Integrand size = 81, antiderivative size = 21 \[ \int \frac {e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \left (\left (-5+e^{x+x^2}\right ) \log \left (5-e^{x+x^2}\right )+e^{x+x^2} \left (x+2 x^2\right ) \log \left (-\frac {x}{4}\right )\right )}{-5 x+e^{x+x^2} x} \, dx=e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \] Output:
exp(ln(-1/4*x)*ln(-exp(x^2+x)+5))
Time = 0.48 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \left (\left (-5+e^{x+x^2}\right ) \log \left (5-e^{x+x^2}\right )+e^{x+x^2} \left (x+2 x^2\right ) \log \left (-\frac {x}{4}\right )\right )}{-5 x+e^{x+x^2} x} \, dx=\left (5-e^{x+x^2}\right )^{\log \left (-\frac {x}{4}\right )} \] Input:
Integrate[(E^(Log[5 - E^(x + x^2)]*Log[-1/4*x])*((-5 + E^(x + x^2))*Log[5 - E^(x + x^2)] + E^(x + x^2)*(x + 2*x^2)*Log[-1/4*x]))/(-5*x + E^(x + x^2) *x),x]
Output:
(5 - E^(x + x^2))^Log[-1/4*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.
\(\displaystyle \int \frac {e^{\log \left (5-e^{x^2+x}\right ) \log \left (-\frac {x}{4}\right )} \left (\left (e^{x^2+x}-5\right ) \log \left (5-e^{x^2+x}\right )+e^{x^2+x} \left (2 x^2+x\right ) \log \left (-\frac {x}{4}\right )\right )}{e^{x^2+x} x-5 x} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {5 (2 x+1) e^{\log \left (5-e^{x^2+x}\right ) \log \left (-\frac {x}{4}\right )} \log \left (-\frac {x}{4}\right )}{e^{x^2+x}-5}+\frac {e^{\log \left (5-e^{x^2+x}\right ) \log \left (-\frac {x}{4}\right )} \left (2 x^2 \log \left (-\frac {x}{4}\right )+\log \left (5-e^{x^2+x}\right )+x \log \left (-\frac {x}{4}\right )\right )}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \text {Subst}\left (\int e^{\log \left (5-e^{4 x (4 x+1)}\right ) \log (-x)} \log (-x)dx,x,\frac {x}{4}\right )+20 \text {Subst}\left (\int \frac {e^{\log \left (5-e^{4 x (4 x+1)}\right ) \log (-x)} \log (-x)}{-5+e^{4 x (4 x+1)}}dx,x,\frac {x}{4}\right )+32 \text {Subst}\left (\int e^{\log \left (5-e^{4 x (4 x+1)}\right ) \log (-x)} x \log (-x)dx,x,\frac {x}{4}\right )+160 \text {Subst}\left (\int \frac {e^{\log \left (5-e^{4 x (4 x+1)}\right ) \log (-x)} x \log (-x)}{-5+e^{4 x (4 x+1)}}dx,x,\frac {x}{4}\right )+\int \frac {e^{\log \left (5-e^{x^2+x}\right ) \log \left (-\frac {x}{4}\right )} \log \left (5-e^{x^2+x}\right )}{x}dx\) |
Input:
Int[(E^(Log[5 - E^(x + x^2)]*Log[-1/4*x])*((-5 + E^(x + x^2))*Log[5 - E^(x + x^2)] + E^(x + x^2)*(x + 2*x^2)*Log[-1/4*x]))/(-5*x + E^(x + x^2)*x),x]
Output:
$Aborted
Time = 4.89 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76
method | result | size |
risch | \(\left (-\frac {x}{4}\right )^{\ln \left (-{\mathrm e}^{\left (1+x \right ) x}+5\right )}\) | \(16\) |
parallelrisch | \({\mathrm e}^{\ln \left (-\frac {x}{4}\right ) \ln \left (-{\mathrm e}^{x^{2}+x}+5\right )}\) | \(18\) |
Input:
int(((exp(x^2+x)-5)*ln(-exp(x^2+x)+5)+(2*x^2+x)*exp(x^2+x)*ln(-1/4*x))*exp (ln(-1/4*x)*ln(-exp(x^2+x)+5))/(x*exp(x^2+x)-5*x),x,method=_RETURNVERBOSE)
Output:
(-1/4*x)^ln(-exp((1+x)*x)+5)
Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \left (\left (-5+e^{x+x^2}\right ) \log \left (5-e^{x+x^2}\right )+e^{x+x^2} \left (x+2 x^2\right ) \log \left (-\frac {x}{4}\right )\right )}{-5 x+e^{x+x^2} x} \, dx=e^{\left (\log \left (-\frac {1}{4} \, x\right ) \log \left (-e^{\left (x^{2} + x\right )} + 5\right )\right )} \] Input:
integrate(((exp(x^2+x)-5)*log(-exp(x^2+x)+5)+(2*x^2+x)*exp(x^2+x)*log(-1/4 *x))*exp(log(-1/4*x)*log(-exp(x^2+x)+5))/(x*exp(x^2+x)-5*x),x, algorithm=" fricas")
Output:
e^(log(-1/4*x)*log(-e^(x^2 + x) + 5))
Timed out. \[ \int \frac {e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \left (\left (-5+e^{x+x^2}\right ) \log \left (5-e^{x+x^2}\right )+e^{x+x^2} \left (x+2 x^2\right ) \log \left (-\frac {x}{4}\right )\right )}{-5 x+e^{x+x^2} x} \, dx=\text {Timed out} \] Input:
integrate(((exp(x**2+x)-5)*ln(-exp(x**2+x)+5)+(2*x**2+x)*exp(x**2+x)*ln(-1 /4*x))*exp(ln(-1/4*x)*ln(-exp(x**2+x)+5))/(x*exp(x**2+x)-5*x),x)
Output:
Timed out
Time = 0.33 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.57 \[ \int \frac {e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \left (\left (-5+e^{x+x^2}\right ) \log \left (5-e^{x+x^2}\right )+e^{x+x^2} \left (x+2 x^2\right ) \log \left (-\frac {x}{4}\right )\right )}{-5 x+e^{x+x^2} x} \, dx=e^{\left (-2 \, \log \left (2\right ) \log \left (-e^{\left (x^{2} + x\right )} + 5\right ) + \log \left (-x\right ) \log \left (-e^{\left (x^{2} + x\right )} + 5\right )\right )} \] Input:
integrate(((exp(x^2+x)-5)*log(-exp(x^2+x)+5)+(2*x^2+x)*exp(x^2+x)*log(-1/4 *x))*exp(log(-1/4*x)*log(-exp(x^2+x)+5))/(x*exp(x^2+x)-5*x),x, algorithm=" maxima")
Output:
e^(-2*log(2)*log(-e^(x^2 + x) + 5) + log(-x)*log(-e^(x^2 + x) + 5))
Time = 0.15 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \left (\left (-5+e^{x+x^2}\right ) \log \left (5-e^{x+x^2}\right )+e^{x+x^2} \left (x+2 x^2\right ) \log \left (-\frac {x}{4}\right )\right )}{-5 x+e^{x+x^2} x} \, dx=e^{\left (\log \left (-\frac {1}{4} \, x\right ) \log \left (-e^{\left (x^{2} + x\right )} + 5\right )\right )} \] Input:
integrate(((exp(x^2+x)-5)*log(-exp(x^2+x)+5)+(2*x^2+x)*exp(x^2+x)*log(-1/4 *x))*exp(log(-1/4*x)*log(-exp(x^2+x)+5))/(x*exp(x^2+x)-5*x),x, algorithm=" giac")
Output:
e^(log(-1/4*x)*log(-e^(x^2 + x) + 5))
Time = 2.68 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.71 \[ \int \frac {e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \left (\left (-5+e^{x+x^2}\right ) \log \left (5-e^{x+x^2}\right )+e^{x+x^2} \left (x+2 x^2\right ) \log \left (-\frac {x}{4}\right )\right )}{-5 x+e^{x+x^2} x} \, dx={\left (5-{\mathrm {e}}^{x^2+x}\right )}^{\ln \left (-\frac {x}{4}\right )} \] Input:
int(-(exp(log(5 - exp(x + x^2))*log(-x/4))*(log(5 - exp(x + x^2))*(exp(x + x^2) - 5) + log(-x/4)*exp(x + x^2)*(x + 2*x^2)))/(5*x - x*exp(x + x^2)),x )
Output:
(5 - exp(x + x^2))^log(-x/4)
\[ \int \frac {e^{\log \left (5-e^{x+x^2}\right ) \log \left (-\frac {x}{4}\right )} \left (\left (-5+e^{x+x^2}\right ) \log \left (5-e^{x+x^2}\right )+e^{x+x^2} \left (x+2 x^2\right ) \log \left (-\frac {x}{4}\right )\right )}{-5 x+e^{x+x^2} x} \, dx=\int \frac {\left (\left ({\mathrm e}^{x^{2}+x}-5\right ) \mathrm {log}\left (-{\mathrm e}^{x^{2}+x}+5\right )+\left (2 x^{2}+x \right ) {\mathrm e}^{x^{2}+x} \mathrm {log}\left (-\frac {x}{4}\right )\right ) {\mathrm e}^{\mathrm {log}\left (-\frac {x}{4}\right ) \mathrm {log}\left (-{\mathrm e}^{x^{2}+x}+5\right )}}{x \,{\mathrm e}^{x^{2}+x}-5 x}d x \] Input:
int(((exp(x^2+x)-5)*log(-exp(x^2+x)+5)+(2*x^2+x)*exp(x^2+x)*log(-1/4*x))*e xp(log(-1/4*x)*log(-exp(x^2+x)+5))/(x*exp(x^2+x)-5*x),x)
Output:
int(((exp(x^2+x)-5)*log(-exp(x^2+x)+5)+(2*x^2+x)*exp(x^2+x)*log(-1/4*x))*e xp(log(-1/4*x)*log(-exp(x^2+x)+5))/(x*exp(x^2+x)-5*x),x)