Integrand size = 112, antiderivative size = 27 \[ \int \frac {40-44 x+12 x^2+e^{2 x^2} x^2+e^{x^2} \left (-13 x+7 x^2\right )+\left (15-10 x+3 x^2+e^{x^2} \left (x^2+6 x^3-2 x^4\right )\right ) \log \left (\frac {x}{2}\right )}{25-30 x+9 x^2+e^{2 x^2} x^2+e^{x^2} \left (-10 x+6 x^2\right )} \, dx=-2+x+\frac {(-3+x) \log \left (\frac {x}{2}\right )}{3+e^{x^2}-\frac {5}{x}} \]
Time = 0.10 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int \frac {40-44 x+12 x^2+e^{2 x^2} x^2+e^{x^2} \left (-13 x+7 x^2\right )+\left (15-10 x+3 x^2+e^{x^2} \left (x^2+6 x^3-2 x^4\right )\right ) \log \left (\frac {x}{2}\right )}{25-30 x+9 x^2+e^{2 x^2} x^2+e^{x^2} \left (-10 x+6 x^2\right )} \, dx=x+\frac {(-3+x) x \log \left (\frac {x}{2}\right )}{-5+3 x+e^{x^2} x} \]
Integrate[(40 - 44*x + 12*x^2 + E^(2*x^2)*x^2 + E^x^2*(-13*x + 7*x^2) + (1 5 - 10*x + 3*x^2 + E^x^2*(x^2 + 6*x^3 - 2*x^4))*Log[x/2])/(25 - 30*x + 9*x ^2 + E^(2*x^2)*x^2 + E^x^2*(-10*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.
\(\displaystyle \int \frac {e^{2 x^2} x^2+12 x^2+e^{x^2} \left (7 x^2-13 x\right )+\left (3 x^2+e^{x^2} \left (-2 x^4+6 x^3+x^2\right )-10 x+15\right ) \log \left (\frac {x}{2}\right )-44 x+40}{e^{2 x^2} x^2+9 x^2+e^{x^2} \left (6 x^2-10 x\right )-30 x+25} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 x^2} x^2+12 x^2+e^{x^2} \left (7 x^2-13 x\right )+\left (3 x^2+e^{x^2} \left (-2 x^4+6 x^3+x^2\right )-10 x+15\right ) \log \left (\frac {x}{2}\right )-44 x+40}{\left (-e^{x^2} x-3 x+5\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {2 x^3 \log \left (\frac {x}{2}\right )-6 x^2 \log \left (\frac {x}{2}\right )-x-x \log \left (\frac {x}{2}\right )+3}{e^{x^2} x+3 x-5}+\frac {\left (6 x^4-28 x^3+30 x^2-5 x+15\right ) \log \left (\frac {x}{2}\right )}{\left (e^{x^2} x+3 x-5\right )^2}+1\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -3 \int \frac {1}{e^{x^2} x+3 x-5}dx+\int \frac {x}{e^{x^2} x+3 x-5}dx-15 \int \frac {\int \frac {1}{\left (\left (3+e^{x^2}\right ) x-5\right )^2}dx}{x}dx+5 \int \frac {\int \frac {x}{\left (\left (3+e^{x^2}\right ) x-5\right )^2}dx}{x}dx-30 \int \frac {\int \frac {x^2}{\left (\left (3+e^{x^2}\right ) x-5\right )^2}dx}{x}dx-\int \frac {\int \frac {x}{\left (3+e^{x^2}\right ) x-5}dx}{x}dx-6 \int \frac {\int \frac {x^2}{\left (3+e^{x^2}\right ) x-5}dx}{x}dx+15 \log \left (\frac {x}{2}\right ) \int \frac {1}{\left (e^{x^2} x+3 x-5\right )^2}dx-5 \log \left (\frac {x}{2}\right ) \int \frac {x}{\left (e^{x^2} x+3 x-5\right )^2}dx+30 \log \left (\frac {x}{2}\right ) \int \frac {x^2}{\left (e^{x^2} x+3 x-5\right )^2}dx+\log \left (\frac {x}{2}\right ) \int \frac {x}{e^{x^2} x+3 x-5}dx+6 \log \left (\frac {x}{2}\right ) \int \frac {x^2}{e^{x^2} x+3 x-5}dx-6 \int \frac {\int \frac {x^4}{\left (\left (3+e^{x^2}\right ) x-5\right )^2}dx}{x}dx+6 \log \left (\frac {x}{2}\right ) \int \frac {x^4}{\left (e^{x^2} x+3 x-5\right )^2}dx+28 \int \frac {\int \frac {x^3}{\left (\left (3+e^{x^2}\right ) x-5\right )^2}dx}{x}dx+2 \int \frac {\int \frac {x^3}{\left (3+e^{x^2}\right ) x-5}dx}{x}dx-28 \log \left (\frac {x}{2}\right ) \int \frac {x^3}{\left (e^{x^2} x+3 x-5\right )^2}dx-2 \log \left (\frac {x}{2}\right ) \int \frac {x^3}{e^{x^2} x+3 x-5}dx+x\) |
Int[(40 - 44*x + 12*x^2 + E^(2*x^2)*x^2 + E^x^2*(-13*x + 7*x^2) + (15 - 10 *x + 3*x^2 + E^x^2*(x^2 + 6*x^3 - 2*x^4))*Log[x/2])/(25 - 30*x + 9*x^2 + E ^(2*x^2)*x^2 + E^x^2*(-10*x + 6*x^2)),x]
3.28.25.3.1 Defintions of rubi rules used
Time = 0.22 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93
method | result | size |
risch | \(\frac {x \left (-3+x \right ) \ln \left (\frac {x}{2}\right )}{{\mathrm e}^{x^{2}} x +3 x -5}+x\) | \(25\) |
parallelrisch | \(-\frac {-x^{2} {\mathrm e}^{x^{2}}-x^{2} \ln \left (\frac {x}{2}\right )-3 x^{2}+3 x \ln \left (\frac {x}{2}\right )+5 x}{{\mathrm e}^{x^{2}} x +3 x -5}\) | \(50\) |
int((((-2*x^4+6*x^3+x^2)*exp(x^2)+3*x^2-10*x+15)*ln(1/2*x)+x^2*exp(x^2)^2+ (7*x^2-13*x)*exp(x^2)+12*x^2-44*x+40)/(x^2*exp(x^2)^2+(6*x^2-10*x)*exp(x^2 )+9*x^2-30*x+25),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.59 \[ \int \frac {40-44 x+12 x^2+e^{2 x^2} x^2+e^{x^2} \left (-13 x+7 x^2\right )+\left (15-10 x+3 x^2+e^{x^2} \left (x^2+6 x^3-2 x^4\right )\right ) \log \left (\frac {x}{2}\right )}{25-30 x+9 x^2+e^{2 x^2} x^2+e^{x^2} \left (-10 x+6 x^2\right )} \, dx=\frac {x^{2} e^{\left (x^{2}\right )} + 3 \, x^{2} + {\left (x^{2} - 3 \, x\right )} \log \left (\frac {1}{2} \, x\right ) - 5 \, x}{x e^{\left (x^{2}\right )} + 3 \, x - 5} \]
integrate((((-2*x^4+6*x^3+x^2)*exp(x^2)+3*x^2-10*x+15)*log(1/2*x)+x^2*exp( x^2)^2+(7*x^2-13*x)*exp(x^2)+12*x^2-44*x+40)/(x^2*exp(x^2)^2+(6*x^2-10*x)* exp(x^2)+9*x^2-30*x+25),x, algorithm=\
Time = 0.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {40-44 x+12 x^2+e^{2 x^2} x^2+e^{x^2} \left (-13 x+7 x^2\right )+\left (15-10 x+3 x^2+e^{x^2} \left (x^2+6 x^3-2 x^4\right )\right ) \log \left (\frac {x}{2}\right )}{25-30 x+9 x^2+e^{2 x^2} x^2+e^{x^2} \left (-10 x+6 x^2\right )} \, dx=x + \frac {x^{2} \log {\left (\frac {x}{2} \right )} - 3 x \log {\left (\frac {x}{2} \right )}}{x e^{x^{2}} + 3 x - 5} \]
integrate((((-2*x**4+6*x**3+x**2)*exp(x**2)+3*x**2-10*x+15)*ln(1/2*x)+x**2 *exp(x**2)**2+(7*x**2-13*x)*exp(x**2)+12*x**2-44*x+40)/(x**2*exp(x**2)**2+ (6*x**2-10*x)*exp(x**2)+9*x**2-30*x+25),x)
Time = 0.30 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.96 \[ \int \frac {40-44 x+12 x^2+e^{2 x^2} x^2+e^{x^2} \left (-13 x+7 x^2\right )+\left (15-10 x+3 x^2+e^{x^2} \left (x^2+6 x^3-2 x^4\right )\right ) \log \left (\frac {x}{2}\right )}{25-30 x+9 x^2+e^{2 x^2} x^2+e^{x^2} \left (-10 x+6 x^2\right )} \, dx=-\frac {x^{2} {\left (\log \left (2\right ) - 3\right )} - x^{2} e^{\left (x^{2}\right )} - x {\left (3 \, \log \left (2\right ) - 5\right )} - {\left (x^{2} - 3 \, x\right )} \log \left (x\right )}{x e^{\left (x^{2}\right )} + 3 \, x - 5} \]
integrate((((-2*x^4+6*x^3+x^2)*exp(x^2)+3*x^2-10*x+15)*log(1/2*x)+x^2*exp( x^2)^2+(7*x^2-13*x)*exp(x^2)+12*x^2-44*x+40)/(x^2*exp(x^2)^2+(6*x^2-10*x)* exp(x^2)+9*x^2-30*x+25),x, algorithm=\
-(x^2*(log(2) - 3) - x^2*e^(x^2) - x*(3*log(2) - 5) - (x^2 - 3*x)*log(x))/ (x*e^(x^2) + 3*x - 5)
Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.70 \[ \int \frac {40-44 x+12 x^2+e^{2 x^2} x^2+e^{x^2} \left (-13 x+7 x^2\right )+\left (15-10 x+3 x^2+e^{x^2} \left (x^2+6 x^3-2 x^4\right )\right ) \log \left (\frac {x}{2}\right )}{25-30 x+9 x^2+e^{2 x^2} x^2+e^{x^2} \left (-10 x+6 x^2\right )} \, dx=\frac {x^{2} e^{\left (x^{2}\right )} + x^{2} \log \left (\frac {1}{2} \, x\right ) + 3 \, x^{2} - 3 \, x \log \left (\frac {1}{2} \, x\right ) - 5 \, x}{x e^{\left (x^{2}\right )} + 3 \, x - 5} \]
integrate((((-2*x^4+6*x^3+x^2)*exp(x^2)+3*x^2-10*x+15)*log(1/2*x)+x^2*exp( x^2)^2+(7*x^2-13*x)*exp(x^2)+12*x^2-44*x+40)/(x^2*exp(x^2)^2+(6*x^2-10*x)* exp(x^2)+9*x^2-30*x+25),x, algorithm=\
Time = 11.25 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {40-44 x+12 x^2+e^{2 x^2} x^2+e^{x^2} \left (-13 x+7 x^2\right )+\left (15-10 x+3 x^2+e^{x^2} \left (x^2+6 x^3-2 x^4\right )\right ) \log \left (\frac {x}{2}\right )}{25-30 x+9 x^2+e^{2 x^2} x^2+e^{x^2} \left (-10 x+6 x^2\right )} \, dx=x-\frac {\ln \left (\frac {x}{2}\right )\,\left (3\,x-x^2\right )}{3\,x+x\,{\mathrm {e}}^{x^2}-5} \]