\(\int \frac {-2+e^x (1+(-2-x) \log (5))+(-2 x+e^x x \log (5)) \log (x)+(2+e^x (-1+(2+x) \log (5))) \log (x) \log (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))})}{(2 x^2+e^x (-x^2+(2 x^2+x^3) \log (5))) \log (x)+(4 x+e^x (-2 x+(4 x+2 x^2) \log (5))) \log (x) \log (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))})+(2+e^x (-1+(2+x) \log (5))) \log (x) \log ^2(\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))})} \, dx\) [6675]

Optimal result

Integrand size = 203, antiderivative size = 32 \[ \int \frac {-2+e^x (1+(-2-x) \log (5))+\left (-2 x+e^x x \log (5)\right ) \log (x)+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}{\left (2 x^2+e^x \left (-x^2+\left (2 x^2+x^3\right ) \log (5)\right )\right ) \log (x)+\left (4 x+e^x \left (-2 x+\left (4 x+2 x^2\right ) \log (5)\right )\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log ^2\left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \, dx=2+\frac {x}{x+\log \left (\frac {\log (x)}{2+x-\frac {1-2 e^{-x}}{\log (5)}}\right )} \]

[Out]

x/(x+ln(ln(x)/(x-(1-2/exp(x))/ln(5)+2)))+2

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {6820, 6843, 32} \[ \int \frac {-2+e^x (1+(-2-x) \log (5))+\left (-2 x+e^x x \log (5)\right ) \log (x)+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}{\left (2 x^2+e^x \left (-x^2+\left (2 x^2+x^3\right ) \log (5)\right )\right ) \log (x)+\left (4 x+e^x \left (-2 x+\left (4 x+2 x^2\right ) \log (5)\right )\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log ^2\left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \, dx=-\frac {1}{\frac {x}{\log \left (\frac {e^x \log (5) \log (x)}{e^x ((x+2) \log (5)-1)+2}\right )}+1} \]

[In]

Int[(-2 + E^x*(1 + (-2 - x)*Log[5]) + (-2*x + E^x*x*Log[5])*Log[x] + (2 + E^x*(-1 + (2 + x)*Log[5]))*Log[x]*Lo
g[(E^x*Log[5]*Log[x])/(2 + E^x*(-1 + (2 + x)*Log[5]))])/((2*x^2 + E^x*(-x^2 + (2*x^2 + x^3)*Log[5]))*Log[x] +
(4*x + E^x*(-2*x + (4*x + 2*x^2)*Log[5]))*Log[x]*Log[(E^x*Log[5]*Log[x])/(2 + E^x*(-1 + (2 + x)*Log[5]))] + (2
+ E^x*(-1 + (2 + x)*Log[5]))*Log[x]*Log[(E^x*Log[5]*Log[x])/(2 + E^x*(-1 + (2 + x)*Log[5]))]^2),x]

[Out]

-(1 + x/Log[(E^x*Log[5]*Log[x])/(2 + E^x*(-1 + (2 + x)*Log[5]))])^(-1)

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 6820

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

Rule 6843

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{align*} \text {integral}& = \int \frac {-2-e^x (-1+x \log (5)+\log (25))+\log (x) \left (x \left (-2+e^x \log (5)\right )+\left (2+e^x (-1+x \log (5)+\log (25))\right ) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )\right )}{\left (2+e^x (-1+x \log (5)+\log (25))\right ) \log (x) \left (x+\log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )\right )^2} \, dx \\ & = \text {Subst}\left (\int \frac {1}{(1+x)^2} \, dx,x,\frac {x}{\log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}\right ) \\ & = -\frac {1}{1+\frac {x}{\log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {-2+e^x (1+(-2-x) \log (5))+\left (-2 x+e^x x \log (5)\right ) \log (x)+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}{\left (2 x^2+e^x \left (-x^2+\left (2 x^2+x^3\right ) \log (5)\right )\right ) \log (x)+\left (4 x+e^x \left (-2 x+\left (4 x+2 x^2\right ) \log (5)\right )\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log ^2\left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \, dx=\frac {x}{x+\log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \]

[In]

Integrate[(-2 + E^x*(1 + (-2 - x)*Log[5]) + (-2*x + E^x*x*Log[5])*Log[x] + (2 + E^x*(-1 + (2 + x)*Log[5]))*Log
[x]*Log[(E^x*Log[5]*Log[x])/(2 + E^x*(-1 + (2 + x)*Log[5]))])/((2*x^2 + E^x*(-x^2 + (2*x^2 + x^3)*Log[5]))*Log
[x] + (4*x + E^x*(-2*x + (4*x + 2*x^2)*Log[5]))*Log[x]*Log[(E^x*Log[5]*Log[x])/(2 + E^x*(-1 + (2 + x)*Log[5]))
] + (2 + E^x*(-1 + (2 + x)*Log[5]))*Log[x]*Log[(E^x*Log[5]*Log[x])/(2 + E^x*(-1 + (2 + x)*Log[5]))]^2),x]

[Out]

x/(x + Log[(E^x*Log[5]*Log[x])/(2 + E^x*(-1 + (2 + x)*Log[5]))])

Maple [A] (verified)

Time = 63.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09

method	result	size
parallelrisch	\(\frac {x}{\ln \left (\frac {\ln \left (5\right ) {\mathrm e}^{x} \ln \left (x \right )}{x \,{\mathrm e}^{x} \ln \left (5\right )+2 \,{\mathrm e}^{x} \ln \left (5\right )-{\mathrm e}^{x}+2}\right )+x}\)	\(35\)
risch	\(\frac {2 x}{i \pi \,\operatorname {csgn}\left (\frac {i \ln \left (x \right )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \left (5\right )-{\mathrm e}^{x}+2}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{x} \ln \left (x \right )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \left (5\right )-{\mathrm e}^{x}+2}\right )^{2}-i \pi \,\operatorname {csgn}\left (\frac {i}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \left (5\right )-{\mathrm e}^{x}+2}\right ) \operatorname {csgn}\left (\frac {i \ln \left (x \right )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \left (5\right )-{\mathrm e}^{x}+2}\right ) \operatorname {csgn}\left (i \ln \left (x \right )\right )+i \pi \,\operatorname {csgn}\left (\frac {i}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \left (5\right )-{\mathrm e}^{x}+2}\right ) {\operatorname {csgn}\left (\frac {i \ln \left (x \right )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \left (5\right )-{\mathrm e}^{x}+2}\right )}^{2}-i \pi \,\operatorname {csgn}\left (\frac {i \ln \left (x \right )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \left (5\right )-{\mathrm e}^{x}+2}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{x} \ln \left (x \right )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \left (5\right )-{\mathrm e}^{x}+2}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )-i \pi {\operatorname {csgn}\left (\frac {i \ln \left (x \right )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \left (5\right )-{\mathrm e}^{x}+2}\right )}^{3}-i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{x} \ln \left (x \right )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \left (5\right )-{\mathrm e}^{x}+2}\right )^{3}+i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{x} \ln \left (x \right )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \left (5\right )-{\mathrm e}^{x}+2}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )+i \pi {\operatorname {csgn}\left (\frac {i \ln \left (x \right )}{\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \left (5\right )-{\mathrm e}^{x}+2}\right )}^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )+2 \ln \left (\ln \left (5\right )\right )+2 x +2 \ln \left (\ln \left (x \right )\right )-2 \ln \left (\left ({\mathrm e}^{x} x +2 \,{\mathrm e}^{x}\right ) \ln \left (5\right )-{\mathrm e}^{x}+2\right )+2 \ln \left ({\mathrm e}^{x}\right )}\)	\(430\)

[In]

int((((ln(5)*(2+x)-1)*exp(x)+2)*ln(x)*ln(ln(5)*exp(x)*ln(x)/((ln(5)*(2+x)-1)*exp(x)+2))+(x*exp(x)*ln(5)-2*x)*l
n(x)+((-2-x)*ln(5)+1)*exp(x)-2)/(((ln(5)*(2+x)-1)*exp(x)+2)*ln(x)*ln(ln(5)*exp(x)*ln(x)/((ln(5)*(2+x)-1)*exp(x
)+2))^2+(((2*x^2+4*x)*ln(5)-2*x)*exp(x)+4*x)*ln(x)*ln(ln(5)*exp(x)*ln(x)/((ln(5)*(2+x)-1)*exp(x)+2))+(((x^3+2*
x^2)*ln(5)-x^2)*exp(x)+2*x^2)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

x/(ln(ln(5)*exp(x)*ln(x)/(x*exp(x)*ln(5)+2*exp(x)*ln(5)-exp(x)+2))+x)

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {-2+e^x (1+(-2-x) \log (5))+\left (-2 x+e^x x \log (5)\right ) \log (x)+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}{\left (2 x^2+e^x \left (-x^2+\left (2 x^2+x^3\right ) \log (5)\right )\right ) \log (x)+\left (4 x+e^x \left (-2 x+\left (4 x+2 x^2\right ) \log (5)\right )\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log ^2\left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \, dx=\frac {x}{x + \log \left (\frac {e^{x} \log \left (5\right ) \log \left (x\right )}{{\left ({\left (x + 2\right )} \log \left (5\right ) - 1\right )} e^{x} + 2}\right )} \]

[In]

integrate((((log(5)*(2+x)-1)*exp(x)+2)*log(x)*log(log(5)*exp(x)*log(x)/((log(5)*(2+x)-1)*exp(x)+2))+(x*exp(x)*
log(5)-2*x)*log(x)+((-2-x)*log(5)+1)*exp(x)-2)/(((log(5)*(2+x)-1)*exp(x)+2)*log(x)*log(log(5)*exp(x)*log(x)/((
log(5)*(2+x)-1)*exp(x)+2))^2+(((2*x^2+4*x)*log(5)-2*x)*exp(x)+4*x)*log(x)*log(log(5)*exp(x)*log(x)/((log(5)*(2
+x)-1)*exp(x)+2))+(((x^3+2*x^2)*log(5)-x^2)*exp(x)+2*x^2)*log(x)),x, algorithm="fricas")

[Out]

x/(x + log(e^x*log(5)*log(x)/(((x + 2)*log(5) - 1)*e^x + 2)))

Sympy [A] (verification not implemented)

Time = 0.94 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {-2+e^x (1+(-2-x) \log (5))+\left (-2 x+e^x x \log (5)\right ) \log (x)+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}{\left (2 x^2+e^x \left (-x^2+\left (2 x^2+x^3\right ) \log (5)\right )\right ) \log (x)+\left (4 x+e^x \left (-2 x+\left (4 x+2 x^2\right ) \log (5)\right )\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log ^2\left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \, dx=\frac {x}{x + \log {\left (\frac {e^{x} \log {\left (5 \right )} \log {\left (x \right )}}{\left (\left (x + 2\right ) \log {\left (5 \right )} - 1\right ) e^{x} + 2} \right )}} \]

[In]

integrate((((ln(5)*(2+x)-1)*exp(x)+2)*ln(x)*ln(ln(5)*exp(x)*ln(x)/((ln(5)*(2+x)-1)*exp(x)+2))+(x*exp(x)*ln(5)-
2*x)*ln(x)+((-2-x)*ln(5)+1)*exp(x)-2)/(((ln(5)*(2+x)-1)*exp(x)+2)*ln(x)*ln(ln(5)*exp(x)*ln(x)/((ln(5)*(2+x)-1)
*exp(x)+2))**2+(((2*x**2+4*x)*ln(5)-2*x)*exp(x)+4*x)*ln(x)*ln(ln(5)*exp(x)*ln(x)/((ln(5)*(2+x)-1)*exp(x)+2))+(
((x**3+2*x**2)*ln(5)-x**2)*exp(x)+2*x**2)*ln(x)),x)

[Out]

x/(x + log(exp(x)*log(5)*log(x)/(((x + 2)*log(5) - 1)*exp(x) + 2)))

Maxima [A] (verification not implemented)

none

Time = 0.76 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {-2+e^x (1+(-2-x) \log (5))+\left (-2 x+e^x x \log (5)\right ) \log (x)+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}{\left (2 x^2+e^x \left (-x^2+\left (2 x^2+x^3\right ) \log (5)\right )\right ) \log (x)+\left (4 x+e^x \left (-2 x+\left (4 x+2 x^2\right ) \log (5)\right )\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log ^2\left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \, dx=\frac {x}{2 \, x - \log \left ({\left (x \log \left (5\right ) + 2 \, \log \left (5\right ) - 1\right )} e^{x} + 2\right ) + \log \left (\log \left (5\right )\right ) + \log \left (\log \left (x\right )\right )} \]

[In]

integrate((((log(5)*(2+x)-1)*exp(x)+2)*log(x)*log(log(5)*exp(x)*log(x)/((log(5)*(2+x)-1)*exp(x)+2))+(x*exp(x)*
log(5)-2*x)*log(x)+((-2-x)*log(5)+1)*exp(x)-2)/(((log(5)*(2+x)-1)*exp(x)+2)*log(x)*log(log(5)*exp(x)*log(x)/((
log(5)*(2+x)-1)*exp(x)+2))^2+(((2*x^2+4*x)*log(5)-2*x)*exp(x)+4*x)*log(x)*log(log(5)*exp(x)*log(x)/((log(5)*(2
+x)-1)*exp(x)+2))+(((x^3+2*x^2)*log(5)-x^2)*exp(x)+2*x^2)*log(x)),x, algorithm="maxima")

[Out]

x/(2*x - log((x*log(5) + 2*log(5) - 1)*e^x + 2) + log(log(5)) + log(log(x)))

Giac [F(-1)]

Timed out. \[ \int \frac {-2+e^x (1+(-2-x) \log (5))+\left (-2 x+e^x x \log (5)\right ) \log (x)+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}{\left (2 x^2+e^x \left (-x^2+\left (2 x^2+x^3\right ) \log (5)\right )\right ) \log (x)+\left (4 x+e^x \left (-2 x+\left (4 x+2 x^2\right ) \log (5)\right )\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log ^2\left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \, dx=\text {Timed out} \]

[In]

integrate((((log(5)*(2+x)-1)*exp(x)+2)*log(x)*log(log(5)*exp(x)*log(x)/((log(5)*(2+x)-1)*exp(x)+2))+(x*exp(x)*
log(5)-2*x)*log(x)+((-2-x)*log(5)+1)*exp(x)-2)/(((log(5)*(2+x)-1)*exp(x)+2)*log(x)*log(log(5)*exp(x)*log(x)/((
log(5)*(2+x)-1)*exp(x)+2))^2+(((2*x^2+4*x)*log(5)-2*x)*exp(x)+4*x)*log(x)*log(log(5)*exp(x)*log(x)/((log(5)*(2
+x)-1)*exp(x)+2))+(((x^3+2*x^2)*log(5)-x^2)*exp(x)+2*x^2)*log(x)),x, algorithm="giac")

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {-2+e^x (1+(-2-x) \log (5))+\left (-2 x+e^x x \log (5)\right ) \log (x)+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}{\left (2 x^2+e^x \left (-x^2+\left (2 x^2+x^3\right ) \log (5)\right )\right ) \log (x)+\left (4 x+e^x \left (-2 x+\left (4 x+2 x^2\right ) \log (5)\right )\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log ^2\left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \, dx=-\int \frac {{\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (x+2\right )-1\right )+\ln \left (x\right )\,\left (2\,x-x\,{\mathrm {e}}^x\,\ln \left (5\right )\right )-\ln \left (x\right )\,\ln \left (\frac {{\mathrm {e}}^x\,\ln \left (5\right )\,\ln \left (x\right )}{{\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (x+2\right )-1\right )+2}\right )\,\left ({\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (x+2\right )-1\right )+2\right )+2}{\ln \left (x\right )\,\left ({\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (x+2\right )-1\right )+2\right )\,{\ln \left (\frac {{\mathrm {e}}^x\,\ln \left (5\right )\,\ln \left (x\right )}{{\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (x+2\right )-1\right )+2}\right )}^2+\ln \left (x\right )\,\left (4\,x-{\mathrm {e}}^x\,\left (2\,x-\ln \left (5\right )\,\left (2\,x^2+4\,x\right )\right )\right )\,\ln \left (\frac {{\mathrm {e}}^x\,\ln \left (5\right )\,\ln \left (x\right )}{{\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (x+2\right )-1\right )+2}\right )+\ln \left (x\right )\,\left ({\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (x^3+2\,x^2\right )-x^2\right )+2\,x^2\right )} \,d x \]

[In]

int(-(exp(x)*(log(5)*(x + 2) - 1) + log(x)*(2*x - x*exp(x)*log(5)) - log(x)*log((exp(x)*log(5)*log(x))/(exp(x)
*(log(5)*(x + 2) - 1) + 2))*(exp(x)*(log(5)*(x + 2) - 1) + 2) + 2)/(log(x)*(exp(x)*(log(5)*(2*x^2 + x^3) - x^2
) + 2*x^2) + log(x)*log((exp(x)*log(5)*log(x))/(exp(x)*(log(5)*(x + 2) - 1) + 2))*(4*x - exp(x)*(2*x - log(5)*
(4*x + 2*x^2))) + log(x)*log((exp(x)*log(5)*log(x))/(exp(x)*(log(5)*(x + 2) - 1) + 2))^2*(exp(x)*(log(5)*(x +
2) - 1) + 2)),x)

[Out]

-int((exp(x)*(log(5)*(x + 2) - 1) + log(x)*(2*x - x*exp(x)*log(5)) - log(x)*log((exp(x)*log(5)*log(x))/(exp(x)
*(log(5)*(x + 2) - 1) + 2))*(exp(x)*(log(5)*(x + 2) - 1) + 2) + 2)/(log(x)*(exp(x)*(log(5)*(2*x^2 + x^3) - x^2
) + 2*x^2) + log(x)*log((exp(x)*log(5)*log(x))/(exp(x)*(log(5)*(x + 2) - 1) + 2))*(4*x - exp(x)*(2*x - log(5)*
(4*x + 2*x^2))) + log(x)*log((exp(x)*log(5)*log(x))/(exp(x)*(log(5)*(x + 2) - 1) + 2))^2*(exp(x)*(log(5)*(x +
2) - 1) + 2)), x)