Integrand size = 91, antiderivative size = 22 \[ \int \frac {-9-18 x+e^{x^2} \left (1+2 x-8 x^2-2 x^3-2 x^4\right )+\left (-36-9 x-9 x^2+e^{x^2} \left (4+x+x^2\right )\right ) \log \left (\frac {2}{-9+e^{x^2}}\right )}{-36-9 x-9 x^2+e^{x^2} \left (4+x+x^2\right )} \, dx=x \log \left (\frac {2}{-9+e^{x^2}}\right )+\log \left (4+x+x^2\right ) \] Output:
ln(2/(exp(x^2)-9))*x+ln(x^2+x+4)
Time = 0.38 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-9-18 x+e^{x^2} \left (1+2 x-8 x^2-2 x^3-2 x^4\right )+\left (-36-9 x-9 x^2+e^{x^2} \left (4+x+x^2\right )\right ) \log \left (\frac {2}{-9+e^{x^2}}\right )}{-36-9 x-9 x^2+e^{x^2} \left (4+x+x^2\right )} \, dx=x \log \left (\frac {2}{-9+e^{x^2}}\right )+\log \left (4+x+x^2\right ) \] Input:
Integrate[(-9 - 18*x + E^x^2*(1 + 2*x - 8*x^2 - 2*x^3 - 2*x^4) + (-36 - 9* x - 9*x^2 + E^x^2*(4 + x + x^2))*Log[2/(-9 + E^x^2)])/(-36 - 9*x - 9*x^2 + E^x^2*(4 + x + x^2)),x]
Output:
x*Log[2/(-9 + E^x^2)] + Log[4 + x + x^2]
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.
\(\displaystyle \int \frac {\left (-9 x^2+e^{x^2} \left (x^2+x+4\right )-9 x-36\right ) \log \left (\frac {2}{e^{x^2}-9}\right )+e^{x^2} \left (-2 x^4-2 x^3-8 x^2+2 x+1\right )-18 x-9}{-9 x^2+e^{x^2} \left (x^2+x+4\right )-9 x-36} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-\left (-9 x^2+e^{x^2} \left (x^2+x+4\right )-9 x-36\right ) \log \left (\frac {2}{e^{x^2}-9}\right )-e^{x^2} \left (-2 x^4-2 x^3-8 x^2+2 x+1\right )+18 x+9}{\left (9-e^{x^2}\right ) \left (x^2+x+4\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {-2 x^4-2 x^3-8 x^2+x^2 \log \left (\frac {2}{e^{x^2}-9}\right )+x \log \left (\frac {2}{e^{x^2}-9}\right )+4 \log \left (\frac {2}{e^{x^2}-9}\right )+2 x+1}{x^2+x+4}-\frac {18 x^2}{e^{x^2}-9}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -18 \int \frac {x^2}{-9+e^{x^2}}dx+2 \int \frac {e^{x^2} x^2}{-9+e^{x^2}}dx-\frac {2 x^3}{3}+x \log \left (-\frac {2}{9-e^{x^2}}\right )+\log \left (x^2+x+4\right )\) |
Input:
Int[(-9 - 18*x + E^x^2*(1 + 2*x - 8*x^2 - 2*x^3 - 2*x^4) + (-36 - 9*x - 9* x^2 + E^x^2*(4 + x + x^2))*Log[2/(-9 + E^x^2)])/(-36 - 9*x - 9*x^2 + E^x^2 *(4 + x + x^2)),x]
Output:
$Aborted
Time = 0.30 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00
method | result | size |
norman | \(\ln \left (\frac {2}{{\mathrm e}^{x^{2}}-9}\right ) x +\ln \left (x^{2}+x +4\right )\) | \(22\) |
parallelrisch | \(\ln \left (\frac {2}{{\mathrm e}^{x^{2}}-9}\right ) x +\ln \left (x^{2}+x +4\right )\) | \(22\) |
risch | \(-\ln \left ({\mathrm e}^{x^{2}}-9\right ) x +x \ln \left (2\right )+\ln \left (x^{2}+x +4\right )\) | \(23\) |
default | \(x \left (\ln \left (\frac {2}{{\mathrm e}^{x^{2}}-9}\right )+\ln \left ({\mathrm e}^{x^{2}}-9\right )\right )-\ln \left ({\mathrm e}^{x^{2}}-9\right ) x +\ln \left (x^{2}+x +4\right )\) | \(40\) |
Input:
int((((x^2+x+4)*exp(x^2)-9*x^2-9*x-36)*ln(2/(exp(x^2)-9))+(-2*x^4-2*x^3-8* x^2+2*x+1)*exp(x^2)-18*x-9)/((x^2+x+4)*exp(x^2)-9*x^2-9*x-36),x,method=_RE TURNVERBOSE)
Output:
ln(2/(exp(x^2)-9))*x+ln(x^2+x+4)
Time = 0.11 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-9-18 x+e^{x^2} \left (1+2 x-8 x^2-2 x^3-2 x^4\right )+\left (-36-9 x-9 x^2+e^{x^2} \left (4+x+x^2\right )\right ) \log \left (\frac {2}{-9+e^{x^2}}\right )}{-36-9 x-9 x^2+e^{x^2} \left (4+x+x^2\right )} \, dx=x \log \left (\frac {2}{e^{\left (x^{2}\right )} - 9}\right ) + \log \left (x^{2} + x + 4\right ) \] Input:
integrate((((x^2+x+4)*exp(x^2)-9*x^2-9*x-36)*log(2/(exp(x^2)-9))+(-2*x^4-2 *x^3-8*x^2+2*x+1)*exp(x^2)-18*x-9)/((x^2+x+4)*exp(x^2)-9*x^2-9*x-36),x, al gorithm="fricas")
Output:
x*log(2/(e^(x^2) - 9)) + log(x^2 + x + 4)
Time = 0.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {-9-18 x+e^{x^2} \left (1+2 x-8 x^2-2 x^3-2 x^4\right )+\left (-36-9 x-9 x^2+e^{x^2} \left (4+x+x^2\right )\right ) \log \left (\frac {2}{-9+e^{x^2}}\right )}{-36-9 x-9 x^2+e^{x^2} \left (4+x+x^2\right )} \, dx=x \log {\left (\frac {2}{e^{x^{2}} - 9} \right )} + \log {\left (x^{2} + x + 4 \right )} \] Input:
integrate((((x**2+x+4)*exp(x**2)-9*x**2-9*x-36)*ln(2/(exp(x**2)-9))+(-2*x* *4-2*x**3-8*x**2+2*x+1)*exp(x**2)-18*x-9)/((x**2+x+4)*exp(x**2)-9*x**2-9*x -36),x)
Output:
x*log(2/(exp(x**2) - 9)) + log(x**2 + x + 4)
Time = 0.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-9-18 x+e^{x^2} \left (1+2 x-8 x^2-2 x^3-2 x^4\right )+\left (-36-9 x-9 x^2+e^{x^2} \left (4+x+x^2\right )\right ) \log \left (\frac {2}{-9+e^{x^2}}\right )}{-36-9 x-9 x^2+e^{x^2} \left (4+x+x^2\right )} \, dx=x \log \left (2\right ) - x \log \left (e^{\left (x^{2}\right )} - 9\right ) + \log \left (x^{2} + x + 4\right ) \] Input:
integrate((((x^2+x+4)*exp(x^2)-9*x^2-9*x-36)*log(2/(exp(x^2)-9))+(-2*x^4-2 *x^3-8*x^2+2*x+1)*exp(x^2)-18*x-9)/((x^2+x+4)*exp(x^2)-9*x^2-9*x-36),x, al gorithm="maxima")
Output:
x*log(2) - x*log(e^(x^2) - 9) + log(x^2 + x + 4)
Time = 0.14 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-9-18 x+e^{x^2} \left (1+2 x-8 x^2-2 x^3-2 x^4\right )+\left (-36-9 x-9 x^2+e^{x^2} \left (4+x+x^2\right )\right ) \log \left (\frac {2}{-9+e^{x^2}}\right )}{-36-9 x-9 x^2+e^{x^2} \left (4+x+x^2\right )} \, dx=x \log \left (\frac {2}{e^{\left (x^{2}\right )} - 9}\right ) + \log \left (x^{2} + x + 4\right ) \] Input:
integrate((((x^2+x+4)*exp(x^2)-9*x^2-9*x-36)*log(2/(exp(x^2)-9))+(-2*x^4-2 *x^3-8*x^2+2*x+1)*exp(x^2)-18*x-9)/((x^2+x+4)*exp(x^2)-9*x^2-9*x-36),x, al gorithm="giac")
Output:
x*log(2/(e^(x^2) - 9)) + log(x^2 + x + 4)
Time = 4.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {-9-18 x+e^{x^2} \left (1+2 x-8 x^2-2 x^3-2 x^4\right )+\left (-36-9 x-9 x^2+e^{x^2} \left (4+x+x^2\right )\right ) \log \left (\frac {2}{-9+e^{x^2}}\right )}{-36-9 x-9 x^2+e^{x^2} \left (4+x+x^2\right )} \, dx=\ln \left (x^2+x+4\right )+x\,\ln \left (\frac {2}{{\mathrm {e}}^{x^2}-9}\right ) \] Input:
int((18*x + exp(x^2)*(8*x^2 - 2*x + 2*x^3 + 2*x^4 - 1) + log(2/(exp(x^2) - 9))*(9*x - exp(x^2)*(x + x^2 + 4) + 9*x^2 + 36) + 9)/(9*x - exp(x^2)*(x + x^2 + 4) + 9*x^2 + 36),x)
Output:
log(x + x^2 + 4) + x*log(2/(exp(x^2) - 9))
Time = 0.36 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {-9-18 x+e^{x^2} \left (1+2 x-8 x^2-2 x^3-2 x^4\right )+\left (-36-9 x-9 x^2+e^{x^2} \left (4+x+x^2\right )\right ) \log \left (\frac {2}{-9+e^{x^2}}\right )}{-36-9 x-9 x^2+e^{x^2} \left (4+x+x^2\right )} \, dx=\mathrm {log}\left (x^{2}+x +4\right )+\mathrm {log}\left (\frac {2}{e^{x^{2}}-9}\right ) x \] Input:
int((((x^2+x+4)*exp(x^2)-9*x^2-9*x-36)*log(2/(exp(x^2)-9))+(-2*x^4-2*x^3-8 *x^2+2*x+1)*exp(x^2)-18*x-9)/((x^2+x+4)*exp(x^2)-9*x^2-9*x-36),x)
Output:
log(x**2 + x + 4) + log(2/(e**(x**2) - 9))*x