Integrand size = 68, antiderivative size = 23 \[ \int \frac {1}{5} e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \left (6+24 x+75 e^x x^2+\left (18 x-24 x^2+150 e^x x^2\right ) \log (x)\right ) \, dx=x^{3 x \left (\frac {2}{5} e^{-x} \left (4+\frac {1}{x}\right )+5 x\right )} \] Output:
exp(3*x*ln(x)*(5*x+2/5/exp(x)*(1/x+4)))
Time = 4.21 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{5} e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \left (6+24 x+75 e^x x^2+\left (18 x-24 x^2+150 e^x x^2\right ) \log (x)\right ) \, dx=x^{\frac {3}{5} e^{-x} \left (2+8 x+25 e^x x^2\right )} \] Input:
Integrate[(x^(-1 + (6 + 24*x + 75*E^x*x^2)/(5*E^x))*(6 + 24*x + 75*E^x*x^2 + (18*x - 24*x^2 + 150*E^x*x^2)*Log[x]))/(5*E^x),x]
Output:
x^((3*(2 + 8*x + 25*E^x*x^2))/(5*E^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.
\(\displaystyle \int \frac {1}{5} e^{-x} x^{\frac {1}{5} e^{-x} \left (75 e^x x^2+24 x+6\right )-1} \left (75 e^x x^2+\left (150 e^x x^2-24 x^2+18 x\right ) \log (x)+24 x+6\right ) \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int 3 e^{-x} x^{\frac {3}{5} e^{-x} \left (25 e^x x^2+8 x+2\right )-1} \left (25 e^x x^2+8 x+2 \left (25 e^x x^2-4 x^2+3 x\right ) \log (x)+2\right )dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3}{5} \int e^{-x} x^{\frac {3}{5} e^{-x} \left (25 e^x x^2+8 x+2\right )-1} \left (25 e^x x^2+8 x+2 \left (25 e^x x^2-4 x^2+3 x\right ) \log (x)+2\right )dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {3}{5} \int \left (8 e^{-x} x^{\frac {3}{5} e^{-x} \left (25 e^x x^2+8 x+2\right )}+2 e^{-x} \left (25 e^x x-4 x+3\right ) \log (x) x^{\frac {3}{5} e^{-x} \left (25 e^x x^2+8 x+2\right )}+2 e^{-x} x^{\frac {3}{5} e^{-x} \left (25 e^x x^2+8 x+2\right )-1}+25 x^{\frac {3}{5} e^{-x} \left (25 e^x x^2+8 x+2\right )+1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3}{5} \left (8 \int e^{-x} x^{\frac {3}{5} e^{-x} \left (25 e^x x^2+8 x+2\right )}dx+2 \int e^{-x} x^{\frac {3}{5} e^{-x} \left (25 e^x x^2+8 x+2\right )-1}dx+25 \int x^{\frac {3}{5} e^{-x} \left (25 e^x x^2+8 x+2\right )+1}dx-6 \int \frac {\int e^{-x} x^{\frac {3}{5} e^{-x} \left (25 e^x x^2+8 x+2\right )}dx}{x}dx-50 \int \frac {\int x^{15 x^2+\frac {6}{5} e^{-x} (4 x+1)+1}dx}{x}dx+8 \int \frac {\int e^{-x} x^{15 x^2+\frac {6}{5} e^{-x} (4 x+1)+1}dx}{x}dx+6 \log (x) \int e^{-x} x^{\frac {3}{5} e^{-x} \left (25 e^x x^2+8 x+2\right )}dx+50 \log (x) \int x^{\frac {3}{5} e^{-x} \left (25 e^x x^2+8 x+2\right )+1}dx-8 \log (x) \int e^{-x} x^{\frac {3}{5} e^{-x} \left (25 e^x x^2+8 x+2\right )+1}dx\right )\) |
Input:
Int[(x^(-1 + (6 + 24*x + 75*E^x*x^2)/(5*E^x))*(6 + 24*x + 75*E^x*x^2 + (18 *x - 24*x^2 + 150*E^x*x^2)*Log[x]))/(5*E^x),x]
Output:
$Aborted
Time = 0.83 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96
method | result | size |
risch | \(x^{15 x^{2}+\frac {24 x \,{\mathrm e}^{-x}}{5}+\frac {6 \,{\mathrm e}^{-x}}{5}}\) | \(22\) |
parallelrisch | \({\mathrm e}^{\frac {\left (75 \,{\mathrm e}^{x} x^{2}+24 x +6\right ) \ln \left (x \right ) {\mathrm e}^{-x}}{5}}\) | \(22\) |
Input:
int(1/5*((150*exp(x)*x^2-24*x^2+18*x)*ln(x)+75*exp(x)*x^2+24*x+6)*exp(1/5* (75*exp(x)*x^2+24*x+6)*ln(x)/exp(x))/exp(x)/x,x,method=_RETURNVERBOSE)
Output:
x^(15*x^2+24/5*x*exp(-x)+6/5*exp(-x))
Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {1}{5} e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \left (6+24 x+75 e^x x^2+\left (18 x-24 x^2+150 e^x x^2\right ) \log (x)\right ) \, dx=x^{\frac {3}{5} \, {\left (25 \, x^{2} e^{x} + 8 \, x + 2\right )} e^{\left (-x\right )}} \] Input:
integrate(1/5*((150*exp(x)*x^2-24*x^2+18*x)*log(x)+75*exp(x)*x^2+24*x+6)*e xp(1/5*(75*exp(x)*x^2+24*x+6)*log(x)/exp(x))/exp(x)/x,x, algorithm="fricas ")
Output:
x^(3/5*(25*x^2*e^x + 8*x + 2)*e^(-x))
Time = 0.34 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {1}{5} e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \left (6+24 x+75 e^x x^2+\left (18 x-24 x^2+150 e^x x^2\right ) \log (x)\right ) \, dx=e^{\left (15 x^{2} e^{x} + \frac {24 x}{5} + \frac {6}{5}\right ) e^{- x} \log {\left (x \right )}} \] Input:
integrate(1/5*((150*exp(x)*x**2-24*x**2+18*x)*ln(x)+75*exp(x)*x**2+24*x+6) *exp(1/5*(75*exp(x)*x**2+24*x+6)*ln(x)/exp(x))/exp(x)/x,x)
Output:
exp((15*x**2*exp(x) + 24*x/5 + 6/5)*exp(-x)*log(x))
Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {1}{5} e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \left (6+24 x+75 e^x x^2+\left (18 x-24 x^2+150 e^x x^2\right ) \log (x)\right ) \, dx=e^{\left (15 \, x^{2} \log \left (x\right ) + \frac {24}{5} \, x e^{\left (-x\right )} \log \left (x\right ) + \frac {6}{5} \, e^{\left (-x\right )} \log \left (x\right )\right )} \] Input:
integrate(1/5*((150*exp(x)*x^2-24*x^2+18*x)*log(x)+75*exp(x)*x^2+24*x+6)*e xp(1/5*(75*exp(x)*x^2+24*x+6)*log(x)/exp(x))/exp(x)/x,x, algorithm="maxima ")
Output:
e^(15*x^2*log(x) + 24/5*x*e^(-x)*log(x) + 6/5*e^(-x)*log(x))
Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09 \[ \int \frac {1}{5} e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \left (6+24 x+75 e^x x^2+\left (18 x-24 x^2+150 e^x x^2\right ) \log (x)\right ) \, dx=x^{15 \, x^{2}} x^{\frac {24}{5} \, x e^{\left (-x\right )}} x^{\frac {6}{5} \, e^{\left (-x\right )}} \] Input:
integrate(1/5*((150*exp(x)*x^2-24*x^2+18*x)*log(x)+75*exp(x)*x^2+24*x+6)*e xp(1/5*(75*exp(x)*x^2+24*x+6)*log(x)/exp(x))/exp(x)/x,x, algorithm="giac")
Output:
x^(15*x^2)*x^(24/5*x*e^(-x))*x^(6/5*e^(-x))
Time = 2.35 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.87 \[ \int \frac {1}{5} e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \left (6+24 x+75 e^x x^2+\left (18 x-24 x^2+150 e^x x^2\right ) \log (x)\right ) \, dx=x^{\frac {3\,{\mathrm {e}}^{-x}\,\left (8\,x+25\,x^2\,{\mathrm {e}}^x+2\right )}{5}} \] Input:
int((exp((exp(-x)*log(x)*(24*x + 75*x^2*exp(x) + 6))/5)*exp(-x)*(24*x + 75 *x^2*exp(x) + log(x)*(18*x + 150*x^2*exp(x) - 24*x^2) + 6))/(5*x),x)
Output:
x^((3*exp(-x)*(8*x + 25*x^2*exp(x) + 2))/5)
Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {1}{5} e^{-x} x^{-1+\frac {1}{5} e^{-x} \left (6+24 x+75 e^x x^2\right )} \left (6+24 x+75 e^x x^2+\left (18 x-24 x^2+150 e^x x^2\right ) \log (x)\right ) \, dx=x^{15 x^{2}} e^{\frac {24 \,\mathrm {log}\left (x \right ) x +6 \,\mathrm {log}\left (x \right )}{5 e^{x}}} \] Input:
int(1/5*((150*exp(x)*x^2-24*x^2+18*x)*log(x)+75*exp(x)*x^2+24*x+6)*exp(1/5 *(75*exp(x)*x^2+24*x+6)*log(x)/exp(x))/exp(x)/x,x)
Output:
x**(15*x**2)*e**((24*log(x)*x + 6*log(x))/(5*e**x))