3.81.89 \(\int \frac {(-4-4 x+2 e^x x^2) \log (\frac {e^x x}{2+e^x (x+x^2)})}{90 e^4 x+e^{4+x} (45 x^2+45 x^3)+(60 e^4 x+e^{4+x} (30 x^2+30 x^3)) \log ^2(\frac {e^x x}{2+e^x (x+x^2)})+(10 e^4 x+e^{4+x} (5 x^2+5 x^3)) \log ^4(\frac {e^x x}{2+e^x (x+x^2)})} \, dx\)
Optimal. Leaf size=30 \[ \frac {1}{5 e^4 \left (3+\log ^2\left (\frac {x}{2 e^{-x}+x+x^2}\right )\right )} \]
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Rubi [A] time = 9.45, antiderivative size = 33, normalized size of antiderivative = 1.10,
number of steps used = 4, number of rules used = 5, integrand size = 151, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used
= {6688, 12, 6742, 6711, 32} \begin {gather*} -\frac {1}{15 e^4 \left (\frac {3}{\log ^2\left (\frac {e^x x}{e^x x (x+1)+2}\right )}+1\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[((-4 - 4*x + 2*E^x*x^2)*Log[(E^x*x)/(2 + E^x*(x + x^2))])/(90*E^4*x + E^(4 + x)*(45*x^2 + 45*x^3) + (60*E^
4*x + E^(4 + x)*(30*x^2 + 30*x^3))*Log[(E^x*x)/(2 + E^x*(x + x^2))]^2 + (10*E^4*x + E^(4 + x)*(5*x^2 + 5*x^3))
*Log[(E^x*x)/(2 + E^x*(x + x^2))]^4),x]
[Out]
-1/15*1/(E^4*(1 + 3/Log[(E^x*x)/(2 + E^x*x*(1 + 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]
Rule 6742
Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]
Rubi steps
\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (-2-2 x+e^x x^2\right ) \log \left (\frac {e^x x}{2+e^x x (1+x)}\right )}{5 e^4 x \left (2+e^x x (1+x)\right ) \left (3+\log ^2\left (\frac {e^x x}{2+e^x x (1+x)}\right )\right )^2} \, dx\\ &=\frac {2 \int \frac {\left (-2-2 x+e^x x^2\right ) \log \left (\frac {e^x x}{2+e^x x (1+x)}\right )}{x \left (2+e^x x (1+x)\right ) \left (3+\log ^2\left (\frac {e^x x}{2+e^x x (1+x)}\right )\right )^2} \, dx}{5 e^4}\\ &=\frac {\operatorname {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,\frac {3}{\log ^2\left (\frac {e^x x}{2+e^x x (1+x)}\right )}\right )}{15 e^4}\\ &=-\frac {1}{15 e^4 \left (1+\frac {3}{\log ^2\left (\frac {e^x x}{2+e^x x (1+x)}\right )}\right )}\\ \end {aligned} \end {gather*}
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Mathematica [A] time = 0.34, size = 31, normalized size = 1.03 \begin {gather*} \frac {1}{5 e^4 \left (3+\log ^2\left (\frac {e^x x}{2+e^x x (1+x)}\right )\right )} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((-4 - 4*x + 2*E^x*x^2)*Log[(E^x*x)/(2 + E^x*(x + x^2))])/(90*E^4*x + E^(4 + x)*(45*x^2 + 45*x^3) +
(60*E^4*x + E^(4 + x)*(30*x^2 + 30*x^3))*Log[(E^x*x)/(2 + E^x*(x + x^2))]^2 + (10*E^4*x + E^(4 + x)*(5*x^2 + 5
*x^3))*Log[(E^x*x)/(2 + E^x*(x + x^2))]^4),x]
[Out]
1/(5*E^4*(3 + Log[(E^x*x)/(2 + E^x*x*(1 + x))]^2))
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fricas [A] time = 0.59, size = 38, normalized size = 1.27 \begin {gather*} \frac {1}{5 \, {\left (e^{4} \log \left (\frac {x e^{\left (x + 4\right )}}{{\left (x^{2} + x\right )} e^{\left (x + 4\right )} + 2 \, e^{4}}\right )^{2} + 3 \, e^{4}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((2*exp(x)*x^2-4*x-4)*log(x*exp(x)/((x^2+x)*exp(x)+2))/(((5*x^3+5*x^2)*exp(4)*exp(x)+10*x*exp(4))*log
(x*exp(x)/((x^2+x)*exp(x)+2))^4+((30*x^3+30*x^2)*exp(4)*exp(x)+60*x*exp(4))*log(x*exp(x)/((x^2+x)*exp(x)+2))^2
+(45*x^3+45*x^2)*exp(4)*exp(x)+90*x*exp(4)),x, algorithm="fricas")
[Out]
1/5/(e^4*log(x*e^(x + 4)/((x^2 + x)*e^(x + 4) + 2*e^4))^2 + 3*e^4)
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giac [B] time = 76.37, size = 59, normalized size = 1.97 \begin {gather*} \frac {2}{5 \, {\left (x^{2} e^{4} + 2 \, x e^{4} \log \left (\frac {x}{x^{2} e^{x} + x e^{x} + 2}\right ) + e^{4} \log \left (\frac {x}{x^{2} e^{x} + x e^{x} + 2}\right )^{2} + 3 \, e^{4}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((2*exp(x)*x^2-4*x-4)*log(x*exp(x)/((x^2+x)*exp(x)+2))/(((5*x^3+5*x^2)*exp(4)*exp(x)+10*x*exp(4))*log
(x*exp(x)/((x^2+x)*exp(x)+2))^4+((30*x^3+30*x^2)*exp(4)*exp(x)+60*x*exp(4))*log(x*exp(x)/((x^2+x)*exp(x)+2))^2
+(45*x^3+45*x^2)*exp(4)*exp(x)+90*x*exp(4)),x, algorithm="giac")
[Out]
2/5/(x^2*e^4 + 2*x*e^4*log(x/(x^2*e^x + x*e^x + 2)) + e^4*log(x/(x^2*e^x + x*e^x + 2))^2 + 3*e^4)
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maple [C] time = 2.24, size = 2968, normalized size = 98.93
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((2*exp(x)*x^2-4*x-4)*ln(x*exp(x)/((x^2+x)*exp(x)+2))/(((5*x^3+5*x^2)*exp(4)*exp(x)+10*x*exp(4))*ln(x*exp(x
)/((x^2+x)*exp(x)+2))^4+((30*x^3+30*x^2)*exp(4)*exp(x)+60*x*exp(4))*ln(x*exp(x)/((x^2+x)*exp(x)+2))^2+(45*x^3+
45*x^2)*exp(4)*exp(x)+90*x*exp(4)),x,method=_RETURNVERBOSE)
[Out]
-4/5*exp(-4)/(-12-4*ln(x)^2-2*Pi^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp
(x))^5-2*Pi^2*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^5*csgn(I*x)+Pi^2*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x
))^4*csgn(I*x)^2+Pi^2*csgn(I/(exp(x)*x^2+exp(x)*x+2))^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^4-2*Pi^2*csgn(I
/(exp(x)*x^2+exp(x)*x+2))*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^5-2*Pi^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2
))^5*csgn(I*exp(x))-2*Pi^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^4*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^2
+2*Pi^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^3*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^3+Pi^2*csgn(I*exp(x)
/(exp(x)*x^2+exp(x)*x+2))^4*csgn(I*exp(x))^2+Pi^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^2*csgn(I*x/(exp(x)*x^
2+exp(x)*x+2)*exp(x))^4+4*Pi^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))
^4*csgn(I*x)+Pi^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^2*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^2*csgn(I*x
)^2-2*Pi^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^3*csgn(I*x)^2-2*Pi^
2*csgn(I/(exp(x)*x^2+exp(x)*x+2))^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^3*csgn(I*exp(x))+4*Pi^2*csgn(I/(exp
(x)*x^2+exp(x)*x+2))*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^4*csgn(I*exp(x))+2*Pi^2*csgn(I/(exp(x)*x^2+exp(x)*
x+2))*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^3*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^2-2*Pi^2*csgn(I/(exp(x
)*x^2+exp(x)*x+2))*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^2*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^3+Pi^2*cs
gn(I/(exp(x)*x^2+exp(x)*x+2))^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^2*csgn(I*exp(x))^2-2*Pi^2*csgn(I/(exp(x
)*x^2+exp(x)*x+2))*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^3*csgn(I*exp(x))^2+2*Pi^2*csgn(I*exp(x)/(exp(x)*x^2+
exp(x)*x+2))^4*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))*csgn(I*x)-2*Pi^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2)
)^3*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^2*csgn(I*x)+2*Pi^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^3*csgn(
I*exp(x))*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^2-2*Pi^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^2*csgn(I*ex
p(x))*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^3+4*I*ln(exp(x))*Pi*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^3-
4*I*ln(exp(x)*x^2+exp(x)*x+2)*Pi*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^3-4*I*ln(exp(x)*x^2+exp(x)*x+2)*Pi*csg
n(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^3+4*I*ln(x)*Pi*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^3+4*I*ln(x)*Pi*csg
n(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^3+4*I*ln(exp(x))*Pi*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^3-4*ln(exp(x)
)^2+Pi^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^6+Pi^2*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^6-4*ln(exp(x)*
x^2+exp(x)*x+2)^2+8*ln(x)*ln(exp(x)*x^2+exp(x)*x+2)-8*ln(x)*ln(exp(x))+8*ln(exp(x))*ln(exp(x)*x^2+exp(x)*x+2)-
2*Pi^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^2*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^3*csgn(I*x)+2*Pi^2*cs
gn(I/(exp(x)*x^2+exp(x)*x+2))*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^2*csgn(I*exp(x))*csgn(I*x/(exp(x)*x^2+exp
(x)*x+2)*exp(x))*csgn(I*x)-2*Pi^2*csgn(I/(exp(x)*x^2+exp(x)*x+2))*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))*csgn(
I*exp(x))*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^2*csgn(I*x)+4*I*ln(x)*Pi*csgn(I/(exp(x)*x^2+exp(x)*x+2))*cs
gn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))*csgn(I*exp(x))+4*I*ln(x)*Pi*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))*csgn(I
*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))*csgn(I*x)+4*I*ln(exp(x))*Pi*csgn(I/(exp(x)*x^2+exp(x)*x+2))*csgn(I*exp(x)/(
exp(x)*x^2+exp(x)*x+2))*csgn(I*exp(x))+4*I*ln(exp(x))*Pi*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))*csgn(I*x/(exp(
x)*x^2+exp(x)*x+2)*exp(x))*csgn(I*x)-4*I*ln(exp(x)*x^2+exp(x)*x+2)*Pi*csgn(I/(exp(x)*x^2+exp(x)*x+2))*csgn(I*e
xp(x)/(exp(x)*x^2+exp(x)*x+2))*csgn(I*exp(x))-4*I*ln(exp(x)*x^2+exp(x)*x+2)*Pi*csgn(I*exp(x)/(exp(x)*x^2+exp(x
)*x+2))*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))*csgn(I*x)+4*I*ln(exp(x)*x^2+exp(x)*x+2)*Pi*csgn(I*exp(x)/(exp
(x)*x^2+exp(x)*x+2))^2*csgn(I*exp(x))+4*I*ln(exp(x)*x^2+exp(x)*x+2)*Pi*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))*
csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^2+4*I*ln(exp(x)*x^2+exp(x)*x+2)*Pi*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*e
xp(x))^2*csgn(I*x)-4*I*ln(x)*Pi*csgn(I/(exp(x)*x^2+exp(x)*x+2))*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^2-4*I*l
n(x)*Pi*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^2*csgn(I*exp(x))-4*I*ln(x)*Pi*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*
x+2))*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^2-4*I*ln(x)*Pi*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^2*csgn(
I*x)-4*I*ln(exp(x))*Pi*csgn(I/(exp(x)*x^2+exp(x)*x+2))*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^2-4*I*ln(exp(x))
*Pi*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^2*csgn(I*exp(x))-4*I*ln(exp(x))*Pi*csgn(I*exp(x)/(exp(x)*x^2+exp(x)
*x+2))*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^2-4*I*ln(exp(x))*Pi*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^2
*csgn(I*x)+4*I*ln(exp(x)*x^2+exp(x)*x+2)*Pi*csgn(I/(exp(x)*x^2+exp(x)*x+2))*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x
+2))^2-2*Pi^2*csgn(I/(exp(x)*x^2+exp(x)*x+2))*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^3*csgn(I*x/(exp(x)*x^2+ex
p(x)*x+2)*exp(x))*csgn(I*x)+2*Pi^2*csgn(I/(exp(x)*x^2+exp(x)*x+2))*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^2*cs
gn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^2*csgn(I*x)-2*Pi^2*csgn(I/(exp(x)*x^2+exp(x)*x+2))*csgn(I*exp(x)/(exp(x
)*x^2+exp(x)*x+2))^2*csgn(I*exp(x))*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^2+2*Pi^2*csgn(I/(exp(x)*x^2+exp(x
)*x+2))*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))*csgn(I*exp(x))*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^3-2*Pi^
2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^3*csgn(I*exp(x))*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))*csgn(I*x)+2
*Pi^2*csgn(I*exp(x)/(exp(x)*x^2+exp(x)*x+2))^2*csgn(I*exp(x))*csgn(I*x/(exp(x)*x^2+exp(x)*x+2)*exp(x))^2*csgn(
I*x))
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maxima [B] time = 0.64, size = 68, normalized size = 2.27 \begin {gather*} \frac {1}{5 \, {\left (x^{2} e^{4} + e^{4} \log \left ({\left (x^{2} + x\right )} e^{x} + 2\right )^{2} + 2 \, x e^{4} \log \relax (x) + e^{4} \log \relax (x)^{2} - 2 \, {\left (x e^{4} + e^{4} \log \relax (x)\right )} \log \left ({\left (x^{2} + x\right )} e^{x} + 2\right ) + 3 \, e^{4}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((2*exp(x)*x^2-4*x-4)*log(x*exp(x)/((x^2+x)*exp(x)+2))/(((5*x^3+5*x^2)*exp(4)*exp(x)+10*x*exp(4))*log
(x*exp(x)/((x^2+x)*exp(x)+2))^4+((30*x^3+30*x^2)*exp(4)*exp(x)+60*x*exp(4))*log(x*exp(x)/((x^2+x)*exp(x)+2))^2
+(45*x^3+45*x^2)*exp(4)*exp(x)+90*x*exp(4)),x, algorithm="maxima")
[Out]
1/5/(x^2*e^4 + e^4*log((x^2 + x)*e^x + 2)^2 + 2*x*e^4*log(x) + e^4*log(x)^2 - 2*(x*e^4 + e^4*log(x))*log((x^2
+ x)*e^x + 2) + 3*e^4)
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mupad [B] time = 6.03, size = 28, normalized size = 0.93 \begin {gather*} \frac {{\mathrm {e}}^{-4}}{5\,\left ({\ln \left (\frac {x\,{\mathrm {e}}^x}{{\mathrm {e}}^x\,\left (x^2+x\right )+2}\right )}^2+3\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(-(log((x*exp(x))/(exp(x)*(x + x^2) + 2))*(4*x - 2*x^2*exp(x) + 4))/(90*x*exp(4) + log((x*exp(x))/(exp(x)*(
x + x^2) + 2))^4*(10*x*exp(4) + exp(4)*exp(x)*(5*x^2 + 5*x^3)) + log((x*exp(x))/(exp(x)*(x + x^2) + 2))^2*(60*
x*exp(4) + exp(4)*exp(x)*(30*x^2 + 30*x^3)) + exp(4)*exp(x)*(45*x^2 + 45*x^3)),x)
[Out]
exp(-4)/(5*(log((x*exp(x))/(exp(x)*(x + x^2) + 2))^2 + 3))
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sympy [A] time = 0.31, size = 29, normalized size = 0.97 \begin {gather*} \frac {1}{5 e^{4} \log {\left (\frac {x e^{x}}{\left (x^{2} + x\right ) e^{x} + 2} \right )}^{2} + 15 e^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((2*exp(x)*x**2-4*x-4)*ln(x*exp(x)/((x**2+x)*exp(x)+2))/(((5*x**3+5*x**2)*exp(4)*exp(x)+10*x*exp(4))*
ln(x*exp(x)/((x**2+x)*exp(x)+2))**4+((30*x**3+30*x**2)*exp(4)*exp(x)+60*x*exp(4))*ln(x*exp(x)/((x**2+x)*exp(x)
+2))**2+(45*x**3+45*x**2)*exp(4)*exp(x)+90*x*exp(4)),x)
[Out]
1/(5*exp(4)*log(x*exp(x)/((x**2 + x)*exp(x) + 2))**2 + 15*exp(4))
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