Integrand size = 48, antiderivative size = 25 \[ \int \frac {5-20 x-20 x \log (x)+e^x x \left (2 x^2+5 x^3+4 x^3 \log (x)\right )}{-20 x+4 e^x x^4} \, dx=5+x \log (x)+\frac {1}{4} \log \left (\frac {5}{x}-e^x x^2\right ) \] Output:
1/4*ln(5/x-exp(x+ln(x))*x)+x*ln(x)+5
Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {5-20 x-20 x \log (x)+e^x x \left (2 x^2+5 x^3+4 x^3 \log (x)\right )}{-20 x+4 e^x x^4} \, dx=\frac {1}{4} \left (-\log (x)+4 x \log (x)+\log \left (5-e^x x^3\right )\right ) \] Input:
Integrate[(5 - 20*x - 20*x*Log[x] + E^x*x*(2*x^2 + 5*x^3 + 4*x^3*Log[x]))/ (-20*x + 4*E^x*x^4),x]
Output:
(-Log[x] + 4*x*Log[x] + Log[5 - E^x*x^3])/4
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.
\(\displaystyle \int \frac {e^x x \left (5 x^3+4 x^3 \log (x)+2 x^2\right )-20 x-20 x \log (x)+5}{4 e^x x^4-20 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-e^x x \left (5 x^3+4 x^3 \log (x)+2 x^2\right )+20 x+20 x \log (x)-5}{4 x \left (5-e^x x^3\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \int -\frac {-20 \log (x) x+e^x \left (4 \log (x) x^3+5 x^3+2 x^2\right ) x-20 x+5}{x \left (5-e^x x^3\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{4} \int \frac {-20 \log (x) x+e^x \left (4 \log (x) x^3+5 x^3+2 x^2\right ) x-20 x+5}{x \left (5-e^x x^3\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{4} \int \left (\frac {-4 \log (x) x-5 x-2}{x}-\frac {5 (x+3)}{x \left (e^x x^3-5\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{4} \left (5 \int \frac {1}{e^x x^3-5}dx+15 \int \frac {1}{x \left (e^x x^3-5\right )}dx+x+4 x \log (x)+2 \log (x)\right )\) |
Input:
Int[(5 - 20*x - 20*x*Log[x] + E^x*x*(2*x^2 + 5*x^3 + 4*x^3*Log[x]))/(-20*x + 4*E^x*x^4),x]
Output:
$Aborted
Time = 0.23 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92
method | result | size |
risch | \(x \ln \left (x \right )+\frac {\ln \left (x \right )}{4}+\frac {\ln \left ({\mathrm e}^{x} x -\frac {5}{x^{2}}\right )}{4}\) | \(23\) |
norman | \(x \ln \left (x \right )-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (x^{2} {\mathrm e}^{x +\ln \left (x \right )}-5\right )}{4}\) | \(24\) |
parallelrisch | \(x \ln \left (x \right )-\frac {\ln \left (x \right )}{4}+\frac {\ln \left (x^{2} {\mathrm e}^{x +\ln \left (x \right )}-5\right )}{4}\) | \(24\) |
Input:
int(((4*x^3*ln(x)+5*x^3+2*x^2)*exp(x+ln(x))-20*x*ln(x)-20*x+5)/(4*x^3*exp( x+ln(x))-20*x),x,method=_RETURNVERBOSE)
Output:
x*ln(x)+1/4*ln(x)+1/4*ln(exp(x)*x-5/x^2)
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {5-20 x-20 x \log (x)+e^x x \left (2 x^2+5 x^3+4 x^3 \log (x)\right )}{-20 x+4 e^x x^4} \, dx=\frac {1}{4} \, {\left (4 \, x + 1\right )} \log \left (x\right ) + \frac {1}{4} \, \log \left (\frac {x^{2} e^{\left (x + \log \left (x\right )\right )} - 5}{x^{2}}\right ) \] Input:
integrate(((4*x^3*log(x)+5*x^3+2*x^2)*exp(x+log(x))-20*x*log(x)-20*x+5)/(4 *x^3*exp(x+log(x))-20*x),x, algorithm="fricas")
Output:
1/4*(4*x + 1)*log(x) + 1/4*log((x^2*e^(x + log(x)) - 5)/x^2)
Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {5-20 x-20 x \log (x)+e^x x \left (2 x^2+5 x^3+4 x^3 \log (x)\right )}{-20 x+4 e^x x^4} \, dx=x \log {\left (x \right )} + \frac {\log {\left (x \right )}}{2} + \frac {\log {\left (e^{x} - \frac {5}{x^{3}} \right )}}{4} \] Input:
integrate(((4*x**3*ln(x)+5*x**3+2*x**2)*exp(x+ln(x))-20*x*ln(x)-20*x+5)/(4 *x**3*exp(x+ln(x))-20*x),x)
Output:
x*log(x) + log(x)/2 + log(exp(x) - 5/x**3)/4
Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {5-20 x-20 x \log (x)+e^x x \left (2 x^2+5 x^3+4 x^3 \log (x)\right )}{-20 x+4 e^x x^4} \, dx=\frac {1}{2} \, {\left (2 \, x + 1\right )} \log \left (x\right ) + \frac {1}{4} \, \log \left (\frac {x^{3} e^{x} - 5}{x^{3}}\right ) \] Input:
integrate(((4*x^3*log(x)+5*x^3+2*x^2)*exp(x+log(x))-20*x*log(x)-20*x+5)/(4 *x^3*exp(x+log(x))-20*x),x, algorithm="maxima")
Output:
1/2*(2*x + 1)*log(x) + 1/4*log((x^3*e^x - 5)/x^3)
Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {5-20 x-20 x \log (x)+e^x x \left (2 x^2+5 x^3+4 x^3 \log (x)\right )}{-20 x+4 e^x x^4} \, dx=x \log \left (x\right ) + \frac {1}{4} \, \log \left (x^{3} e^{x} - 5\right ) - \frac {1}{4} \, \log \left (x\right ) \] Input:
integrate(((4*x^3*log(x)+5*x^3+2*x^2)*exp(x+log(x))-20*x*log(x)-20*x+5)/(4 *x^3*exp(x+log(x))-20*x),x, algorithm="giac")
Output:
x*log(x) + 1/4*log(x^3*e^x - 5) - 1/4*log(x)
Time = 1.86 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {5-20 x-20 x \log (x)+e^x x \left (2 x^2+5 x^3+4 x^3 \log (x)\right )}{-20 x+4 e^x x^4} \, dx=\frac {\ln \left (\frac {1}{x^2}\right )}{4}+\frac {\ln \left (x^3\,{\mathrm {e}}^x-5\right )}{4}+\frac {\ln \left (x\right )}{4}+x\,\ln \left (x\right ) \] Input:
int((20*x - exp(x + log(x))*(4*x^3*log(x) + 2*x^2 + 5*x^3) + 20*x*log(x) - 5)/(20*x - 4*x^3*exp(x + log(x))),x)
Output:
log(1/x^2)/4 + log(x^3*exp(x) - 5)/4 + log(x)/4 + x*log(x)
\[ \int \frac {5-20 x-20 x \log (x)+e^x x \left (2 x^2+5 x^3+4 x^3 \log (x)\right )}{-20 x+4 e^x x^4} \, dx=-5 \left (\int \frac {\mathrm {log}\left (x \right )}{e^{x} x^{3}-5}d x \right )+\frac {5 \left (\int \frac {e^{x} x^{3}}{e^{x} x^{3}-5}d x \right )}{4}+\frac {\left (\int \frac {e^{x} x^{2}}{e^{x} x^{3}-5}d x \right )}{2}+\int \frac {e^{x} \mathrm {log}\left (x \right ) x^{3}}{e^{x} x^{3}-5}d x +\frac {5 \left (\int \frac {1}{e^{x} x^{4}-5 x}d x \right )}{4}-5 \left (\int \frac {1}{e^{x} x^{3}-5}d x \right ) \] Input:
int(((4*x^3*log(x)+5*x^3+2*x^2)*exp(x+log(x))-20*x*log(x)-20*x+5)/(4*x^3*e xp(x+log(x))-20*x),x)
Output:
( - 20*int(log(x)/(e**x*x**3 - 5),x) + 5*int((e**x*x**3)/(e**x*x**3 - 5),x ) + 2*int((e**x*x**2)/(e**x*x**3 - 5),x) + 4*int((e**x*log(x)*x**3)/(e**x* x**3 - 5),x) + 5*int(1/(e**x*x**4 - 5*x),x) - 20*int(1/(e**x*x**3 - 5),x)) /4