Integrand size = 108, antiderivative size = 25 \[ \int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx=\log \left (\frac {e^{x^2}}{5 x^2}-\log \left (4+e^{2 x}+x\right )\right ) \] Output:
ln(1/5*exp(x^2)/x^2-ln(exp(x)^2+4+x))
Time = 0.07 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.52 \[ \int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx=-2 x-2 \log (x)+\log \left (e^{2 x+x^2}-5 e^{2 x} x^2 \log \left (4+e^{2 x}+x\right )\right ) \] Input:
Integrate[(5*x^3 + 10*E^(2*x)*x^3 + E^x^2*(8 + 2*x - 8*x^2 - 2*x^3 + E^(2* x)*(2 - 2*x^2)))/(E^x^2*(-4*x - E^(2*x)*x - x^2) + (20*x^3 + 5*E^(2*x)*x^3 + 5*x^4)*Log[4 + E^(2*x) + x]),x]
Output:
-2*x - 2*Log[x] + Log[E^(2*x + x^2) - 5*E^(2*x)*x^2*Log[4 + E^(2*x) + 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.
\(\displaystyle \int \frac {10 e^{2 x} x^3+5 x^3+e^{x^2} \left (-2 x^3-8 x^2+e^{2 x} \left (2-2 x^2\right )+2 x+8\right )}{e^{x^2} \left (-x^2-e^{2 x} x-4 x\right )+\left (5 x^4+5 e^{2 x} x^3+20 x^3\right ) \log \left (x+e^{2 x}+4\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-10 e^{2 x} x^3-5 x^3-e^{x^2} \left (-2 x^3-8 x^2+e^{2 x} \left (2-2 x^2\right )+2 x+8\right )}{x \left (x+e^{2 x}+4\right ) \left (e^{x^2}-5 x^2 \log \left (x+e^{2 x}+4\right )\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {2 \left (x^2-1\right )}{x}-\frac {5 x \left (2 x^3 \log \left (x+e^{2 x}+4\right )+2 e^{2 x} x^2 \log \left (x+e^{2 x}+4\right )+8 x^2 \log \left (x+e^{2 x}+4\right )-2 e^{2 x} x-x-2 x \log \left (x+e^{2 x}+4\right )-2 e^{2 x} \log \left (x+e^{2 x}+4\right )-8 \log \left (x+e^{2 x}+4\right )\right )}{\left (x+e^{2 x}+4\right ) \left (5 x^2 \log \left (x+e^{2 x}+4\right )-e^{x^2}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 \int \frac {x^2}{\left (x+e^{2 x}+4\right ) \left (5 x^2 \log \left (x+e^{2 x}+4\right )-e^{x^2}\right )}dx+10 \int \frac {e^{2 x} x^2}{\left (x+e^{2 x}+4\right ) \left (5 x^2 \log \left (x+e^{2 x}+4\right )-e^{x^2}\right )}dx+40 \int \frac {x \log \left (x+e^{2 x}+4\right )}{\left (x+e^{2 x}+4\right ) \left (5 x^2 \log \left (x+e^{2 x}+4\right )-e^{x^2}\right )}dx+10 \int \frac {e^{2 x} x \log \left (x+e^{2 x}+4\right )}{\left (x+e^{2 x}+4\right ) \left (5 x^2 \log \left (x+e^{2 x}+4\right )-e^{x^2}\right )}dx+10 \int \frac {x^2 \log \left (x+e^{2 x}+4\right )}{\left (x+e^{2 x}+4\right ) \left (5 x^2 \log \left (x+e^{2 x}+4\right )-e^{x^2}\right )}dx-10 \int \frac {x^4 \log \left (x+e^{2 x}+4\right )}{\left (x+e^{2 x}+4\right ) \left (5 x^2 \log \left (x+e^{2 x}+4\right )-e^{x^2}\right )}dx-40 \int \frac {x^3 \log \left (x+e^{2 x}+4\right )}{\left (x+e^{2 x}+4\right ) \left (5 x^2 \log \left (x+e^{2 x}+4\right )-e^{x^2}\right )}dx-10 \int \frac {e^{2 x} x^3 \log \left (x+e^{2 x}+4\right )}{\left (x+e^{2 x}+4\right ) \left (5 x^2 \log \left (x+e^{2 x}+4\right )-e^{x^2}\right )}dx+x^2-2 \log (x)\) |
Input:
Int[(5*x^3 + 10*E^(2*x)*x^3 + E^x^2*(8 + 2*x - 8*x^2 - 2*x^3 + E^(2*x)*(2 - 2*x^2)))/(E^x^2*(-4*x - E^(2*x)*x - x^2) + (20*x^3 + 5*E^(2*x)*x^3 + 5*x ^4)*Log[4 + E^(2*x) + x]),x]
Output:
$Aborted
Time = 4.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80
method | result | size |
risch | \(\ln \left (\ln \left ({\mathrm e}^{2 x}+4+x \right )-\frac {{\mathrm e}^{x^{2}}}{5 x^{2}}\right )\) | \(20\) |
parallelrisch | \(-2 \ln \left (x \right )+\ln \left (\ln \left ({\mathrm e}^{2 x}+4+x \right ) x^{2}-\frac {{\mathrm e}^{x^{2}}}{5}\right )\) | \(26\) |
Input:
int((((-2*x^2+2)*exp(x)^2-2*x^3-8*x^2+2*x+8)*exp(x^2)+10*exp(x)^2*x^3+5*x^ 3)/((5*exp(x)^2*x^3+5*x^4+20*x^3)*ln(exp(x)^2+4+x)+(-x*exp(x)^2-x^2-4*x)*e xp(x^2)),x,method=_RETURNVERBOSE)
Output:
ln(ln(exp(2*x)+4+x)-1/5*exp(x^2)/x^2)
Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx=\log \left (\frac {5 \, x^{2} \log \left (x + e^{\left (2 \, x\right )} + 4\right ) - e^{\left (x^{2}\right )}}{x^{2}}\right ) \] Input:
integrate((((-2*x^2+2)*exp(x)^2-2*x^3-8*x^2+2*x+8)*exp(x^2)+10*exp(x)^2*x^ 3+5*x^3)/((5*exp(x)^2*x^3+5*x^4+20*x^3)*log(exp(x)^2+4+x)+(-x*exp(x)^2-x^2 -4*x)*exp(x^2)),x, algorithm="fricas")
Output:
log((5*x^2*log(x + e^(2*x) + 4) - e^(x^2))/x^2)
Time = 0.41 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx=\log {\left (\log {\left (x + e^{2 x} + 4 \right )} - \frac {e^{x^{2}}}{5 x^{2}} \right )} \] Input:
integrate((((-2*x**2+2)*exp(x)**2-2*x**3-8*x**2+2*x+8)*exp(x**2)+10*exp(x) **2*x**3+5*x**3)/((5*exp(x)**2*x**3+5*x**4+20*x**3)*ln(exp(x)**2+4+x)+(-x* exp(x)**2-x**2-4*x)*exp(x**2)),x)
Output:
log(log(x + exp(2*x) + 4) - exp(x**2)/(5*x**2))
Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx=\log \left (\frac {5 \, x^{2} \log \left (x + e^{\left (2 \, x\right )} + 4\right ) - e^{\left (x^{2}\right )}}{5 \, x^{2}}\right ) \] Input:
integrate((((-2*x^2+2)*exp(x)^2-2*x^3-8*x^2+2*x+8)*exp(x^2)+10*exp(x)^2*x^ 3+5*x^3)/((5*exp(x)^2*x^3+5*x^4+20*x^3)*log(exp(x)^2+4+x)+(-x*exp(x)^2-x^2 -4*x)*exp(x^2)),x, algorithm="maxima")
Output:
log(1/5*(5*x^2*log(x + e^(2*x) + 4) - e^(x^2))/x^2)
\[ \int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx=\int { -\frac {10 \, x^{3} e^{\left (2 \, x\right )} + 5 \, x^{3} - 2 \, {\left (x^{3} + 4 \, x^{2} + {\left (x^{2} - 1\right )} e^{\left (2 \, x\right )} - x - 4\right )} e^{\left (x^{2}\right )}}{{\left (x^{2} + x e^{\left (2 \, x\right )} + 4 \, x\right )} e^{\left (x^{2}\right )} - 5 \, {\left (x^{4} + x^{3} e^{\left (2 \, x\right )} + 4 \, x^{3}\right )} \log \left (x + e^{\left (2 \, x\right )} + 4\right )} \,d x } \] Input:
integrate((((-2*x^2+2)*exp(x)^2-2*x^3-8*x^2+2*x+8)*exp(x^2)+10*exp(x)^2*x^ 3+5*x^3)/((5*exp(x)^2*x^3+5*x^4+20*x^3)*log(exp(x)^2+4+x)+(-x*exp(x)^2-x^2 -4*x)*exp(x^2)),x, algorithm="giac")
Output:
integrate(-(10*x^3*e^(2*x) + 5*x^3 - 2*(x^3 + 4*x^2 + (x^2 - 1)*e^(2*x) - x - 4)*e^(x^2))/((x^2 + x*e^(2*x) + 4*x)*e^(x^2) - 5*(x^4 + x^3*e^(2*x) + 4*x^3)*log(x + e^(2*x) + 4)), x)
Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx=\ln \left (5\,x^2\,\ln \left (x+{\mathrm {e}}^{2\,x}+4\right )-{\mathrm {e}}^{x^2}\right )+\ln \left (\frac {1}{x^2}\right ) \] Input:
int((10*x^3*exp(2*x) - exp(x^2)*(exp(2*x)*(2*x^2 - 2) - 2*x + 8*x^2 + 2*x^ 3 - 8) + 5*x^3)/(log(x + exp(2*x) + 4)*(5*x^3*exp(2*x) + 20*x^3 + 5*x^4) - exp(x^2)*(4*x + x*exp(2*x) + x^2)),x)
Output:
log(5*x^2*log(x + exp(2*x) + 4) - exp(x^2)) + log(1/x^2)
Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {5 x^3+10 e^{2 x} x^3+e^{x^2} \left (8+2 x-8 x^2-2 x^3+e^{2 x} \left (2-2 x^2\right )\right )}{e^{x^2} \left (-4 x-e^{2 x} x-x^2\right )+\left (20 x^3+5 e^{2 x} x^3+5 x^4\right ) \log \left (4+e^{2 x}+x\right )} \, dx=-\mathrm {log}\left (x^{2}\right )+\mathrm {log}\left (e^{x^{2}}-5 \,\mathrm {log}\left (e^{2 x}+x +4\right ) x^{2}\right ) \] Input:
int((((-2*x^2+2)*exp(x)^2-2*x^3-8*x^2+2*x+8)*exp(x^2)+10*exp(x)^2*x^3+5*x^ 3)/((5*exp(x)^2*x^3+5*x^4+20*x^3)*log(exp(x)^2+4+x)+(-x*exp(x)^2-x^2-4*x)* exp(x^2)),x)
Output:
- log(x**2) + log(e**(x**2) - 5*log(e**(2*x) + x + 4)*x**2)