Integrand size = 43, antiderivative size = 24 \[ \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{5 x^3} \, dx=\frac {e \left (11-\frac {e^{\frac {2}{x}+x}}{x}\right )}{5 x} \] Output:
1/5*(11-exp(ln(exp(x)/x)+2/x))/x*exp(1)
Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.88 \[ \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{5 x^3} \, dx=-\frac {e \left (e^{\frac {2}{x}+x}-11 x\right )}{5 x^2} \] Input:
Integrate[(-11*E*x + E^(1 + (2 + x*Log[E^x/x])/x)*(2 + 2*x - x^2))/(5*x^3) ,x]
Output:
-1/5*(E*(E^(2/x + x) - 11*x))/x^2
Time = 0.26 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.71, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.093, Rules used = {27, 25, 2010, 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.
\(\displaystyle \int \frac {\left (-x^2+2 x+2\right ) e^{\frac {x \log \left (\frac {e^x}{x}\right )+2}{x}+1}-11 e x}{5 x^3} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {11 e x-e^{\frac {x \log \left (\frac {e^x}{x}\right )+2}{x}+1} \left (-x^2+2 x+2\right )}{x^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int \frac {11 e x-e^{\frac {x \log \left (\frac {e^x}{x}\right )+2}{x}+1} \left (-x^2+2 x+2\right )}{x^3}dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle -\frac {1}{5} \int \left (\frac {e^{x+1+\frac {2}{x}} \left (x^2-2 x-2\right )}{x^4}+\frac {11 e}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (\frac {e^{x+\frac {2}{x}+1} \left (2-x^2\right )}{\left (1-\frac {2}{x^2}\right ) x^4}+\frac {11 e}{x}\right )\) |
Input:
Int[(-11*E*x + E^(1 + (2 + x*Log[E^x/x])/x)*(2 + 2*x - x^2))/(5*x^3),x]
Output:
((11*E)/x + (E^(1 + 2/x + x)*(2 - x^2))/((1 - 2/x^2)*x^4))/5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
Time = 0.77 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29
method | result | size |
parallelrisch | \(\frac {-{\mathrm e} \,{\mathrm e}^{\frac {x \ln \left (\frac {{\mathrm e}^{x}}{x}\right )+2}{x}}+11 \,{\mathrm e}}{5 x}\) | \(31\) |
default | \(\frac {11 \,{\mathrm e}}{5 x}-\frac {{\mathrm e} \,{\mathrm e}^{\frac {x \ln \left (\frac {{\mathrm e}^{x}}{x}\right )+2}{x}}}{5 x}\) | \(32\) |
parts | \(\frac {11 \,{\mathrm e}}{5 x}-\frac {{\mathrm e} \,{\mathrm e}^{\frac {x \ln \left (\frac {{\mathrm e}^{x}}{x}\right )+2}{x}}}{5 x}\) | \(32\) |
orering | \(-\frac {\left (x^{4}-4 x^{3}+12 x +4\right ) \left (\left (-x^{2}+2 x +2\right ) {\mathrm e} \,{\mathrm e}^{\frac {x \ln \left (\frac {{\mathrm e}^{x}}{x}\right )+2}{x}}-11 x \,{\mathrm e}\right )}{5 x^{2} \left (x^{4}-2 x^{3}-2 x^{2}+8 x +4\right )}+\frac {\left (x^{2}-x -2\right ) x^{3} \left (\frac {\left (2-2 x \right ) {\mathrm e} \,{\mathrm e}^{\frac {x \ln \left (\frac {{\mathrm e}^{x}}{x}\right )+2}{x}}+\left (-x^{2}+2 x +2\right ) {\mathrm e} \left (\frac {\ln \left (\frac {{\mathrm e}^{x}}{x}\right )+x^{2} \left (\frac {{\mathrm e}^{x}}{x}-\frac {{\mathrm e}^{x}}{x^{2}}\right ) {\mathrm e}^{-x}}{x}-\frac {x \ln \left (\frac {{\mathrm e}^{x}}{x}\right )+2}{x^{2}}\right ) {\mathrm e}^{\frac {x \ln \left (\frac {{\mathrm e}^{x}}{x}\right )+2}{x}}-11 \,{\mathrm e}}{5 x^{3}}-\frac {3 \left (\left (-x^{2}+2 x +2\right ) {\mathrm e} \,{\mathrm e}^{\frac {x \ln \left (\frac {{\mathrm e}^{x}}{x}\right )+2}{x}}-11 x \,{\mathrm e}\right )}{5 x^{4}}\right )}{x^{4}-2 x^{3}-2 x^{2}+8 x +4}\) | \(262\) |
Input:
int(1/5*((-x^2+2*x+2)*exp(1)*exp((x*ln(exp(x)/x)+2)/x)-11*x*exp(1))/x^3,x, method=_RETURNVERBOSE)
Output:
1/5/x*(-exp(1)*exp((x*ln(exp(x)/x)+2)/x)+11*exp(1))
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{5 x^3} \, dx=\frac {11 \, e - e^{\left (\frac {x \log \left (\frac {e^{x}}{x}\right ) + x + 2}{x}\right )}}{5 \, x} \] Input:
integrate(1/5*((-x^2+2*x+2)*exp(1)*exp((x*log(exp(x)/x)+2)/x)-11*exp(1)*x) /x^3,x, algorithm="fricas")
Output:
1/5*(11*e - e^((x*log(e^x/x) + x + 2)/x))/x
\[ \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{5 x^3} \, dx=- \frac {e \left (\int \frac {11}{x^{2}}\, dx + \int \left (- \frac {2 e^{\frac {2}{x}} e^{x}}{x^{4}}\right )\, dx + \int \left (- \frac {2 e^{\frac {2}{x}} e^{x}}{x^{3}}\right )\, dx + \int \frac {e^{\frac {2}{x}} e^{x}}{x^{2}}\, dx\right )}{5} \] Input:
integrate(1/5*((-x**2+2*x+2)*exp(1)*exp((x*ln(exp(x)/x)+2)/x)-11*exp(1)*x) /x**3,x)
Output:
-E*(Integral(11/x**2, x) + Integral(-2*exp(2/x)*exp(x)/x**4, x) + Integral (-2*exp(2/x)*exp(x)/x**3, x) + Integral(exp(2/x)*exp(x)/x**2, x))/5
Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{5 x^3} \, dx=\frac {11 \, e}{5 \, x} - \frac {e^{\left (x + \frac {2}{x} + 1\right )}}{5 \, x^{2}} \] Input:
integrate(1/5*((-x^2+2*x+2)*exp(1)*exp((x*log(exp(x)/x)+2)/x)-11*exp(1)*x) /x^3,x, algorithm="maxima")
Output:
11/5*e/x - 1/5*e^(x + 2/x + 1)/x^2
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{5 x^3} \, dx=\frac {11 \, x e - e^{\left (\frac {x^{2} + x + 2}{x}\right )}}{5 \, x^{2}} \] Input:
integrate(1/5*((-x^2+2*x+2)*exp(1)*exp((x*log(exp(x)/x)+2)/x)-11*exp(1)*x) /x^3,x, algorithm="giac")
Output:
1/5*(11*x*e - e^((x^2 + x + 2)/x))/x^2
Timed out. \[ \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{5 x^3} \, dx=\int -\frac {\frac {11\,x\,\mathrm {e}}{5}-\frac {{\mathrm {e}}^{\frac {x\,\ln \left (\frac {{\mathrm {e}}^x}{x}\right )+2}{x}}\,\mathrm {e}\,\left (-x^2+2\,x+2\right )}{5}}{x^3} \,d x \] Input:
int(-((11*x*exp(1))/5 - (exp((x*log(exp(x)/x) + 2)/x)*exp(1)*(2*x - x^2 + 2))/5)/x^3,x)
Output:
int(-((11*x*exp(1))/5 - (exp((x*log(exp(x)/x) + 2)/x)*exp(1)*(2*x - x^2 + 2))/5)/x^3, x)
Time = 1.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {-11 e x+e^{1+\frac {2+x \log \left (\frac {e^x}{x}\right )}{x}} \left (2+2 x-x^2\right )}{5 x^3} \, dx=\frac {e \left (-e^{\frac {x^{2}+2}{x}}+11 x \right )}{5 x^{2}} \] Input:
int(1/5*((-x^2+2*x+2)*exp(1)*exp((x*log(exp(x)/x)+2)/x)-11*exp(1)*x)/x^3,x )
Output:
(e*( - e**((x**2 + 2)/x) + 11*x))/(5*x**2)