Integrand size = 44, antiderivative size = 22 \[ \int \frac {1}{36} e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx=-\frac {x}{4}+\left (2-\frac {5}{6} e^{-x} x^2\right )^2 \] Output:
(2-5/2/exp(ln(x)+ln(3/x^2)+x)*x)^2-1/4*x
Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {1}{36} e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx=\frac {1}{36} \left (-9 x-120 e^{-x} x^2+25 e^{-2 x} x^4\right ) \] Input:
Integrate[(x^2*((-9*E^(2*x))/x^2 + 100*x - 50*x^2 + (3*E^x*(-80 + 40*x))/x ))/(36*E^(2*x)),x]
Output:
(-9*x - (120*x^2)/E^x + (25*x^4)/E^(2*x))/36
Time = 0.64 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {27, 25, 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.
\(\displaystyle \int \frac {1}{36} e^{-2 x} x^2 \left (-50 x^2-\frac {9 e^{2 x}}{x^2}+100 x+\frac {3 e^x (40 x-80)}{x}\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{36} \int -e^{-2 x} x^2 \left (50 x^2-100 x+\frac {120 e^x (2-x)}{x}+\frac {9 e^{2 x}}{x^2}\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{36} \int e^{-2 x} x^2 \left (50 x^2-100 x+\frac {120 e^x (2-x)}{x}+\frac {9 e^{2 x}}{x^2}\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{36} \int \left (10 e^{-2 x} (x-2) x \left (5 x^2-12 e^x\right )+9\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{36} \left (25 e^{-2 x} x^4-120 e^{-x} x^2-9 x\right )\) |
Input:
Int[(x^2*((-9*E^(2*x))/x^2 + 100*x - 50*x^2 + (3*E^x*(-80 + 40*x))/x))/(36 *E^(2*x)),x]
Output:
(-9*x - (120*x^2)/E^x + (25*x^4)/E^(2*x))/36
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 0.31 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77
method | result | size |
default | \(-\frac {x}{4}-\frac {10 x^{2} {\mathrm e}^{-x}}{3}+\frac {25 x^{4} {\mathrm e}^{-2 x}}{36}\) | \(39\) |
parts | \(-\frac {x}{4}-\frac {10 x^{2} {\mathrm e}^{-x}}{3}+\frac {25 x^{4} {\mathrm e}^{-2 x}}{36}\) | \(39\) |
parallelrisch | \(-\frac {\left (\frac {9 \,{\mathrm e}^{2 x}}{x}-25 x^{2}+40 \,{\mathrm e}^{\ln \left (x \right )+\ln \left (\frac {3}{x^{2}}\right )+x} x \right ) x^{2} {\mathrm e}^{-2 x}}{36}\) | \(51\) |
risch | \(-\frac {x}{4}+\frac {10 x^{2} {\mathrm e}^{-x} {\mathrm e}^{-\frac {i \pi \operatorname {csgn}\left (i x^{2}\right )^{3}}{2}} {\mathrm e}^{-\frac {i \pi \,\operatorname {csgn}\left (i x^{2}\right ) \operatorname {csgn}\left (i x \right )^{2}}{2}}}{3}+\frac {25 x^{4} {\mathrm e}^{-2 x}}{36}\) | \(56\) |
orering | \(\frac {\left (x +\frac {9}{2}\right ) \left (-\frac {9 \,{\mathrm e}^{2 x}}{x^{2}}+\left (40 x -80\right ) {\mathrm e}^{\ln \left (x \right )+\ln \left (\frac {3}{x^{2}}\right )+x}-50 x^{2}+100 x \right ) x^{2} {\mathrm e}^{-2 x}}{36}+\frac {x \left (3 x^{5}-7 x^{4}-63 x^{3}+258 x^{2}-312 x +108\right ) \left (\frac {\left (-\frac {18 \,{\mathrm e}^{2 x} \left (-\frac {1}{x}+1\right )}{x^{2}}+40 \,{\mathrm e}^{\ln \left (x \right )+\ln \left (\frac {3}{x^{2}}\right )+x}+\left (40 x -80\right ) \left (-\frac {1}{x}+1\right ) {\mathrm e}^{\ln \left (x \right )+\ln \left (\frac {3}{x^{2}}\right )+x}-100 x +100\right ) x^{2} {\mathrm e}^{-2 x}}{36}-\frac {\left (-\frac {9 \,{\mathrm e}^{2 x}}{x^{2}}+\left (40 x -80\right ) {\mathrm e}^{\ln \left (x \right )+\ln \left (\frac {3}{x^{2}}\right )+x}-50 x^{2}+100 x \right ) x^{2} {\mathrm e}^{-2 x} \left (-\frac {1}{x}+1\right )}{18}\right )}{2 x^{5}-20 x^{4}+74 x^{3}-120 x^{2}+84 x -24}+\frac {x^{2} \left (x^{4}-21 x^{2}+51 x -36\right ) \left (\frac {\left (-\frac {36 \,{\mathrm e}^{2 x} \left (-\frac {1}{x}+1\right )^{2}}{x^{2}}-\frac {18 \,{\mathrm e}^{2 x}}{x^{4}}+80 \left (-\frac {1}{x}+1\right ) {\mathrm e}^{\ln \left (x \right )+\ln \left (\frac {3}{x^{2}}\right )+x}+\frac {\left (40 x -80\right ) {\mathrm e}^{\ln \left (x \right )+\ln \left (\frac {3}{x^{2}}\right )+x}}{x^{2}}+\left (40 x -80\right ) \left (-\frac {1}{x}+1\right )^{2} {\mathrm e}^{\ln \left (x \right )+\ln \left (\frac {3}{x^{2}}\right )+x}-100\right ) x^{2} {\mathrm e}^{-2 x}}{36}-\frac {\left (-\frac {18 \,{\mathrm e}^{2 x} \left (-\frac {1}{x}+1\right )}{x^{2}}+40 \,{\mathrm e}^{\ln \left (x \right )+\ln \left (\frac {3}{x^{2}}\right )+x}+\left (40 x -80\right ) \left (-\frac {1}{x}+1\right ) {\mathrm e}^{\ln \left (x \right )+\ln \left (\frac {3}{x^{2}}\right )+x}-100 x +100\right ) x^{2} {\mathrm e}^{-2 x} \left (-\frac {1}{x}+1\right )}{9}+\frac {\left (-\frac {9 \,{\mathrm e}^{2 x}}{x^{2}}+\left (40 x -80\right ) {\mathrm e}^{\ln \left (x \right )+\ln \left (\frac {3}{x^{2}}\right )+x}-50 x^{2}+100 x \right ) x^{2} {\mathrm e}^{-2 x} \left (-\frac {1}{x}+1\right )^{2}}{9}-\frac {\left (-\frac {9 \,{\mathrm e}^{2 x}}{x^{2}}+\left (40 x -80\right ) {\mathrm e}^{\ln \left (x \right )+\ln \left (\frac {3}{x^{2}}\right )+x}-50 x^{2}+100 x \right ) {\mathrm e}^{-2 x}}{18}\right )}{2 \left (-2+x \right ) \left (x^{4}-8 x^{3}+21 x^{2}-18 x +6\right )}\) | \(635\) |
Input:
int(1/4*(-exp(ln(x)+ln(3/x^2)+x)^2+(40*x-80)*exp(ln(x)+ln(3/x^2)+x)-50*x^2 +100*x)/exp(ln(x)+ln(3/x^2)+x)^2,x,method=_RETURNVERBOSE)
Output:
-1/4*x-10/exp(ln(x)+ln(3/x^2)+x)*x+25/4*x^2/exp(ln(x)+ln(3/x^2)+x)^2
Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.68 \[ \int \frac {1}{36} e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx=\frac {1}{4} \, {\left (25 \, x^{2} - x e^{\left (2 \, x + \log \left (3\right ) + \log \left (\frac {3}{x^{2}}\right )\right )} - 40 \, x e^{\left (x + \frac {1}{2} \, \log \left (3\right ) + \frac {1}{2} \, \log \left (\frac {3}{x^{2}}\right )\right )}\right )} e^{\left (-2 \, x - \log \left (3\right ) - \log \left (\frac {3}{x^{2}}\right )\right )} \] Input:
integrate(1/4*(-exp(log(x)+log(3/x^2)+x)^2+(40*x-80)*exp(log(x)+log(3/x^2) +x)-50*x^2+100*x)/exp(log(x)+log(3/x^2)+x)^2,x, algorithm="fricas")
Output:
1/4*(25*x^2 - x*e^(2*x + log(3) + log(3/x^2)) - 40*x*e^(x + 1/2*log(3) + 1 /2*log(3/x^2)))*e^(-2*x - log(3) - log(3/x^2))
Time = 0.06 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.09 \[ \int \frac {1}{36} e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx=\frac {25 x^{4} e^{- 2 x}}{36} - \frac {10 x^{2} e^{- x}}{3} - \frac {x}{4} \] Input:
integrate(1/4*(-exp(ln(x)+ln(3/x**2)+x)**2+(40*x-80)*exp(ln(x)+ln(3/x**2)+ x)-50*x**2+100*x)/exp(ln(x)+ln(3/x**2)+x)**2,x)
Output:
25*x**4*exp(-2*x)/36 - 10*x**2*exp(-x)/3 - x/4
Leaf count of result is larger than twice the leaf count of optimal. 74 vs. \(2 (30) = 60\).
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 3.36 \[ \int \frac {1}{36} e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx=-\frac {10}{3} \, {\left (x^{2} + 2 \, x + 2\right )} e^{\left (-x\right )} + \frac {20}{3} \, {\left (x + 1\right )} e^{\left (-x\right )} + \frac {25}{72} \, {\left (2 \, x^{4} + 4 \, x^{3} + 6 \, x^{2} + 6 \, x + 3\right )} e^{\left (-2 \, x\right )} - \frac {25}{72} \, {\left (4 \, x^{3} + 6 \, x^{2} + 6 \, x + 3\right )} e^{\left (-2 \, x\right )} - \frac {1}{4} \, x \] Input:
integrate(1/4*(-exp(log(x)+log(3/x^2)+x)^2+(40*x-80)*exp(log(x)+log(3/x^2) +x)-50*x^2+100*x)/exp(log(x)+log(3/x^2)+x)^2,x, algorithm="maxima")
Output:
-10/3*(x^2 + 2*x + 2)*e^(-x) + 20/3*(x + 1)*e^(-x) + 25/72*(2*x^4 + 4*x^3 + 6*x^2 + 6*x + 3)*e^(-2*x) - 25/72*(4*x^3 + 6*x^2 + 6*x + 3)*e^(-2*x) - 1 /4*x
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {1}{36} e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx=\frac {1}{36} \, {\left (25 \, x^{4} - 120 \, x^{2} e^{x} - 9 \, x e^{\left (2 \, x\right )}\right )} e^{\left (-2 \, x\right )} \] Input:
integrate(1/4*(-exp(log(x)+log(3/x^2)+x)^2+(40*x-80)*exp(log(x)+log(3/x^2) +x)-50*x^2+100*x)/exp(log(x)+log(3/x^2)+x)^2,x, algorithm="giac")
Output:
1/36*(25*x^4 - 120*x^2*e^x - 9*x*e^(2*x))*e^(-2*x)
Time = 3.40 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {1}{36} e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx=-\frac {x\,\left (120\,x\,{\mathrm {e}}^{-x}-25\,x^3\,{\mathrm {e}}^{-2\,x}+9\right )}{36} \] Input:
int(exp(- 2*x - 2*log(3/x^2) - 2*log(x))*(25*x - exp(2*x + 2*log(3/x^2) + 2*log(x))/4 + (exp(x + log(3/x^2) + log(x))*(40*x - 80))/4 - (25*x^2)/2),x )
Output:
-(x*(120*x*exp(-x) - 25*x^3*exp(-2*x) + 9))/36
Time = 0.15 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {1}{36} e^{-2 x} x^2 \left (-\frac {9 e^{2 x}}{x^2}+100 x-50 x^2+\frac {3 e^x (-80+40 x)}{x}\right ) \, dx=\frac {x \left (-9 e^{2 x}-120 e^{x} x +25 x^{3}\right )}{36 e^{2 x}} \] Input:
int(1/4*(-exp(log(x)+log(3/x^2)+x)^2+(40*x-80)*exp(log(x)+log(3/x^2)+x)-50 *x^2+100*x)/exp(log(x)+log(3/x^2)+x)^2,x)
Output:
(x*( - 9*e**(2*x) - 120*e**x*x + 25*x**3))/(36*e**(2*x))