Integrand size = 68, antiderivative size = 33 \[ \int \frac {e^{-x} \left (-1+x+x^2+x^2 \log (2)+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{\left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )} \, dx=x+\frac {e^{-x} \left (1+\log (2)+\log \left (\frac {2}{x}\right )\right )}{(1-x) \log \left (5 e^2\right )} \]
Time = 5.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.18 \[ \int \frac {e^{-x} \left (-1+x+x^2+x^2 \log (2)+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{\left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )} \, dx=\frac {e^{-x} \left (-1+e^x (-1+x) x (2+\log (5))-\log \left (\frac {4}{x}\right )\right )}{(-1+x) (2+\log (5))} \]
Integrate[(-1 + x + x^2 + x^2*Log[2] + E^x*(x - 2*x^2 + x^3)*Log[5*E^2] + x^2*Log[2/x])/(E^x*(x - 2*x^2 + x^3)*Log[5*E^2]),x]
Time = 2.36 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.64, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6, 27, 25, 2026, 7277, 27, 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.
\(\displaystyle \int \frac {e^{-x} \left (x^2+x^2 \log \left (\frac {2}{x}\right )+x^2 \log (2)+e^x \left (x^3-2 x^2+x\right ) \log \left (5 e^2\right )+x-1\right )}{\left (x^3-2 x^2+x\right ) \log \left (5 e^2\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^{-x} \left (x^2 \log \left (\frac {2}{x}\right )+x^2 (1+\log (2))+e^x \left (x^3-2 x^2+x\right ) \log \left (5 e^2\right )+x-1\right )}{\left (x^3-2 x^2+x\right ) \log \left (5 e^2\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int -\frac {e^{-x} \left (-\log \left (\frac {2}{x}\right ) x^2-(1+\log (2)) x^2-x-e^x \left (x^3-2 x^2+x\right ) (2+\log (5))+1\right )}{x^3-2 x^2+x}dx}{2+\log (5)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {e^{-x} \left (-\log \left (\frac {2}{x}\right ) x^2-(1+\log (2)) x^2-x-e^x \left (x^3-2 x^2+x\right ) (2+\log (5))+1\right )}{x^3-2 x^2+x}dx}{2+\log (5)}\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle -\frac {\int \frac {e^{-x} \left (-\log \left (\frac {2}{x}\right ) x^2-(1+\log (2)) x^2-x-e^x \left (x^3-2 x^2+x\right ) (2+\log (5))+1\right )}{x \left (x^2-2 x+1\right )}dx}{2+\log (5)}\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle -\frac {4 \int \frac {e^{-x} \left (-\log \left (\frac {2}{x}\right ) x^2-(1+\log (2)) x^2-x-e^x \left (x^3-2 x^2+x\right ) (2+\log (5))+1\right )}{4 (1-x)^2 x}dx}{2+\log (5)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {e^{-x} \left (-\log \left (\frac {2}{x}\right ) x^2-(1+\log (2)) x^2-x-e^x \left (x^3-2 x^2+x\right ) (2+\log (5))+1\right )}{(1-x)^2 x}dx}{2+\log (5)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {\int \left (-\frac {e^{-x} \log \left (\frac {2}{x}\right ) x}{(x-1)^2}-\frac {e^{-x} (1+\log (2)) x}{(x-1)^2}-\frac {e^{-x}}{(x-1)^2}-\log (5)-2+\frac {e^{-x}}{(x-1)^2 x}\right )dx}{2+\log (5)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-(x (2+\log (5)))-\frac {e^{-x} \log \left (\frac {2}{x}\right )}{1-x}-\frac {e^{-x} (1+\log (2))}{1-x}}{2+\log (5)}\) |
Int[(-1 + x + x^2 + x^2*Log[2] + E^x*(x - 2*x^2 + x^3)*Log[5*E^2] + x^2*Lo g[2/x])/(E^x*(x - 2*x^2 + x^3)*Log[5*E^2]),x]
3.26.47.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
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]
Time = 0.43 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24
method | result | size |
default | \(\frac {\frac {\left (-1-\ln \left (2\right )-\ln \left (\frac {2}{x}\right )\right ) {\mathrm e}^{-x}}{-1+x}+x \ln \left (5 \,{\mathrm e}^{2}\right )}{\ln \left (5 \,{\mathrm e}^{2}\right )}\) | \(41\) |
parts | \(x +\frac {\left (\frac {\ln \left (x \right )}{\ln \left (5 \,{\mathrm e}^{2}\right )}-\frac {\ln \left (2\right )+\ln \left (\frac {2}{x}\right )+\ln \left (x \right )+1}{\ln \left (5 \,{\mathrm e}^{2}\right )}\right ) {\mathrm e}^{-x}}{-1+x}\) | \(45\) |
norman | \(\frac {\left (-{\mathrm e}^{x}+{\mathrm e}^{x} x^{2}-\frac {1+\ln \left (2\right )}{2+\ln \left (5\right )}-\frac {\ln \left (\frac {2}{x}\right )}{2+\ln \left (5\right )}\right ) {\mathrm e}^{-x}}{-1+x}\) | \(48\) |
parallelrisch | \(\frac {\left (\ln \left (5 \,{\mathrm e}^{2}\right ) {\mathrm e}^{x} x^{2}-1-{\mathrm e}^{x} \ln \left (5 \,{\mathrm e}^{2}\right )-\ln \left (2\right )-\ln \left (\frac {2}{x}\right )\right ) {\mathrm e}^{-x}}{\ln \left (5 \,{\mathrm e}^{2}\right ) \left (-1+x \right )}\) | \(52\) |
risch | \(\frac {{\mathrm e}^{-x} \ln \left (x \right )}{\left (2+\ln \left (5\right )\right ) \left (-1+x \right )}-\frac {\left (2-2 x^{2} \ln \left (5\right ) {\mathrm e}^{x}+2 x \,{\mathrm e}^{x} \ln \left (5\right )-4 \,{\mathrm e}^{x} x^{2}+4 \,{\mathrm e}^{x} x +4 \ln \left (2\right )\right ) {\mathrm e}^{-x}}{2 \left (2+\ln \left (5\right )\right ) \left (-1+x \right )}\) | \(71\) |
int(((x^3-2*x^2+x)*exp(x)*ln(5*exp(2))+x^2*ln(2/x)+x^2*ln(2)+x^2+x-1)/(x^3 -2*x^2+x)/exp(x)/ln(5*exp(2)),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.64 \[ \int \frac {e^{-x} \left (-1+x+x^2+x^2 \log (2)+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{\left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )} \, dx=\frac {{\left ({\left (2 \, x^{2} + {\left (x^{2} - x\right )} \log \left (5\right ) - 2 \, x\right )} e^{x} - \log \left (2\right ) - \log \left (\frac {2}{x}\right ) - 1\right )} e^{\left (-x\right )}}{{\left (x - 1\right )} \log \left (5\right ) + 2 \, x - 2} \]
integrate(((x^3-2*x^2+x)*exp(x)*log(5*exp(2))+x^2*log(2/x)+x^2*log(2)+x^2+ x-1)/(x^3-2*x^2+x)/exp(x)/log(5*exp(2)),x, algorithm=\
((2*x^2 + (x^2 - x)*log(5) - 2*x)*e^x - log(2) - log(2/x) - 1)*e^(-x)/((x - 1)*log(5) + 2*x - 2)
Time = 0.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-x} \left (-1+x+x^2+x^2 \log (2)+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{\left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )} \, dx=x + \frac {\left (- \log {\left (\frac {2}{x} \right )} - 1 - \log {\left (2 \right )}\right ) e^{- x}}{x \log {\left (5 \right )} + 2 x - 2 - \log {\left (5 \right )}} \]
integrate(((x**3-2*x**2+x)*exp(x)*ln(5*exp(2))+x**2*ln(2/x)+x**2*ln(2)+x** 2+x-1)/(x**3-2*x**2+x)/exp(x)/ln(5*exp(2)),x)
Time = 0.32 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.36 \[ \int \frac {e^{-x} \left (-1+x+x^2+x^2 \log (2)+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{\left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )} \, dx=\frac {x^{2} {\left (\log \left (5\right ) + 2\right )} - x {\left (\log \left (5\right ) + 2\right )} - {\left (2 \, \log \left (2\right ) - \log \left (x\right ) + 1\right )} e^{\left (-x\right )}}{{\left (x - 1\right )} \log \left (5 \, e^{2}\right )} \]
integrate(((x^3-2*x^2+x)*exp(x)*log(5*exp(2))+x^2*log(2/x)+x^2*log(2)+x^2+ x-1)/(x^3-2*x^2+x)/exp(x)/log(5*exp(2)),x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.64 \[ \int \frac {e^{-x} \left (-1+x+x^2+x^2 \log (2)+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{\left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )} \, dx=\frac {x^{2} \log \left (5\right ) + 2 \, x^{2} - x \log \left (5\right ) - 2 \, e^{\left (-x\right )} \log \left (2\right ) + e^{\left (-x\right )} \log \left (x\right ) - 2 \, x - e^{\left (-x\right )}}{{\left (x - 1\right )} \log \left (5 \, e^{2}\right )} \]
integrate(((x^3-2*x^2+x)*exp(x)*log(5*exp(2))+x^2*log(2/x)+x^2*log(2)+x^2+ x-1)/(x^3-2*x^2+x)/exp(x)/log(5*exp(2)),x, algorithm=\
(x^2*log(5) + 2*x^2 - x*log(5) - 2*e^(-x)*log(2) + e^(-x)*log(x) - 2*x - e ^(-x))/((x - 1)*log(5*e^2))
Timed out. \[ \int \frac {e^{-x} \left (-1+x+x^2+x^2 \log (2)+e^x \left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )+x^2 \log \left (\frac {2}{x}\right )\right )}{\left (x-2 x^2+x^3\right ) \log \left (5 e^2\right )} \, dx=\int \frac {{\mathrm {e}}^{-x}\,\left (x+x^2\,\ln \left (2\right )+x^2+x^2\,\ln \left (\frac {2}{x}\right )+\ln \left (5\,{\mathrm {e}}^2\right )\,{\mathrm {e}}^x\,\left (x^3-2\,x^2+x\right )-1\right )}{\ln \left (5\,{\mathrm {e}}^2\right )\,\left (x^3-2\,x^2+x\right )} \,d x \]
int((exp(-x)*(x + x^2*log(2) + x^2 + x^2*log(2/x) + log(5*exp(2))*exp(x)*( x - 2*x^2 + x^3) - 1))/(log(5*exp(2))*(x - 2*x^2 + x^3)),x)