Integrand size = 96, antiderivative size = 30 \[ \int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx=-5-e^{-x+\frac {25 \log ^2(3)}{(4-x)^2}}+x+\log \left (\frac {3 x}{2}\right ) \] Output:
ln(3/2*x)+x-5-exp(25*ln(3)^2/(4-x)^2-x)
Time = 5.07 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.77 \[ \int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx=-e^{-x+\frac {25 \log ^2(3)}{(-4+x)^2}}+x+\log (x) \] Input:
Integrate[(-64 - 16*x + 36*x^2 - 11*x^3 + x^4 + E^((-16*x + 8*x^2 - x^3 + 25*Log[3]^2)/(16 - 8*x + x^2))*(-64*x + 48*x^2 - 12*x^3 + x^4 + 50*x*Log[3 ]^2))/(-64*x + 48*x^2 - 12*x^3 + x^4),x]
Output:
-E^(-x + (25*Log[3]^2)/(-4 + x)^2) + x + Log[x]
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.
\(\displaystyle \int \frac {\left (x^4-12 x^3+48 x^2-64 x+50 x \log ^2(3)\right ) \exp \left (\frac {-x^3+8 x^2-16 x+25 \log ^2(3)}{x^2-8 x+16}\right )+x^4-11 x^3+36 x^2-16 x-64}{x^4-12 x^3+48 x^2-64 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (x^4-12 x^3+48 x^2-64 x+50 x \log ^2(3)\right ) \exp \left (\frac {-x^3+8 x^2-16 x+25 \log ^2(3)}{x^2-8 x+16}\right )+x^4-11 x^3+36 x^2-16 x-64}{x \left (x^3-12 x^2+48 x-64\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {\left (x^4-12 x^3+48 x^2-64 x+50 x \log ^2(3)\right ) \exp \left (\frac {-x^3+8 x^2-16 x+25 \log ^2(3)}{x^2-8 x+16}\right )+x^4-11 x^3+36 x^2-16 x-64}{(x-4)^3 x}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {x^3}{(x-4)^3}-\frac {11 x^2}{(x-4)^3}+\frac {e^{\frac {-x^3+8 x^2-16 x+25 \log ^2(3)}{(x-4)^2}} \left (-x^3+12 x^2-48 x+2 \left (32-25 \log ^2(3)\right )\right )}{(4-x)^3}+\frac {36 x}{(x-4)^3}-\frac {16}{(x-4)^3}-\frac {64}{(x-4)^3 x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int e^{\frac {-x^3+8 x^2-16 x+25 \log ^2(3)}{(x-4)^2}}dx+50 \log ^2(3) \int \frac {e^{\frac {-x^3+8 x^2-16 x+25 \log ^2(3)}{(x-4)^2}}}{(x-4)^3}dx-\frac {9 x^2}{2 (4-x)^2}+x-\frac {36}{4-x}+\frac {72}{(4-x)^2}+\log (x)\) |
Input:
Int[(-64 - 16*x + 36*x^2 - 11*x^3 + x^4 + E^((-16*x + 8*x^2 - x^3 + 25*Log [3]^2)/(16 - 8*x + x^2))*(-64*x + 48*x^2 - 12*x^3 + x^4 + 50*x*Log[3]^2))/ (-64*x + 48*x^2 - 12*x^3 + x^4),x]
Output:
$Aborted
Time = 3.43 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.13
method | result | size |
risch | \(x +\ln \left (x \right )-{\mathrm e}^{\frac {25 \ln \left (3\right )^{2}-x^{3}+8 x^{2}-16 x}{\left (x -4\right )^{2}}}\) | \(34\) |
parallelrisch | \(x +\ln \left (x \right )-{\mathrm e}^{\frac {25 \ln \left (3\right )^{2}-x^{3}+8 x^{2}-16 x}{x^{2}-8 x +16}}+24\) | \(40\) |
parts | \(x +\ln \left (x \right )+\frac {8 x \,{\mathrm e}^{\frac {25 \ln \left (3\right )^{2}-x^{3}+8 x^{2}-16 x}{x^{2}-8 x +16}}-{\mathrm e}^{\frac {25 \ln \left (3\right )^{2}-x^{3}+8 x^{2}-16 x}{x^{2}-8 x +16}} x^{2}-16 \,{\mathrm e}^{\frac {25 \ln \left (3\right )^{2}-x^{3}+8 x^{2}-16 x}{x^{2}-8 x +16}}}{\left (x -4\right )^{2}}\) | \(118\) |
norman | \(\frac {x^{3}-48 x +8 x \,{\mathrm e}^{\frac {25 \ln \left (3\right )^{2}-x^{3}+8 x^{2}-16 x}{x^{2}-8 x +16}}-{\mathrm e}^{\frac {25 \ln \left (3\right )^{2}-x^{3}+8 x^{2}-16 x}{x^{2}-8 x +16}} x^{2}-16 \,{\mathrm e}^{\frac {25 \ln \left (3\right )^{2}-x^{3}+8 x^{2}-16 x}{x^{2}-8 x +16}}+128}{\left (x -4\right )^{2}}+\ln \left (x \right )\) | \(124\) |
Input:
int(((50*x*ln(3)^2+x^4-12*x^3+48*x^2-64*x)*exp((25*ln(3)^2-x^3+8*x^2-16*x) /(x^2-8*x+16))+x^4-11*x^3+36*x^2-16*x-64)/(x^4-12*x^3+48*x^2-64*x),x,metho d=_RETURNVERBOSE)
Output:
x+ln(x)-exp((25*ln(3)^2-x^3+8*x^2-16*x)/(x-4)^2)
Time = 0.09 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.23 \[ \int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx=x - e^{\left (-\frac {x^{3} - 8 \, x^{2} - 25 \, \log \left (3\right )^{2} + 16 \, x}{x^{2} - 8 \, x + 16}\right )} + \log \left (x\right ) \] Input:
integrate(((50*x*log(3)^2+x^4-12*x^3+48*x^2-64*x)*exp((25*log(3)^2-x^3+8*x ^2-16*x)/(x^2-8*x+16))+x^4-11*x^3+36*x^2-16*x-64)/(x^4-12*x^3+48*x^2-64*x) ,x, algorithm="fricas")
Output:
x - e^(-(x^3 - 8*x^2 - 25*log(3)^2 + 16*x)/(x^2 - 8*x + 16)) + log(x)
Time = 0.16 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.07 \[ \int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx=x - e^{\frac {- x^{3} + 8 x^{2} - 16 x + 25 \log {\left (3 \right )}^{2}}{x^{2} - 8 x + 16}} + \log {\left (x \right )} \] Input:
integrate(((50*x*ln(3)**2+x**4-12*x**3+48*x**2-64*x)*exp((25*ln(3)**2-x**3 +8*x**2-16*x)/(x**2-8*x+16))+x**4-11*x**3+36*x**2-16*x-64)/(x**4-12*x**3+4 8*x**2-64*x),x)
Output:
x - exp((-x**3 + 8*x**2 - 16*x + 25*log(3)**2)/(x**2 - 8*x + 16)) + log(x)
Leaf count of result is larger than twice the leaf count of optimal. 101 vs. \(2 (25) = 50\).
Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 3.37 \[ \int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx=x - \frac {16 \, {\left (3 \, x - 10\right )}}{x^{2} - 8 \, x + 16} - \frac {36 \, {\left (x - 2\right )}}{x^{2} - 8 \, x + 16} + \frac {88 \, {\left (x - 3\right )}}{x^{2} - 8 \, x + 16} - \frac {4 \, {\left (x - 6\right )}}{x^{2} - 8 \, x + 16} + \frac {8}{x^{2} - 8 \, x + 16} - e^{\left (-x + \frac {25 \, \log \left (3\right )^{2}}{x^{2} - 8 \, x + 16}\right )} + \log \left (x\right ) \] Input:
integrate(((50*x*log(3)^2+x^4-12*x^3+48*x^2-64*x)*exp((25*log(3)^2-x^3+8*x ^2-16*x)/(x^2-8*x+16))+x^4-11*x^3+36*x^2-16*x-64)/(x^4-12*x^3+48*x^2-64*x) ,x, algorithm="maxima")
Output:
x - 16*(3*x - 10)/(x^2 - 8*x + 16) - 36*(x - 2)/(x^2 - 8*x + 16) + 88*(x - 3)/(x^2 - 8*x + 16) - 4*(x - 6)/(x^2 - 8*x + 16) + 8/(x^2 - 8*x + 16) - e ^(-x + 25*log(3)^2/(x^2 - 8*x + 16)) + log(x)
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (25) = 50\).
Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.87 \[ \int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx=x - e^{\left (\frac {25}{16} \, \log \left (3\right )^{2} - \frac {25 \, x^{2} \log \left (3\right )^{2} + 16 \, x^{3} - 200 \, x \log \left (3\right )^{2} - 128 \, x^{2} + 256 \, x}{16 \, {\left (x^{2} - 8 \, x + 16\right )}}\right )} + \log \left (x\right ) \] Input:
integrate(((50*x*log(3)^2+x^4-12*x^3+48*x^2-64*x)*exp((25*log(3)^2-x^3+8*x ^2-16*x)/(x^2-8*x+16))+x^4-11*x^3+36*x^2-16*x-64)/(x^4-12*x^3+48*x^2-64*x) ,x, algorithm="giac")
Output:
x - e^(25/16*log(3)^2 - 1/16*(25*x^2*log(3)^2 + 16*x^3 - 200*x*log(3)^2 - 128*x^2 + 256*x)/(x^2 - 8*x + 16)) + log(x)
Time = 3.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 2.30 \[ \int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx=x+\ln \left (x\right )-{\mathrm {e}}^{-\frac {x^3}{x^2-8\,x+16}}\,{\mathrm {e}}^{\frac {8\,x^2}{x^2-8\,x+16}}\,{\mathrm {e}}^{-\frac {16\,x}{x^2-8\,x+16}}\,{\mathrm {e}}^{\frac {25\,{\ln \left (3\right )}^2}{x^2-8\,x+16}} \] Input:
int((16*x - exp(-(16*x - 25*log(3)^2 - 8*x^2 + x^3)/(x^2 - 8*x + 16))*(50* x*log(3)^2 - 64*x + 48*x^2 - 12*x^3 + x^4) - 36*x^2 + 11*x^3 - x^4 + 64)/( 64*x - 48*x^2 + 12*x^3 - x^4),x)
Output:
x + log(x) - exp(-x^3/(x^2 - 8*x + 16))*exp((8*x^2)/(x^2 - 8*x + 16))*exp( -(16*x)/(x^2 - 8*x + 16))*exp((25*log(3)^2)/(x^2 - 8*x + 16))
Time = 0.17 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.27 \[ \int \frac {-64-16 x+36 x^2-11 x^3+x^4+e^{\frac {-16 x+8 x^2-x^3+25 \log ^2(3)}{16-8 x+x^2}} \left (-64 x+48 x^2-12 x^3+x^4+50 x \log ^2(3)\right )}{-64 x+48 x^2-12 x^3+x^4} \, dx=\frac {-e^{\frac {25 \mathrm {log}\left (3\right )^{2}}{x^{2}-8 x +16}}+e^{x} \mathrm {log}\left (x \right )+e^{x} x}{e^{x}} \] Input:
int(((50*x*log(3)^2+x^4-12*x^3+48*x^2-64*x)*exp((25*log(3)^2-x^3+8*x^2-16* x)/(x^2-8*x+16))+x^4-11*x^3+36*x^2-16*x-64)/(x^4-12*x^3+48*x^2-64*x),x)
Output:
( - e**((25*log(3)**2)/(x**2 - 8*x + 16)) + e**x*log(x) + e**x*x)/e**x