Integrand size = 64, antiderivative size = 28 \[ \int \frac {e^{8-2 x} \left (1+x^2\right )+e^{4-x} (-2-x) \log (4)+\log ^2(4)}{e^{8-2 x} x^2-2 e^{4-x} x \log (4)+\log ^2(4)} \, dx=-4+\frac {1-x}{-x+e^{-4+x} \log (4)}+\log \left (e^{2+x}\right ) \] Output:
ln(exp(2)*exp(x))-4+(1-x)/(2*ln(2)/exp(4-x)-x)
Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.86 \[ \int \frac {e^{8-2 x} \left (1+x^2\right )+e^{4-x} (-2-x) \log (4)+\log ^2(4)}{e^{8-2 x} x^2-2 e^{4-x} x \log (4)+\log ^2(4)} \, dx=x+\frac {e^4 (-1+x)}{e^4 x-e^x \log (4)} \] Input:
Integrate[(E^(8 - 2*x)*(1 + x^2) + E^(4 - x)*(-2 - x)*Log[4] + Log[4]^2)/( E^(8 - 2*x)*x^2 - 2*E^(4 - x)*x*Log[4] + Log[4]^2),x]
Output:
x + (E^4*(-1 + x))/(E^4*x - E^x*Log[4])
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.
\(\displaystyle \int \frac {e^{8-2 x} \left (x^2+1\right )+e^{4-x} (-x-2) \log (4)+\log ^2(4)}{e^{8-2 x} x^2-2 e^{4-x} x \log (4)+\log ^2(4)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 x} \left (e^{8-2 x} \left (x^2+1\right )+e^{4-x} (-x-2) \log (4)+\log ^2(4)\right )}{\left (e^4 x-e^x \log (4)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^2+1}{x^2}+\frac {e^{2 x} (x-1)^2 \log ^2(4)}{x^2 \left (e^4 x-e^x \log (4)\right )^2}+\frac {e^{2 x-4} \left (x^2-2 x+2\right ) \log ^2(4)}{x^3 \left (e^4 x-e^x \log (4)\right )}+\frac {e^{x-4} \left (x^2-2 x+2\right ) \log (4)}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \log ^2(4) \int \frac {e^{2 x-4}}{x^3 \left (e^4 x-e^x \log (4)\right )}dx+\log ^2(4) \int \frac {e^{2 x}}{x^2 \left (e^4 x-e^x \log (4)\right )^2}dx-2 \log ^2(4) \int \frac {e^{2 x-4}}{x^2 \left (e^4 x-e^x \log (4)\right )}dx+\log ^2(4) \int \frac {e^{2 x}}{\left (e^4 x-e^x \log (4)\right )^2}dx-2 \log ^2(4) \int \frac {e^{2 x}}{x \left (e^4 x-e^x \log (4)\right )^2}dx+\log ^2(4) \int \frac {e^{2 x-4}}{x \left (e^4 x-e^x \log (4)\right )}dx-\frac {e^{x-4} \log (4)}{x^2}+x-\frac {1}{x}+\frac {e^{x-4} \log (4)}{x}\) |
Input:
Int[(E^(8 - 2*x)*(1 + x^2) + E^(4 - x)*(-2 - x)*Log[4] + Log[4]^2)/(E^(8 - 2*x)*x^2 - 2*E^(4 - x)*x*Log[4] + Log[4]^2),x]
Output:
$Aborted
Time = 0.30 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21
method | result | size |
risch | \(x -\frac {1}{x}-\frac {2 \left (-1+x \right ) \ln \left (2\right )}{x \left (-x \,{\mathrm e}^{-x +4}+2 \ln \left (2\right )\right )}\) | \(34\) |
norman | \(\frac {2 x \ln \left (2\right )-x^{2} {\mathrm e}^{-x +4}-2 \ln \left (2\right )+{\mathrm e}^{-x +4}}{-x \,{\mathrm e}^{-x +4}+2 \ln \left (2\right )}\) | \(45\) |
parallelrisch | \(\frac {2 x \ln \left (2\right )-x^{2} {\mathrm e}^{-x +4}-2 \ln \left (2\right )+{\mathrm e}^{-x +4}}{-x \,{\mathrm e}^{-x +4}+2 \ln \left (2\right )}\) | \(45\) |
Input:
int(((x^2+1)*exp(-x+4)^2+2*(-2-x)*ln(2)*exp(-x+4)+4*ln(2)^2)/(x^2*exp(-x+4 )^2-4*x*ln(2)*exp(-x+4)+4*ln(2)^2),x,method=_RETURNVERBOSE)
Output:
x-1/x-2*(-1+x)*ln(2)/x/(-x*exp(-x+4)+2*ln(2))
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^{8-2 x} \left (1+x^2\right )+e^{4-x} (-2-x) \log (4)+\log ^2(4)}{e^{8-2 x} x^2-2 e^{4-x} x \log (4)+\log ^2(4)} \, dx=\frac {{\left (x^{2} - 1\right )} e^{\left (-x + 4\right )} - 2 \, {\left (x - 1\right )} \log \left (2\right )}{x e^{\left (-x + 4\right )} - 2 \, \log \left (2\right )} \] Input:
integrate(((x^2+1)*exp(-x+4)^2+2*(-2-x)*log(2)*exp(-x+4)+4*log(2)^2)/(x^2* exp(-x+4)^2-4*x*log(2)*exp(-x+4)+4*log(2)^2),x, algorithm="fricas")
Output:
((x^2 - 1)*e^(-x + 4) - 2*(x - 1)*log(2))/(x*e^(-x + 4) - 2*log(2))
Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11 \[ \int \frac {e^{8-2 x} \left (1+x^2\right )+e^{4-x} (-2-x) \log (4)+\log ^2(4)}{e^{8-2 x} x^2-2 e^{4-x} x \log (4)+\log ^2(4)} \, dx=x + \frac {2 x \log {\left (2 \right )} - 2 \log {\left (2 \right )}}{x^{2} e^{4 - x} - 2 x \log {\left (2 \right )}} - \frac {1}{x} \] Input:
integrate(((x**2+1)*exp(-x+4)**2+2*(-2-x)*ln(2)*exp(-x+4)+4*ln(2)**2)/(x** 2*exp(-x+4)**2-4*x*ln(2)*exp(-x+4)+4*ln(2)**2),x)
Output:
x + (2*x*log(2) - 2*log(2))/(x**2*exp(4 - x) - 2*x*log(2)) - 1/x
Time = 0.16 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.29 \[ \int \frac {e^{8-2 x} \left (1+x^2\right )+e^{4-x} (-2-x) \log (4)+\log ^2(4)}{e^{8-2 x} x^2-2 e^{4-x} x \log (4)+\log ^2(4)} \, dx=\frac {x^{2} e^{4} - 2 \, x e^{x} \log \left (2\right ) + x e^{4} - e^{4}}{x e^{4} - 2 \, e^{x} \log \left (2\right )} \] Input:
integrate(((x^2+1)*exp(-x+4)^2+2*(-2-x)*log(2)*exp(-x+4)+4*log(2)^2)/(x^2* exp(-x+4)^2-4*x*log(2)*exp(-x+4)+4*log(2)^2),x, algorithm="maxima")
Output:
(x^2*e^4 - 2*x*e^x*log(2) + x*e^4 - e^4)/(x*e^4 - 2*e^x*log(2))
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {e^{8-2 x} \left (1+x^2\right )+e^{4-x} (-2-x) \log (4)+\log ^2(4)}{e^{8-2 x} x^2-2 e^{4-x} x \log (4)+\log ^2(4)} \, dx=\frac {x^{2} e^{\left (-x + 4\right )} - 2 \, x \log \left (2\right ) - e^{\left (-x + 4\right )} + 2 \, \log \left (2\right )}{x e^{\left (-x + 4\right )} - 2 \, \log \left (2\right )} \] Input:
integrate(((x^2+1)*exp(-x+4)^2+2*(-2-x)*log(2)*exp(-x+4)+4*log(2)^2)/(x^2* exp(-x+4)^2-4*x*log(2)*exp(-x+4)+4*log(2)^2),x, algorithm="giac")
Output:
(x^2*e^(-x + 4) - 2*x*log(2) - e^(-x + 4) + 2*log(2))/(x*e^(-x + 4) - 2*lo g(2))
Time = 0.11 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {e^{8-2 x} \left (1+x^2\right )+e^{4-x} (-2-x) \log (4)+\log ^2(4)}{e^{8-2 x} x^2-2 e^{4-x} x \log (4)+\log ^2(4)} \, dx=\frac {x+x^2-x\,{\mathrm {e}}^{x-4}\,\ln \left (4\right )-1}{x-2\,{\mathrm {e}}^{x-4}\,\ln \left (2\right )} \] Input:
int((exp(8 - 2*x)*(x^2 + 1) + 4*log(2)^2 - 2*exp(4 - x)*log(2)*(x + 2))/(4 *log(2)^2 + x^2*exp(8 - 2*x) - 4*x*exp(4 - x)*log(2)),x)
Output:
(x + x^2 - x*exp(x - 4)*log(4) - 1)/(x - 2*exp(x - 4)*log(2))
Time = 0.17 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.57 \[ \int \frac {e^{8-2 x} \left (1+x^2\right )+e^{4-x} (-2-x) \log (4)+\log ^2(4)}{e^{8-2 x} x^2-2 e^{4-x} x \log (4)+\log ^2(4)} \, dx=\frac {2 e^{x} \mathrm {log}\left (2\right ) x -2 e^{x} \mathrm {log}\left (2\right )-e^{4} x^{2}+e^{4}}{2 e^{x} \mathrm {log}\left (2\right )-e^{4} x} \] Input:
int(((x^2+1)*exp(-x+4)^2+2*(-2-x)*log(2)*exp(-x+4)+4*log(2)^2)/(x^2*exp(-x +4)^2-4*x*log(2)*exp(-x+4)+4*log(2)^2),x)
Output:
(2*e**x*log(2)*x - 2*e**x*log(2) - e**4*x**2 + e**4)/(2*e**x*log(2) - e**4 *x)