Integrand size = 104, antiderivative size = 34 \[ \int \frac {-15 x^2+e^{\frac {2-x^3+x^4}{x}} \left (10+10 x^3-15 x^4\right )}{12 x^2+3 e^{\frac {2 \left (2-x^3+x^4\right )}{x}} x^2-36 x^3+27 x^4+e^{\frac {2-x^3+x^4}{x}} \left (-12 x^2+18 x^3\right )} \, dx=\frac {5}{\left (9+\frac {3 \left (-2+e^{\frac {2}{x}-x \left (x-x^2\right )}\right )}{x}\right ) x} \]
Time = 1.22 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {-15 x^2+e^{\frac {2-x^3+x^4}{x}} \left (10+10 x^3-15 x^4\right )}{12 x^2+3 e^{\frac {2 \left (2-x^3+x^4\right )}{x}} x^2-36 x^3+27 x^4+e^{\frac {2-x^3+x^4}{x}} \left (-12 x^2+18 x^3\right )} \, dx=\frac {5 e^{x^2}}{3 \left (e^{\frac {2}{x}+x^3}+e^{x^2} (-2+3 x)\right )} \]
Integrate[(-15*x^2 + E^((2 - x^3 + x^4)/x)*(10 + 10*x^3 - 15*x^4))/(12*x^2 + 3*E^((2*(2 - x^3 + x^4))/x)*x^2 - 36*x^3 + 27*x^4 + E^((2 - x^3 + x^4)/ x)*(-12*x^2 + 18*x^3)),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.
\(\displaystyle \int \frac {e^{\frac {x^4-x^3+2}{x}} \left (-15 x^4+10 x^3+10\right )-15 x^2}{27 x^4-36 x^3+12 x^2+3 e^{\frac {2 \left (x^4-x^3+2\right )}{x}} x^2+e^{\frac {x^4-x^3+2}{x}} \left (18 x^3-12 x^2\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{2 x^2} \left (e^{\frac {x^4-x^3+2}{x}} \left (-15 x^4+10 x^3+10\right )-15 x^2\right )}{3 x^2 \left (-e^{x^3+\frac {2}{x}}-3 e^{x^2} x+2 e^{x^2}\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \int -\frac {5 e^{2 x^2} \left (3 x^2-e^{\frac {x^4-x^3+2}{x}} \left (-3 x^4+2 x^3+2\right )\right )}{x^2 \left (-3 e^{x^2} x+2 e^{x^2}-e^{x^3+\frac {2}{x}}\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {5}{3} \int \frac {e^{2 x^2} \left (3 x^2-e^{\frac {x^4-x^3+2}{x}} \left (-3 x^4+2 x^3+2\right )\right )}{x^2 \left (-3 e^{x^2} x+2 e^{x^2}-e^{x^3+\frac {2}{x}}\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {5}{3} \int \left (\frac {e^{x^2} \left (3 x^4-2 x^3-2\right )}{x^2 \left (3 e^{x^2} x-2 e^{x^2}+e^{x^3+\frac {2}{x}}\right )}-\frac {e^{2 x^2} \left (9 x^5-12 x^4+4 x^3-3 x^2-6 x+4\right )}{x^2 \left (3 e^{x^2} x-2 e^{x^2}+e^{x^3+\frac {2}{x}}\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {5}{3} \left (3 \int \frac {e^{2 x^2}}{\left (3 e^{x^2} x-2 e^{x^2}+e^{x^3+\frac {2}{x}}\right )^2}dx-4 \int \frac {e^{2 x^2}}{x^2 \left (3 e^{x^2} x-2 e^{x^2}+e^{x^3+\frac {2}{x}}\right )^2}dx+6 \int \frac {e^{2 x^2}}{x \left (3 e^{x^2} x-2 e^{x^2}+e^{x^3+\frac {2}{x}}\right )^2}dx-4 \int \frac {e^{2 x^2} x}{\left (3 e^{x^2} x-2 e^{x^2}+e^{x^3+\frac {2}{x}}\right )^2}dx+12 \int \frac {e^{2 x^2} x^2}{\left (3 e^{x^2} x-2 e^{x^2}+e^{x^3+\frac {2}{x}}\right )^2}dx-9 \int \frac {e^{2 x^2} x^3}{\left (3 e^{x^2} x-2 e^{x^2}+e^{x^3+\frac {2}{x}}\right )^2}dx-2 \int \frac {e^{x^2}}{x^2 \left (3 e^{x^2} x-2 e^{x^2}+e^{x^3+\frac {2}{x}}\right )}dx-2 \int \frac {e^{x^2} x}{3 e^{x^2} x-2 e^{x^2}+e^{x^3+\frac {2}{x}}}dx+3 \int \frac {e^{x^2} x^2}{3 e^{x^2} x-2 e^{x^2}+e^{x^3+\frac {2}{x}}}dx\right )\) |
Int[(-15*x^2 + E^((2 - x^3 + x^4)/x)*(10 + 10*x^3 - 15*x^4))/(12*x^2 + 3*E ^((2*(2 - x^3 + x^4))/x)*x^2 - 36*x^3 + 27*x^4 + E^((2 - x^3 + x^4)/x)*(-1 2*x^2 + 18*x^3)),x]
3.28.57.3.1 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 = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.74
method | result | size |
norman | \(\frac {5}{3 \left (-2+{\mathrm e}^{\frac {x^{4}-x^{3}+2}{x}}+3 x \right )}\) | \(25\) |
risch | \(\frac {5}{3 \left (-2+{\mathrm e}^{\frac {x^{4}-x^{3}+2}{x}}+3 x \right )}\) | \(25\) |
parallelrisch | \(\frac {5}{3 \left (-2+{\mathrm e}^{\frac {x^{4}-x^{3}+2}{x}}+3 x \right )}\) | \(25\) |
int(((-15*x^4+10*x^3+10)*exp((x^4-x^3+2)/x)-15*x^2)/(3*x^2*exp((x^4-x^3+2) /x)^2+(18*x^3-12*x^2)*exp((x^4-x^3+2)/x)+27*x^4-36*x^3+12*x^2),x,method=_R ETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {-15 x^2+e^{\frac {2-x^3+x^4}{x}} \left (10+10 x^3-15 x^4\right )}{12 x^2+3 e^{\frac {2 \left (2-x^3+x^4\right )}{x}} x^2-36 x^3+27 x^4+e^{\frac {2-x^3+x^4}{x}} \left (-12 x^2+18 x^3\right )} \, dx=\frac {5}{3 \, {\left (3 \, x + e^{\left (\frac {x^{4} - x^{3} + 2}{x}\right )} - 2\right )}} \]
integrate(((-15*x^4+10*x^3+10)*exp((x^4-x^3+2)/x)-15*x^2)/(3*x^2*exp((x^4- x^3+2)/x)^2+(18*x^3-12*x^2)*exp((x^4-x^3+2)/x)+27*x^4-36*x^3+12*x^2),x, al gorithm=\
Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.56 \[ \int \frac {-15 x^2+e^{\frac {2-x^3+x^4}{x}} \left (10+10 x^3-15 x^4\right )}{12 x^2+3 e^{\frac {2 \left (2-x^3+x^4\right )}{x}} x^2-36 x^3+27 x^4+e^{\frac {2-x^3+x^4}{x}} \left (-12 x^2+18 x^3\right )} \, dx=\frac {5}{9 x + 3 e^{\frac {x^{4} - x^{3} + 2}{x}} - 6} \]
integrate(((-15*x**4+10*x**3+10)*exp((x**4-x**3+2)/x)-15*x**2)/(3*x**2*exp ((x**4-x**3+2)/x)**2+(18*x**3-12*x**2)*exp((x**4-x**3+2)/x)+27*x**4-36*x** 3+12*x**2),x)
Time = 0.24 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {-15 x^2+e^{\frac {2-x^3+x^4}{x}} \left (10+10 x^3-15 x^4\right )}{12 x^2+3 e^{\frac {2 \left (2-x^3+x^4\right )}{x}} x^2-36 x^3+27 x^4+e^{\frac {2-x^3+x^4}{x}} \left (-12 x^2+18 x^3\right )} \, dx=\frac {5 \, e^{\left (x^{2}\right )}}{3 \, {\left ({\left (3 \, x - 2\right )} e^{\left (x^{2}\right )} + e^{\left (x^{3} + \frac {2}{x}\right )}\right )}} \]
integrate(((-15*x^4+10*x^3+10)*exp((x^4-x^3+2)/x)-15*x^2)/(3*x^2*exp((x^4- x^3+2)/x)^2+(18*x^3-12*x^2)*exp((x^4-x^3+2)/x)+27*x^4-36*x^3+12*x^2),x, al gorithm=\
Time = 0.30 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.71 \[ \int \frac {-15 x^2+e^{\frac {2-x^3+x^4}{x}} \left (10+10 x^3-15 x^4\right )}{12 x^2+3 e^{\frac {2 \left (2-x^3+x^4\right )}{x}} x^2-36 x^3+27 x^4+e^{\frac {2-x^3+x^4}{x}} \left (-12 x^2+18 x^3\right )} \, dx=\frac {5}{3 \, {\left (3 \, x + e^{\left (\frac {x^{4} - x^{3} + 2}{x}\right )} - 2\right )}} \]
integrate(((-15*x^4+10*x^3+10)*exp((x^4-x^3+2)/x)-15*x^2)/(3*x^2*exp((x^4- x^3+2)/x)^2+(18*x^3-12*x^2)*exp((x^4-x^3+2)/x)+27*x^4-36*x^3+12*x^2),x, al gorithm=\
Time = 13.63 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.79 \[ \int \frac {-15 x^2+e^{\frac {2-x^3+x^4}{x}} \left (10+10 x^3-15 x^4\right )}{12 x^2+3 e^{\frac {2 \left (2-x^3+x^4\right )}{x}} x^2-36 x^3+27 x^4+e^{\frac {2-x^3+x^4}{x}} \left (-12 x^2+18 x^3\right )} \, dx=\frac {5}{3\,\left (3\,x+{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{2/x}\,{\mathrm {e}}^{-x^2}-2\right )} \]