Integrand size = 120, antiderivative size = 38 \[ \int \frac {-x^3-e x^3+(-2-2 e-3 x) \log (5)+e^{2 x^2} \left (-4 x^6+\left (3 x+4 x^3\right ) \log (5)\right )}{x^3+e^2 x^3+2 x^4+x^5+e^{4 x^2} x^5+e \left (2 x^3+2 x^4\right )+e^{2 x^2} \left (-2 x^4-2 e x^4-2 x^5\right )} \, dx=\frac {x^2-\frac {\log (5)}{x}}{-x-e x-x^2+e^{2 x^2} x^2} \] Output:
(x^2-ln(5)/x)/(x^2*exp(x^2)^2-x-x*exp(1)-x^2)
Time = 0.66 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.84 \[ \int \frac {-x^3-e x^3+(-2-2 e-3 x) \log (5)+e^{2 x^2} \left (-4 x^6+\left (3 x+4 x^3\right ) \log (5)\right )}{x^3+e^2 x^3+2 x^4+x^5+e^{4 x^2} x^5+e \left (2 x^3+2 x^4\right )+e^{2 x^2} \left (-2 x^4-2 e x^4-2 x^5\right )} \, dx=-\frac {-x^3+\log (5)}{x^2 \left (-1-e-x+e^{2 x^2} x\right )} \] Input:
Integrate[(-x^3 - E*x^3 + (-2 - 2*E - 3*x)*Log[5] + E^(2*x^2)*(-4*x^6 + (3 *x + 4*x^3)*Log[5]))/(x^3 + E^2*x^3 + 2*x^4 + x^5 + E^(4*x^2)*x^5 + E*(2*x ^3 + 2*x^4) + E^(2*x^2)*(-2*x^4 - 2*E*x^4 - 2*x^5)),x]
Output:
-((-x^3 + Log[5])/(x^2*(-1 - E - x + E^(2*x^2)*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.
\(\displaystyle \int \frac {-e x^3-x^3+e^{2 x^2} \left (\left (4 x^3+3 x\right ) \log (5)-4 x^6\right )+(-3 x-2 e-2) \log (5)}{x^5+2 x^4+e^2 x^3+x^3+e^{4 x^2} x^5+e \left (2 x^4+2 x^3\right )+e^{2 x^2} \left (-2 x^5-2 e x^4-2 x^4\right )} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {-e x^3-x^3+e^{2 x^2} \left (\left (4 x^3+3 x\right ) \log (5)-4 x^6\right )+(-3 x-2 e-2) \log (5)}{x^5+2 x^4+\left (1+e^2\right ) x^3+e^{4 x^2} x^5+e \left (2 x^4+2 x^3\right )+e^{2 x^2} \left (-2 x^5-2 e x^4-2 x^4\right )}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {(-1-e) x^3+e^{2 x^2} \left (\left (4 x^3+3 x\right ) \log (5)-4 x^6\right )+(-3 x-2 e-2) \log (5)}{x^5+2 x^4+\left (1+e^2\right ) x^3+e^{4 x^2} x^5+e \left (2 x^4+2 x^3\right )+e^{2 x^2} \left (-2 x^5-2 e x^4-2 x^4\right )}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {(-1-e) x^3+e^{2 x^2} \left (\left (4 x^3+3 x\right ) \log (5)-4 x^6\right )+(-3 x-2 e-2) \log (5)}{x^3 \left (-e^{2 x^2} x+x+e+1\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\left (4 x^3+4 (1+e) x^2+e+1\right ) \left (\log (5)-x^3\right )}{x^3 \left (-e^{2 x^2} x+x+e+1\right )^2}-\frac {4 x^5-4 x^2 \log (5)-\log (125)}{x^3 \left (e^{2 x^2} x-x-e-1\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -4 (1+e) \int \frac {x^2}{\left (-e^{2 x^2} x+x+e+1\right )^2}dx-4 \int \frac {x^2}{e^{2 x^2} x-x-e-1}dx-\left ((1+e-4 \log (5)) \int \frac {1}{\left (-e^{2 x^2} x+x+e+1\right )^2}dx\right )+4 (1+e) \log (5) \int \frac {1}{x \left (-e^{2 x^2} x+x+e+1\right )^2}dx+4 \log (5) \int \frac {1}{x \left (e^{2 x^2} x-x-e-1\right )}dx-4 \int \frac {x^3}{\left (e^{2 x^2} x-x-e-1\right )^2}dx+(1+e) \log (5) \int \frac {1}{x^3 \left (-e^{2 x^2} x+x+e+1\right )^2}dx+\log (125) \int \frac {1}{x^3 \left (e^{2 x^2} x-x-e-1\right )}dx\) |
Input:
Int[(-x^3 - E*x^3 + (-2 - 2*E - 3*x)*Log[5] + E^(2*x^2)*(-4*x^6 + (3*x + 4 *x^3)*Log[5]))/(x^3 + E^2*x^3 + 2*x^4 + x^5 + E^(4*x^2)*x^5 + E*(2*x^3 + 2 *x^4) + E^(2*x^2)*(-2*x^4 - 2*E*x^4 - 2*x^5)),x]
Output:
$Aborted
Time = 0.98 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76
method | result | size |
risch | \(\frac {-x^{3}+\ln \left (5\right )}{x^{2} \left (-x \,{\mathrm e}^{2 x^{2}}+{\mathrm e}+x +1\right )}\) | \(29\) |
parallelrisch | \(\frac {-x^{3}+\ln \left (5\right )}{x^{2} \left (-x \,{\mathrm e}^{2 x^{2}}+{\mathrm e}+x +1\right )}\) | \(29\) |
Input:
int((((4*x^3+3*x)*ln(5)-4*x^6)*exp(x^2)^2+(-2*exp(1)-2-3*x)*ln(5)-x^3*exp( 1)-x^3)/(x^5*exp(x^2)^4+(-2*x^4*exp(1)-2*x^5-2*x^4)*exp(x^2)^2+x^3*exp(1)^ 2+(2*x^4+2*x^3)*exp(1)+x^5+2*x^4+x^3),x,method=_RETURNVERBOSE)
Output:
(-x^3+ln(5))/x^2/(-x*exp(2*x^2)+exp(1)+x+1)
Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.03 \[ \int \frac {-x^3-e x^3+(-2-2 e-3 x) \log (5)+e^{2 x^2} \left (-4 x^6+\left (3 x+4 x^3\right ) \log (5)\right )}{x^3+e^2 x^3+2 x^4+x^5+e^{4 x^2} x^5+e \left (2 x^3+2 x^4\right )+e^{2 x^2} \left (-2 x^4-2 e x^4-2 x^5\right )} \, dx=\frac {x^{3} - \log \left (5\right )}{x^{3} e^{\left (2 \, x^{2}\right )} - x^{3} - x^{2} e - x^{2}} \] Input:
integrate((((4*x^3+3*x)*log(5)-4*x^6)*exp(x^2)^2+(-2*exp(1)-2-3*x)*log(5)- x^3*exp(1)-x^3)/(x^5*exp(x^2)^4+(-2*x^4*exp(1)-2*x^5-2*x^4)*exp(x^2)^2+x^3 *exp(1)^2+(2*x^4+2*x^3)*exp(1)+x^5+2*x^4+x^3),x, algorithm="fricas")
Output:
(x^3 - log(5))/(x^3*e^(2*x^2) - x^3 - x^2*e - x^2)
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76 \[ \int \frac {-x^3-e x^3+(-2-2 e-3 x) \log (5)+e^{2 x^2} \left (-4 x^6+\left (3 x+4 x^3\right ) \log (5)\right )}{x^3+e^2 x^3+2 x^4+x^5+e^{4 x^2} x^5+e \left (2 x^3+2 x^4\right )+e^{2 x^2} \left (-2 x^4-2 e x^4-2 x^5\right )} \, dx=\frac {x^{3} - \log {\left (5 \right )}}{x^{3} e^{2 x^{2}} - x^{3} - e x^{2} - x^{2}} \] Input:
integrate((((4*x**3+3*x)*ln(5)-4*x**6)*exp(x**2)**2+(-2*exp(1)-2-3*x)*ln(5 )-x**3*exp(1)-x**3)/(x**5*exp(x**2)**4+(-2*x**4*exp(1)-2*x**5-2*x**4)*exp( x**2)**2+x**3*exp(1)**2+(2*x**4+2*x**3)*exp(1)+x**5+2*x**4+x**3),x)
Output:
(x**3 - log(5))/(x**3*exp(2*x**2) - x**3 - E*x**2 - x**2)
Time = 0.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.95 \[ \int \frac {-x^3-e x^3+(-2-2 e-3 x) \log (5)+e^{2 x^2} \left (-4 x^6+\left (3 x+4 x^3\right ) \log (5)\right )}{x^3+e^2 x^3+2 x^4+x^5+e^{4 x^2} x^5+e \left (2 x^3+2 x^4\right )+e^{2 x^2} \left (-2 x^4-2 e x^4-2 x^5\right )} \, dx=\frac {x^{3} - \log \left (5\right )}{x^{3} e^{\left (2 \, x^{2}\right )} - x^{3} - x^{2} {\left (e + 1\right )}} \] Input:
integrate((((4*x^3+3*x)*log(5)-4*x^6)*exp(x^2)^2+(-2*exp(1)-2-3*x)*log(5)- x^3*exp(1)-x^3)/(x^5*exp(x^2)^4+(-2*x^4*exp(1)-2*x^5-2*x^4)*exp(x^2)^2+x^3 *exp(1)^2+(2*x^4+2*x^3)*exp(1)+x^5+2*x^4+x^3),x, algorithm="maxima")
Output:
(x^3 - log(5))/(x^3*e^(2*x^2) - x^3 - x^2*(e + 1))
Time = 0.15 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.03 \[ \int \frac {-x^3-e x^3+(-2-2 e-3 x) \log (5)+e^{2 x^2} \left (-4 x^6+\left (3 x+4 x^3\right ) \log (5)\right )}{x^3+e^2 x^3+2 x^4+x^5+e^{4 x^2} x^5+e \left (2 x^3+2 x^4\right )+e^{2 x^2} \left (-2 x^4-2 e x^4-2 x^5\right )} \, dx=\frac {x^{3} - \log \left (5\right )}{x^{3} e^{\left (2 \, x^{2}\right )} - x^{3} - x^{2} e - x^{2}} \] Input:
integrate((((4*x^3+3*x)*log(5)-4*x^6)*exp(x^2)^2+(-2*exp(1)-2-3*x)*log(5)- x^3*exp(1)-x^3)/(x^5*exp(x^2)^4+(-2*x^4*exp(1)-2*x^5-2*x^4)*exp(x^2)^2+x^3 *exp(1)^2+(2*x^4+2*x^3)*exp(1)+x^5+2*x^4+x^3),x, algorithm="giac")
Output:
(x^3 - log(5))/(x^3*e^(2*x^2) - x^3 - x^2*e - x^2)
Time = 2.75 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.92 \[ \int \frac {-x^3-e x^3+(-2-2 e-3 x) \log (5)+e^{2 x^2} \left (-4 x^6+\left (3 x+4 x^3\right ) \log (5)\right )}{x^3+e^2 x^3+2 x^4+x^5+e^{4 x^2} x^5+e \left (2 x^3+2 x^4\right )+e^{2 x^2} \left (-2 x^4-2 e x^4-2 x^5\right )} \, dx=\frac {\ln \left (5\right )-x^3}{x^2\,\mathrm {e}-x^3\,{\mathrm {e}}^{2\,x^2}+x^2+x^3} \] Input:
int(-(x^3*exp(1) + log(5)*(3*x + 2*exp(1) + 2) - exp(2*x^2)*(log(5)*(3*x + 4*x^3) - 4*x^6) + x^3)/(exp(1)*(2*x^3 + 2*x^4) - exp(2*x^2)*(2*x^4*exp(1) + 2*x^4 + 2*x^5) + x^3*exp(2) + x^5*exp(4*x^2) + x^3 + 2*x^4 + x^5),x)
Output:
(log(5) - x^3)/(x^2*exp(1) - x^3*exp(2*x^2) + x^2 + x^3)
Time = 0.18 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {-x^3-e x^3+(-2-2 e-3 x) \log (5)+e^{2 x^2} \left (-4 x^6+\left (3 x+4 x^3\right ) \log (5)\right )}{x^3+e^2 x^3+2 x^4+x^5+e^{4 x^2} x^5+e \left (2 x^3+2 x^4\right )+e^{2 x^2} \left (-2 x^4-2 e x^4-2 x^5\right )} \, dx=\frac {-\mathrm {log}\left (5\right )+x^{3}}{x^{2} \left (e^{2 x^{2}} x -e -x -1\right )} \] Input:
int((((4*x^3+3*x)*log(5)-4*x^6)*exp(x^2)^2+(-2*exp(1)-2-3*x)*log(5)-x^3*ex p(1)-x^3)/(x^5*exp(x^2)^4+(-2*x^4*exp(1)-2*x^5-2*x^4)*exp(x^2)^2+x^3*exp(1 )^2+(2*x^4+2*x^3)*exp(1)+x^5+2*x^4+x^3),x)
Output:
( - log(5) + x**3)/(x**2*(e**(2*x**2)*x - e - x - 1))