Integrand size = 101, antiderivative size = 15 \[ \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \left (-11 x-x^2+\left (-96+16 x+2 x^2+(-24-2 x) \log (12+x)\right ) \log (-4+x-\log (12+x))\right )}{48 x^3-8 x^4-x^5+\left (12 x^3+x^4\right ) \log (12+x)} \, dx=e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} \]
Time = 5.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \left (-11 x-x^2+\left (-96+16 x+2 x^2+(-24-2 x) \log (12+x)\right ) \log (-4+x-\log (12+x))\right )}{48 x^3-8 x^4-x^5+\left (12 x^3+x^4\right ) \log (12+x)} \, dx=e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} \]
Integrate[(E^(-4 + x - Log[12 + x])^x^(-2)*(-4 + x - Log[12 + x])^x^(-2)*( -11*x - x^2 + (-96 + 16*x + 2*x^2 + (-24 - 2*x)*Log[12 + x])*Log[-4 + x - Log[12 + x]]))/(48*x^3 - 8*x^4 - x^5 + (12*x^3 + x^4)*Log[12 + x]),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.
\(\displaystyle \int \frac {e^{(x-\log (x+12)-4)^{\frac {1}{x^2}}} (x-\log (x+12)-4)^{\frac {1}{x^2}} \left (-x^2+\left (2 x^2+16 x+(-2 x-24) \log (x+12)-96\right ) \log (x-\log (x+12)-4)-11 x\right )}{-x^5-8 x^4+48 x^3+\left (x^4+12 x^3\right ) \log (x+12)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{(x-\log (x+12)-4)^{\frac {1}{x^2}}} (x-\log (x+12)-4)^{\frac {1}{x^2}-1} \left (x^2-\left (2 x^2+16 x+(-2 x-24) \log (x+12)-96\right ) \log (x-\log (x+12)-4)+11 x\right )}{x^3 (x+12)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {(x+11) e^{(x-\log (x+12)-4)^{\frac {1}{x^2}}} (x-\log (x+12)-4)^{\frac {1}{x^2}-1}}{x^2 (x+12)}-\frac {2 e^{(x-\log (x+12)-4)^{\frac {1}{x^2}}} (x-\log (x+12)-4)^{\frac {1}{x^2}} \log (x-\log (x+12)-4)}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {11}{12} \int \frac {e^{(x-\log (x+12)-4)^{\frac {1}{x^2}}} (x-\log (x+12)-4)^{\frac {1}{x^2}-1}}{x^2}dx+\frac {1}{144} \int \frac {e^{(x-\log (x+12)-4)^{\frac {1}{x^2}}} (x-\log (x+12)-4)^{\frac {1}{x^2}-1}}{x}dx-\frac {1}{144} \int \frac {e^{(x-\log (x+12)-4)^{\frac {1}{x^2}}} (x-\log (x+12)-4)^{\frac {1}{x^2}-1}}{x+12}dx-2 \int \frac {e^{(x-\log (x+12)-4)^{\frac {1}{x^2}}} (x-\log (x+12)-4)^{\frac {1}{x^2}} \log (x-\log (x+12)-4)}{x^3}dx\) |
Int[(E^(-4 + x - Log[12 + x])^x^(-2)*(-4 + x - Log[12 + x])^x^(-2)*(-11*x - x^2 + (-96 + 16*x + 2*x^2 + (-24 - 2*x)*Log[12 + x])*Log[-4 + x - Log[12 + x]]))/(48*x^3 - 8*x^4 - x^5 + (12*x^3 + x^4)*Log[12 + x]),x]
3.5.15.3.1 Defintions of rubi rules used
Time = 247.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00
method | result | size |
risch | \({\mathrm e}^{\left (-\ln \left (x +12\right )+x -4\right )^{\frac {1}{x^{2}}}}\) | \(15\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{\frac {\ln \left (-\ln \left (x +12\right )+x -4\right )}{x^{2}}}}\) | \(17\) |
int((((-2*x-24)*ln(x+12)+2*x^2+16*x-96)*ln(-ln(x+12)+x-4)-x^2-11*x)*exp(ln (-ln(x+12)+x-4)/x^2)*exp(exp(ln(-ln(x+12)+x-4)/x^2))/((x^4+12*x^3)*ln(x+12 )-x^5-8*x^4+48*x^3),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \left (-11 x-x^2+\left (-96+16 x+2 x^2+(-24-2 x) \log (12+x)\right ) \log (-4+x-\log (12+x))\right )}{48 x^3-8 x^4-x^5+\left (12 x^3+x^4\right ) \log (12+x)} \, dx=e^{\left ({\left (x - \log \left (x + 12\right ) - 4\right )}^{\left (\frac {1}{x^{2}}\right )}\right )} \]
integrate((((-2*x-24)*log(x+12)+2*x^2+16*x-96)*log(-log(x+12)+x-4)-x^2-11* x)*exp(log(-log(x+12)+x-4)/x^2)*exp(exp(log(-log(x+12)+x-4)/x^2))/((x^4+12 *x^3)*log(x+12)-x^5-8*x^4+48*x^3),x, algorithm=\
Time = 171.53 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \left (-11 x-x^2+\left (-96+16 x+2 x^2+(-24-2 x) \log (12+x)\right ) \log (-4+x-\log (12+x))\right )}{48 x^3-8 x^4-x^5+\left (12 x^3+x^4\right ) \log (12+x)} \, dx=e^{e^{\frac {\log {\left (x - \log {\left (x + 12 \right )} - 4 \right )}}{x^{2}}}} \]
integrate((((-2*x-24)*ln(x+12)+2*x**2+16*x-96)*ln(-ln(x+12)+x-4)-x**2-11*x )*exp(ln(-ln(x+12)+x-4)/x**2)*exp(exp(ln(-ln(x+12)+x-4)/x**2))/((x**4+12*x **3)*ln(x+12)-x**5-8*x**4+48*x**3),x)
Time = 0.27 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \left (-11 x-x^2+\left (-96+16 x+2 x^2+(-24-2 x) \log (12+x)\right ) \log (-4+x-\log (12+x))\right )}{48 x^3-8 x^4-x^5+\left (12 x^3+x^4\right ) \log (12+x)} \, dx=e^{\left ({\left (x - \log \left (x + 12\right ) - 4\right )}^{\left (\frac {1}{x^{2}}\right )}\right )} \]
integrate((((-2*x-24)*log(x+12)+2*x^2+16*x-96)*log(-log(x+12)+x-4)-x^2-11* x)*exp(log(-log(x+12)+x-4)/x^2)*exp(exp(log(-log(x+12)+x-4)/x^2))/((x^4+12 *x^3)*log(x+12)-x^5-8*x^4+48*x^3),x, algorithm=\
\[ \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \left (-11 x-x^2+\left (-96+16 x+2 x^2+(-24-2 x) \log (12+x)\right ) \log (-4+x-\log (12+x))\right )}{48 x^3-8 x^4-x^5+\left (12 x^3+x^4\right ) \log (12+x)} \, dx=\int { \frac {{\left (x^{2} - 2 \, {\left (x^{2} - {\left (x + 12\right )} \log \left (x + 12\right ) + 8 \, x - 48\right )} \log \left (x - \log \left (x + 12\right ) - 4\right ) + 11 \, x\right )} {\left (x - \log \left (x + 12\right ) - 4\right )}^{\left (\frac {1}{x^{2}}\right )} e^{\left ({\left (x - \log \left (x + 12\right ) - 4\right )}^{\left (\frac {1}{x^{2}}\right )}\right )}}{x^{5} + 8 \, x^{4} - 48 \, x^{3} - {\left (x^{4} + 12 \, x^{3}\right )} \log \left (x + 12\right )} \,d x } \]
integrate((((-2*x-24)*log(x+12)+2*x^2+16*x-96)*log(-log(x+12)+x-4)-x^2-11* x)*exp(log(-log(x+12)+x-4)/x^2)*exp(exp(log(-log(x+12)+x-4)/x^2))/((x^4+12 *x^3)*log(x+12)-x^5-8*x^4+48*x^3),x, algorithm=\
integrate((x^2 - 2*(x^2 - (x + 12)*log(x + 12) + 8*x - 48)*log(x - log(x + 12) - 4) + 11*x)*(x - log(x + 12) - 4)^(x^(-2))*e^((x - log(x + 12) - 4)^ (x^(-2)))/(x^5 + 8*x^4 - 48*x^3 - (x^4 + 12*x^3)*log(x + 12)), x)
Time = 14.47 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93 \[ \int \frac {e^{(-4+x-\log (12+x))^{\frac {1}{x^2}}} (-4+x-\log (12+x))^{\frac {1}{x^2}} \left (-11 x-x^2+\left (-96+16 x+2 x^2+(-24-2 x) \log (12+x)\right ) \log (-4+x-\log (12+x))\right )}{48 x^3-8 x^4-x^5+\left (12 x^3+x^4\right ) \log (12+x)} \, dx={\mathrm {e}}^{{\left (x-\ln \left (x+12\right )-4\right )}^{\frac {1}{x^2}}} \]