Integrand size = 97, antiderivative size = 20 \[ \int \frac {\left (-4 e^x x+e^{x+x^2} \left (4 x+8 x^2\right )\right ) \log ^3\left (e^x-e^{x+x^2}\right )+\left (4 e^x-4 e^{x+x^2}\right ) \log ^4\left (e^x-e^{x+x^2}\right )}{-e^x x^5+e^{x+x^2} x^5} \, dx=\frac {\log ^4\left (e^x-e^{x+x^2}\right )}{x^4} \] Output:
ln(-exp(x^2+x)+exp(1/2*x)^2)^4/x^4
Time = 0.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {\left (-4 e^x x+e^{x+x^2} \left (4 x+8 x^2\right )\right ) \log ^3\left (e^x-e^{x+x^2}\right )+\left (4 e^x-4 e^{x+x^2}\right ) \log ^4\left (e^x-e^{x+x^2}\right )}{-e^x x^5+e^{x+x^2} x^5} \, dx=-1+\frac {\log ^4\left (-e^x \left (-1+e^{x^2}\right )\right )}{x^4} \] Input:
Integrate[((-4*E^x*x + E^(x + x^2)*(4*x + 8*x^2))*Log[E^x - E^(x + x^2)]^3 + (4*E^x - 4*E^(x + x^2))*Log[E^x - E^(x + x^2)]^4)/(-(E^x*x^5) + E^(x + x^2)*x^5),x]
Output:
-1 + Log[-(E^x*(-1 + E^x^2))]^4/x^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.
\(\displaystyle \int \frac {\left (4 e^x-4 e^{x^2+x}\right ) \log ^4\left (e^x-e^{x^2+x}\right )+\left (e^{x^2+x} \left (8 x^2+4 x\right )-4 e^x x\right ) \log ^3\left (e^x-e^{x^2+x}\right )}{e^{x^2+x} x^5-e^x x^5} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {4 \log ^3\left (-e^x \left (e^{x^2}-1\right )\right ) \left (-2 e^{x^2} x^2-e^{x^2} x+e^{x^2} \log \left (-e^x \left (e^{x^2}-1\right )\right )-\log \left (-e^x \left (e^{x^2}-1\right )\right )+x\right )}{\left (1-e^{x^2}\right ) x^5}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int \frac {\log ^3\left (e^x \left (1-e^{x^2}\right )\right ) \left (-2 e^{x^2} x^2-e^{x^2} x+x+e^{x^2} \log \left (e^x \left (1-e^{x^2}\right )\right )-\log \left (e^x \left (1-e^{x^2}\right )\right )\right )}{\left (1-e^{x^2}\right ) x^5}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 4 \int \left (\frac {\left (2 x^2+x-\log \left (-e^x \left (-1+e^{x^2}\right )\right )\right ) \log ^3\left (-e^x \left (-1+e^{x^2}\right )\right )}{x^5}+\frac {2 \log ^3\left (-e^x \left (-1+e^{x^2}\right )\right )}{\left (-1+e^{x^2}\right ) x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \left (-\int \frac {\log ^4\left (-e^x \left (-1+e^{x^2}\right )\right )}{x^5}dx+\int \frac {\log ^3\left (-e^x \left (-1+e^{x^2}\right )\right )}{x^4}dx+2 \int \frac {\log ^3\left (-e^x \left (-1+e^{x^2}\right )\right )}{x^3}dx+2 \int \frac {\log ^3\left (-e^x \left (-1+e^{x^2}\right )\right )}{\left (-1+e^{x^2}\right ) x^3}dx\right )\) |
Input:
Int[((-4*E^x*x + E^(x + x^2)*(4*x + 8*x^2))*Log[E^x - E^(x + x^2)]^3 + (4* E^x - 4*E^(x + x^2))*Log[E^x - E^(x + x^2)]^4)/(-(E^x*x^5) + E^(x + x^2)*x ^5),x]
Output:
$Aborted
Time = 2.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95
method | result | size |
risch | \(\frac {\ln \left (-{\mathrm e}^{\left (1+x \right ) x}+{\mathrm e}^{x}\right )^{4}}{x^{4}}\) | \(19\) |
parallelrisch | \(\frac {\ln \left (-{\mathrm e}^{x^{2}+x}+{\mathrm e}^{x}\right )^{4}}{x^{4}}\) | \(23\) |
Input:
int(((-4*exp(x^2+x)+4*exp(1/2*x)^2)*ln(-exp(x^2+x)+exp(1/2*x)^2)^4+((8*x^2 +4*x)*exp(x^2+x)-4*x*exp(1/2*x)^2)*ln(-exp(x^2+x)+exp(1/2*x)^2)^3)/(x^5*ex p(x^2+x)-x^5*exp(1/2*x)^2),x,method=_RETURNVERBOSE)
Output:
ln(-exp((1+x)*x)+exp(x))^4/x^4
Time = 0.09 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {\left (-4 e^x x+e^{x+x^2} \left (4 x+8 x^2\right )\right ) \log ^3\left (e^x-e^{x+x^2}\right )+\left (4 e^x-4 e^{x+x^2}\right ) \log ^4\left (e^x-e^{x+x^2}\right )}{-e^x x^5+e^{x+x^2} x^5} \, dx=\frac {\log \left (-e^{\left (x^{2} + x\right )} + e^{x}\right )^{4}}{x^{4}} \] Input:
integrate(((-4*exp(x^2+x)+4*exp(1/2*x)^2)*log(-exp(x^2+x)+exp(1/2*x)^2)^4+ ((8*x^2+4*x)*exp(x^2+x)-4*x*exp(1/2*x)^2)*log(-exp(x^2+x)+exp(1/2*x)^2)^3) /(x^5*exp(x^2+x)-x^5*exp(1/2*x)^2),x, algorithm="fricas")
Output:
log(-e^(x^2 + x) + e^x)^4/x^4
Time = 0.22 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75 \[ \int \frac {\left (-4 e^x x+e^{x+x^2} \left (4 x+8 x^2\right )\right ) \log ^3\left (e^x-e^{x+x^2}\right )+\left (4 e^x-4 e^{x+x^2}\right ) \log ^4\left (e^x-e^{x+x^2}\right )}{-e^x x^5+e^{x+x^2} x^5} \, dx=\frac {\log {\left (e^{x} - e^{x^{2} + x} \right )}^{4}}{x^{4}} \] Input:
integrate(((-4*exp(x**2+x)+4*exp(1/2*x)**2)*ln(-exp(x**2+x)+exp(1/2*x)**2) **4+((8*x**2+4*x)*exp(x**2+x)-4*x*exp(1/2*x)**2)*ln(-exp(x**2+x)+exp(1/2*x )**2)**3)/(x**5*exp(x**2+x)-x**5*exp(1/2*x)**2),x)
Output:
log(exp(x) - exp(x**2 + x))**4/x**4
Leaf count of result is larger than twice the leaf count of optimal. 60 vs. \(2 (18) = 36\).
Time = 0.09 (sec) , antiderivative size = 60, normalized size of antiderivative = 3.00 \[ \int \frac {\left (-4 e^x x+e^{x+x^2} \left (4 x+8 x^2\right )\right ) \log ^3\left (e^x-e^{x+x^2}\right )+\left (4 e^x-4 e^{x+x^2}\right ) \log ^4\left (e^x-e^{x+x^2}\right )}{-e^x x^5+e^{x+x^2} x^5} \, dx=\frac {4 \, x^{3} \log \left (-e^{\left (x^{2}\right )} + 1\right ) + 6 \, x^{2} \log \left (-e^{\left (x^{2}\right )} + 1\right )^{2} + 4 \, x \log \left (-e^{\left (x^{2}\right )} + 1\right )^{3} + \log \left (-e^{\left (x^{2}\right )} + 1\right )^{4}}{x^{4}} \] Input:
integrate(((-4*exp(x^2+x)+4*exp(1/2*x)^2)*log(-exp(x^2+x)+exp(1/2*x)^2)^4+ ((8*x^2+4*x)*exp(x^2+x)-4*x*exp(1/2*x)^2)*log(-exp(x^2+x)+exp(1/2*x)^2)^3) /(x^5*exp(x^2+x)-x^5*exp(1/2*x)^2),x, algorithm="maxima")
Output:
(4*x^3*log(-e^(x^2) + 1) + 6*x^2*log(-e^(x^2) + 1)^2 + 4*x*log(-e^(x^2) + 1)^3 + log(-e^(x^2) + 1)^4)/x^4
Time = 1.13 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {\left (-4 e^x x+e^{x+x^2} \left (4 x+8 x^2\right )\right ) \log ^3\left (e^x-e^{x+x^2}\right )+\left (4 e^x-4 e^{x+x^2}\right ) \log ^4\left (e^x-e^{x+x^2}\right )}{-e^x x^5+e^{x+x^2} x^5} \, dx=\frac {\log \left (-e^{\left (x^{2} + x\right )} + e^{x}\right )^{4}}{x^{4}} \] Input:
integrate(((-4*exp(x^2+x)+4*exp(1/2*x)^2)*log(-exp(x^2+x)+exp(1/2*x)^2)^4+ ((8*x^2+4*x)*exp(x^2+x)-4*x*exp(1/2*x)^2)*log(-exp(x^2+x)+exp(1/2*x)^2)^3) /(x^5*exp(x^2+x)-x^5*exp(1/2*x)^2),x, algorithm="giac")
Output:
log(-e^(x^2 + x) + e^x)^4/x^4
Time = 7.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int \frac {\left (-4 e^x x+e^{x+x^2} \left (4 x+8 x^2\right )\right ) \log ^3\left (e^x-e^{x+x^2}\right )+\left (4 e^x-4 e^{x+x^2}\right ) \log ^4\left (e^x-e^{x+x^2}\right )}{-e^x x^5+e^{x+x^2} x^5} \, dx=\frac {{\ln \left ({\mathrm {e}}^x-{\mathrm {e}}^{x^2}\,{\mathrm {e}}^x\right )}^4}{x^4} \] Input:
int((log(exp(x) - exp(x + x^2))^3*(4*x*exp(x) - exp(x + x^2)*(4*x + 8*x^2) ) + log(exp(x) - exp(x + x^2))^4*(4*exp(x + x^2) - 4*exp(x)))/(x^5*exp(x) - x^5*exp(x + x^2)),x)
Output:
log(exp(x) - exp(x^2)*exp(x))^4/x^4
Time = 0.64 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-4 e^x x+e^{x+x^2} \left (4 x+8 x^2\right )\right ) \log ^3\left (e^x-e^{x+x^2}\right )+\left (4 e^x-4 e^{x+x^2}\right ) \log ^4\left (e^x-e^{x+x^2}\right )}{-e^x x^5+e^{x+x^2} x^5} \, dx=\frac {\mathrm {log}\left (-e^{x^{2}+x}+e^{x}\right )^{4}}{x^{4}} \] Input:
int(((-4*exp(x^2+x)+4*exp(1/2*x)^2)*log(-exp(x^2+x)+exp(1/2*x)^2)^4+((8*x^ 2+4*x)*exp(x^2+x)-4*x*exp(1/2*x)^2)*log(-exp(x^2+x)+exp(1/2*x)^2)^3)/(x^5* exp(x^2+x)-x^5*exp(1/2*x)^2),x)
Output:
log( - e**(x**2 + x) + e**x)**4/x**4