Integrand size = 77, antiderivative size = 29 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=-4 x+\frac {4 e^{3-x} \log (6)}{\left (e^x-\frac {10}{x}\right ) x} \] Output:
4*ln(6)/x/(exp(x)-10/x)/exp(-3+x)-4*x
Time = 1.48 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=-\frac {4 \left (10 x-e^x x^2+e^{3-x} \log (6)\right )}{10-e^x x} \] Input:
Integrate[(-400*E^(-3 + x) - 4*E^(-3 + 3*x)*x^2 + 40*Log[6] + E^x*(80*E^(- 3 + x)*x + (-4 - 8*x)*Log[6]))/(100*E^(-3 + x) - 20*E^(-3 + 2*x)*x + E^(-3 + 3*x)*x^2),x]
Output:
(-4*(10*x - E^x*x^2 + E^(3 - x)*Log[6]))/(10 - E^x*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.
\(\displaystyle \int \frac {-4 e^{3 x-3} x^2-400 e^{x-3}+e^x \left (80 e^{x-3} x+(-8 x-4) \log (6)\right )+40 \log (6)}{e^{3 x-3} x^2-20 e^{2 x-3} x+100 e^{x-3}} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{3-x} \left (-4 e^{3 x-3} x^2-400 e^{x-3}+e^x \left (80 e^{x-3} x+(-8 x-4) \log (6)\right )+40 \log (6)\right )}{\left (10-e^x x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {40 e^{3-x} (x+1) \log (6)}{x \left (e^x x-10\right )^2}-\frac {4 e^{3-x} (2 x+1) \log (6)}{x \left (e^x x-10\right )}-4\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -40 \log (6) \int \frac {e^{3-x}}{\left (e^x x-10\right )^2}dx-40 \log (6) \int \frac {e^{3-x}}{x \left (e^x x-10\right )^2}dx-8 \log (6) \int \frac {e^{3-x}}{e^x x-10}dx-4 \log (6) \int \frac {e^{3-x}}{x \left (e^x x-10\right )}dx-4 x\) |
Input:
Int[(-400*E^(-3 + x) - 4*E^(-3 + 3*x)*x^2 + 40*Log[6] + E^x*(80*E^(-3 + x) *x + (-4 - 8*x)*Log[6]))/(100*E^(-3 + x) - 20*E^(-3 + 2*x)*x + E^(-3 + 3*x )*x^2),x]
Output:
$Aborted
Time = 0.41 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21
method | result | size |
norman | \(\frac {\left (4 \,{\mathrm e}^{3} \ln \left (6\right )+40 \,{\mathrm e}^{x} x -4 \,{\mathrm e}^{2 x} x^{2}\right ) {\mathrm e}^{-x}}{{\mathrm e}^{x} x -10}\) | \(35\) |
parallelrisch | \(\frac {\left (-4 \,{\mathrm e}^{-3+x} x^{2} {\mathrm e}^{x}+40 x \,{\mathrm e}^{-3+x}+4 \ln \left (6\right )\right ) {\mathrm e}^{-x +3}}{{\mathrm e}^{x} x -10}\) | \(39\) |
risch | \(-4 x -\frac {2 \ln \left (2\right ) {\mathrm e}^{-x +3}}{5}-\frac {2 \ln \left (3\right ) {\mathrm e}^{-x +3}}{5}+\frac {2 x \left (\ln \left (2\right )+\ln \left (3\right )\right ) {\mathrm e}^{3}}{5 \left ({\mathrm e}^{x} x -10\right )}\) | \(43\) |
Input:
int((-4*x^2*exp(-3+x)*exp(x)^2+(80*x*exp(-3+x)+(-8*x-4)*ln(6))*exp(x)-400* exp(-3+x)+40*ln(6))/(x^2*exp(-3+x)*exp(x)^2-20*x*exp(-3+x)*exp(x)+100*exp( -3+x)),x,method=_RETURNVERBOSE)
Output:
(4*exp(3)*ln(6)+40*exp(x)*x-4*exp(x)^2*x^2)/exp(x)/(exp(x)*x-10)
Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.21 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=-\frac {4 \, {\left (x^{2} e^{\left (2 \, x\right )} - 10 \, x e^{x} - e^{3} \log \left (6\right )\right )}}{x e^{\left (2 \, x\right )} - 10 \, e^{x}} \] Input:
integrate((-4*x^2*exp(-3+x)*exp(x)^2+(80*x*exp(-3+x)+(-8*x-4)*log(6))*exp( x)-400*exp(-3+x)+40*log(6))/(x^2*exp(-3+x)*exp(x)^2-20*x*exp(-3+x)*exp(x)+ 100*exp(-3+x)),x, algorithm="fricas")
Output:
-4*(x^2*e^(2*x) - 10*x*e^x - e^3*log(6))/(x*e^(2*x) - 10*e^x)
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.17 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=- 4 x + \frac {2 x e^{3} \log {\left (6 \right )}}{5 x e^{x} - 50} - \frac {2 e^{3} e^{- x} \log {\left (6 \right )}}{5} \] Input:
integrate((-4*x**2*exp(-3+x)*exp(x)**2+(80*x*exp(-3+x)+(-8*x-4)*ln(6))*exp (x)-400*exp(-3+x)+40*ln(6))/(x**2*exp(-3+x)*exp(x)**2-20*x*exp(-3+x)*exp(x )+100*exp(-3+x)),x)
Output:
-4*x + 2*x*exp(3)*log(6)/(5*x*exp(x) - 50) - 2*exp(3)*exp(-x)*log(6)/5
Time = 0.17 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=-\frac {4 \, {\left (x^{2} e^{\left (2 \, x\right )} - {\left (\log \left (3\right ) + \log \left (2\right )\right )} e^{3} - 10 \, x e^{x}\right )}}{x e^{\left (2 \, x\right )} - 10 \, e^{x}} \] Input:
integrate((-4*x^2*exp(-3+x)*exp(x)^2+(80*x*exp(-3+x)+(-8*x-4)*log(6))*exp( x)-400*exp(-3+x)+40*log(6))/(x^2*exp(-3+x)*exp(x)^2-20*x*exp(-3+x)*exp(x)+ 100*exp(-3+x)),x, algorithm="maxima")
Output:
-4*(x^2*e^(2*x) - (log(3) + log(2))*e^3 - 10*x*e^x)/(x*e^(2*x) - 10*e^x)
Leaf count of result is larger than twice the leaf count of optimal. 66 vs. \(2 (29) = 58\).
Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 2.28 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=-\frac {4 \, {\left ({\left (x - 3\right )}^{2} e^{\left (2 \, x - 3\right )} + 3 \, {\left (x - 3\right )} e^{\left (2 \, x - 3\right )} - 10 \, {\left (x - 3\right )} e^{\left (x - 3\right )} - \log \left (6\right )\right )}}{{\left (x - 3\right )} e^{\left (2 \, x - 3\right )} + 3 \, e^{\left (2 \, x - 3\right )} - 10 \, e^{\left (x - 3\right )}} \] Input:
integrate((-4*x^2*exp(-3+x)*exp(x)^2+(80*x*exp(-3+x)+(-8*x-4)*log(6))*exp( x)-400*exp(-3+x)+40*log(6))/(x^2*exp(-3+x)*exp(x)^2-20*x*exp(-3+x)*exp(x)+ 100*exp(-3+x)),x, algorithm="giac")
Output:
-4*((x - 3)^2*e^(2*x - 3) + 3*(x - 3)*e^(2*x - 3) - 10*(x - 3)*e^(x - 3) - log(6))/((x - 3)*e^(2*x - 3) + 3*e^(2*x - 3) - 10*e^(x - 3))
Time = 4.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=\frac {4\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^3\,\ln \left (6\right )}{x\,{\mathrm {e}}^x-10}-4\,x \] Input:
int(-(400*exp(x - 3) - 40*log(6) + exp(x)*(log(6)*(8*x + 4) - 80*x*exp(x - 3)) + 4*x^2*exp(2*x)*exp(x - 3))/(100*exp(x - 3) + x^2*exp(2*x)*exp(x - 3 ) - 20*x*exp(x - 3)*exp(x)),x)
Output:
(4*exp(-x)*exp(3)*log(6))/(x*exp(x) - 10) - 4*x
Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {-400 e^{-3+x}-4 e^{-3+3 x} x^2+40 \log (6)+e^x \left (80 e^{-3+x} x+(-4-8 x) \log (6)\right )}{100 e^{-3+x}-20 e^{-3+2 x} x+e^{-3+3 x} x^2} \, dx=\frac {-4 e^{2 x} x^{2}+40 e^{x} x +4 \,\mathrm {log}\left (6\right ) e^{3}}{e^{x} \left (e^{x} x -10\right )} \] Input:
int((-4*x^2*exp(-3+x)*exp(x)^2+(80*x*exp(-3+x)+(-8*x-4)*log(6))*exp(x)-400 *exp(-3+x)+40*log(6))/(x^2*exp(-3+x)*exp(x)^2-20*x*exp(-3+x)*exp(x)+100*ex p(-3+x)),x)
Output:
(4*( - e**(2*x)*x**2 + 10*e**x*x + log(6)*e**3))/(e**x*(e**x*x - 10))