Integrand size = 75, antiderivative size = 24 \[ \int \frac {7 x+e^{\log ^2\left (x^2\right )} \left (3 x+(4-12 x) \log \left (x^2\right )\right )}{12 x+3 e^{2 \log ^2\left (x^2\right )} x+12 x^2+3 x^3+e^{\log ^2\left (x^2\right )} \left (12 x+6 x^2\right )} \, dx=\frac {-1+3 x}{3 \left (2+e^{\log ^2\left (x^2\right )}\right )+3 x} \]
Time = 0.48 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {7 x+e^{\log ^2\left (x^2\right )} \left (3 x+(4-12 x) \log \left (x^2\right )\right )}{12 x+3 e^{2 \log ^2\left (x^2\right )} x+12 x^2+3 x^3+e^{\log ^2\left (x^2\right )} \left (12 x+6 x^2\right )} \, dx=\frac {-1+3 x}{3 \left (2+e^{\log ^2\left (x^2\right )}+x\right )} \]
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]
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 |
\(\displaystyle \frac {1}{3} \left (\int \frac {1}{\left (x+e^{\log ^2\left (x^2\right )}+2\right )^2}dx-3 \int \frac {x}{\left (x+e^{\log ^2\left (x^2\right )}+2\right )^2}dx+3 \int \frac {1}{x+e^{\log ^2\left (x^2\right )}+2}dx+20 \int \frac {\log \left (x^2\right )}{\left (x+e^{\log ^2\left (x^2\right )}+2\right )^2}dx-8 \int \frac {\log \left (x^2\right )}{x \left (x+e^{\log ^2\left (x^2\right )}+2\right )^2}dx+12 \int \frac {x \log \left (x^2\right )}{\left (x+e^{\log ^2\left (x^2\right )}+2\right )^2}dx-12 \int \frac {\log \left (x^2\right )}{x+e^{\log ^2\left (x^2\right )}+2}dx+4 \int \frac {\log \left (x^2\right )}{x \left (x+e^{\log ^2\left (x^2\right )}+2\right )}dx\right )\) |
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]
3.9.67.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 1.22 (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 | \(\frac {6 x -2}{12+6 \,{\mathrm e}^{\ln \left (x^{2}\right )^{2}}+6 x}\) | \(20\) |
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)
Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {7 x+e^{\log ^2\left (x^2\right )} \left (3 x+(4-12 x) \log \left (x^2\right )\right )}{12 x+3 e^{2 \log ^2\left (x^2\right )} x+12 x^2+3 x^3+e^{\log ^2\left (x^2\right )} \left (12 x+6 x^2\right )} \, dx=\frac {3 \, x - 1}{3 \, {\left (x + e^{\left (\log \left (x^{2}\right )^{2}\right )} + 2\right )}} \]
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=\
Time = 0.09 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {7 x+e^{\log ^2\left (x^2\right )} \left (3 x+(4-12 x) \log \left (x^2\right )\right )}{12 x+3 e^{2 \log ^2\left (x^2\right )} x+12 x^2+3 x^3+e^{\log ^2\left (x^2\right )} \left (12 x+6 x^2\right )} \, dx=\frac {3 x - 1}{3 x + 3 e^{\log {\left (x^{2} \right )}^{2}} + 6} \]
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)
Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {7 x+e^{\log ^2\left (x^2\right )} \left (3 x+(4-12 x) \log \left (x^2\right )\right )}{12 x+3 e^{2 \log ^2\left (x^2\right )} x+12 x^2+3 x^3+e^{\log ^2\left (x^2\right )} \left (12 x+6 x^2\right )} \, dx=\frac {3 \, x - 1}{3 \, {\left (x + e^{\left (4 \, \log \left (x\right )^{2}\right )} + 2\right )}} \]
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=\
Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {7 x+e^{\log ^2\left (x^2\right )} \left (3 x+(4-12 x) \log \left (x^2\right )\right )}{12 x+3 e^{2 \log ^2\left (x^2\right )} x+12 x^2+3 x^3+e^{\log ^2\left (x^2\right )} \left (12 x+6 x^2\right )} \, dx=\frac {3 \, x - 1}{3 \, {\left (x + e^{\left (\log \left (x^{2}\right )^{2}\right )} + 2\right )}} \]
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=\
Time = 9.68 (sec) , antiderivative size = 72, normalized size of antiderivative = 3.00 \[ \int \frac {7 x+e^{\log ^2\left (x^2\right )} \left (3 x+(4-12 x) \log \left (x^2\right )\right )}{12 x+3 e^{2 \log ^2\left (x^2\right )} x+12 x^2+3 x^3+e^{\log ^2\left (x^2\right )} \left (12 x+6 x^2\right )} \, dx=\frac {20\,x^2\,\ln \left (x^2\right )-8\,x\,\ln \left (x^2\right )+12\,x^3\,\ln \left (x^2\right )+x^2-3\,x^3}{3\,\left (8\,x\,\ln \left (x^2\right )+4\,x^2\,\ln \left (x^2\right )-x^2\right )\,\left (x+{\mathrm {e}}^{{\ln \left (x^2\right )}^2}+2\right )} \]
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)