Integrand size = 59, antiderivative size = 22 \[ \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx=\log \left (\log \left (e^{-1+x \left (4+\frac {3}{4} e^{-x^2} x\right )}\right )\right ) \] Output:
ln(ln(exp(x*(3/4*x/exp(x^2)+4))/exp(1)))
Time = 0.10 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx=\log \left (\log \left (e^{-1+4 x+\frac {3}{4} e^{-x^2} x^2}\right )\right ) \] Input:
Integrate[(8*E^x^2 + 3*x - 3*x^3)/(2*E^x^2*Log[E^(-1 + (16*E^x^2*x + 3*x^2 )/(4*E^x^2))]),x]
Output:
Log[Log[E^(-1 + 4*x + (3*x^2)/(4*E^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.
\(\displaystyle \int \frac {e^{-x^2} \left (-3 x^3+8 e^{x^2}+3 x\right )}{2 \log \left (e^{\frac {1}{4} e^{-x^2} \left (3 x^2+16 e^{x^2} x\right )-1}\right )} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {e^{-x^2} \left (-3 x^3+3 x+8 e^{x^2}\right )}{\log \left (e^{\frac {1}{4} e^{-x^2} \left (3 x^2+16 e^{x^2} x\right )-1}\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{2} \int \left (\frac {8}{\log \left (e^{\frac {3}{4} e^{-x^2} x^2+4 x-1}\right )}-\frac {3 e^{-x^2} x \left (x^2-1\right )}{\log \left (e^{\frac {3}{4} e^{-x^2} x^2+4 x-1}\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (8 \int \frac {1}{\log \left (e^{\frac {3}{4} e^{-x^2} x^2+4 x-1}\right )}dx+3 \int \frac {e^{-x^2} x}{\log \left (e^{\frac {3}{4} e^{-x^2} x^2+4 x-1}\right )}dx-3 \int \frac {e^{-x^2} x^3}{\log \left (e^{\frac {3}{4} e^{-x^2} x^2+4 x-1}\right )}dx\right )\) |
Input:
Int[(8*E^x^2 + 3*x - 3*x^3)/(2*E^x^2*Log[E^(-1 + (16*E^x^2*x + 3*x^2)/(4*E ^x^2))]),x]
Output:
$Aborted
Time = 0.38 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14
method | result | size |
risch | \(\ln \left (\ln \left ({\mathrm e}^{\frac {x \left (16 \,{\mathrm e}^{x^{2}}+3 x \right ) {\mathrm e}^{-x^{2}}}{4}}\right )-1\right )\) | \(25\) |
parallelrisch | \(\ln \left (\ln \left ({\mathrm e}^{\frac {x \left (16 \,{\mathrm e}^{x^{2}}+3 x \right ) {\mathrm e}^{-x^{2}}}{4}} {\mathrm e}^{-1}\right )\right )\) | \(28\) |
derivativedivides | \(\ln \left (\ln \left ({\mathrm e}^{\frac {\left (16 \,{\mathrm e}^{x^{2}} x +3 x^{2}\right ) {\mathrm e}^{-x^{2}}}{4}} {\mathrm e}^{-1}\right )\right )\) | \(30\) |
default | \(\ln \left (\ln \left ({\mathrm e}^{\frac {\left (16 \,{\mathrm e}^{x^{2}} x +3 x^{2}\right ) {\mathrm e}^{-x^{2}}}{4}} {\mathrm e}^{-1}\right )\right )\) | \(30\) |
Input:
int(1/2*(8*exp(x^2)-3*x^3+3*x)/exp(x^2)/ln(exp(1/4*(16*exp(x^2)*x+3*x^2)/e xp(x^2))/exp(1)),x,method=_RETURNVERBOSE)
Output:
ln(ln(exp(1/4*x*(16*exp(x^2)+3*x)*exp(-x^2)))-1)
Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.73 \[ \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx=-x^{2} + \log \left (4 \, x - 1\right ) + \log \left (\frac {3 \, x^{2} + 4 \, {\left (4 \, x - 1\right )} e^{\left (x^{2}\right )}}{4 \, x - 1}\right ) \] Input:
integrate(1/2*(8*exp(x^2)-3*x^3+3*x)/exp(x^2)/log(exp(1/4*(16*exp(x^2)*x+3 *x^2)/exp(x^2))/exp(1)),x, algorithm="fricas")
Output:
-x^2 + log(4*x - 1) + log((3*x^2 + 4*(4*x - 1)*e^(x^2))/(4*x - 1))
Time = 0.25 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx=2 \log {\left (x \right )} + \log {\left (e^{- x^{2}} + \frac {16 x - 4}{3 x^{2}} \right )} \] Input:
integrate(1/2*(8*exp(x**2)-3*x**3+3*x)/exp(x**2)/ln(exp(1/4*(16*exp(x**2)* x+3*x**2)/exp(x**2))/exp(1)),x)
Output:
2*log(x) + log(exp(-x**2) + (16*x - 4)/(3*x**2))
Leaf count of result is larger than twice the leaf count of optimal. 39 vs. \(2 (19) = 38\).
Time = 0.06 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.77 \[ \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx=-x^{2} + \log \left (4 \, x - 1\right ) + \log \left (\frac {3 \, x^{2} + 4 \, {\left (4 \, x - 1\right )} e^{\left (x^{2}\right )}}{4 \, {\left (4 \, x - 1\right )}}\right ) \] Input:
integrate(1/2*(8*exp(x^2)-3*x^3+3*x)/exp(x^2)/log(exp(1/4*(16*exp(x^2)*x+3 *x^2)/exp(x^2))/exp(1)),x, algorithm="maxima")
Output:
-x^2 + log(4*x - 1) + log(1/4*(3*x^2 + 4*(4*x - 1)*e^(x^2))/(4*x - 1))
Time = 0.12 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx=\log \left (3 \, x^{2} e^{\left (-x^{2}\right )} + 16 \, x - 4\right ) \] Input:
integrate(1/2*(8*exp(x^2)-3*x^3+3*x)/exp(x^2)/log(exp(1/4*(16*exp(x^2)*x+3 *x^2)/exp(x^2))/exp(1)),x, algorithm="giac")
Output:
log(3*x^2*e^(-x^2) + 16*x - 4)
Time = 3.14 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx=\ln \left (16\,x+3\,x^2\,{\mathrm {e}}^{-x^2}-4\right ) \] Input:
int((exp(-x^2)*((3*x)/2 + 4*exp(x^2) - (3*x^3)/2))/log(exp(exp(-x^2)*(4*x* exp(x^2) + (3*x^2)/4))*exp(-1)),x)
Output:
log(16*x + 3*x^2*exp(-x^2) - 4)
\[ \int \frac {e^{-x^2} \left (8 e^{x^2}+3 x-3 x^3\right )}{2 \log \left (e^{-1+\frac {1}{4} e^{-x^2} \left (16 e^{x^2} x+3 x^2\right )}\right )} \, dx=-\frac {3 \left (\int \frac {x^{3}}{e^{x^{2}} \mathrm {log}\left (\frac {e^{\frac {16 e^{x^{2}} x +3 x^{2}}{4 e^{x^{2}}}}}{e}\right )}d x \right )}{2}+\frac {3 \left (\int \frac {x}{e^{x^{2}} \mathrm {log}\left (\frac {e^{\frac {16 e^{x^{2}} x +3 x^{2}}{4 e^{x^{2}}}}}{e}\right )}d x \right )}{2}+4 \left (\int \frac {1}{\mathrm {log}\left (\frac {e^{\frac {16 e^{x^{2}} x +3 x^{2}}{4 e^{x^{2}}}}}{e}\right )}d x \right ) \] Input:
int(1/2*(8*exp(x^2)-3*x^3+3*x)/exp(x^2)/log(exp(1/4*(16*exp(x^2)*x+3*x^2)/ exp(x^2))/exp(1)),x)
Output:
( - 3*int(x**3/(e**(x**2)*log(e**((16*e**(x**2)*x + 3*x**2)/(4*e**(x**2))) /e)),x) + 3*int(x/(e**(x**2)*log(e**((16*e**(x**2)*x + 3*x**2)/(4*e**(x**2 )))/e)),x) + 8*int(1/log(e**((16*e**(x**2)*x + 3*x**2)/(4*e**(x**2)))/e),x ))/2