Integrand size = 50, antiderivative size = 19 \[ \int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx=e^{\frac {1}{25} x^2 (x+\log (3)-25 \log (5 x))} \] Output:
exp(1/25*x^2*(ln(3)+x-25*ln(5*x)))
\[ \int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx=\int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx \] Input:
Integrate[(E^((x^3 + x^2*Log[3] - 25*x^2*Log[5*x])/25)*(-25*x + 3*x^2 + 2* x*Log[3] - 50*x*Log[5*x]))/25,x]
Output:
Integrate[E^((x^3 + x^2*Log[3] - 25*x^2*Log[5*x])/25)*(-25*x + 3*x^2 + 2*x *Log[3] - 50*x*Log[5*x]), x]/25
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.
\(\displaystyle \int \frac {1}{25} e^{\frac {1}{25} \left (x^3-25 x^2 \log (5 x)+x^2 \log (3)\right )} \left (3 x^2-25 x-50 x \log (5 x)+2 x \log (3)\right ) \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {1}{25} e^{\frac {1}{25} \left (x^3-25 x^2 \log (5 x)+x^2 \log (3)\right )} \left (3 x^2-50 x \log (5 x)+x (2 \log (3)-25)\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{25} \int 3^{\frac {x^2}{25}} 5^{-x^2} e^{\frac {x^3}{25}} x^{-x^2} \left (3 x^2-50 \log (5 x) x-(25-\log (9)) x\right )dx\) |
\(\Big \downarrow \) 2725 |
\(\displaystyle \frac {1}{25} \int e^{\frac {x^3}{25}+\frac {1}{25} (\log (3)-25 \log (5)) x^2} x^{-x^2} \left (3 x^2-50 \log (5 x) x-(25-\log (9)) x\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{25} \int \left (-50 e^{\frac {x^3}{25}+\frac {1}{25} (\log (3)-25 \log (5)) x^2} \log (5 x) x^{1-x^2}+e^{\frac {x^3}{25}+\frac {1}{25} (\log (3)-25 \log (5)) x^2} (-25+\log (9)) x^{1-x^2}+3 e^{\frac {x^3}{25}+\frac {1}{25} (\log (3)-25 \log (5)) x^2} x^{2-x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{25} \left (50 \int \frac {\int e^{\frac {1}{25} x^2 (x-25 \log (5)+\log (3))} x^{1-x^2}dx}{x}dx-50 \log (5 x) \int e^{\frac {x^3}{25}+\frac {1}{25} (\log (3)-25 \log (5)) x^2} x^{1-x^2}dx-(25-\log (9)) \int e^{\frac {x^3}{25}+\frac {1}{25} (\log (3)-25 \log (5)) x^2} x^{1-x^2}dx+3 \int e^{\frac {x^3}{25}+\frac {1}{25} (\log (3)-25 \log (5)) x^2} x^{2-x^2}dx\right )\) |
Input:
Int[(E^((x^3 + x^2*Log[3] - 25*x^2*Log[5*x])/25)*(-25*x + 3*x^2 + 2*x*Log[ 3] - 50*x*Log[5*x]))/25,x]
Output:
$Aborted
Time = 0.20 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.89
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {x^{2} \left (\ln \left (3\right )+x -25 \ln \left (5 x \right )\right )}{25}}\) | \(17\) |
norman | \({\mathrm e}^{-x^{2} \ln \left (5 x \right )+\frac {x^{2} \ln \left (3\right )}{25}+\frac {x^{3}}{25}}\) | \(24\) |
risch | \(\left (5 x \right )^{-x^{2}} 3^{\frac {x^{2}}{25}} {\mathrm e}^{\frac {x^{3}}{25}}\) | \(24\) |
Input:
int(1/25*(-50*x*ln(5*x)+2*x*ln(3)+3*x^2-25*x)*exp(-x^2*ln(5*x)+1/25*x^2*ln (3)+1/25*x^3),x,method=_RETURNVERBOSE)
Output:
exp(1/25*x^2*(ln(3)+x-25*ln(5*x)))
Time = 0.08 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx=e^{\left (\frac {1}{25} \, x^{3} + \frac {1}{25} \, x^{2} \log \left (3\right ) - x^{2} \log \left (5 \, x\right )\right )} \] Input:
integrate(1/25*(-50*x*log(5*x)+2*x*log(3)+3*x^2-25*x)*exp(-x^2*log(5*x)+1/ 25*x^2*log(3)+1/25*x^3),x, algorithm="fricas")
Output:
e^(1/25*x^3 + 1/25*x^2*log(3) - x^2*log(5*x))
Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx=e^{\frac {x^{3}}{25} - x^{2} \log {\left (5 x \right )} + \frac {x^{2} \log {\left (3 \right )}}{25}} \] Input:
integrate(1/25*(-50*x*ln(5*x)+2*x*ln(3)+3*x**2-25*x)*exp(-x**2*ln(5*x)+1/2 5*x**2*ln(3)+1/25*x**3),x)
Output:
exp(x**3/25 - x**2*log(5*x) + x**2*log(3)/25)
Time = 0.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx=e^{\left (\frac {1}{25} \, x^{3} + \frac {1}{25} \, x^{2} \log \left (3\right ) - x^{2} \log \left (5 \, x\right )\right )} \] Input:
integrate(1/25*(-50*x*log(5*x)+2*x*log(3)+3*x^2-25*x)*exp(-x^2*log(5*x)+1/ 25*x^2*log(3)+1/25*x^3),x, algorithm="maxima")
Output:
e^(1/25*x^3 + 1/25*x^2*log(3) - x^2*log(5*x))
Time = 0.13 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.21 \[ \int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx=e^{\left (\frac {1}{25} \, x^{3} + \frac {1}{25} \, x^{2} \log \left (3\right ) - x^{2} \log \left (5 \, x\right )\right )} \] Input:
integrate(1/25*(-50*x*log(5*x)+2*x*log(3)+3*x^2-25*x)*exp(-x^2*log(5*x)+1/ 25*x^2*log(3)+1/25*x^3),x, algorithm="giac")
Output:
e^(1/25*x^3 + 1/25*x^2*log(3) - x^2*log(5*x))
Time = 3.78 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.37 \[ \int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx=\frac {{\left (\frac {1}{5}\right )}^{x^2}\,3^{\frac {x^2}{25}}\,{\mathrm {e}}^{\frac {x^3}{25}}}{x^{x^2}} \] Input:
int(-(exp((x^2*log(3))/25 - x^2*log(5*x) + x^3/25)*(25*x + 50*x*log(5*x) - 2*x*log(3) - 3*x^2))/25,x)
Output:
((1/5)^(x^2)*3^(x^2/25)*exp(x^3/25))/x^(x^2)
\[ \int \frac {1}{25} e^{\frac {1}{25} \left (x^3+x^2 \log (3)-25 x^2 \log (5 x)\right )} \left (-25 x+3 x^2+2 x \log (3)-50 x \log (5 x)\right ) \, dx=\frac {3 \left (\int \frac {e^{\frac {x^{3}}{25}} 3^{\frac {x^{2}}{25}} x^{2}}{x^{x^{2}} 5^{x^{2}}}d x \right )}{25}-2 \left (\int \frac {e^{\frac {x^{3}}{25}} 3^{\frac {x^{2}}{25}} \mathrm {log}\left (5 x \right ) x}{x^{x^{2}} 5^{x^{2}}}d x \right )+\frac {2 \left (\int \frac {e^{\frac {x^{3}}{25}} 3^{\frac {x^{2}}{25}} x}{x^{x^{2}} 5^{x^{2}}}d x \right ) \mathrm {log}\left (3\right )}{25}-\left (\int \frac {e^{\frac {x^{3}}{25}} 3^{\frac {x^{2}}{25}} x}{x^{x^{2}} 5^{x^{2}}}d x \right ) \] Input:
int(1/25*(-50*x*log(5*x)+2*x*log(3)+3*x^2-25*x)*exp(-x^2*log(5*x)+1/25*x^2 *log(3)+1/25*x^3),x)
Output:
(3*int((e**(x**3/25)*3**(x**2/25)*x**2)/(x**(x**2)*5**(x**2)),x) - 50*int( (e**(x**3/25)*3**(x**2/25)*log(5*x)*x)/(x**(x**2)*5**(x**2)),x) + 2*int((e **(x**3/25)*3**(x**2/25)*x)/(x**(x**2)*5**(x**2)),x)*log(3) - 25*int((e**( x**3/25)*3**(x**2/25)*x)/(x**(x**2)*5**(x**2)),x))/25