Integrand size = 118, antiderivative size = 26 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=5 (4-x) \log \left (x+e^x x \left (2 x+\frac {2}{\log (\log (2))}\right )\right ) \]
Time = 0.96 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=20 \log (x)-5 x \log \left (x+2 e^x x \left (x+\frac {1}{\log (\log (2))}\right )\right )+20 \log \left (2 e^x+\log (\log (2))+2 e^x x \log (\log (2))\right ) \]
Integrate[(E^x*(40 + 30*x - 10*x^2) + (20 - 5*x + E^x*(80*x + 20*x^2 - 10* x^3))*Log[Log[2]] + (-10*E^x*x + (-5*x - 10*E^x*x^2)*Log[Log[2]])*Log[(2*E ^x*x + (x + 2*E^x*x^2)*Log[Log[2]])/Log[Log[2]]])/(2*E^x*x + (x + 2*E^x*x^ 2)*Log[Log[2]]),x]
20*Log[x] - 5*x*Log[x + 2*E^x*x*(x + Log[Log[2]]^(-1))] + 20*Log[2*E^x + L og[Log[2]] + 2*E^x*x*Log[Log[2]]]
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.
| \(\displaystyle \int \frac {e^x \left (-10 x^2+30 x+40\right )+\left (\left (-10 e^x x^2-5 x\right ) \log (\log (2))-10 e^x x\right ) \log \left (\frac {\left (2 e^x x^2+x\right ) \log (\log (2))+2 e^x x}{\log (\log (2))}\right )+\left (e^x \left (-10 x^3+20 x^2+80 x\right )-5 x+20\right ) \log (\log (2))}{\left (2 e^x x^2+x\right ) \log (\log (2))+2 e^x x} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {\left (x^4-11 x^3+35 x^2-5 x-50\right ) \exp \left (\frac {2 x^2-10 x-10}{x^2-5 x}+\frac {e^{\frac {2 x^2-10 x-10}{x^2-5 x}+x-5}}{x}+x-5\right )}{x^3 \left (x^2-10 x+25\right )}dx\) |
| \(\Big \downarrow \) 7277 |
| \(\displaystyle 4 \int -\frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right ) \left (-x^4+11 x^3-35 x^2+5 x+50\right )}{4 (5-x)^2 x^3}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right ) \left (-x^4+11 x^3-35 x^2+5 x+50\right )}{(5-x)^2 x^3}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\int \left (-\frac {27 \exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{25 x}+\frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x^2}+\frac {2 \exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x^3}+\frac {2 \exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{25 (x-5)}-\frac {2 \exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{5 (x-5)^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2}{5} \int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{(x-5)^2}dx-\frac {2}{25} \int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x-5}dx-\int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x^2}dx+\frac {27}{25} \int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x}dx-2 \int \frac {\exp \left (x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5+\frac {e^{x+\frac {2 \left (-x^2+5 x+5\right )}{5 x-x^2}-5}}{x}\right )}{x^3}dx\) |
Int[(E^x*(40 + 30*x - 10*x^2) + (20 - 5*x + E^x*(80*x + 20*x^2 - 10*x^3))* Log[Log[2]] + (-10*E^x*x + (-5*x - 10*E^x*x^2)*Log[Log[2]])*Log[(2*E^x*x + (x + 2*E^x*x^2)*Log[Log[2]])/Log[Log[2]]])/(2*E^x*x + (x + 2*E^x*x^2)*Log [Log[2]]),x]
3.13.56.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Leaf count of result is larger than twice the leaf count of optimal. \(55\) vs. \(2(25)=50\).
Time = 3.94 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15
| method | result | size |
| default | \(20 \ln \left (-x \left (2 \,{\mathrm e}^{x} \ln \left (\ln \left (2\right )\right ) x +2 \,{\mathrm e}^{x}+\ln \left (\ln \left (2\right )\right )\right )\right )+5 \ln \left (-\ln \left (\ln \left (2\right )\right )\right ) x -5 x \ln \left (-x \left (2 \,{\mathrm e}^{x} \ln \left (\ln \left (2\right )\right ) x +2 \,{\mathrm e}^{x}+\ln \left (\ln \left (2\right )\right )\right )\right )\) | \(56\) |
| norman | \(20 \ln \left (\frac {\left (2 \,{\mathrm e}^{x} x^{2}+x \right ) \ln \left (\ln \left (2\right )\right )+2 \,{\mathrm e}^{x} x}{\ln \left (\ln \left (2\right )\right )}\right )-5 x \ln \left (\frac {\left (2 \,{\mathrm e}^{x} x^{2}+x \right ) \ln \left (\ln \left (2\right )\right )+2 \,{\mathrm e}^{x} x}{\ln \left (\ln \left (2\right )\right )}\right )\) | \(59\) |
| parallelrisch | \(20 \ln \left (\frac {\left (2 \,{\mathrm e}^{x} x^{2}+x \right ) \ln \left (\ln \left (2\right )\right )+2 \,{\mathrm e}^{x} x}{\ln \left (\ln \left (2\right )\right )}\right )-5 x \ln \left (\frac {\left (2 \,{\mathrm e}^{x} x^{2}+x \right ) \ln \left (\ln \left (2\right )\right )+2 \,{\mathrm e}^{x} x}{\ln \left (\ln \left (2\right )\right )}\right )\) | \(59\) |
| risch | \(-5 x \ln \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )-5 x \ln \left (x \right )+\frac {5 i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (i \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right ) \operatorname {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right )}{2}-\frac {5 i \pi x \,\operatorname {csgn}\left (i x \right ) {\operatorname {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right )}^{2}}{2}-\frac {5 i \pi x \,\operatorname {csgn}\left (i \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right ) {\operatorname {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right )}^{2}}{2}+5 i \pi x {\operatorname {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right )}^{2}-\frac {5 i \pi x {\operatorname {csgn}\left (i x \left (\left ({\mathrm e}^{x} x +\frac {1}{2}\right ) \ln \left (\ln \left (2\right )\right )+{\mathrm e}^{x}\right )\right )}^{3}}{2}-5 i \pi x -5 x \ln \left (2\right )+5 \ln \left (-\ln \left (\ln \left (2\right )\right )\right ) x +20 \ln \left (x^{2} \ln \left (\ln \left (2\right )\right )+x \right )+20 \ln \left ({\mathrm e}^{x}+\frac {\ln \left (\ln \left (2\right )\right )}{2+2 x \ln \left (\ln \left (2\right )\right )}\right )\) | \(242\) |
int((((-10*exp(x)*x^2-5*x)*ln(ln(2))-10*exp(x)*x)*ln(((2*exp(x)*x^2+x)*ln( ln(2))+2*exp(x)*x)/ln(ln(2)))+((-10*x^3+20*x^2+80*x)*exp(x)-5*x+20)*ln(ln( 2))+(-10*x^2+30*x+40)*exp(x))/((2*exp(x)*x^2+x)*ln(ln(2))+2*exp(x)*x),x,me thod=_RETURNVERBOSE)
20*ln(-x*(2*exp(x)*ln(ln(2))*x+2*exp(x)+ln(ln(2))))+5*ln(-ln(ln(2)))*x-5*x *ln(-x*(2*exp(x)*ln(ln(2))*x+2*exp(x)+ln(ln(2))))
Time = 0.25 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=-5 \, {\left (x - 4\right )} \log \left (\frac {2 \, x e^{x} + {\left (2 \, x^{2} e^{x} + x\right )} \log \left (\log \left (2\right )\right )}{\log \left (\log \left (2\right )\right )}\right ) \]
integrate((((-10*exp(x)*x^2-5*x)*log(log(2))-10*exp(x)*x)*log(((2*exp(x)*x ^2+x)*log(log(2))+2*exp(x)*x)/log(log(2)))+((-10*x^3+20*x^2+80*x)*exp(x)-5 *x+20)*log(log(2))+(-10*x^2+30*x+40)*exp(x))/((2*exp(x)*x^2+x)*log(log(2)) +2*exp(x)*x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (22) = 44\).
Time = 0.42 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.54 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=- 5 x \log {\left (\frac {2 x e^{x} + \left (2 x^{2} e^{x} + x\right ) \log {\left (\log {\left (2 \right )} \right )}}{\log {\left (\log {\left (2 \right )} \right )}} \right )} + 20 \log {\left (x^{2} \log {\left (\log {\left (2 \right )} \right )} + x \right )} + 20 \log {\left (e^{x} + \frac {\log {\left (\log {\left (2 \right )} \right )}}{2 x \log {\left (\log {\left (2 \right )} \right )} + 2} \right )} \]
integrate((((-10*exp(x)*x**2-5*x)*ln(ln(2))-10*exp(x)*x)*ln(((2*exp(x)*x** 2+x)*ln(ln(2))+2*exp(x)*x)/ln(ln(2)))+((-10*x**3+20*x**2+80*x)*exp(x)-5*x+ 20)*ln(ln(2))+(-10*x**2+30*x+40)*exp(x))/((2*exp(x)*x**2+x)*ln(ln(2))+2*ex p(x)*x),x)
-5*x*log((2*x*exp(x) + (2*x**2*exp(x) + x)*log(log(2)))/log(log(2))) + 20* log(x**2*log(log(2)) + x) + 20*log(exp(x) + log(log(2))/(2*x*log(log(2)) + 2))
Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (20) = 40\).
Time = 0.32 (sec) , antiderivative size = 75, normalized size of antiderivative = 2.88 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=-5 \, x \log \left (2 \, {\left (x \log \left (\log \left (2\right )\right ) + 1\right )} e^{x} + \log \left (\log \left (2\right )\right )\right ) - 5 \, x \log \left (x\right ) + 5 \, x \log \left (\log \left (\log \left (2\right )\right )\right ) + 20 \, \log \left (x \log \left (\log \left (2\right )\right ) + 1\right ) + 20 \, \log \left (x\right ) + 20 \, \log \left (\frac {2 \, {\left (x \log \left (\log \left (2\right )\right ) + 1\right )} e^{x} + \log \left (\log \left (2\right )\right )}{2 \, {\left (x \log \left (\log \left (2\right )\right ) + 1\right )}}\right ) \]
integrate((((-10*exp(x)*x^2-5*x)*log(log(2))-10*exp(x)*x)*log(((2*exp(x)*x ^2+x)*log(log(2))+2*exp(x)*x)/log(log(2)))+((-10*x^3+20*x^2+80*x)*exp(x)-5 *x+20)*log(log(2))+(-10*x^2+30*x+40)*exp(x))/((2*exp(x)*x^2+x)*log(log(2)) +2*exp(x)*x),x, algorithm=\
-5*x*log(2*(x*log(log(2)) + 1)*e^x + log(log(2))) - 5*x*log(x) + 5*x*log(l og(log(2))) + 20*log(x*log(log(2)) + 1) + 20*log(x) + 20*log(1/2*(2*(x*log (log(2)) + 1)*e^x + log(log(2)))/(x*log(log(2)) + 1))
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (20) = 40\).
Time = 0.30 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=-5 \, x \log \left (2 \, x^{2} e^{x} \log \left (\log \left (2\right )\right ) + 2 \, x e^{x} + x \log \left (\log \left (2\right )\right )\right ) + 5 \, x \log \left (\log \left (\log \left (2\right )\right )\right ) + 20 \, \log \left (2 \, x e^{x} \log \left (\log \left (2\right )\right ) + 2 \, e^{x} + \log \left (\log \left (2\right )\right )\right ) + 20 \, \log \left (x\right ) \]
integrate((((-10*exp(x)*x^2-5*x)*log(log(2))-10*exp(x)*x)*log(((2*exp(x)*x ^2+x)*log(log(2))+2*exp(x)*x)/log(log(2)))+((-10*x^3+20*x^2+80*x)*exp(x)-5 *x+20)*log(log(2))+(-10*x^2+30*x+40)*exp(x))/((2*exp(x)*x^2+x)*log(log(2)) +2*exp(x)*x),x, algorithm=\
-5*x*log(2*x^2*e^x*log(log(2)) + 2*x*e^x + x*log(log(2))) + 5*x*log(log(lo g(2))) + 20*log(2*x*e^x*log(log(2)) + 2*e^x + log(log(2))) + 20*log(x)
Time = 10.06 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.73 \[ \int \frac {e^x \left (40+30 x-10 x^2\right )+\left (20-5 x+e^x \left (80 x+20 x^2-10 x^3\right )\right ) \log (\log (2))+\left (-10 e^x x+\left (-5 x-10 e^x x^2\right ) \log (\log (2))\right ) \log \left (\frac {2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))}{\log (\log (2))}\right )}{2 e^x x+\left (x+2 e^x x^2\right ) \log (\log (2))} \, dx=20\,\ln \left (\frac {\ln \left (\ln \left (2\right )\right )+2\,{\mathrm {e}}^x+2\,x\,{\mathrm {e}}^x\,\ln \left (\ln \left (2\right )\right )}{x\,\ln \left (\ln \left (2\right )\right )+1}\right )+20\,\ln \left (\ln \left (\ln \left (2\right )\right )\,x^2+x\right )-5\,x\,\ln \left (\frac {\ln \left (\ln \left (2\right )\right )\,\left (x+2\,x^2\,{\mathrm {e}}^x\right )+2\,x\,{\mathrm {e}}^x}{\ln \left (\ln \left (2\right )\right )}\right ) \]
int((log(log(2))*(exp(x)*(80*x + 20*x^2 - 10*x^3) - 5*x + 20) - log((log(l og(2))*(x + 2*x^2*exp(x)) + 2*x*exp(x))/log(log(2)))*(log(log(2))*(5*x + 1 0*x^2*exp(x)) + 10*x*exp(x)) + exp(x)*(30*x - 10*x^2 + 40))/(log(log(2))*( x + 2*x^2*exp(x)) + 2*x*exp(x)),x)