Integrand size = 115, antiderivative size = 38 \[ \int \frac {e^x \left (4 x-8 x^2\right )+e \left (-2 x^2-x^3\right )+\left (-4 e^x x+e \left (-4 x-2 x^2\right )\right ) \log \left (\frac {2}{x}\right )+e (-2-x) \log ^2\left (\frac {2}{x}\right )}{5 e^{1+x} x^5+10 e^{1+x} x^4 \log \left (\frac {2}{x}\right )+5 e^{1+x} x^3 \log ^2\left (\frac {2}{x}\right )} \, dx=\frac {2+\frac {\frac {e^{1-x}}{x}+x+\frac {4}{x+\log \left (\frac {2}{x}\right )}}{5 x}}{e} \] Output:
(2+1/5*(x+4/(x+ln(2/x))+exp(1)/exp(x)/x)/x)/exp(1)
Time = 5.05 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.84 \[ \int \frac {e^x \left (4 x-8 x^2\right )+e \left (-2 x^2-x^3\right )+\left (-4 e^x x+e \left (-4 x-2 x^2\right )\right ) \log \left (\frac {2}{x}\right )+e (-2-x) \log ^2\left (\frac {2}{x}\right )}{5 e^{1+x} x^5+10 e^{1+x} x^4 \log \left (\frac {2}{x}\right )+5 e^{1+x} x^3 \log ^2\left (\frac {2}{x}\right )} \, dx=\frac {1}{5} \left (\frac {e^{-x}}{x^2}+\frac {4}{e x \left (x+\log \left (\frac {2}{x}\right )\right )}\right ) \] Input:
Integrate[(E^x*(4*x - 8*x^2) + E*(-2*x^2 - x^3) + (-4*E^x*x + E*(-4*x - 2* x^2))*Log[2/x] + E*(-2 - x)*Log[2/x]^2)/(5*E^(1 + x)*x^5 + 10*E^(1 + x)*x^ 4*Log[2/x] + 5*E^(1 + x)*x^3*Log[2/x]^2),x]
Output:
(1/(E^x*x^2) + 4/(E*x*(x + Log[2/x])))/5
Time = 2.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.84, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {7292, 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.
\(\displaystyle \int \frac {e^x \left (4 x-8 x^2\right )+\left (e \left (-2 x^2-4 x\right )-4 e^x x\right ) \log \left (\frac {2}{x}\right )+e \left (-x^3-2 x^2\right )+e (-x-2) \log ^2\left (\frac {2}{x}\right )}{5 e^{x+1} x^5+10 e^{x+1} x^4 \log \left (\frac {2}{x}\right )+5 e^{x+1} x^3 \log ^2\left (\frac {2}{x}\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-x-1} \left (e^x \left (4 x-8 x^2\right )+\left (e \left (-2 x^2-4 x\right )-4 e^x x\right ) \log \left (\frac {2}{x}\right )+e \left (-x^3-2 x^2\right )+e (-x-2) \log ^2\left (\frac {2}{x}\right )\right )}{5 x^3 \left (x+\log \left (\frac {2}{x}\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int \frac {e^{-x-1} \left (-e (x+2) \log ^2\left (\frac {2}{x}\right )-2 \left (2 e^x x+e \left (x^2+2 x\right )\right ) \log \left (\frac {2}{x}\right )+4 e^x \left (x-2 x^2\right )-e \left (x^3+2 x^2\right )\right )}{x^3 \left (x+\log \left (\frac {2}{x}\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{5} \int \left (-\frac {e^{-x} (x+2)}{x^3}-\frac {4 \left (2 x+\log \left (\frac {2}{x}\right )-1\right )}{e x^2 \left (x+\log \left (\frac {2}{x}\right )\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} \left (\frac {e^{-x}}{x^2}+\frac {4}{e x \left (x+\log \left (\frac {2}{x}\right )\right )}\right )\) |
Input:
Int[(E^x*(4*x - 8*x^2) + E*(-2*x^2 - x^3) + (-4*E^x*x + E*(-4*x - 2*x^2))* Log[2/x] + E*(-2 - x)*Log[2/x]^2)/(5*E^(1 + x)*x^5 + 10*E^(1 + x)*x^4*Log[ 2/x] + 5*E^(1 + x)*x^3*Log[2/x]^2),x]
Output:
(1/(E^x*x^2) + 4/(E*x*(x + Log[2/x])))/5
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 3.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.84
method | result | size |
risch | \(\frac {{\mathrm e}^{-x}}{5 x^{2}}+\frac {8 \,{\mathrm e}^{-1}}{5 x \left (2 \ln \left (2\right )+2 x -2 \ln \left (x \right )\right )}\) | \(32\) |
parallelrisch | \(\frac {\left (x \,{\mathrm e}+{\mathrm e} \ln \left (\frac {2}{x}\right )+4 \,{\mathrm e}^{x} x \right ) {\mathrm e}^{-x} {\mathrm e}^{-1}}{5 x^{2} \left (x +\ln \left (\frac {2}{x}\right )\right )}\) | \(43\) |
Input:
int(((-2-x)*exp(1)*ln(2/x)^2+(-4*exp(x)*x+(-2*x^2-4*x)*exp(1))*ln(2/x)+(-8 *x^2+4*x)*exp(x)+(-x^3-2*x^2)*exp(1))/(5*x^3*exp(1)*exp(x)*ln(2/x)^2+10*x^ 4*exp(1)*exp(x)*ln(2/x)+5*x^5*exp(1)*exp(x)),x,method=_RETURNVERBOSE)
Output:
1/5*exp(-x)/x^2+8/5/x*exp(-1)/(2*ln(2)+2*x-2*ln(x))
Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int \frac {e^x \left (4 x-8 x^2\right )+e \left (-2 x^2-x^3\right )+\left (-4 e^x x+e \left (-4 x-2 x^2\right )\right ) \log \left (\frac {2}{x}\right )+e (-2-x) \log ^2\left (\frac {2}{x}\right )}{5 e^{1+x} x^5+10 e^{1+x} x^4 \log \left (\frac {2}{x}\right )+5 e^{1+x} x^3 \log ^2\left (\frac {2}{x}\right )} \, dx=\frac {x e^{2} + 4 \, x e^{\left (x + 1\right )} + e^{2} \log \left (\frac {2}{x}\right )}{5 \, {\left (x^{3} e^{\left (x + 2\right )} + x^{2} e^{\left (x + 2\right )} \log \left (\frac {2}{x}\right )\right )}} \] Input:
integrate(((-2-x)*exp(1)*log(2/x)^2+(-4*exp(x)*x+(-2*x^2-4*x)*exp(1))*log( 2/x)+(-8*x^2+4*x)*exp(x)+(-x^3-2*x^2)*exp(1))/(5*x^3*exp(1)*exp(x)*log(2/x )^2+10*x^4*exp(1)*exp(x)*log(2/x)+5*x^5*exp(1)*exp(x)),x, algorithm="frica s")
Output:
1/5*(x*e^2 + 4*x*e^(x + 1) + e^2*log(2/x))/(x^3*e^(x + 2) + x^2*e^(x + 2)* log(2/x))
Time = 0.13 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {e^x \left (4 x-8 x^2\right )+e \left (-2 x^2-x^3\right )+\left (-4 e^x x+e \left (-4 x-2 x^2\right )\right ) \log \left (\frac {2}{x}\right )+e (-2-x) \log ^2\left (\frac {2}{x}\right )}{5 e^{1+x} x^5+10 e^{1+x} x^4 \log \left (\frac {2}{x}\right )+5 e^{1+x} x^3 \log ^2\left (\frac {2}{x}\right )} \, dx=\frac {4}{5 e x^{2} + 5 e x \log {\left (\frac {2}{x} \right )}} + \frac {e^{- x}}{5 x^{2}} \] Input:
integrate(((-2-x)*exp(1)*ln(2/x)**2+(-4*exp(x)*x+(-2*x**2-4*x)*exp(1))*ln( 2/x)+(-8*x**2+4*x)*exp(x)+(-x**3-2*x**2)*exp(1))/(5*x**3*exp(1)*exp(x)*ln( 2/x)**2+10*x**4*exp(1)*exp(x)*ln(2/x)+5*x**5*exp(1)*exp(x)),x)
Output:
4/(5*E*x**2 + 5*E*x*log(2/x)) + exp(-x)/(5*x**2)
Time = 0.23 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.39 \[ \int \frac {e^x \left (4 x-8 x^2\right )+e \left (-2 x^2-x^3\right )+\left (-4 e^x x+e \left (-4 x-2 x^2\right )\right ) \log \left (\frac {2}{x}\right )+e (-2-x) \log ^2\left (\frac {2}{x}\right )}{5 e^{1+x} x^5+10 e^{1+x} x^4 \log \left (\frac {2}{x}\right )+5 e^{1+x} x^3 \log ^2\left (\frac {2}{x}\right )} \, dx=\frac {{\left (x e + e \log \left (2\right ) - e \log \left (x\right )\right )} e^{\left (-x\right )} + 4 \, x}{5 \, {\left (x^{3} e + x^{2} e \log \left (2\right ) - x^{2} e \log \left (x\right )\right )}} \] Input:
integrate(((-2-x)*exp(1)*log(2/x)^2+(-4*exp(x)*x+(-2*x^2-4*x)*exp(1))*log( 2/x)+(-8*x^2+4*x)*exp(x)+(-x^3-2*x^2)*exp(1))/(5*x^3*exp(1)*exp(x)*log(2/x )^2+10*x^4*exp(1)*exp(x)*log(2/x)+5*x^5*exp(1)*exp(x)),x, algorithm="maxim a")
Output:
1/5*((x*e + e*log(2) - e*log(x))*e^(-x) + 4*x)/(x^3*e + x^2*e*log(2) - x^2 *e*log(x))
Time = 0.13 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.55 \[ \int \frac {e^x \left (4 x-8 x^2\right )+e \left (-2 x^2-x^3\right )+\left (-4 e^x x+e \left (-4 x-2 x^2\right )\right ) \log \left (\frac {2}{x}\right )+e (-2-x) \log ^2\left (\frac {2}{x}\right )}{5 e^{1+x} x^5+10 e^{1+x} x^4 \log \left (\frac {2}{x}\right )+5 e^{1+x} x^3 \log ^2\left (\frac {2}{x}\right )} \, dx=\frac {x e^{\left (-x + 1\right )} + e^{\left (-x + 1\right )} \log \left (2\right ) - e^{\left (-x + 1\right )} \log \left (x\right ) + 4 \, x}{5 \, {\left (x^{3} e + x^{2} e \log \left (2\right ) - x^{2} e \log \left (x\right )\right )}} \] Input:
integrate(((-2-x)*exp(1)*log(2/x)^2+(-4*exp(x)*x+(-2*x^2-4*x)*exp(1))*log( 2/x)+(-8*x^2+4*x)*exp(x)+(-x^3-2*x^2)*exp(1))/(5*x^3*exp(1)*exp(x)*log(2/x )^2+10*x^4*exp(1)*exp(x)*log(2/x)+5*x^5*exp(1)*exp(x)),x, algorithm="giac" )
Output:
1/5*(x*e^(-x + 1) + e^(-x + 1)*log(2) - e^(-x + 1)*log(x) + 4*x)/(x^3*e + x^2*e*log(2) - x^2*e*log(x))
Time = 4.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.32 \[ \int \frac {e^x \left (4 x-8 x^2\right )+e \left (-2 x^2-x^3\right )+\left (-4 e^x x+e \left (-4 x-2 x^2\right )\right ) \log \left (\frac {2}{x}\right )+e (-2-x) \log ^2\left (\frac {2}{x}\right )}{5 e^{1+x} x^5+10 e^{1+x} x^4 \log \left (\frac {2}{x}\right )+5 e^{1+x} x^3 \log ^2\left (\frac {2}{x}\right )} \, dx=\frac {x^2\,\left (\mathrm {e}+4\,{\mathrm {e}}^x\right )+x\,\mathrm {e}\,\ln \left (\frac {2}{x}\right )}{5\,x^4\,{\mathrm {e}}^{x+1}+5\,x^3\,{\mathrm {e}}^{x+1}\,\ln \left (\frac {2}{x}\right )} \] Input:
int(-(exp(1)*(2*x^2 + x^3) + log(2/x)*(exp(1)*(4*x + 2*x^2) + 4*x*exp(x)) - exp(x)*(4*x - 8*x^2) + exp(1)*log(2/x)^2*(x + 2))/(5*x^5*exp(1)*exp(x) + 10*x^4*exp(1)*exp(x)*log(2/x) + 5*x^3*exp(1)*exp(x)*log(2/x)^2),x)
Output:
(x^2*(exp(1) + 4*exp(x)) + x*exp(1)*log(2/x))/(5*x^4*exp(x + 1) + 5*x^3*ex p(x + 1)*log(2/x))
Time = 0.17 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.08 \[ \int \frac {e^x \left (4 x-8 x^2\right )+e \left (-2 x^2-x^3\right )+\left (-4 e^x x+e \left (-4 x-2 x^2\right )\right ) \log \left (\frac {2}{x}\right )+e (-2-x) \log ^2\left (\frac {2}{x}\right )}{5 e^{1+x} x^5+10 e^{1+x} x^4 \log \left (\frac {2}{x}\right )+5 e^{1+x} x^3 \log ^2\left (\frac {2}{x}\right )} \, dx=\frac {4 e^{x} x +\mathrm {log}\left (\frac {2}{x}\right ) e +e x}{5 e^{x} e \,x^{2} \left (\mathrm {log}\left (\frac {2}{x}\right )+x \right )} \] Input:
int(((-2-x)*exp(1)*log(2/x)^2+(-4*exp(x)*x+(-2*x^2-4*x)*exp(1))*log(2/x)+( -8*x^2+4*x)*exp(x)+(-x^3-2*x^2)*exp(1))/(5*x^3*exp(1)*exp(x)*log(2/x)^2+10 *x^4*exp(1)*exp(x)*log(2/x)+5*x^5*exp(1)*exp(x)),x)
Output:
(4*e**x*x + log(2/x)*e + e*x)/(5*e**x*e*x**2*(log(2/x) + x))