4*x)*exp(x)-8
*x**2+20*x+2)*ln(x))*ln(ln(ln(x)+2*exp(x)*x-2*x**2+10*x))+((-12*x**2-12*x)*exp(x)+24*x**2-60*x-6)*ln(x)**2)/(x
*ln(x)**4+(2*exp(x)*x**2-2*x**3+10*x**2)*ln(x)**3)/ln(ln(x)+2*exp(x)*x-2*x**2+10*x),x)

[Out]

Exception raised: TypeError >> '>' not supported between instances of 'Poly' and 'int'

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \frac {\left (-6-60 x+24 x^2+e^x \left (-12 x-12 x^2\right )\right ) \log ^2(x)+\left (\left (2+20 x-8 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (\left (60 x+12 e^x x-12 x^2\right ) \log (x)+6 \log ^2(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right ) \log \left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )+\left (-20 x-4 e^x x+4 x^2-2 \log (x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right ) \log ^2\left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )}{\left (\left (10 x^2+2 e^x x^2-2 x^3\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )} \, dx=-\frac {6 \, \log \left (x\right ) \log \left (\log \left (-2 \, x^{2} + 2 \, x e^{x} + 10 \, x + \log \left (x\right )\right )\right ) - \log \left (\log \left (-2 \, x^{2} + 2 \, x e^{x} + 10 \, x + \log \left (x\right )\right )\right )^{2}}{\log \left (x\right )^{2}} \]

[In]

integrate(((-2*log(x)-4*exp(x)*x+4*x^2-20*x)*log(log(x)+2*exp(x)*x-2*x^2+10*x)*log(log(log(x)+2*exp(x)*x-2*x^2
+10*x))^2+((6*log(x)^2+(12*exp(x)*x-12*x^2+60*x)*log(x))*log(log(x)+2*exp(x)*x-2*x^2+10*x)+((4*x^2+4*x)*exp(x)
-8*x^2+20*x+2)*log(x))*log(log(log(x)+2*exp(x)*x-2*x^2+10*x))+((-12*x^2-12*x)*exp(x)+24*x^2-60*x-6)*log(x)^2)/
(x*log(x)^4+(2*exp(x)*x^2-2*x^3+10*x^2)*log(x)^3)/log(log(x)+2*exp(x)*x-2*x^2+10*x),x, algorithm="maxima")

[Out]

-(6*log(x)*log(log(-2*x^2 + 2*x*e^x + 10*x + log(x))) - log(log(-2*x^2 + 2*x*e^x + 10*x + log(x)))^2)/log(x)^2

Giac [F(-2)]

Exception generated. \[ \int \frac {\left (-6-60 x+24 x^2+e^x \left (-12 x-12 x^2\right )\right ) \log ^2(x)+\left (\left (2+20 x-8 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (\left (60 x+12 e^x x-12 x^2\right ) \log (x)+6 \log ^2(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right ) \log \left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )+\left (-20 x-4 e^x x+4 x^2-2 \log (x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right ) \log ^2\left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )}{\left (\left (10 x^2+2 e^x x^2-2 x^3\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(((-2*log(x)-4*exp(x)*x+4*x^2-20*x)*log(log(x)+2*exp(x)*x-2*x^2+10*x)*log(log(log(x)+2*exp(x)*x-2*x^2
+10*x))^2+((6*log(x)^2+(12*exp(x)*x-12*x^2+60*x)*log(x))*log(log(x)+2*exp(x)*x-2*x^2+10*x)+((4*x^2+4*x)*exp(x)
-8*x^2+20*x+2)*log(x))*log(log(log(x)+2*exp(x)*x-2*x^2+10*x))+((-12*x^2-12*x)*exp(x)+24*x^2-60*x-6)*log(x)^2)/
(x*log(x)^4+(2*exp(x)*x^2-2*x^3+10*x^2)*log(x)^3)/log(log(x)+2*exp(x)*x-2*x^2+10*x),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Sign error %%%{ln(w),0%%%}Sign error %%%{ln(w),0%%%}Sign error %%%{ln(w),0%%%}Done

Mupad [B] (verification not implemented)

Time = 13.81 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.48 \[ \int \frac {\left (-6-60 x+24 x^2+e^x \left (-12 x-12 x^2\right )\right ) \log ^2(x)+\left (\left (2+20 x-8 x^2+e^x \left (4 x+4 x^2\right )\right ) \log (x)+\left (\left (60 x+12 e^x x-12 x^2\right ) \log (x)+6 \log ^2(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right ) \log \left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )+\left (-20 x-4 e^x x+4 x^2-2 \log (x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right ) \log ^2\left (\log \left (10 x+2 e^x x-2 x^2+\log (x)\right )\right )}{\left (\left (10 x^2+2 e^x x^2-2 x^3\right ) \log ^3(x)+x \log ^4(x)\right ) \log \left (10 x+2 e^x x-2 x^2+\log (x)\right )} \, dx=\frac {\ln \left (\ln \left (10\,x+\ln \left (x\right )+2\,x\,{\mathrm {e}}^x-2\,x^2\right )\right )\,\left (\ln \left (\ln \left (10\,x+\ln \left (x\right )+2\,x\,{\mathrm {e}}^x-2\,x^2\right )\right )-6\,\ln \left (x\right )\right )}{{\ln \left (x\right )}^2} \]

[In]

int(-(log(x)^2*(60*x + exp(x)*(12*x + 12*x^2) - 24*x^2 + 6) - log(log(10*x + log(x) + 2*x*exp(x) - 2*x^2))*(lo
g(10*x + log(x) + 2*x*exp(x) - 2*x^2)*(6*log(x)^2 + log(x)*(60*x + 12*x*exp(x) - 12*x^2)) + log(x)*(20*x + exp
(x)*(4*x + 4*x^2) - 8*x^2 + 2)) + log(10*x + log(x) + 2*x*exp(x) - 2*x^2)*log(log(10*x + log(x) + 2*x*exp(x) -
2*x^2))^2*(20*x + 2*log(x) + 4*x*exp(x) - 4*x^2))/(log(10*x + log(x) + 2*x*exp(x) - 2*x^2)*(x*log(x)^4 + log(
x)^3*(2*x^2*exp(x) + 10*x^2 - 2*x^3))),x)

[Out]

(log(log(10*x + log(x) + 2*x*exp(x) - 2*x^2))*(log(log(10*x + log(x) + 2*x*exp(x) - 2*x^2)) - 6*log(x)))/log(x
)^2