Integrand size = 76, antiderivative size = 25 \[ \int \frac {e^{\frac {2-\log \left (\frac {(-4+x) \log (4)}{16 x}\right )}{-2+x}} \left (8+4 x-2 x^2+\left (-4 x+x^2\right ) \log \left (\frac {(-4+x) \log (4)}{16 x}\right )\right )}{-16 x+20 x^2-8 x^3+x^4} \, dx=e^{\frac {2-\log \left (\frac {(-4+x) \log (4)}{16 x}\right )}{-2+x}} \] Output:
exp((-ln(1/8*(-4+x)*ln(2)/x)+2)/(-2+x))
Time = 5.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {e^{\frac {2-\log \left (\frac {(-4+x) \log (4)}{16 x}\right )}{-2+x}} \left (8+4 x-2 x^2+\left (-4 x+x^2\right ) \log \left (\frac {(-4+x) \log (4)}{16 x}\right )\right )}{-16 x+20 x^2-8 x^3+x^4} \, dx=e^{\frac {2}{-2+x}} \left (\frac {-4+x}{x}\right )^{-\frac {1}{-2+x}} \left (\frac {16}{\log (4)}\right )^{\frac {1}{-2+x}} \] Input:
Integrate[(E^((2 - Log[((-4 + x)*Log[4])/(16*x)])/(-2 + x))*(8 + 4*x - 2*x ^2 + (-4*x + x^2)*Log[((-4 + x)*Log[4])/(16*x)]))/(-16*x + 20*x^2 - 8*x^3 + x^4),x]
Output:
(E^(2/(-2 + x))*(16/Log[4])^(-2 + x)^(-1))/((-4 + x)/x)^(-2 + x)^(-1)
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.
\(\displaystyle \int \frac {e^{\frac {2-\log \left (\frac {(x-4) \log (4)}{16 x}\right )}{x-2}} \left (-2 x^2+\left (x^2-4 x\right ) \log \left (\frac {(x-4) \log (4)}{16 x}\right )+4 x+8\right )}{x^4-8 x^3+20 x^2-16 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{\frac {2-\log \left (\frac {(x-4) \log (4)}{16 x}\right )}{x-2}} \left (-2 x^2+\left (x^2-4 x\right ) \log \left (\frac {(x-4) \log (4)}{16 x}\right )+4 x+8\right )}{x \left (x^3-8 x^2+20 x-16\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {e^{\frac {2-\log \left (\frac {(x-4) \log (4)}{16 x}\right )}{x-2}} \left (-2 x^2+\left (x^2-4 x\right ) \log \left (\frac {(x-4) \log (4)}{16 x}\right )+4 x+8\right )}{4 (x-4) x}-\frac {e^{\frac {2-\log \left (\frac {(x-4) \log (4)}{16 x}\right )}{x-2}} \left (-2 x^2+\left (x^2-4 x\right ) \log \left (\frac {(x-4) \log (4)}{16 x}\right )+4 x+8\right )}{4 (x-2) x}-\frac {e^{\frac {2-\log \left (\frac {(x-4) \log (4)}{16 x}\right )}{x-2}} \left (-2 x^2+\left (x^2-4 x\right ) \log \left (\frac {(x-4) \log (4)}{16 x}\right )+4 x+8\right )}{2 (x-2)^2 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{2} \int \frac {e^{\frac {2-\log \left (\frac {(x-4) \log (4)}{16 x}\right )}{x-2}}}{x-4}dx-2 \int \frac {e^{\frac {2-\log \left (\frac {(x-4) \log (4)}{16 x}\right )}{x-2}}}{(x-2)^2}dx+\int \frac {e^{\frac {2-\log \left (\frac {(x-4) \log (4)}{16 x}\right )}{x-2}}}{x-2}dx-\frac {1}{2} \int \frac {e^{\frac {2-\log \left (\frac {(x-4) \log (4)}{16 x}\right )}{x-2}}}{x}dx+\int \frac {e^{\frac {2-\log \left (\frac {(x-4) \log (4)}{16 x}\right )}{x-2}} \log \left (\frac {\log (4)}{16}-\frac {\log (4)}{4 x}\right )}{(x-2)^2}dx\) |
Input:
Int[(E^((2 - Log[((-4 + x)*Log[4])/(16*x)])/(-2 + x))*(8 + 4*x - 2*x^2 + ( -4*x + x^2)*Log[((-4 + x)*Log[4])/(16*x)]))/(-16*x + 20*x^2 - 8*x^3 + x^4) ,x]
Output:
$Aborted
Time = 0.54 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
risch | \({\mathrm e}^{-\frac {\ln \left (\frac {\left (x -4\right ) \ln \left (2\right )}{8 x}\right )-2}{-2+x}}\) | \(22\) |
parallelrisch | \({\mathrm e}^{-\frac {\ln \left (\frac {\left (x -4\right ) \ln \left (2\right )}{8 x}\right )-2}{-2+x}}\) | \(22\) |
norman | \(\frac {x \,{\mathrm e}^{\frac {-\ln \left (\frac {\left (x -4\right ) \ln \left (2\right )}{8 x}\right )+2}{-2+x}}-2 \,{\mathrm e}^{\frac {-\ln \left (\frac {\left (x -4\right ) \ln \left (2\right )}{8 x}\right )+2}{-2+x}}}{-2+x}\) | \(56\) |
Input:
int(((x^2-4*x)*ln(1/8*(x-4)*ln(2)/x)-2*x^2+4*x+8)*exp((-ln(1/8*(x-4)*ln(2) /x)+2)/(-2+x))/(x^4-8*x^3+20*x^2-16*x),x,method=_RETURNVERBOSE)
Output:
exp(-(ln(1/8*(x-4)*ln(2)/x)-2)/(-2+x))
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{\frac {2-\log \left (\frac {(-4+x) \log (4)}{16 x}\right )}{-2+x}} \left (8+4 x-2 x^2+\left (-4 x+x^2\right ) \log \left (\frac {(-4+x) \log (4)}{16 x}\right )\right )}{-16 x+20 x^2-8 x^3+x^4} \, dx=e^{\left (-\frac {\log \left (\frac {{\left (x - 4\right )} \log \left (2\right )}{8 \, x}\right ) - 2}{x - 2}\right )} \] Input:
integrate(((x^2-4*x)*log(1/8*(-4+x)*log(2)/x)-2*x^2+4*x+8)*exp((-log(1/8*( -4+x)*log(2)/x)+2)/(-2+x))/(x^4-8*x^3+20*x^2-16*x),x, algorithm="fricas")
Output:
e^(-(log(1/8*(x - 4)*log(2)/x) - 2)/(x - 2))
Time = 0.21 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {2-\log \left (\frac {(-4+x) \log (4)}{16 x}\right )}{-2+x}} \left (8+4 x-2 x^2+\left (-4 x+x^2\right ) \log \left (\frac {(-4+x) \log (4)}{16 x}\right )\right )}{-16 x+20 x^2-8 x^3+x^4} \, dx=e^{\frac {2 - \log {\left (\frac {\left (\frac {x}{8} - \frac {1}{2}\right ) \log {\left (2 \right )}}{x} \right )}}{x - 2}} \] Input:
integrate(((x**2-4*x)*ln(1/8*(-4+x)*ln(2)/x)-2*x**2+4*x+8)*exp((-ln(1/8*(- 4+x)*ln(2)/x)+2)/(-2+x))/(x**4-8*x**3+20*x**2-16*x),x)
Output:
exp((2 - log((x/8 - 1/2)*log(2)/x))/(x - 2))
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (21) = 42\).
Time = 0.22 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88 \[ \int \frac {e^{\frac {2-\log \left (\frac {(-4+x) \log (4)}{16 x}\right )}{-2+x}} \left (8+4 x-2 x^2+\left (-4 x+x^2\right ) \log \left (\frac {(-4+x) \log (4)}{16 x}\right )\right )}{-16 x+20 x^2-8 x^3+x^4} \, dx=e^{\left (\frac {3 \, \log \left (2\right )}{x - 2} - \frac {\log \left (x - 4\right )}{x - 2} + \frac {\log \left (x\right )}{x - 2} - \frac {\log \left (\log \left (2\right )\right )}{x - 2} + \frac {2}{x - 2}\right )} \] Input:
integrate(((x^2-4*x)*log(1/8*(-4+x)*log(2)/x)-2*x^2+4*x+8)*exp((-log(1/8*( -4+x)*log(2)/x)+2)/(-2+x))/(x^4-8*x^3+20*x^2-16*x),x, algorithm="maxima")
Output:
e^(3*log(2)/(x - 2) - log(x - 4)/(x - 2) + log(x)/(x - 2) - log(log(2))/(x - 2) + 2/(x - 2))
Time = 0.17 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {2-\log \left (\frac {(-4+x) \log (4)}{16 x}\right )}{-2+x}} \left (8+4 x-2 x^2+\left (-4 x+x^2\right ) \log \left (\frac {(-4+x) \log (4)}{16 x}\right )\right )}{-16 x+20 x^2-8 x^3+x^4} \, dx=e^{\left (-\frac {\log \left (-\frac {\log \left (2\right )}{2 \, x} + \frac {1}{8} \, \log \left (2\right )\right )}{x - 2} + \frac {2}{x - 2}\right )} \] Input:
integrate(((x^2-4*x)*log(1/8*(-4+x)*log(2)/x)-2*x^2+4*x+8)*exp((-log(1/8*( -4+x)*log(2)/x)+2)/(-2+x))/(x^4-8*x^3+20*x^2-16*x),x, algorithm="giac")
Output:
e^(-log(-1/2*log(2)/x + 1/8*log(2))/(x - 2) + 2/(x - 2))
Time = 2.73 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{\frac {2-\log \left (\frac {(-4+x) \log (4)}{16 x}\right )}{-2+x}} \left (8+4 x-2 x^2+\left (-4 x+x^2\right ) \log \left (\frac {(-4+x) \log (4)}{16 x}\right )\right )}{-16 x+20 x^2-8 x^3+x^4} \, dx=\frac {{\mathrm {e}}^{\frac {2}{x-2}}}{{\left (\frac {\ln \left (2\right )}{8}-\frac {\ln \left (2\right )}{2\,x}\right )}^{\frac {1}{x-2}}} \] Input:
int(-(exp(-(log((log(2)*(x - 4))/(8*x)) - 2)/(x - 2))*(4*x - log((log(2)*( x - 4))/(8*x))*(4*x - x^2) - 2*x^2 + 8))/(16*x - 20*x^2 + 8*x^3 - x^4),x)
Output:
exp(2/(x - 2))/(log(2)/8 - log(2)/(2*x))^(1/(x - 2))
Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {e^{\frac {2-\log \left (\frac {(-4+x) \log (4)}{16 x}\right )}{-2+x}} \left (8+4 x-2 x^2+\left (-4 x+x^2\right ) \log \left (\frac {(-4+x) \log (4)}{16 x}\right )\right )}{-16 x+20 x^2-8 x^3+x^4} \, dx=\frac {e^{\frac {2}{x -2}}}{e^{\frac {\mathrm {log}\left (\frac {\mathrm {log}\left (2\right ) x -4 \,\mathrm {log}\left (2\right )}{8 x}\right )}{x -2}}} \] Input:
int(((x^2-4*x)*log(1/8*(-4+x)*log(2)/x)-2*x^2+4*x+8)*exp((-log(1/8*(-4+x)* log(2)/x)+2)/(-2+x))/(x^4-8*x^3+20*x^2-16*x),x)
Output:
e**(2/(x - 2))/e**(log((log(2)*x - 4*log(2))/(8*x))/(x - 2))