Integrand size = 69, antiderivative size = 26 \[ \int \left (36+18 x+\frac {25 e^{-8-x+2 x^2} \left (-9 x^2+36 x^3\right )}{x^2}+e^{\frac {1}{2} \left (-8-x+2 x^2+2 \log \left (\frac {5}{x}\right )\right )} \left (63 x^2+36 x^3\right )\right ) \, dx=9 \left (2+5 e^{-\frac {x}{2}+x \left (-\frac {4}{x}+x\right )}+x\right )^2 \] Output:
3*(2+x+exp((x-4/x)*x-1/2*x+ln(5/x))*x)*(6+3*x+3*exp((x-4/x)*x-1/2*x+ln(5/x ))*x)
Time = 0.27 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.85 \[ \int \left (36+18 x+\frac {25 e^{-8-x+2 x^2} \left (-9 x^2+36 x^3\right )}{x^2}+e^{\frac {1}{2} \left (-8-x+2 x^2+2 \log \left (\frac {5}{x}\right )\right )} \left (63 x^2+36 x^3\right )\right ) \, dx=9 \left (25 e^{-8-x+2 x^2}+4 x+x^2+5 e^{-\frac {x}{2}+x^2} \left (\frac {4}{e^4}+\frac {2 x}{e^4}\right )\right ) \] Input:
Integrate[36 + 18*x + (25*E^(-8 - x + 2*x^2)*(-9*x^2 + 36*x^3))/x^2 + E^(( -8 - x + 2*x^2 + 2*Log[5/x])/2)*(63*x^2 + 36*x^3),x]
Output:
9*(25*E^(-8 - x + 2*x^2) + 4*x + x^2 + 5*E^(-1/2*x + x^2)*(4/E^4 + (2*x)/E ^4))
Time = 0.46 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {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.
\(\displaystyle \int \left (\frac {25 e^{2 x^2-x-8} \left (36 x^3-9 x^2\right )}{x^2}+\left (36 x^3+63 x^2\right ) e^{\frac {1}{2} \left (2 x^2-x+2 \log \left (\frac {5}{x}\right )-8\right )}+18 x+36\right ) \, dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 9 x^2+90 e^{x^2-\frac {x}{2}-4} x+180 e^{x^2-\frac {x}{2}-4}+225 e^{2 x^2-x-8}+36 x\) |
Input:
Int[36 + 18*x + (25*E^(-8 - x + 2*x^2)*(-9*x^2 + 36*x^3))/x^2 + E^((-8 - x + 2*x^2 + 2*Log[5/x])/2)*(63*x^2 + 36*x^3),x]
Output:
180*E^(-4 - x/2 + x^2) + 225*E^(-8 - x + 2*x^2) + 36*x + 90*E^(-4 - x/2 + x^2)*x + 9*x^2
Time = 0.41 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.50
method | result | size |
risch | \(225 \,{\mathrm e}^{2 x^{2}-x -8}+5 \left (36+18 x \right ) {\mathrm e}^{-4+x^{2}-\frac {1}{2} x}+9 x^{2}+36 x\) | \(39\) |
default | \(36 x +225 \,{\mathrm e}^{2 x^{2}-x -8}+18 x^{2} {\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}+36 x \,{\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}+9 x^{2}\) | \(70\) |
norman | \(36 x +225 \,{\mathrm e}^{2 x^{2}-x -8}+18 x^{2} {\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}+36 x \,{\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}+9 x^{2}\) | \(70\) |
parallelrisch | \(36 x +225 \,{\mathrm e}^{2 x^{2}-x -8}+18 x^{2} {\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}+36 x \,{\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}+9 x^{2}\) | \(70\) |
parts | \(36 x +225 \,{\mathrm e}^{2 x^{2}-x -8}+18 x^{2} {\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}+36 x \,{\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}+9 x^{2}\) | \(70\) |
orering | \(\frac {\left (16384 x^{8}+77824 x^{7}+3072 x^{6}-42752 x^{5}+102848 x^{4}-82864 x^{3}+67540 x^{2}-74097 x +20224\right ) \left (\frac {25 \left (36 x^{3}-9 x^{2}\right ) {\mathrm e}^{2 x^{2}-x -8}}{x^{2}}+\left (36 x^{3}+63 x^{2}\right ) {\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}+18 x +36\right )}{32 \left (4 x^{2}+7 x -4\right ) \left (256 x^{5}+256 x^{4}-160 x^{3}+416 x^{2}-7 x +193\right )}-\frac {3 \left (4096 x^{7}+20480 x^{6}-1280 x^{5}-34816 x^{4}+21104 x^{3}-15952 x^{2}-10099 x +4476\right ) \left (\frac {25 \left (108 x^{2}-18 x \right ) {\mathrm e}^{2 x^{2}-x -8}}{x^{2}}+\frac {50 \left (36 x^{3}-9 x^{2}\right ) {\mathrm e}^{2 x^{2}-x -8} \left (-\frac {1}{x}+2 x -\frac {1}{2}\right )}{x^{2}}+\left (108 x^{2}+126 x \right ) {\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}+\left (36 x^{3}+63 x^{2}\right ) \left (-\frac {1}{x}+2 x -\frac {1}{2}\right ) {\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}+18\right )}{32 \left (4 x^{2}+7 x -4\right ) \left (256 x^{5}+256 x^{4}-160 x^{3}+416 x^{2}-7 x +193\right )}+\frac {\left (1024 x^{6}+5376 x^{5}-13184 x^{3}+1980 x^{2}+3099 x -2410\right ) \left (\frac {25 \left (216 x -18\right ) {\mathrm e}^{2 x^{2}-x -8}}{x^{2}}+\frac {100 \left (108 x^{2}-18 x \right ) {\mathrm e}^{2 x^{2}-x -8} \left (-\frac {1}{x}+2 x -\frac {1}{2}\right )}{x^{2}}+\frac {100 \left (36 x^{3}-9 x^{2}\right ) {\mathrm e}^{2 x^{2}-x -8} \left (-\frac {1}{x}+2 x -\frac {1}{2}\right )^{2}}{x^{2}}+\frac {50 \left (36 x^{3}-9 x^{2}\right ) {\mathrm e}^{2 x^{2}-x -8} \left (\frac {1}{x^{2}}+2\right )}{x^{2}}+\left (216 x +126\right ) {\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}+2 \left (108 x^{2}+126 x \right ) \left (-\frac {1}{x}+2 x -\frac {1}{2}\right ) {\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}+\left (36 x^{3}+63 x^{2}\right ) \left (\frac {1}{x^{2}}+2\right ) {\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}+\left (36 x^{3}+63 x^{2}\right ) \left (-\frac {1}{x}+2 x -\frac {1}{2}\right )^{2} {\mathrm e}^{\ln \left (\frac {5}{x}\right )+x^{2}-\frac {x}{2}-4}\right )}{16 \left (4 x^{2}+7 x -4\right ) \left (256 x^{5}+256 x^{4}-160 x^{3}+416 x^{2}-7 x +193\right )}\) | \(684\) |
Input:
int((36*x^3-9*x^2)*exp(ln(5/x)+x^2-1/2*x-4)^2+(36*x^3+63*x^2)*exp(ln(5/x)+ x^2-1/2*x-4)+18*x+36,x,method=_RETURNVERBOSE)
Output:
225*exp(2*x^2-x-8)+5*(36+18*x)*exp(-4+x^2-1/2*x)+9*x^2+36*x
Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 2.19 \[ \int \left (36+18 x+\frac {25 e^{-8-x+2 x^2} \left (-9 x^2+36 x^3\right )}{x^2}+e^{\frac {1}{2} \left (-8-x+2 x^2+2 \log \left (\frac {5}{x}\right )\right )} \left (63 x^2+36 x^3\right )\right ) \, dx=9 \, x^{2} e^{\left (2 \, x^{2} - x + 2 \, \log \left (\frac {5}{x}\right ) - 8\right )} + 9 \, x^{2} + 18 \, {\left (x^{2} + 2 \, x\right )} e^{\left (x^{2} - \frac {1}{2} \, x + \log \left (\frac {5}{x}\right ) - 4\right )} + 36 \, x \] Input:
integrate((36*x^3-9*x^2)*exp(log(5/x)+x^2-1/2*x-4)^2+(36*x^3+63*x^2)*exp(l og(5/x)+x^2-1/2*x-4)+18*x+36,x, algorithm="fricas")
Output:
9*x^2*e^(2*x^2 - x + 2*log(5/x) - 8) + 9*x^2 + 18*(x^2 + 2*x)*e^(x^2 - 1/2 *x + log(5/x) - 4) + 36*x
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.31 \[ \int \left (36+18 x+\frac {25 e^{-8-x+2 x^2} \left (-9 x^2+36 x^3\right )}{x^2}+e^{\frac {1}{2} \left (-8-x+2 x^2+2 \log \left (\frac {5}{x}\right )\right )} \left (63 x^2+36 x^3\right )\right ) \, dx=9 x^{2} + 36 x + \left (90 x + 180\right ) e^{x^{2} - \frac {x}{2} - 4} + 225 e^{2 x^{2} - x - 8} \] Input:
integrate((36*x**3-9*x**2)*exp(ln(5/x)+x**2-1/2*x-4)**2+(36*x**3+63*x**2)* exp(ln(5/x)+x**2-1/2*x-4)+18*x+36,x)
Output:
9*x**2 + 36*x + (90*x + 180)*exp(x**2 - x/2 - 4) + 225*exp(2*x**2 - x - 8)
Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.38 \[ \int \left (36+18 x+\frac {25 e^{-8-x+2 x^2} \left (-9 x^2+36 x^3\right )}{x^2}+e^{\frac {1}{2} \left (-8-x+2 x^2+2 \log \left (\frac {5}{x}\right )\right )} \left (63 x^2+36 x^3\right )\right ) \, dx=9 \, x^{2} + 90 \, {\left (x + 2\right )} e^{\left (x^{2} - \frac {1}{2} \, x - 4\right )} + 36 \, x + 225 \, e^{\left (2 \, x^{2} - x - 8\right )} \] Input:
integrate((36*x^3-9*x^2)*exp(log(5/x)+x^2-1/2*x-4)^2+(36*x^3+63*x^2)*exp(l og(5/x)+x^2-1/2*x-4)+18*x+36,x, algorithm="maxima")
Output:
9*x^2 + 90*(x + 2)*e^(x^2 - 1/2*x - 4) + 36*x + 225*e^(2*x^2 - x - 8)
Time = 0.13 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.81 \[ \int \left (36+18 x+\frac {25 e^{-8-x+2 x^2} \left (-9 x^2+36 x^3\right )}{x^2}+e^{\frac {1}{2} \left (-8-x+2 x^2+2 \log \left (\frac {5}{x}\right )\right )} \left (63 x^2+36 x^3\right )\right ) \, dx=9 \, x^{2} + 90 \, {\left (x e^{\left (x^{2} - \frac {1}{2} \, x\right )} + 2 \, e^{\left (x^{2} - \frac {1}{2} \, x\right )}\right )} e^{\left (-4\right )} + 36 \, x + 225 \, e^{\left (2 \, x^{2} - x - 8\right )} \] Input:
integrate((36*x^3-9*x^2)*exp(log(5/x)+x^2-1/2*x-4)^2+(36*x^3+63*x^2)*exp(l og(5/x)+x^2-1/2*x-4)+18*x+36,x, algorithm="giac")
Output:
9*x^2 + 90*(x*e^(x^2 - 1/2*x) + 2*e^(x^2 - 1/2*x))*e^(-4) + 36*x + 225*e^( 2*x^2 - x - 8)
Time = 0.52 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \left (36+18 x+\frac {25 e^{-8-x+2 x^2} \left (-9 x^2+36 x^3\right )}{x^2}+e^{\frac {1}{2} \left (-8-x+2 x^2+2 \log \left (\frac {5}{x}\right )\right )} \left (63 x^2+36 x^3\right )\right ) \, dx=36\,x+180\,{\mathrm {e}}^{x^2-\frac {x}{2}-4}+225\,{\mathrm {e}}^{2\,x^2-x-8}+90\,x\,{\mathrm {e}}^{x^2-\frac {x}{2}-4}+9\,x^2 \] Input:
int(18*x - exp(2*log(5/x) - x + 2*x^2 - 8)*(9*x^2 - 36*x^3) + exp(log(5/x) - x/2 + x^2 - 4)*(63*x^2 + 36*x^3) + 36,x)
Output:
36*x + 180*exp(x^2 - x/2 - 4) + 225*exp(2*x^2 - x - 8) + 90*x*exp(x^2 - x/ 2 - 4) + 9*x^2
Time = 0.24 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.62 \[ \int \left (36+18 x+\frac {25 e^{-8-x+2 x^2} \left (-9 x^2+36 x^3\right )}{x^2}+e^{\frac {1}{2} \left (-8-x+2 x^2+2 \log \left (\frac {5}{x}\right )\right )} \left (63 x^2+36 x^3\right )\right ) \, dx=\frac {225 e^{2 x^{2}}+90 e^{x^{2}+\frac {1}{2} x} e^{4} x +180 e^{x^{2}+\frac {1}{2} x} e^{4}+9 e^{x} e^{8} x^{2}+36 e^{x} e^{8} x}{e^{x} e^{8}} \] Input:
int((36*x^3-9*x^2)*exp(log(5/x)+x^2-1/2*x-4)^2+(36*x^3+63*x^2)*exp(log(5/x )+x^2-1/2*x-4)+18*x+36,x)
Output:
(9*(25*e**(2*x**2) + 10*e**((2*x**2 + x)/2)*e**4*x + 20*e**((2*x**2 + x)/2 )*e**4 + e**x*e**8*x**2 + 4*e**x*e**8*x))/(e**x*e**8)