Integrand size = 68, antiderivative size = 23 \[ \int \frac {-30 e^{2 \log ^2(x)}+e^{\log ^2(x)} (-20+40 \log (x))}{4 x^2+e^{2 \log ^2(x)} \left (1-6 x+9 x^2\right )+e^{\log ^2(x)} \left (-4 x+12 x^2\right )} \, dx=\frac {5}{\frac {1}{2} (-1+x)+x+e^{-\log ^2(x)} x} \] Output:
5/(x/exp(ln(x)^2)+3/2*x-1/2)
Time = 0.33 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-30 e^{2 \log ^2(x)}+e^{\log ^2(x)} (-20+40 \log (x))}{4 x^2+e^{2 \log ^2(x)} \left (1-6 x+9 x^2\right )+e^{\log ^2(x)} \left (-4 x+12 x^2\right )} \, dx=\frac {10 e^{\log ^2(x)}}{2 x+e^{\log ^2(x)} (-1+3 x)} \] Input:
Integrate[(-30*E^(2*Log[x]^2) + E^Log[x]^2*(-20 + 40*Log[x]))/(4*x^2 + E^( 2*Log[x]^2)*(1 - 6*x + 9*x^2) + E^Log[x]^2*(-4*x + 12*x^2)),x]
Output:
(10*E^Log[x]^2)/(2*x + E^Log[x]^2*(-1 + 3*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.
\(\displaystyle \int \frac {e^{\log ^2(x)} (40 \log (x)-20)-30 e^{2 \log ^2(x)}}{4 x^2+\left (9 x^2-6 x+1\right ) e^{2 \log ^2(x)}+\left (12 x^2-4 x\right ) e^{\log ^2(x)}} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {10 e^{\log ^2(x)} \left (-3 e^{\log ^2(x)}+4 \log (x)-2\right )}{\left (2 x+(3 x-1) e^{\log ^2(x)}\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 10 \int -\frac {e^{\log ^2(x)} \left (-4 \log (x)+3 e^{\log ^2(x)}+2\right )}{\left (e^{\log ^2(x)} (1-3 x)-2 x\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -10 \int \frac {e^{\log ^2(x)} \left (-4 \log (x)+3 e^{\log ^2(x)}+2\right )}{\left (e^{\log ^2(x)} (1-3 x)-2 x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -10 \int \left (\frac {3 e^{\log ^2(x)}}{(3 x-1) \left (3 e^{\log ^2(x)} x+2 x-e^{\log ^2(x)}\right )}-\frac {2 e^{\log ^2(x)} (6 x \log (x)-2 \log (x)+1)}{(3 x-1) \left (3 e^{\log ^2(x)} x+2 x-e^{\log ^2(x)}\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -10 \left (-2 \int \frac {e^{\log ^2(x)}}{(3 x-1) \left (3 e^{\log ^2(x)} x+2 x-e^{\log ^2(x)}\right )^2}dx+3 \int \frac {e^{\log ^2(x)}}{(3 x-1) \left (3 e^{\log ^2(x)} x+2 x-e^{\log ^2(x)}\right )}dx-4 \int \frac {e^{\log ^2(x)} \log (x)}{\left (3 e^{\log ^2(x)} x+2 x-e^{\log ^2(x)}\right )^2}dx\right )\) |
Input:
Int[(-30*E^(2*Log[x]^2) + E^Log[x]^2*(-20 + 40*Log[x]))/(4*x^2 + E^(2*Log[ x]^2)*(1 - 6*x + 9*x^2) + E^Log[x]^2*(-4*x + 12*x^2)),x]
Output:
$Aborted
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26
method | result | size |
norman | \(\frac {10 \,{\mathrm e}^{\ln \left (x \right )^{2}}}{3 \,{\mathrm e}^{\ln \left (x \right )^{2}} x -{\mathrm e}^{\ln \left (x \right )^{2}}+2 x}\) | \(29\) |
parallelrisch | \(\frac {10 \,{\mathrm e}^{\ln \left (x \right )^{2}}}{3 \,{\mathrm e}^{\ln \left (x \right )^{2}} x -{\mathrm e}^{\ln \left (x \right )^{2}}+2 x}\) | \(29\) |
risch | \(\frac {10}{3 \left (x -\frac {1}{3}\right )}-\frac {20 x}{\left (-1+3 x \right ) \left (3 \,{\mathrm e}^{\ln \left (x \right )^{2}} x -{\mathrm e}^{\ln \left (x \right )^{2}}+2 x \right )}\) | \(40\) |
Input:
int((-30*exp(ln(x)^2)^2+(40*ln(x)-20)*exp(ln(x)^2))/((9*x^2-6*x+1)*exp(ln( x)^2)^2+(12*x^2-4*x)*exp(ln(x)^2)+4*x^2),x,method=_RETURNVERBOSE)
Output:
10*exp(ln(x)^2)/(3*exp(ln(x)^2)*x-exp(ln(x)^2)+2*x)
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-30 e^{2 \log ^2(x)}+e^{\log ^2(x)} (-20+40 \log (x))}{4 x^2+e^{2 \log ^2(x)} \left (1-6 x+9 x^2\right )+e^{\log ^2(x)} \left (-4 x+12 x^2\right )} \, dx=\frac {10 \, e^{\left (\log \left (x\right )^{2}\right )}}{{\left (3 \, x - 1\right )} e^{\left (\log \left (x\right )^{2}\right )} + 2 \, x} \] Input:
integrate((-30*exp(log(x)^2)^2+(40*log(x)-20)*exp(log(x)^2))/((9*x^2-6*x+1 )*exp(log(x)^2)^2+(12*x^2-4*x)*exp(log(x)^2)+4*x^2),x, algorithm="fricas")
Output:
10*e^(log(x)^2)/((3*x - 1)*e^(log(x)^2) + 2*x)
Time = 0.13 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.48 \[ \int \frac {-30 e^{2 \log ^2(x)}+e^{\log ^2(x)} (-20+40 \log (x))}{4 x^2+e^{2 \log ^2(x)} \left (1-6 x+9 x^2\right )+e^{\log ^2(x)} \left (-4 x+12 x^2\right )} \, dx=- \frac {20 x}{6 x^{2} - 2 x + \left (9 x^{2} - 6 x + 1\right ) e^{\log {\left (x \right )}^{2}}} + \frac {30}{9 x - 3} \] Input:
integrate((-30*exp(ln(x)**2)**2+(40*ln(x)-20)*exp(ln(x)**2))/((9*x**2-6*x+ 1)*exp(ln(x)**2)**2+(12*x**2-4*x)*exp(ln(x)**2)+4*x**2),x)
Output:
-20*x/(6*x**2 - 2*x + (9*x**2 - 6*x + 1)*exp(log(x)**2)) + 30/(9*x - 3)
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-30 e^{2 \log ^2(x)}+e^{\log ^2(x)} (-20+40 \log (x))}{4 x^2+e^{2 \log ^2(x)} \left (1-6 x+9 x^2\right )+e^{\log ^2(x)} \left (-4 x+12 x^2\right )} \, dx=\frac {10 \, e^{\left (\log \left (x\right )^{2}\right )}}{{\left (3 \, x - 1\right )} e^{\left (\log \left (x\right )^{2}\right )} + 2 \, x} \] Input:
integrate((-30*exp(log(x)^2)^2+(40*log(x)-20)*exp(log(x)^2))/((9*x^2-6*x+1 )*exp(log(x)^2)^2+(12*x^2-4*x)*exp(log(x)^2)+4*x^2),x, algorithm="maxima")
Output:
10*e^(log(x)^2)/((3*x - 1)*e^(log(x)^2) + 2*x)
Time = 0.15 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-30 e^{2 \log ^2(x)}+e^{\log ^2(x)} (-20+40 \log (x))}{4 x^2+e^{2 \log ^2(x)} \left (1-6 x+9 x^2\right )+e^{\log ^2(x)} \left (-4 x+12 x^2\right )} \, dx=\frac {10 \, e^{\left (\log \left (x\right )^{2}\right )}}{3 \, x e^{\left (\log \left (x\right )^{2}\right )} + 2 \, x - e^{\left (\log \left (x\right )^{2}\right )}} \] Input:
integrate((-30*exp(log(x)^2)^2+(40*log(x)-20)*exp(log(x)^2))/((9*x^2-6*x+1 )*exp(log(x)^2)^2+(12*x^2-4*x)*exp(log(x)^2)+4*x^2),x, algorithm="giac")
Output:
10*e^(log(x)^2)/(3*x*e^(log(x)^2) + 2*x - e^(log(x)^2))
Time = 2.39 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-30 e^{2 \log ^2(x)}+e^{\log ^2(x)} (-20+40 \log (x))}{4 x^2+e^{2 \log ^2(x)} \left (1-6 x+9 x^2\right )+e^{\log ^2(x)} \left (-4 x+12 x^2\right )} \, dx=\frac {10\,{\mathrm {e}}^{{\ln \left (x\right )}^2}}{2\,x-{\mathrm {e}}^{{\ln \left (x\right )}^2}+3\,x\,{\mathrm {e}}^{{\ln \left (x\right )}^2}} \] Input:
int(-(30*exp(2*log(x)^2) - exp(log(x)^2)*(40*log(x) - 20))/(exp(2*log(x)^2 )*(9*x^2 - 6*x + 1) - exp(log(x)^2)*(4*x - 12*x^2) + 4*x^2),x)
Output:
(10*exp(log(x)^2))/(2*x - exp(log(x)^2) + 3*x*exp(log(x)^2))
Time = 0.22 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {-30 e^{2 \log ^2(x)}+e^{\log ^2(x)} (-20+40 \log (x))}{4 x^2+e^{2 \log ^2(x)} \left (1-6 x+9 x^2\right )+e^{\log ^2(x)} \left (-4 x+12 x^2\right )} \, dx=\frac {10 e^{\mathrm {log}\left (x \right )^{2}}}{3 e^{\mathrm {log}\left (x \right )^{2}} x -e^{\mathrm {log}\left (x \right )^{2}}+2 x} \] Input:
int((-30*exp(log(x)^2)^2+(40*log(x)-20)*exp(log(x)^2))/((9*x^2-6*x+1)*exp( log(x)^2)^2+(12*x^2-4*x)*exp(log(x)^2)+4*x^2),x)
Output:
(10*e**(log(x)**2))/(3*e**(log(x)**2)*x - e**(log(x)**2) + 2*x)