Integrand size = 64, antiderivative size = 24 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )} \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx=2 e^{-3 \left (4-(e+x) \left (x-\frac {\log (x)}{5}\right )\right )} x^2 \] Output:
2*x^2/exp(12-3*(x+exp(1))*(x-1/5*ln(x)))
Time = 0.63 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )} \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx=2 e^{3 \left (-4+e x+x^2\right )} x^{2-\frac {3 e}{5}-\frac {3 x}{5}} \] Input:
Integrate[(E^((-60 + 15*E*x + 15*x^2 - (3*E + 3*x)*Log[x])/5)*(20*x - 6*x^ 2 + 60*x^3 + E*(-6*x + 30*x^2) - 6*x^2*Log[x]))/5,x]
Output:
2*E^(3*(-4 + E*x + x^2))*x^(2 - (3*E)/5 - (3*x)/5)
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.
\(\displaystyle \int \frac {1}{5} \left (60 x^3-6 x^2+e \left (30 x^2-6 x\right )-6 x^2 \log (x)+20 x\right ) \exp \left (\frac {1}{5} \left (15 x^2+15 e x-(3 x+3 e) \log (x)-60\right )\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int 2 e^{-3 \left (-x^2-e x+4\right )} x^{\frac {1}{5} (-3 x-3 e)} \left (30 x^3-3 \log (x) x^2-3 x^2+10 x-3 e \left (x-5 x^2\right )\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{5} \int e^{-3 \left (-x^2-e x+4\right )} x^{-\frac {3}{5} (x+e)} \left (30 x^3-3 \log (x) x^2-3 x^2+10 x-3 e \left (x-5 x^2\right )\right )dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \frac {2}{5} \int e^{-3 \left (-x^2-e x+4\right )} x^{-\frac {3 x}{5}-\frac {3 e}{5}} \left (30 x^3-3 \log (x) x^2-3 x^2+10 x-3 e \left (x-5 x^2\right )\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {2}{5} \int \left (10 e^{-3 \left (-x^2-e x+4\right )} x^{-\frac {3 x}{5}-\frac {3 e}{5}+1}+3 e^{1-3 \left (-x^2-e x+4\right )} (5 x-1) x^{-\frac {3 x}{5}-\frac {3 e}{5}+1}-3 e^{-3 \left (-x^2-e x+4\right )} x^{-\frac {3 x}{5}-\frac {3 e}{5}+2}-3 e^{-3 \left (-x^2-e x+4\right )} \log (x) x^{-\frac {3 x}{5}-\frac {3 e}{5}+2}+30 e^{-3 \left (-x^2-e x+4\right )} x^{-\frac {3 x}{5}-\frac {3 e}{5}+3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{5} \left (10 \int e^{-3 \left (-x^2-e x+4\right )} x^{-\frac {3 x}{5}-\frac {3 e}{5}+1}dx-3 \int e^{3 x^2+3 e x-11} x^{-\frac {3 x}{5}-\frac {3 e}{5}+1}dx-3 \int e^{-3 \left (-x^2-e x+4\right )} x^{-\frac {3 x}{5}-\frac {3 e}{5}+2}dx+15 \int e^{3 x^2+3 e x-11} x^{-\frac {3 x}{5}-\frac {3 e}{5}+2}dx+30 \int e^{-3 \left (-x^2-e x+4\right )} x^{-\frac {3 x}{5}-\frac {3 e}{5}+3}dx+3 \int \frac {\int e^{3 \left (x^2+e x-4\right )} x^{-\frac {3 x}{5}-\frac {3 e}{5}+2}dx}{x}dx-3 \log (x) \int e^{-3 \left (-x^2-e x+4\right )} x^{-\frac {3 x}{5}-\frac {3 e}{5}+2}dx\right )\) |
Input:
Int[(E^((-60 + 15*E*x + 15*x^2 - (3*E + 3*x)*Log[x])/5)*(20*x - 6*x^2 + 60 *x^3 + E*(-6*x + 30*x^2) - 6*x^2*Log[x]))/5,x]
Output:
$Aborted
Time = 0.26 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29
method | result | size |
risch | \(2 x^{2} x^{-\frac {3 x}{5}-\frac {3 \,{\mathrm e}}{5}} {\mathrm e}^{-12+3 x \,{\mathrm e}+3 x^{2}}\) | \(31\) |
norman | \(2 x^{2} {\mathrm e}^{\left (-\frac {3 x}{5}-\frac {3 \,{\mathrm e}}{5}\right ) \ln \left (x \right )+3 x \,{\mathrm e}+3 x^{2}-12}\) | \(33\) |
parallelrisch | \(2 x^{2} {\mathrm e}^{\left (-\frac {3 x}{5}-\frac {3 \,{\mathrm e}}{5}\right ) \ln \left (x \right )+3 x \,{\mathrm e}+3 x^{2}-12}\) | \(33\) |
Input:
int(1/5*(-6*x^2*ln(x)+(30*x^2-6*x)*exp(1)+60*x^3-6*x^2+20*x)/exp(1/5*(3*ex p(1)+3*x)*ln(x)-3*x*exp(1)-3*x^2+12),x,method=_RETURNVERBOSE)
Output:
2*x^2/(x^(3/5*x+3/5*exp(1)))*exp(-12+3*x*exp(1)+3*x^2)
Time = 0.11 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )} \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx=2 \, x^{2} e^{\left (3 \, x^{2} + 3 \, x e - \frac {3}{5} \, {\left (x + e\right )} \log \left (x\right ) - 12\right )} \] Input:
integrate(1/5*(-6*x^2*log(x)+(30*x^2-6*x)*exp(1)+60*x^3-6*x^2+20*x)/exp(1/ 5*(3*exp(1)+3*x)*log(x)-3*exp(1)*x-3*x^2+12),x, algorithm="fricas")
Output:
2*x^2*e^(3*x^2 + 3*x*e - 3/5*(x + e)*log(x) - 12)
Time = 0.14 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.42 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )} \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx=2 x^{2} e^{3 x^{2} + 3 e x - \left (\frac {3 x}{5} + \frac {3 e}{5}\right ) \log {\left (x \right )} - 12} \] Input:
integrate(1/5*(-6*x**2*ln(x)+(30*x**2-6*x)*exp(1)+60*x**3-6*x**2+20*x)/exp (1/5*(3*exp(1)+3*x)*ln(x)-3*exp(1)*x-3*x**2+12),x)
Output:
2*x**2*exp(3*x**2 + 3*E*x - (3*x/5 + 3*E/5)*log(x) - 12)
Time = 0.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )} \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx=2 \, x^{2} e^{\left (3 \, x^{2} + 3 \, x e - \frac {3}{5} \, x \log \left (x\right ) - \frac {3}{5} \, e \log \left (x\right ) - 12\right )} \] Input:
integrate(1/5*(-6*x^2*log(x)+(30*x^2-6*x)*exp(1)+60*x^3-6*x^2+20*x)/exp(1/ 5*(3*exp(1)+3*x)*log(x)-3*exp(1)*x-3*x^2+12),x, algorithm="maxima")
Output:
2*x^2*e^(3*x^2 + 3*x*e - 3/5*x*log(x) - 3/5*e*log(x) - 12)
Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )} \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx=2 \, x^{2} e^{\left (3 \, x^{2} + 3 \, x e - \frac {3}{5} \, x \log \left (x\right ) - \frac {3}{5} \, e \log \left (x\right ) - 12\right )} \] Input:
integrate(1/5*(-6*x^2*log(x)+(30*x^2-6*x)*exp(1)+60*x^3-6*x^2+20*x)/exp(1/ 5*(3*exp(1)+3*x)*log(x)-3*exp(1)*x-3*x^2+12),x, algorithm="giac")
Output:
2*x^2*e^(3*x^2 + 3*x*e - 3/5*x*log(x) - 3/5*e*log(x) - 12)
Time = 3.55 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )} \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx=2\,x^{2-\frac {3\,\mathrm {e}}{5}-\frac {3\,x}{5}}\,{\mathrm {e}}^{-12}\,{\mathrm {e}}^{3\,x^2}\,{\mathrm {e}}^{3\,x\,\mathrm {e}} \] Input:
int(-exp(3*x*exp(1) - (log(x)*(3*x + 3*exp(1)))/5 + 3*x^2 - 12)*((6*x^2*lo g(x))/5 - 4*x + (exp(1)*(6*x - 30*x^2))/5 + (6*x^2)/5 - 12*x^3),x)
Output:
2*x^(2 - (3*exp(1))/5 - (3*x)/5)*exp(-12)*exp(3*x^2)*exp(3*x*exp(1))
Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {1}{5} e^{\frac {1}{5} \left (-60+15 e x+15 x^2-(3 e+3 x) \log (x)\right )} \left (20 x-6 x^2+60 x^3+e \left (-6 x+30 x^2\right )-6 x^2 \log (x)\right ) \, dx=\frac {2 e^{3 e x +3 x^{2}} x^{2}}{x^{\frac {3 e}{5}+\frac {3 x}{5}} e^{12}} \] Input:
int(1/5*(-6*x^2*log(x)+(30*x^2-6*x)*exp(1)+60*x^3-6*x^2+20*x)/exp(1/5*(3*e xp(1)+3*x)*log(x)-3*exp(1)*x-3*x^2+12),x)
Output:
(2*e**(3*e*x + 3*x**2)*x**2)/(x**((3*e + 3*x)/5)*e**12)