Integrand size = 75, antiderivative size = 24 \[ \int \frac {7 x+e^{\log ^2\left (x^2\right )} \left (3 x+(4-12 x) \log \left (x^2\right )\right )}{12 x+3 e^{2 \log ^2\left (x^2\right )} x+12 x^2+3 x^3+e^{\log ^2\left (x^2\right )} \left (12 x+6 x^2\right )} \, dx=\frac {-1+3 x}{3 \left (2+e^{\log ^2\left (x^2\right )}\right )+3 x} \] Output:
(-1+3*x)/(3*exp(ln(x^2)^2)+6+3*x)
Time = 0.29 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {7 x+e^{\log ^2\left (x^2\right )} \left (3 x+(4-12 x) \log \left (x^2\right )\right )}{12 x+3 e^{2 \log ^2\left (x^2\right )} x+12 x^2+3 x^3+e^{\log ^2\left (x^2\right )} \left (12 x+6 x^2\right )} \, dx=\frac {-1+3 x}{3 \left (2+e^{\log ^2\left (x^2\right )}+x\right )} \] Input:
Integrate[(7*x + E^Log[x^2]^2*(3*x + (4 - 12*x)*Log[x^2]))/(12*x + 3*E^(2* Log[x^2]^2)*x + 12*x^2 + 3*x^3 + E^Log[x^2]^2*(12*x + 6*x^2)),x]
Output:
(-1 + 3*x)/(3*(2 + E^Log[x^2]^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.
\(\displaystyle \int \frac {e^{\log ^2\left (x^2\right )} \left ((4-12 x) \log \left (x^2\right )+3 x\right )+7 x}{3 x^3+12 x^2+3 x e^{2 \log ^2\left (x^2\right )}+\left (6 x^2+12 x\right ) e^{\log ^2\left (x^2\right )}+12 x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\log ^2\left (x^2\right )} \left ((4-12 x) \log \left (x^2\right )+3 x\right )+7 x}{3 x \left (e^{\log ^2\left (x^2\right )}+x+2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int \frac {7 x+e^{\log ^2\left (x^2\right )} \left (3 x+4 (1-3 x) \log \left (x^2\right )\right )}{x \left (x+e^{\log ^2\left (x^2\right )}+2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{3} \int \left (\frac {(3 x-1) \left (4 \log \left (x^2\right ) x-x+8 \log \left (x^2\right )\right )}{x \left (x+e^{\log ^2\left (x^2\right )}+2\right )^2}-\frac {12 \log \left (x^2\right ) x-3 x-4 \log \left (x^2\right )}{x \left (x+e^{\log ^2\left (x^2\right )}+2\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} \left (\int \frac {1}{\left (x+e^{\log ^2\left (x^2\right )}+2\right )^2}dx-3 \int \frac {x}{\left (x+e^{\log ^2\left (x^2\right )}+2\right )^2}dx+3 \int \frac {1}{x+e^{\log ^2\left (x^2\right )}+2}dx+20 \int \frac {\log \left (x^2\right )}{\left (x+e^{\log ^2\left (x^2\right )}+2\right )^2}dx-8 \int \frac {\log \left (x^2\right )}{x \left (x+e^{\log ^2\left (x^2\right )}+2\right )^2}dx+12 \int \frac {x \log \left (x^2\right )}{\left (x+e^{\log ^2\left (x^2\right )}+2\right )^2}dx-12 \int \frac {\log \left (x^2\right )}{x+e^{\log ^2\left (x^2\right )}+2}dx+4 \int \frac {\log \left (x^2\right )}{x \left (x+e^{\log ^2\left (x^2\right )}+2\right )}dx\right )\) |
Input:
Int[(7*x + E^Log[x^2]^2*(3*x + (4 - 12*x)*Log[x^2]))/(12*x + 3*E^(2*Log[x^ 2]^2)*x + 12*x^2 + 3*x^3 + E^Log[x^2]^2*(12*x + 6*x^2)),x]
Output:
$Aborted
Time = 0.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83
method | result | size |
risch | \(\frac {-1+3 x}{3 \,{\mathrm e}^{\ln \left (x^{2}\right )^{2}}+6+3 x}\) | \(20\) |
parallelrisch | \(\frac {-2+6 x}{12+6 \,{\mathrm e}^{\ln \left (x^{2}\right )^{2}}+6 x}\) | \(20\) |
Input:
int((((-12*x+4)*ln(x^2)+3*x)*exp(ln(x^2)^2)+7*x)/(3*x*exp(ln(x^2)^2)^2+(6* x^2+12*x)*exp(ln(x^2)^2)+3*x^3+12*x^2+12*x),x,method=_RETURNVERBOSE)
Output:
1/3*(-1+3*x)/(2+exp(ln(x^2)^2)+x)
Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {7 x+e^{\log ^2\left (x^2\right )} \left (3 x+(4-12 x) \log \left (x^2\right )\right )}{12 x+3 e^{2 \log ^2\left (x^2\right )} x+12 x^2+3 x^3+e^{\log ^2\left (x^2\right )} \left (12 x+6 x^2\right )} \, dx=\frac {3 \, x - 1}{3 \, {\left (x + e^{\left (\log \left (x^{2}\right )^{2}\right )} + 2\right )}} \] Input:
integrate((((-12*x+4)*log(x^2)+3*x)*exp(log(x^2)^2)+7*x)/(3*x*exp(log(x^2) ^2)^2+(6*x^2+12*x)*exp(log(x^2)^2)+3*x^3+12*x^2+12*x),x, algorithm="fricas ")
Output:
1/3*(3*x - 1)/(x + e^(log(x^2)^2) + 2)
Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {7 x+e^{\log ^2\left (x^2\right )} \left (3 x+(4-12 x) \log \left (x^2\right )\right )}{12 x+3 e^{2 \log ^2\left (x^2\right )} x+12 x^2+3 x^3+e^{\log ^2\left (x^2\right )} \left (12 x+6 x^2\right )} \, dx=\frac {3 x - 1}{3 x + 3 e^{\log {\left (x^{2} \right )}^{2}} + 6} \] Input:
integrate((((-12*x+4)*ln(x**2)+3*x)*exp(ln(x**2)**2)+7*x)/(3*x*exp(ln(x**2 )**2)**2+(6*x**2+12*x)*exp(ln(x**2)**2)+3*x**3+12*x**2+12*x),x)
Output:
(3*x - 1)/(3*x + 3*exp(log(x**2)**2) + 6)
Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {7 x+e^{\log ^2\left (x^2\right )} \left (3 x+(4-12 x) \log \left (x^2\right )\right )}{12 x+3 e^{2 \log ^2\left (x^2\right )} x+12 x^2+3 x^3+e^{\log ^2\left (x^2\right )} \left (12 x+6 x^2\right )} \, dx=\frac {3 \, x - 1}{3 \, {\left (x + e^{\left (4 \, \log \left (x\right )^{2}\right )} + 2\right )}} \] Input:
integrate((((-12*x+4)*log(x^2)+3*x)*exp(log(x^2)^2)+7*x)/(3*x*exp(log(x^2) ^2)^2+(6*x^2+12*x)*exp(log(x^2)^2)+3*x^3+12*x^2+12*x),x, algorithm="maxima ")
Output:
1/3*(3*x - 1)/(x + e^(4*log(x)^2) + 2)
Time = 0.17 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {7 x+e^{\log ^2\left (x^2\right )} \left (3 x+(4-12 x) \log \left (x^2\right )\right )}{12 x+3 e^{2 \log ^2\left (x^2\right )} x+12 x^2+3 x^3+e^{\log ^2\left (x^2\right )} \left (12 x+6 x^2\right )} \, dx=\frac {3 \, x - 1}{3 \, {\left (x + e^{\left (\log \left (x^{2}\right )^{2}\right )} + 2\right )}} \] Input:
integrate((((-12*x+4)*log(x^2)+3*x)*exp(log(x^2)^2)+7*x)/(3*x*exp(log(x^2) ^2)^2+(6*x^2+12*x)*exp(log(x^2)^2)+3*x^3+12*x^2+12*x),x, algorithm="giac")
Output:
1/3*(3*x - 1)/(x + e^(log(x^2)^2) + 2)
Time = 2.88 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.00 \[ \int \frac {7 x+e^{\log ^2\left (x^2\right )} \left (3 x+(4-12 x) \log \left (x^2\right )\right )}{12 x+3 e^{2 \log ^2\left (x^2\right )} x+12 x^2+3 x^3+e^{\log ^2\left (x^2\right )} \left (12 x+6 x^2\right )} \, dx=\frac {20\,x^2\,\ln \left (x^2\right )-8\,x\,\ln \left (x^2\right )+12\,x^3\,\ln \left (x^2\right )+x^2-3\,x^3}{3\,\left (8\,x\,\ln \left (x^2\right )+4\,x^2\,\ln \left (x^2\right )-x^2\right )\,\left (x+{\mathrm {e}}^{{\ln \left (x^2\right )}^2}+2\right )} \] Input:
int((7*x + exp(log(x^2)^2)*(3*x - log(x^2)*(12*x - 4)))/(12*x + 3*x*exp(2* log(x^2)^2) + exp(log(x^2)^2)*(12*x + 6*x^2) + 12*x^2 + 3*x^3),x)
Output:
(20*x^2*log(x^2) - 8*x*log(x^2) + 12*x^3*log(x^2) + x^2 - 3*x^3)/(3*(8*x*l og(x^2) + 4*x^2*log(x^2) - x^2)*(x + exp(log(x^2)^2) + 2))
Time = 0.18 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25 \[ \int \frac {7 x+e^{\log ^2\left (x^2\right )} \left (3 x+(4-12 x) \log \left (x^2\right )\right )}{12 x+3 e^{2 \log ^2\left (x^2\right )} x+12 x^2+3 x^3+e^{\log ^2\left (x^2\right )} \left (12 x+6 x^2\right )} \, dx=\frac {e^{\mathrm {log}\left (x^{2}\right )^{2}}+7 x}{6 e^{\mathrm {log}\left (x^{2}\right )^{2}}+6 x +12} \] Input:
int((((-12*x+4)*log(x^2)+3*x)*exp(log(x^2)^2)+7*x)/(3*x*exp(log(x^2)^2)^2+ (6*x^2+12*x)*exp(log(x^2)^2)+3*x^3+12*x^2+12*x),x)
Output:
(e**(log(x**2)**2) + 7*x)/(6*(e**(log(x**2)**2) + x + 2))