Integrand size = 126, antiderivative size = 31 \[ \int \frac {e^{-\frac {3+e^x}{3 e^{x^2}-6 x}} \left (-6 e^{2 x^2} \log (5)+\left (-6 x-24 x^2\right ) \log (5)+e^x \left (-2 x+2 x^2\right ) \log (5)+e^{x^2} \left (e^x \left (-x+2 x^2\right ) \log (5)+\left (24 x+6 x^2\right ) \log (5)\right )\right )}{48 e^{2 x^2} x^3-192 e^{x^2} x^4+192 x^5} \, dx=\frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}} \log (5)}{16 x^2} \]
Time = 0.15 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-\frac {3+e^x}{3 e^{x^2}-6 x}} \left (-6 e^{2 x^2} \log (5)+\left (-6 x-24 x^2\right ) \log (5)+e^x \left (-2 x+2 x^2\right ) \log (5)+e^{x^2} \left (e^x \left (-x+2 x^2\right ) \log (5)+\left (24 x+6 x^2\right ) \log (5)\right )\right )}{48 e^{2 x^2} x^3-192 e^{x^2} x^4+192 x^5} \, dx=\frac {e^{\frac {-3-e^x}{3 \left (e^{x^2}-2 x\right )}} \log (5)}{16 x^2} \]
Integrate[(-6*E^(2*x^2)*Log[5] + (-6*x - 24*x^2)*Log[5] + E^x*(-2*x + 2*x^ 2)*Log[5] + E^x^2*(E^x*(-x + 2*x^2)*Log[5] + (24*x + 6*x^2)*Log[5]))/(E^(( 3 + E^x)/(3*E^x^2 - 6*x))*(48*E^(2*x^2)*x^3 - 192*E^x^2*x^4 + 192*x^5)),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.
\(\displaystyle \int \frac {e^{-\frac {e^x+3}{3 e^{x^2}-6 x}} \left (\left (-24 x^2-6 x\right ) \log (5)+e^{x^2} \left (e^x \left (2 x^2-x\right ) \log (5)+\left (6 x^2+24 x\right ) \log (5)\right )-6 e^{2 x^2} \log (5)+e^x \left (2 x^2-2 x\right ) \log (5)\right )}{192 x^5-192 e^{x^2} x^4+48 e^{2 x^2} x^3} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{-\frac {e^x+3}{3 \left (e^{x^2}-2 x\right )}} \left (6 e^{x^2} x^2+2 e^x x^2+2 e^{x^2+x} x^2-24 x^2+24 e^{x^2} x-e^{x^2+x} x-6 e^{2 x^2}-2 e^x x-6 x\right ) \log (5)}{48 \left (e^{x^2}-2 x\right )^2 x^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{48} \log (5) \int -\frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}} \left (-6 e^{x^2} x^2-2 e^x x^2-2 e^{x^2+x} x^2+24 x^2-24 e^{x^2} x+2 e^x x+e^{x^2+x} x+6 x+6 e^{2 x^2}\right )}{\left (e^{x^2}-2 x\right )^2 x^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{48} \log (5) \int \frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}} \left (-6 e^{x^2} x^2-2 e^x x^2-2 e^{x^2+x} x^2+24 x^2-24 e^{x^2} x+2 e^x x+e^{x^2+x} x+6 x+6 e^{2 x^2}\right )}{\left (e^{x^2}-2 x\right )^2 x^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{48} \log (5) \int \left (\frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}} \left (2 e^x x+6 x-e^x\right )}{x^2 \left (2 x-e^{x^2}\right )}-\frac {2 e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}} \left (3+e^x\right ) \left (2 x^2-1\right )}{\left (e^{x^2}-2 x\right )^2 x^2}+\frac {6 e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{48} \log (5) \left (-12 \int \frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{\left (e^{x^2}-2 x\right )^2}dx-4 \int \frac {e^{x-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{\left (e^{x^2}-2 x\right )^2}dx+6 \int \frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{\left (e^{x^2}-2 x\right )^2 x^2}dx+2 \int \frac {e^{x-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{\left (e^{x^2}-2 x\right )^2 x^2}dx+\int \frac {e^{x-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{\left (e^{x^2}-2 x\right ) x^2}dx-6 \int \frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{\left (e^{x^2}-2 x\right ) x}dx-2 \int \frac {e^{x-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{\left (e^{x^2}-2 x\right ) x}dx+6 \int \frac {e^{-\frac {3+e^x}{3 \left (e^{x^2}-2 x\right )}}}{x^3}dx\right )\) |
Int[(-6*E^(2*x^2)*Log[5] + (-6*x - 24*x^2)*Log[5] + E^x*(-2*x + 2*x^2)*Log [5] + E^x^2*(E^x*(-x + 2*x^2)*Log[5] + (24*x + 6*x^2)*Log[5]))/(E^((3 + E^ x)/(3*E^x^2 - 6*x))*(48*E^(2*x^2)*x^3 - 192*E^x^2*x^4 + 192*x^5)),x]
3.21.73.3.1 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 = 12.89 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87
method | result | size |
risch | \(\frac {\ln \left (5\right ) {\mathrm e}^{\frac {3+{\mathrm e}^{x}}{-3 \,{\mathrm e}^{x^{2}}+6 x}}}{16 x^{2}}\) | \(27\) |
parallelrisch | \(\frac {\left (6 x \ln \left (5\right )-3 \ln \left (5\right ) {\mathrm e}^{x^{2}}\right ) {\mathrm e}^{-\frac {3+{\mathrm e}^{x}}{3 \left ({\mathrm e}^{x^{2}}-2 x \right )}}}{48 x^{2} \left (-{\mathrm e}^{x^{2}}+2 x \right )}\) | \(51\) |
int((-6*ln(5)*exp(x^2)^2+((2*x^2-x)*ln(5)*exp(x)+(6*x^2+24*x)*ln(5))*exp(x ^2)+(2*x^2-2*x)*ln(5)*exp(x)+(-24*x^2-6*x)*ln(5))/(48*x^3*exp(x^2)^2-192*x ^4*exp(x^2)+192*x^5)/exp((3+exp(x))/(3*exp(x^2)-6*x)),x,method=_RETURNVERB OSE)
Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-\frac {3+e^x}{3 e^{x^2}-6 x}} \left (-6 e^{2 x^2} \log (5)+\left (-6 x-24 x^2\right ) \log (5)+e^x \left (-2 x+2 x^2\right ) \log (5)+e^{x^2} \left (e^x \left (-x+2 x^2\right ) \log (5)+\left (24 x+6 x^2\right ) \log (5)\right )\right )}{48 e^{2 x^2} x^3-192 e^{x^2} x^4+192 x^5} \, dx=\frac {e^{\left (\frac {e^{x} + 3}{3 \, {\left (2 \, x - e^{\left (x^{2}\right )}\right )}}\right )} \log \left (5\right )}{16 \, x^{2}} \]
integrate((-6*log(5)*exp(x^2)^2+((2*x^2-x)*log(5)*exp(x)+(6*x^2+24*x)*log( 5))*exp(x^2)+(2*x^2-2*x)*log(5)*exp(x)+(-24*x^2-6*x)*log(5))/(48*x^3*exp(x ^2)^2-192*x^4*exp(x^2)+192*x^5)/exp((3+exp(x))/(3*exp(x^2)-6*x)),x, algori thm=\
Time = 0.36 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.77 \[ \int \frac {e^{-\frac {3+e^x}{3 e^{x^2}-6 x}} \left (-6 e^{2 x^2} \log (5)+\left (-6 x-24 x^2\right ) \log (5)+e^x \left (-2 x+2 x^2\right ) \log (5)+e^{x^2} \left (e^x \left (-x+2 x^2\right ) \log (5)+\left (24 x+6 x^2\right ) \log (5)\right )\right )}{48 e^{2 x^2} x^3-192 e^{x^2} x^4+192 x^5} \, dx=\frac {e^{- \frac {e^{x} + 3}{- 6 x + 3 e^{x^{2}}}} \log {\left (5 \right )}}{16 x^{2}} \]
integrate((-6*ln(5)*exp(x**2)**2+((2*x**2-x)*ln(5)*exp(x)+(6*x**2+24*x)*ln (5))*exp(x**2)+(2*x**2-2*x)*ln(5)*exp(x)+(-24*x**2-6*x)*ln(5))/(48*x**3*ex p(x**2)**2-192*x**4*exp(x**2)+192*x**5)/exp((3+exp(x))/(3*exp(x**2)-6*x)), x)
\[ \int \frac {e^{-\frac {3+e^x}{3 e^{x^2}-6 x}} \left (-6 e^{2 x^2} \log (5)+\left (-6 x-24 x^2\right ) \log (5)+e^x \left (-2 x+2 x^2\right ) \log (5)+e^{x^2} \left (e^x \left (-x+2 x^2\right ) \log (5)+\left (24 x+6 x^2\right ) \log (5)\right )\right )}{48 e^{2 x^2} x^3-192 e^{x^2} x^4+192 x^5} \, dx=\int { \frac {{\left (2 \, {\left (x^{2} - x\right )} e^{x} \log \left (5\right ) + {\left ({\left (2 \, x^{2} - x\right )} e^{x} \log \left (5\right ) + 6 \, {\left (x^{2} + 4 \, x\right )} \log \left (5\right )\right )} e^{\left (x^{2}\right )} - 6 \, {\left (4 \, x^{2} + x\right )} \log \left (5\right ) - 6 \, e^{\left (2 \, x^{2}\right )} \log \left (5\right )\right )} e^{\left (\frac {e^{x} + 3}{3 \, {\left (2 \, x - e^{\left (x^{2}\right )}\right )}}\right )}}{48 \, {\left (4 \, x^{5} - 4 \, x^{4} e^{\left (x^{2}\right )} + x^{3} e^{\left (2 \, x^{2}\right )}\right )}} \,d x } \]
integrate((-6*log(5)*exp(x^2)^2+((2*x^2-x)*log(5)*exp(x)+(6*x^2+24*x)*log( 5))*exp(x^2)+(2*x^2-2*x)*log(5)*exp(x)+(-24*x^2-6*x)*log(5))/(48*x^3*exp(x ^2)^2-192*x^4*exp(x^2)+192*x^5)/exp((3+exp(x))/(3*exp(x^2)-6*x)),x, algori thm=\
1/48*integrate((2*(x^2 - x)*e^x*log(5) + ((2*x^2 - x)*e^x*log(5) + 6*(x^2 + 4*x)*log(5))*e^(x^2) - 6*(4*x^2 + x)*log(5) - 6*e^(2*x^2)*log(5))*e^(1/3 *(e^x + 3)/(2*x - e^(x^2)))/(4*x^5 - 4*x^4*e^(x^2) + x^3*e^(2*x^2)), x)
Time = 0.42 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int \frac {e^{-\frac {3+e^x}{3 e^{x^2}-6 x}} \left (-6 e^{2 x^2} \log (5)+\left (-6 x-24 x^2\right ) \log (5)+e^x \left (-2 x+2 x^2\right ) \log (5)+e^{x^2} \left (e^x \left (-x+2 x^2\right ) \log (5)+\left (24 x+6 x^2\right ) \log (5)\right )\right )}{48 e^{2 x^2} x^3-192 e^{x^2} x^4+192 x^5} \, dx=\frac {e^{\left (\frac {e^{x} + 3}{3 \, {\left (2 \, x - e^{\left (x^{2}\right )}\right )}}\right )} \log \left (5\right )}{16 \, x^{2}} \]
integrate((-6*log(5)*exp(x^2)^2+((2*x^2-x)*log(5)*exp(x)+(6*x^2+24*x)*log( 5))*exp(x^2)+(2*x^2-2*x)*log(5)*exp(x)+(-24*x^2-6*x)*log(5))/(48*x^3*exp(x ^2)^2-192*x^4*exp(x^2)+192*x^5)/exp((3+exp(x))/(3*exp(x^2)-6*x)),x, algori thm=\
Time = 13.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16 \[ \int \frac {e^{-\frac {3+e^x}{3 e^{x^2}-6 x}} \left (-6 e^{2 x^2} \log (5)+\left (-6 x-24 x^2\right ) \log (5)+e^x \left (-2 x+2 x^2\right ) \log (5)+e^{x^2} \left (e^x \left (-x+2 x^2\right ) \log (5)+\left (24 x+6 x^2\right ) \log (5)\right )\right )}{48 e^{2 x^2} x^3-192 e^{x^2} x^4+192 x^5} \, dx=\frac {{\mathrm {e}}^{\frac {1}{2\,x-{\mathrm {e}}^{x^2}}}\,{\mathrm {e}}^{\frac {{\mathrm {e}}^x}{3\,\left (2\,x-{\mathrm {e}}^{x^2}\right )}}\,\ln \left (5\right )}{16\,x^2} \]