Integrand size = 115, antiderivative size = 32 \[ \int \frac {e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} \left (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} \left (10-5 x^2\right )\right )}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} x^2+2 x^3} \, dx=\frac {x}{e^{\frac {5}{2} \left (e^{\frac {1}{x^2}} x-x^2-x^3\right )}-x} \] Output:
x/(exp(5/2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)-x)
Time = 2.10 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.44 \[ \int \frac {e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} \left (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} \left (10-5 x^2\right )\right )}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} x^2+2 x^3} \, dx=-\frac {e^{\frac {5}{2} x^2 (1+x)} x}{-e^{\frac {5}{2} e^{\frac {1}{x^2}} x}+e^{\frac {5}{2} x^2 (1+x)} x} \] Input:
Integrate[(E^((5*E^x^(-2)*x - 5*x^2 - 5*x^3)/2)*(2*x + 10*x^3 + 15*x^4 + E ^x^(-2)*(10 - 5*x^2)))/(2*E^(5*E^x^(-2)*x - 5*x^2 - 5*x^3)*x - 4*E^((5*E^x ^(-2)*x - 5*x^2 - 5*x^3)/2)*x^2 + 2*x^3),x]
Output:
-((E^((5*x^2*(1 + x))/2)*x)/(-E^((5*E^x^(-2)*x)/2) + E^((5*x^2*(1 + x))/2) *x))
Time = 4.47 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.035, Rules used = {7292, 27, 7262, 17}
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.
\(\displaystyle \int \frac {e^{\frac {1}{2} \left (-5 x^3-5 x^2+5 e^{\frac {1}{x^2}} x\right )} \left (15 x^4+10 x^3+e^{\frac {1}{x^2}} \left (10-5 x^2\right )+2 x\right )}{2 x^3-4 e^{\frac {1}{2} \left (-5 x^3-5 x^2+5 e^{\frac {1}{x^2}} x\right )} x^2+2 e^{-5 x^3-5 x^2+5 e^{\frac {1}{x^2}} x} x} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{5 x^2 (x+1)-\frac {5}{2} x \left (x^2-e^{\frac {1}{x^2}}+x\right )} \left (15 x^4+10 x^3+e^{\frac {1}{x^2}} \left (10-5 x^2\right )+2 x\right )}{2 x \left (e^{\frac {5}{2} e^{\frac {1}{x^2}} x}-e^{\frac {5}{2} x^2 (x+1)} x\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {e^{5 (x+1) x^2+\frac {5}{2} \left (-x^2-x+e^{\frac {1}{x^2}}\right ) x} \left (15 x^4+10 x^3+2 x+5 e^{\frac {1}{x^2}} \left (2-x^2\right )\right )}{x \left (e^{\frac {5}{2} e^{\frac {1}{x^2}} x}-e^{\frac {5}{2} x^2 (x+1)} x\right )^2}dx\) |
\(\Big \downarrow \) 7262 |
\(\displaystyle \int \frac {1}{\left (1-\frac {e^{\frac {5}{2} e^{\frac {1}{x^2}} x-\frac {5}{2} x^2 (x+1)}}{x}\right )^2}d\left (-\frac {e^{\frac {5}{2} e^{\frac {1}{x^2}} x-\frac {5}{2} x^2 (x+1)}}{x}\right )\) |
\(\Big \downarrow \) 17 |
\(\displaystyle -\frac {1}{1-\frac {e^{\frac {5}{2} e^{\frac {1}{x^2}} x-\frac {5}{2} x^2 (x+1)}}{x}}\) |
Input:
Int[(E^((5*E^x^(-2)*x - 5*x^2 - 5*x^3)/2)*(2*x + 10*x^3 + 15*x^4 + E^x^(-2 )*(10 - 5*x^2)))/(2*E^(5*E^x^(-2)*x - 5*x^2 - 5*x^3)*x - 4*E^((5*E^x^(-2)* x - 5*x^2 - 5*x^3)/2)*x^2 + 2*x^3),x]
Output:
-(1 - E^((5*E^x^(-2)*x)/2 - (5*x^2*(1 + x))/2)/x)^(-1)
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((a_.)*(v_)^(p_.) + (b_.)*(w_)^(q_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(p*w*D[v, x] - q*v*D[w, x])]}, Simp[c*p Subst[Int[(b + a*x^p )^m, x], x, v*w^(m*q + 1)], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p, q}, x] && EqQ[p + q*(m*p + 1), 0] && IntegerQ[p] && IntegerQ[m]
Time = 0.48 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.78
method | result | size |
risch | \(-\frac {x}{x -{\mathrm e}^{-\frac {5 x \left (x^{2}-{\mathrm e}^{\frac {1}{x^{2}}}+x \right )}{2}}}\) | \(25\) |
parallelrisch | \(-\frac {{\mathrm e}^{\frac {5 x \left (-x^{2}+{\mathrm e}^{\frac {1}{x^{2}}}-x \right )}{2}}}{x -{\mathrm e}^{\frac {5 x \left (-x^{2}+{\mathrm e}^{\frac {1}{x^{2}}}-x \right )}{2}}}\) | \(43\) |
norman | \(-\frac {{\mathrm e}^{\frac {5 x \,{\mathrm e}^{\frac {1}{x^{2}}}}{2}-\frac {5 x^{3}}{2}-\frac {5 x^{2}}{2}}}{x -{\mathrm e}^{\frac {5 x \,{\mathrm e}^{\frac {1}{x^{2}}}}{2}-\frac {5 x^{3}}{2}-\frac {5 x^{2}}{2}}}\) | \(47\) |
Input:
int(((-5*x^2+10)*exp(1/x^2)+15*x^4+10*x^3+2*x)*exp(5/2*x*exp(1/x^2)-5/2*x^ 3-5/2*x^2)/(2*x*exp(5/2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)^2-4*x^2*exp(5/2*x*ex p(1/x^2)-5/2*x^3-5/2*x^2)+2*x^3),x,method=_RETURNVERBOSE)
Output:
-x/(x-exp(-5/2*x*(x^2-exp(1/x^2)+x)))
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} \left (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} \left (10-5 x^2\right )\right )}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} x^2+2 x^3} \, dx=-\frac {x}{x - e^{\left (-\frac {5}{2} \, x^{3} - \frac {5}{2} \, x^{2} + \frac {5}{2} \, x e^{\left (\frac {1}{x^{2}}\right )}\right )}} \] Input:
integrate(((-5*x^2+10)*exp(1/x^2)+15*x^4+10*x^3+2*x)*exp(5/2*x*exp(1/x^2)- 5/2*x^3-5/2*x^2)/(2*x*exp(5/2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)^2-4*x^2*exp(5/ 2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)+2*x^3),x, algorithm="fricas")
Output:
-x/(x - e^(-5/2*x^3 - 5/2*x^2 + 5/2*x*e^(x^(-2))))
Time = 0.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} \left (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} \left (10-5 x^2\right )\right )}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} x^2+2 x^3} \, dx=\frac {x}{- x + e^{- \frac {5 x^{3}}{2} - \frac {5 x^{2}}{2} + \frac {5 x e^{\frac {1}{x^{2}}}}{2}}} \] Input:
integrate(((-5*x**2+10)*exp(1/x**2)+15*x**4+10*x**3+2*x)*exp(5/2*x*exp(1/x **2)-5/2*x**3-5/2*x**2)/(2*x*exp(5/2*x*exp(1/x**2)-5/2*x**3-5/2*x**2)**2-4 *x**2*exp(5/2*x*exp(1/x**2)-5/2*x**3-5/2*x**2)+2*x**3),x)
Output:
x/(-x + exp(-5*x**3/2 - 5*x**2/2 + 5*x*exp(x**(-2))/2))
Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int \frac {e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} \left (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} \left (10-5 x^2\right )\right )}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} x^2+2 x^3} \, dx=-\frac {x e^{\left (\frac {5}{2} \, x^{3} + \frac {5}{2} \, x^{2}\right )}}{x e^{\left (\frac {5}{2} \, x^{3} + \frac {5}{2} \, x^{2}\right )} - e^{\left (\frac {5}{2} \, x e^{\left (\frac {1}{x^{2}}\right )}\right )}} \] Input:
integrate(((-5*x^2+10)*exp(1/x^2)+15*x^4+10*x^3+2*x)*exp(5/2*x*exp(1/x^2)- 5/2*x^3-5/2*x^2)/(2*x*exp(5/2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)^2-4*x^2*exp(5/ 2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)+2*x^3),x, algorithm="maxima")
Output:
-x*e^(5/2*x^3 + 5/2*x^2)/(x*e^(5/2*x^3 + 5/2*x^2) - e^(5/2*x*e^(x^(-2))))
Time = 0.30 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} \left (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} \left (10-5 x^2\right )\right )}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} x^2+2 x^3} \, dx=-\frac {x}{x - e^{\left (-\frac {5}{2} \, x^{3} - \frac {5}{2} \, x^{2} + \frac {5}{2} \, x e^{\left (\frac {1}{x^{2}}\right )}\right )}} \] Input:
integrate(((-5*x^2+10)*exp(1/x^2)+15*x^4+10*x^3+2*x)*exp(5/2*x*exp(1/x^2)- 5/2*x^3-5/2*x^2)/(2*x*exp(5/2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)^2-4*x^2*exp(5/ 2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)+2*x^3),x, algorithm="giac")
Output:
-x/(x - e^(-5/2*x^3 - 5/2*x^2 + 5/2*x*e^(x^(-2))))
Time = 2.78 (sec) , antiderivative size = 95, normalized size of antiderivative = 2.97 \[ \int \frac {e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} \left (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} \left (10-5 x^2\right )\right )}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} x^2+2 x^3} \, dx=-\frac {x^4\,\left (2\,x+10\,{\mathrm {e}}^{\frac {1}{x^2}}-5\,x^2\,{\mathrm {e}}^{\frac {1}{x^2}}+10\,x^3+15\,x^4\right )}{\left (x-{\mathrm {e}}^{\frac {5\,x\,{\mathrm {e}}^{\frac {1}{x^2}}}{2}-\frac {5\,x^2}{2}-\frac {5\,x^3}{2}}\right )\,\left (10\,x^3\,{\mathrm {e}}^{\frac {1}{x^2}}-5\,x^5\,{\mathrm {e}}^{\frac {1}{x^2}}+2\,x^4+10\,x^6+15\,x^7\right )} \] Input:
int((exp((5*x*exp(1/x^2))/2 - (5*x^2)/2 - (5*x^3)/2)*(2*x - exp(1/x^2)*(5* x^2 - 10) + 10*x^3 + 15*x^4))/(2*x*exp(5*x*exp(1/x^2) - 5*x^2 - 5*x^3) - 4 *x^2*exp((5*x*exp(1/x^2))/2 - (5*x^2)/2 - (5*x^3)/2) + 2*x^3),x)
Output:
-(x^4*(2*x + 10*exp(1/x^2) - 5*x^2*exp(1/x^2) + 10*x^3 + 15*x^4))/((x - ex p((5*x*exp(1/x^2))/2 - (5*x^2)/2 - (5*x^3)/2))*(10*x^3*exp(1/x^2) - 5*x^5* exp(1/x^2) + 2*x^4 + 10*x^6 + 15*x^7))
\[ \int \frac {e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} \left (2 x+10 x^3+15 x^4+e^{\frac {1}{x^2}} \left (10-5 x^2\right )\right )}{2 e^{5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3} x-4 e^{\frac {1}{2} \left (5 e^{\frac {1}{x^2}} x-5 x^2-5 x^3\right )} x^2+2 x^3} \, dx=\int \frac {\left (\left (-5 x^{2}+10\right ) {\mathrm e}^{\frac {1}{x^{2}}}+15 x^{4}+10 x^{3}+2 x \right ) {\mathrm e}^{\frac {5 x \,{\mathrm e}^{\frac {1}{x^{2}}}}{2}-\frac {5 x^{3}}{2}-\frac {5 x^{2}}{2}}}{2 x \left ({\mathrm e}^{\frac {5 x \,{\mathrm e}^{\frac {1}{x^{2}}}}{2}-\frac {5 x^{3}}{2}-\frac {5 x^{2}}{2}}\right )^{2}-4 x^{2} {\mathrm e}^{\frac {5 x \,{\mathrm e}^{\frac {1}{x^{2}}}}{2}-\frac {5 x^{3}}{2}-\frac {5 x^{2}}{2}}+2 x^{3}}d x \] Input:
int(((-5*x^2+10)*exp(1/x^2)+15*x^4+10*x^3+2*x)*exp(5/2*x*exp(1/x^2)-5/2*x^ 3-5/2*x^2)/(2*x*exp(5/2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)^2-4*x^2*exp(5/2*x*ex p(1/x^2)-5/2*x^3-5/2*x^2)+2*x^3),x)
Output:
int(((-5*x^2+10)*exp(1/x^2)+15*x^4+10*x^3+2*x)*exp(5/2*x*exp(1/x^2)-5/2*x^ 3-5/2*x^2)/(2*x*exp(5/2*x*exp(1/x^2)-5/2*x^3-5/2*x^2)^2-4*x^2*exp(5/2*x*ex p(1/x^2)-5/2*x^3-5/2*x^2)+2*x^3),x)