Integrand size = 203, antiderivative size = 32 \[ \int \frac {-2+e^x (1+(-2-x) \log (5))+\left (-2 x+e^x x \log (5)\right ) \log (x)+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}{\left (2 x^2+e^x \left (-x^2+\left (2 x^2+x^3\right ) \log (5)\right )\right ) \log (x)+\left (4 x+e^x \left (-2 x+\left (4 x+2 x^2\right ) \log (5)\right )\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log ^2\left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \, dx=2+\frac {x}{x+\log \left (\frac {\log (x)}{2+x-\frac {1-2 e^{-x}}{\log (5)}}\right )} \]
Time = 0.36 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \frac {-2+e^x (1+(-2-x) \log (5))+\left (-2 x+e^x x \log (5)\right ) \log (x)+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}{\left (2 x^2+e^x \left (-x^2+\left (2 x^2+x^3\right ) \log (5)\right )\right ) \log (x)+\left (4 x+e^x \left (-2 x+\left (4 x+2 x^2\right ) \log (5)\right )\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log ^2\left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \, dx=\frac {x}{x+\log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \]
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]
Time = 1.50 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.16, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.015, Rules used = {7239, 7262, 17}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {e^x ((-x-2) \log (5)+1)+\left (e^x x \log (5)-2 x\right ) \log (x)+\left (e^x ((x+2) \log (5)-1)+2\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{e^x ((x+2) \log (5)-1)+2}\right )-2}{\left (e^x \left (\left (2 x^2+4 x\right ) \log (5)-2 x\right )+4 x\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{e^x ((x+2) \log (5)-1)+2}\right )+\left (2 x^2+e^x \left (\left (x^3+2 x^2\right ) \log (5)-x^2\right )\right ) \log (x)+\left (e^x ((x+2) \log (5)-1)+2\right ) \log (x) \log ^2\left (\frac {e^x \log (5) \log (x)}{e^x ((x+2) \log (5)-1)+2}\right )} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {-e^x (x \log (5)-1+\log (25))+\log (x) \left (x \left (e^x \log (5)-2\right )+\left (e^x (x \log (5)-1+\log (25))+2\right ) \log \left (\frac {e^x \log (5) \log (x)}{e^x ((x+2) \log (5)-1)+2}\right )\right )-2}{\left (e^x (x \log (5)-1+\log (25))+2\right ) \log (x) \left (x+\log \left (\frac {e^x \log (5) \log (x)}{e^x ((x+2) \log (5)-1)+2}\right )\right )^2}dx\) |
| \(\Big \downarrow \) 7262 |
| \(\displaystyle \int \frac {1}{\left (\frac {x}{\log \left (\frac {e^x \log (5) \log (x)}{2-e^x (1-(x+2) \log (5))}\right )}+1\right )^2}d\frac {x}{\log \left (\frac {e^x \log (5) \log (x)}{2-e^x (1-(x+2) \log (5))}\right )}\) |
| \(\Big \downarrow \) 17 |
| \(\displaystyle -\frac {1}{\frac {x}{\log \left (\frac {e^x \log (5) \log (x)}{2-e^x (1-(x+2) \log (5))}\right )}+1}\) |
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]*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]
3.7.75.3.1 Defintions of rubi rules used
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
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])]}, Simp[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]
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 )}\) | \(430\) |
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)*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)/((l n(5)*(2+x)-1)*exp(x)+2))+(((x^3+2*x^2)*ln(5)-x^2)*exp(x)+2*x^2)*ln(x)),x,m ethod=_RETURNVERBOSE)
Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {-2+e^x (1+(-2-x) \log (5))+\left (-2 x+e^x x \log (5)\right ) \log (x)+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}{\left (2 x^2+e^x \left (-x^2+\left (2 x^2+x^3\right ) \log (5)\right )\right ) \log (x)+\left (4 x+e^x \left (-2 x+\left (4 x+2 x^2\right ) \log (5)\right )\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log ^2\left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \, dx=\frac {x}{x + \log \left (\frac {e^{x} \log \left (5\right ) \log \left (x\right )}{{\left ({\left (x + 2\right )} \log \left (5\right ) - 1\right )} e^{x} + 2}\right )} \]
integrate((((log(5)*(2+x)-1)*exp(x)+2)*log(x)*log(log(5)*exp(x)*log(x)/((l og(5)*(2+x)-1)*exp(x)+2))+(x*exp(x)*log(5)-2*x)*log(x)+((-2-x)*log(5)+1)*e xp(x)-2)/(((log(5)*(2+x)-1)*exp(x)+2)*log(x)*log(log(5)*exp(x)*log(x)/((lo g(5)*(2+x)-1)*exp(x)+2))^2+(((2*x^2+4*x)*log(5)-2*x)*exp(x)+4*x)*log(x)*lo g(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=\
Time = 0.94 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {-2+e^x (1+(-2-x) \log (5))+\left (-2 x+e^x x \log (5)\right ) \log (x)+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}{\left (2 x^2+e^x \left (-x^2+\left (2 x^2+x^3\right ) \log (5)\right )\right ) \log (x)+\left (4 x+e^x \left (-2 x+\left (4 x+2 x^2\right ) \log (5)\right )\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log ^2\left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \, dx=\frac {x}{x + \log {\left (\frac {e^{x} \log {\left (5 \right )} \log {\left (x \right )}}{\left (\left (x + 2\right ) \log {\left (5 \right )} - 1\right ) e^{x} + 2} \right )}} \]
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)*e xp(x)+2))**2+(((2*x**2+4*x)*ln(5)-2*x)*exp(x)+4*x)*ln(x)*ln(ln(5)*exp(x)*l n(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)
Time = 0.76 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00 \[ \int \frac {-2+e^x (1+(-2-x) \log (5))+\left (-2 x+e^x x \log (5)\right ) \log (x)+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}{\left (2 x^2+e^x \left (-x^2+\left (2 x^2+x^3\right ) \log (5)\right )\right ) \log (x)+\left (4 x+e^x \left (-2 x+\left (4 x+2 x^2\right ) \log (5)\right )\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log ^2\left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \, dx=\frac {x}{2 \, x - \log \left ({\left (x \log \left (5\right ) + 2 \, \log \left (5\right ) - 1\right )} e^{x} + 2\right ) + \log \left (\log \left (5\right )\right ) + \log \left (\log \left (x\right )\right )} \]
integrate((((log(5)*(2+x)-1)*exp(x)+2)*log(x)*log(log(5)*exp(x)*log(x)/((l og(5)*(2+x)-1)*exp(x)+2))+(x*exp(x)*log(5)-2*x)*log(x)+((-2-x)*log(5)+1)*e xp(x)-2)/(((log(5)*(2+x)-1)*exp(x)+2)*log(x)*log(log(5)*exp(x)*log(x)/((lo g(5)*(2+x)-1)*exp(x)+2))^2+(((2*x^2+4*x)*log(5)-2*x)*exp(x)+4*x)*log(x)*lo g(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=\
Timed out. \[ \int \frac {-2+e^x (1+(-2-x) \log (5))+\left (-2 x+e^x x \log (5)\right ) \log (x)+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}{\left (2 x^2+e^x \left (-x^2+\left (2 x^2+x^3\right ) \log (5)\right )\right ) \log (x)+\left (4 x+e^x \left (-2 x+\left (4 x+2 x^2\right ) \log (5)\right )\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log ^2\left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \, dx=\text {Timed out} \]
integrate((((log(5)*(2+x)-1)*exp(x)+2)*log(x)*log(log(5)*exp(x)*log(x)/((l og(5)*(2+x)-1)*exp(x)+2))+(x*exp(x)*log(5)-2*x)*log(x)+((-2-x)*log(5)+1)*e xp(x)-2)/(((log(5)*(2+x)-1)*exp(x)+2)*log(x)*log(log(5)*exp(x)*log(x)/((lo g(5)*(2+x)-1)*exp(x)+2))^2+(((2*x^2+4*x)*log(5)-2*x)*exp(x)+4*x)*log(x)*lo g(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=\
Timed out. \[ \int \frac {-2+e^x (1+(-2-x) \log (5))+\left (-2 x+e^x x \log (5)\right ) \log (x)+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )}{\left (2 x^2+e^x \left (-x^2+\left (2 x^2+x^3\right ) \log (5)\right )\right ) \log (x)+\left (4 x+e^x \left (-2 x+\left (4 x+2 x^2\right ) \log (5)\right )\right ) \log (x) \log \left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )+\left (2+e^x (-1+(2+x) \log (5))\right ) \log (x) \log ^2\left (\frac {e^x \log (5) \log (x)}{2+e^x (-1+(2+x) \log (5))}\right )} \, dx=-\int \frac {{\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (x+2\right )-1\right )+\ln \left (x\right )\,\left (2\,x-x\,{\mathrm {e}}^x\,\ln \left (5\right )\right )-\ln \left (x\right )\,\ln \left (\frac {{\mathrm {e}}^x\,\ln \left (5\right )\,\ln \left (x\right )}{{\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (x+2\right )-1\right )+2}\right )\,\left ({\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (x+2\right )-1\right )+2\right )+2}{\ln \left (x\right )\,\left ({\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (x+2\right )-1\right )+2\right )\,{\ln \left (\frac {{\mathrm {e}}^x\,\ln \left (5\right )\,\ln \left (x\right )}{{\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (x+2\right )-1\right )+2}\right )}^2+\ln \left (x\right )\,\left (4\,x-{\mathrm {e}}^x\,\left (2\,x-\ln \left (5\right )\,\left (2\,x^2+4\,x\right )\right )\right )\,\ln \left (\frac {{\mathrm {e}}^x\,\ln \left (5\right )\,\ln \left (x\right )}{{\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (x+2\right )-1\right )+2}\right )+\ln \left (x\right )\,\left ({\mathrm {e}}^x\,\left (\ln \left (5\right )\,\left (x^3+2\,x^2\right )-x^2\right )+2\,x^2\right )} \,d x \]
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)*(l og(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)
-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)*(l og(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)