Integrand size = 209, antiderivative size = 29 \[ \int \frac {-45-3 x^2+e^{10} \left (-15-x^2\right )+e^x \left (-15-15 x+x^2+x^3\right )+e^{e^5} \left (-9-3 e^{10}+e^x \left (-3-3 x+x^2\right )\right )+\left (-45+6 x+3 x^2+e^{e^5} \left (-9+e^{10} (-3+x)+e^x (-3+x)+3 x\right )+e^{10} \left (-15+2 x+x^2\right )+e^x \left (-15+2 x+x^2\right )\right ) \log \left (\frac {3+e^{10}+e^x}{-15+e^{e^5} (-3+x)+2 x+x^2}\right )}{-45+6 x+3 x^2+e^{e^5} \left (-9+e^{10} (-3+x)+e^x (-3+x)+3 x\right )+e^{10} \left (-15+2 x+x^2\right )+e^x \left (-15+2 x+x^2\right )} \, dx=x+x \log \left (\frac {3+e^{10}+e^x}{(-3+x) \left (5+e^{e^5}+x\right )}\right ) \]
Time = 0.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {-45-3 x^2+e^{10} \left (-15-x^2\right )+e^x \left (-15-15 x+x^2+x^3\right )+e^{e^5} \left (-9-3 e^{10}+e^x \left (-3-3 x+x^2\right )\right )+\left (-45+6 x+3 x^2+e^{e^5} \left (-9+e^{10} (-3+x)+e^x (-3+x)+3 x\right )+e^{10} \left (-15+2 x+x^2\right )+e^x \left (-15+2 x+x^2\right )\right ) \log \left (\frac {3+e^{10}+e^x}{-15+e^{e^5} (-3+x)+2 x+x^2}\right )}{-45+6 x+3 x^2+e^{e^5} \left (-9+e^{10} (-3+x)+e^x (-3+x)+3 x\right )+e^{10} \left (-15+2 x+x^2\right )+e^x \left (-15+2 x+x^2\right )} \, dx=x+x \log \left (\frac {3+e^{10}+e^x}{(-3+x) \left (5+e^{e^5}+x\right )}\right ) \]
Integrate[(-45 - 3*x^2 + E^10*(-15 - x^2) + E^x*(-15 - 15*x + x^2 + x^3) + E^E^5*(-9 - 3*E^10 + E^x*(-3 - 3*x + x^2)) + (-45 + 6*x + 3*x^2 + E^E^5*( -9 + E^10*(-3 + x) + E^x*(-3 + x) + 3*x) + E^10*(-15 + 2*x + x^2) + E^x*(- 15 + 2*x + x^2))*Log[(3 + E^10 + E^x)/(-15 + E^E^5*(-3 + x) + 2*x + x^2)]) /(-45 + 6*x + 3*x^2 + E^E^5*(-9 + E^10*(-3 + x) + E^x*(-3 + x) + 3*x) + E^ 10*(-15 + 2*x + x^2) + E^x*(-15 + 2*x + x^2)),x]
Leaf count is larger than twice the leaf count of optimal. \(261\) vs. \(2(29)=58\).
Time = 6.24 (sec) , antiderivative size = 261, normalized size of antiderivative = 9.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {7292, 7293, 2009}
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.
\(\displaystyle \int \frac {-3 x^2+e^{10} \left (-x^2-15\right )+e^{e^5} \left (e^x \left (x^2-3 x-3\right )-3 e^{10}-9\right )+\left (3 x^2+e^x \left (x^2+2 x-15\right )+e^{10} \left (x^2+2 x-15\right )+6 x+e^{e^5} \left (e^x (x-3)+e^{10} (x-3)+3 x-9\right )-45\right ) \log \left (\frac {e^x+3+e^{10}}{x^2+2 x+e^{e^5} (x-3)-15}\right )+e^x \left (x^3+x^2-15 x-15\right )-45}{3 x^2+e^x \left (x^2+2 x-15\right )+e^{10} \left (x^2+2 x-15\right )+6 x+e^{e^5} \left (e^x (x-3)+e^{10} (x-3)+3 x-9\right )-45} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {3 x^2-e^{10} \left (-x^2-15\right )-e^{e^5} \left (e^x \left (x^2-3 x-3\right )-3 e^{10}-9\right )-\left (3 x^2+e^x \left (x^2+2 x-15\right )+e^{10} \left (x^2+2 x-15\right )+6 x+e^{e^5} \left (e^x (x-3)+e^{10} (x-3)+3 x-9\right )-45\right ) \log \left (\frac {e^x+3+e^{10}}{x^2+2 x+e^{e^5} (x-3)-15}\right )-e^x \left (x^3+x^2-15 x-15\right )+45}{\left (e^x+3 \left (1+\frac {e^{10}}{3}\right )\right ) (3-x) \left (x+e^{e^5}+5\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {-x^3-\left (1+e^{e^5}\right ) x^2-x^2 \log \left (\frac {e^x+3+e^{10}}{(x-3) \left (x+e^{e^5}+5\right )}\right )+15 \left (1+\frac {e^{e^5}}{5}\right ) x-2 \left (1+\frac {e^{e^5}}{2}\right ) x \log \left (\frac {e^x+3+e^{10}}{(x-3) \left (x+e^{e^5}+5\right )}\right )+15 \left (1+\frac {e^{e^5}}{5}\right ) \log \left (\frac {e^x+3+e^{10}}{(x-3) \left (x+e^{e^5}+5\right )}\right )+15 \left (1+\frac {e^{e^5}}{5}\right )}{(3-x) \left (x+e^{e^5}+5\right )}+\frac {\left (-3-e^{10}\right ) x}{e^x+3 \left (1+\frac {e^{10}}{3}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\left (2+e^{e^5}\right ) x+e^{e^5} x+3 x+x \log \left (-\frac {e^x+3+e^{10}}{(3-x) \left (x+e^{e^5}+5\right )}\right )-\frac {3 e^{e^5} \log (3-x)}{8+e^{e^5}}-\frac {24 \log (3-x)}{8+e^{e^5}}+3 \log (3-x)-\frac {e^{e^5} \left (37+13 e^{e^5}+e^{2 e^5}\right ) \log \left (x+e^{e^5}+5\right )}{8+e^{e^5}}+\frac {\left (5+e^{e^5}\right )^3 \log \left (x+e^{e^5}+5\right )}{8+e^{e^5}}-\frac {\left (5+e^{e^5}\right )^2 \log \left (x+e^{e^5}+5\right )}{8+e^{e^5}}-\frac {15 \left (5+e^{e^5}\right ) \log \left (x+e^{e^5}+5\right )}{8+e^{e^5}}+\frac {15 \log \left (x+e^{e^5}+5\right )}{8+e^{e^5}}-\left (5+e^{e^5}\right ) \log \left (x+e^{e^5}+5\right )\) |
Int[(-45 - 3*x^2 + E^10*(-15 - x^2) + E^x*(-15 - 15*x + x^2 + x^3) + E^E^5 *(-9 - 3*E^10 + E^x*(-3 - 3*x + x^2)) + (-45 + 6*x + 3*x^2 + E^E^5*(-9 + E ^10*(-3 + x) + E^x*(-3 + x) + 3*x) + E^10*(-15 + 2*x + x^2) + E^x*(-15 + 2 *x + x^2))*Log[(3 + E^10 + E^x)/(-15 + E^E^5*(-3 + x) + 2*x + x^2)])/(-45 + 6*x + 3*x^2 + E^E^5*(-9 + E^10*(-3 + x) + E^x*(-3 + x) + 3*x) + E^10*(-1 5 + 2*x + x^2) + E^x*(-15 + 2*x + x^2)),x]
3*x + E^E^5*x - (2 + E^E^5)*x + 3*Log[3 - x] - (24*Log[3 - x])/(8 + E^E^5) - (3*E^E^5*Log[3 - x])/(8 + E^E^5) + x*Log[-((3 + E^10 + E^x)/((3 - x)*(5 + E^E^5 + x)))] - (5 + E^E^5)*Log[5 + E^E^5 + x] + (15*Log[5 + E^E^5 + x] )/(8 + E^E^5) - (15*(5 + E^E^5)*Log[5 + E^E^5 + x])/(8 + E^E^5) - ((5 + E^ E^5)^2*Log[5 + E^E^5 + x])/(8 + E^E^5) + ((5 + E^E^5)^3*Log[5 + E^E^5 + x] )/(8 + E^E^5) - (E^E^5*(37 + 13*E^E^5 + E^(2*E^5))*Log[5 + E^E^5 + x])/(8 + E^E^5)
3.26.60.3.1 Defintions of rubi rules used
Time = 4.06 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10
method | result | size |
norman | \(x +\ln \left (\frac {{\mathrm e}^{x}+{\mathrm e}^{10}+3}{\left (-3+x \right ) {\mathrm e}^{{\mathrm e}^{5}}+x^{2}+2 x -15}\right ) x\) | \(32\) |
parallelrisch | \(-4+\ln \left (\frac {{\mathrm e}^{x}+{\mathrm e}^{10}+3}{x \,{\mathrm e}^{{\mathrm e}^{5}}+x^{2}-3 \,{\mathrm e}^{{\mathrm e}^{5}}+2 x -15}\right ) x -2 \,{\mathrm e}^{{\mathrm e}^{5}}+x\) | \(41\) |
risch | \(x \ln \left ({\mathrm e}^{x}+{\mathrm e}^{10}+3\right )-x \ln \left (x +{\mathrm e}^{{\mathrm e}^{5}}+5\right )-\ln \left (-3+x \right ) x -\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{-3+x}\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+{\mathrm e}^{10}+3\right )}{x +{\mathrm e}^{{\mathrm e}^{5}}+5}\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+{\mathrm e}^{10}+3\right )}{\left (-3+x \right ) \left (x +{\mathrm e}^{{\mathrm e}^{5}}+5\right )}\right )}{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{-3+x}\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+{\mathrm e}^{10}+3\right )}{\left (-3+x \right ) \left (x +{\mathrm e}^{{\mathrm e}^{5}}+5\right )}\right )}^{2}}{2}-\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x +{\mathrm e}^{{\mathrm e}^{5}}+5}\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{x}+{\mathrm e}^{10}+3\right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+{\mathrm e}^{10}+3\right )}{x +{\mathrm e}^{{\mathrm e}^{5}}+5}\right )}{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x +{\mathrm e}^{{\mathrm e}^{5}}+5}\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+{\mathrm e}^{10}+3\right )}{x +{\mathrm e}^{{\mathrm e}^{5}}+5}\right )}^{2}}{2}+\frac {i \pi x \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x}+{\mathrm e}^{10}+3\right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+{\mathrm e}^{10}+3\right )}{x +{\mathrm e}^{{\mathrm e}^{5}}+5}\right )}^{2}}{2}-\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+{\mathrm e}^{10}+3\right )}{x +{\mathrm e}^{{\mathrm e}^{5}}+5}\right )}^{3}}{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+{\mathrm e}^{10}+3\right )}{x +{\mathrm e}^{{\mathrm e}^{5}}+5}\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+{\mathrm e}^{10}+3\right )}{\left (-3+x \right ) \left (x +{\mathrm e}^{{\mathrm e}^{5}}+5\right )}\right )}^{2}}{2}-\frac {i \pi x {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x}+{\mathrm e}^{10}+3\right )}{\left (-3+x \right ) \left (x +{\mathrm e}^{{\mathrm e}^{5}}+5\right )}\right )}^{3}}{2}+x\) | \(343\) |
int(((((-3+x)*exp(x)+(-3+x)*exp(5)^2+3*x-9)*exp(exp(5))+(x^2+2*x-15)*exp(x )+(x^2+2*x-15)*exp(5)^2+3*x^2+6*x-45)*ln((exp(x)+exp(5)^2+3)/((-3+x)*exp(e xp(5))+x^2+2*x-15))+((x^2-3*x-3)*exp(x)-3*exp(5)^2-9)*exp(exp(5))+(x^3+x^2 -15*x-15)*exp(x)+(-x^2-15)*exp(5)^2-3*x^2-45)/(((-3+x)*exp(x)+(-3+x)*exp(5 )^2+3*x-9)*exp(exp(5))+(x^2+2*x-15)*exp(x)+(x^2+2*x-15)*exp(5)^2+3*x^2+6*x -45),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {-45-3 x^2+e^{10} \left (-15-x^2\right )+e^x \left (-15-15 x+x^2+x^3\right )+e^{e^5} \left (-9-3 e^{10}+e^x \left (-3-3 x+x^2\right )\right )+\left (-45+6 x+3 x^2+e^{e^5} \left (-9+e^{10} (-3+x)+e^x (-3+x)+3 x\right )+e^{10} \left (-15+2 x+x^2\right )+e^x \left (-15+2 x+x^2\right )\right ) \log \left (\frac {3+e^{10}+e^x}{-15+e^{e^5} (-3+x)+2 x+x^2}\right )}{-45+6 x+3 x^2+e^{e^5} \left (-9+e^{10} (-3+x)+e^x (-3+x)+3 x\right )+e^{10} \left (-15+2 x+x^2\right )+e^x \left (-15+2 x+x^2\right )} \, dx=x \log \left (\frac {e^{10} + e^{x} + 3}{x^{2} + {\left (x - 3\right )} e^{\left (e^{5}\right )} + 2 \, x - 15}\right ) + x \]
integrate(((((-3+x)*exp(x)+(-3+x)*exp(5)^2+3*x-9)*exp(exp(5))+(x^2+2*x-15) *exp(x)+(x^2+2*x-15)*exp(5)^2+3*x^2+6*x-45)*log((exp(x)+exp(5)^2+3)/((-3+x )*exp(exp(5))+x^2+2*x-15))+((x^2-3*x-3)*exp(x)-3*exp(5)^2-9)*exp(exp(5))+( x^3+x^2-15*x-15)*exp(x)+(-x^2-15)*exp(5)^2-3*x^2-45)/(((-3+x)*exp(x)+(-3+x )*exp(5)^2+3*x-9)*exp(exp(5))+(x^2+2*x-15)*exp(x)+(x^2+2*x-15)*exp(5)^2+3* x^2+6*x-45),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (24) = 48\).
Time = 1.51 (sec) , antiderivative size = 88, normalized size of antiderivative = 3.03 \[ \int \frac {-45-3 x^2+e^{10} \left (-15-x^2\right )+e^x \left (-15-15 x+x^2+x^3\right )+e^{e^5} \left (-9-3 e^{10}+e^x \left (-3-3 x+x^2\right )\right )+\left (-45+6 x+3 x^2+e^{e^5} \left (-9+e^{10} (-3+x)+e^x (-3+x)+3 x\right )+e^{10} \left (-15+2 x+x^2\right )+e^x \left (-15+2 x+x^2\right )\right ) \log \left (\frac {3+e^{10}+e^x}{-15+e^{e^5} (-3+x)+2 x+x^2}\right )}{-45+6 x+3 x^2+e^{e^5} \left (-9+e^{10} (-3+x)+e^x (-3+x)+3 x\right )+e^{10} \left (-15+2 x+x^2\right )+e^x \left (-15+2 x+x^2\right )} \, dx=x + \left (x + \frac {1}{3} + \frac {e^{e^{5}}}{6}\right ) \log {\left (\frac {e^{x} + 3 + e^{10}}{x^{2} + 2 x + \left (x - 3\right ) e^{e^{5}} - 15} \right )} - \frac {\left (2 + e^{e^{5}}\right ) \log {\left (e^{x} + 3 + e^{10} \right )}}{6} + \frac {\left (2 + e^{e^{5}}\right ) \log {\left (x^{2} + x \left (2 + e^{e^{5}}\right ) - 3 e^{e^{5}} - 15 \right )}}{6} \]
integrate(((((-3+x)*exp(x)+(-3+x)*exp(5)**2+3*x-9)*exp(exp(5))+(x**2+2*x-1 5)*exp(x)+(x**2+2*x-15)*exp(5)**2+3*x**2+6*x-45)*ln((exp(x)+exp(5)**2+3)/( (-3+x)*exp(exp(5))+x**2+2*x-15))+((x**2-3*x-3)*exp(x)-3*exp(5)**2-9)*exp(e xp(5))+(x**3+x**2-15*x-15)*exp(x)+(-x**2-15)*exp(5)**2-3*x**2-45)/(((-3+x) *exp(x)+(-3+x)*exp(5)**2+3*x-9)*exp(exp(5))+(x**2+2*x-15)*exp(x)+(x**2+2*x -15)*exp(5)**2+3*x**2+6*x-45),x)
x + (x + 1/3 + exp(exp(5))/6)*log((exp(x) + 3 + exp(10))/(x**2 + 2*x + (x - 3)*exp(exp(5)) - 15)) - (2 + exp(exp(5)))*log(exp(x) + 3 + exp(10))/6 + (2 + exp(exp(5)))*log(x**2 + x*(2 + exp(exp(5))) - 3*exp(exp(5)) - 15)/6
Time = 0.27 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {-45-3 x^2+e^{10} \left (-15-x^2\right )+e^x \left (-15-15 x+x^2+x^3\right )+e^{e^5} \left (-9-3 e^{10}+e^x \left (-3-3 x+x^2\right )\right )+\left (-45+6 x+3 x^2+e^{e^5} \left (-9+e^{10} (-3+x)+e^x (-3+x)+3 x\right )+e^{10} \left (-15+2 x+x^2\right )+e^x \left (-15+2 x+x^2\right )\right ) \log \left (\frac {3+e^{10}+e^x}{-15+e^{e^5} (-3+x)+2 x+x^2}\right )}{-45+6 x+3 x^2+e^{e^5} \left (-9+e^{10} (-3+x)+e^x (-3+x)+3 x\right )+e^{10} \left (-15+2 x+x^2\right )+e^x \left (-15+2 x+x^2\right )} \, dx=-x \log \left (x + e^{\left (e^{5}\right )} + 5\right ) - x \log \left (x - 3\right ) + x \log \left (e^{10} + e^{x} + 3\right ) + x \]
integrate(((((-3+x)*exp(x)+(-3+x)*exp(5)^2+3*x-9)*exp(exp(5))+(x^2+2*x-15) *exp(x)+(x^2+2*x-15)*exp(5)^2+3*x^2+6*x-45)*log((exp(x)+exp(5)^2+3)/((-3+x )*exp(exp(5))+x^2+2*x-15))+((x^2-3*x-3)*exp(x)-3*exp(5)^2-9)*exp(exp(5))+( x^3+x^2-15*x-15)*exp(x)+(-x^2-15)*exp(5)^2-3*x^2-45)/(((-3+x)*exp(x)+(-3+x )*exp(5)^2+3*x-9)*exp(exp(5))+(x^2+2*x-15)*exp(x)+(x^2+2*x-15)*exp(5)^2+3* x^2+6*x-45),x, algorithm=\
Time = 7.17 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.69 \[ \int \frac {-45-3 x^2+e^{10} \left (-15-x^2\right )+e^x \left (-15-15 x+x^2+x^3\right )+e^{e^5} \left (-9-3 e^{10}+e^x \left (-3-3 x+x^2\right )\right )+\left (-45+6 x+3 x^2+e^{e^5} \left (-9+e^{10} (-3+x)+e^x (-3+x)+3 x\right )+e^{10} \left (-15+2 x+x^2\right )+e^x \left (-15+2 x+x^2\right )\right ) \log \left (\frac {3+e^{10}+e^x}{-15+e^{e^5} (-3+x)+2 x+x^2}\right )}{-45+6 x+3 x^2+e^{e^5} \left (-9+e^{10} (-3+x)+e^x (-3+x)+3 x\right )+e^{10} \left (-15+2 x+x^2\right )+e^x \left (-15+2 x+x^2\right )} \, dx=x \log \left (\frac {e^{20} + 3 \, e^{10} + e^{\left (x + 10\right )}}{x^{2} e^{10} + 2 \, x e^{10} + x e^{\left (e^{5} + 10\right )} - 15 \, e^{10} - 3 \, e^{\left (e^{5} + 10\right )}}\right ) + x \]
integrate(((((-3+x)*exp(x)+(-3+x)*exp(5)^2+3*x-9)*exp(exp(5))+(x^2+2*x-15) *exp(x)+(x^2+2*x-15)*exp(5)^2+3*x^2+6*x-45)*log((exp(x)+exp(5)^2+3)/((-3+x )*exp(exp(5))+x^2+2*x-15))+((x^2-3*x-3)*exp(x)-3*exp(5)^2-9)*exp(exp(5))+( x^3+x^2-15*x-15)*exp(x)+(-x^2-15)*exp(5)^2-3*x^2-45)/(((-3+x)*exp(x)+(-3+x )*exp(5)^2+3*x-9)*exp(exp(5))+(x^2+2*x-15)*exp(x)+(x^2+2*x-15)*exp(5)^2+3* x^2+6*x-45),x, algorithm=\
x*log((e^20 + 3*e^10 + e^(x + 10))/(x^2*e^10 + 2*x*e^10 + x*e^(e^5 + 10) - 15*e^10 - 3*e^(e^5 + 10))) + x
Time = 13.47 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {-45-3 x^2+e^{10} \left (-15-x^2\right )+e^x \left (-15-15 x+x^2+x^3\right )+e^{e^5} \left (-9-3 e^{10}+e^x \left (-3-3 x+x^2\right )\right )+\left (-45+6 x+3 x^2+e^{e^5} \left (-9+e^{10} (-3+x)+e^x (-3+x)+3 x\right )+e^{10} \left (-15+2 x+x^2\right )+e^x \left (-15+2 x+x^2\right )\right ) \log \left (\frac {3+e^{10}+e^x}{-15+e^{e^5} (-3+x)+2 x+x^2}\right )}{-45+6 x+3 x^2+e^{e^5} \left (-9+e^{10} (-3+x)+e^x (-3+x)+3 x\right )+e^{10} \left (-15+2 x+x^2\right )+e^x \left (-15+2 x+x^2\right )} \, dx=x\,\left (\ln \left (\frac {{\mathrm {e}}^{10}+{\mathrm {e}}^x+3}{2\,x+{\mathrm {e}}^{{\mathrm {e}}^5}\,\left (x-3\right )+x^2-15}\right )+1\right ) \]
int(-(3*x^2 - log((exp(10) + exp(x) + 3)/(2*x + exp(exp(5))*(x - 3) + x^2 - 15))*(6*x + exp(exp(5))*(3*x + exp(x)*(x - 3) + exp(10)*(x - 3) - 9) + e xp(x)*(2*x + x^2 - 15) + exp(10)*(2*x + x^2 - 15) + 3*x^2 - 45) + exp(x)*( 15*x - x^2 - x^3 + 15) + exp(exp(5))*(3*exp(10) + exp(x)*(3*x - x^2 + 3) + 9) + exp(10)*(x^2 + 15) + 45)/(6*x + exp(exp(5))*(3*x + exp(x)*(x - 3) + exp(10)*(x - 3) - 9) + exp(x)*(2*x + x^2 - 15) + exp(10)*(2*x + x^2 - 15) + 3*x^2 - 45),x)