Integrand size = 76, antiderivative size = 27 \[ \int \frac {e^{-\frac {3-x}{e^2-x}} \left (6-2 e^2+e^{\frac {3-x}{e^2-x}} \left (2 e^4-4 e^2 x+2 x^2\right )\right )}{e^4-2 e^2 x+x^2} \, dx=\log \left (e^{-2 e^{-\frac {3-x}{e^2-x}}+2 x}\right ) \] Output:
ln(exp(2*x-2/exp((3-x)/(exp(2)-x))))
Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {e^{-\frac {3-x}{e^2-x}} \left (6-2 e^2+e^{\frac {3-x}{e^2-x}} \left (2 e^4-4 e^2 x+2 x^2\right )\right )}{e^4-2 e^2 x+x^2} \, dx=-2 e^{\frac {-3+x}{e^2-x}}+2 x \] Input:
Integrate[(6 - 2*E^2 + E^((3 - x)/(E^2 - x))*(2*E^4 - 4*E^2*x + 2*x^2))/(E ^((3 - x)/(E^2 - x))*(E^4 - 2*E^2*x + x^2)),x]
Output:
-2*E^((-3 + x)/(E^2 - x)) + 2*x
Time = 2.47 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {7277, 27, 7292, 7239, 7293, 2009}
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.
\(\displaystyle \int \frac {e^{-\frac {3-x}{e^2-x}} \left (e^{\frac {3-x}{e^2-x}} \left (2 x^2-4 e^2 x+2 e^4\right )-2 e^2+6\right )}{x^2-2 e^2 x+e^4} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 4 \int \frac {e^{-\frac {3-x}{e^2-x}} \left (e^{\frac {3-x}{e^2-x}} \left (x^2-2 e^2 x+e^4\right )-e^2+3\right )}{2 \left (e^2-x\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 2 \int \frac {e^{-\frac {3-x}{e^2-x}} \left (e^{\frac {3-x}{e^2-x}} \left (x^2-2 e^2 x+e^4\right )-e^2+3\right )}{\left (e^2-x\right )^2}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle 2 \int \frac {e^{-\frac {3-x}{e^2-x}} \left (e^{\frac {3-x}{e^2-x}} \left (x^2-2 e^2 x+e^4\right )+3 \left (1-\frac {e^2}{3}\right )\right )}{\left (e^2-x\right )^2}dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle 2 \int \frac {x^2-2 e^2 x+e^{\frac {x-3}{e^2-x}} \left (3-e^2\right )+e^4}{\left (e^2-x\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle 2 \int \left (1+\frac {e^{\frac {x-3}{e^2-x}} \left (3-e^2\right )}{\left (e^2-x\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \left (x-e^{-\frac {3-e^2}{e^2-x}-1}\right )\) |
Input:
Int[(6 - 2*E^2 + E^((3 - x)/(E^2 - x))*(2*E^4 - 4*E^2*x + 2*x^2))/(E^((3 - x)/(E^2 - x))*(E^4 - 2*E^2*x + x^2)),x]
Output:
2*(-E^(-1 - (3 - E^2)/(E^2 - x)) + x)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(u_)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 0.50 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74
method | result | size |
risch | \(2 x -2 \,{\mathrm e}^{\frac {-3+x}{{\mathrm e}^{2}-x}}\) | \(20\) |
parts | \(2 x +\frac {\left (-2 \,{\mathrm e}^{2}+6\right ) \left ({\mathrm e}^{2}-3\right ) {\mathrm e}^{-1-\frac {{\mathrm e}^{2}-3}{-{\mathrm e}^{2}+x}}}{{\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9}\) | \(46\) |
norman | \(\frac {\left (2 \,{\mathrm e}^{4} {\mathrm e}^{\frac {-x +3}{{\mathrm e}^{2}-x}}+2 x -2 x^{2} {\mathrm e}^{\frac {-x +3}{{\mathrm e}^{2}-x}}-2 \,{\mathrm e}^{2}\right ) {\mathrm e}^{-\frac {-x +3}{{\mathrm e}^{2}-x}}}{{\mathrm e}^{2}-x}\) | \(76\) |
parallelrisch | \(\frac {\left (8 \,{\mathrm e}^{4} {\mathrm e}^{-\frac {-3+x}{{\mathrm e}^{2}-x}}-6 \,{\mathrm e}^{2} {\mathrm e}^{-\frac {-3+x}{{\mathrm e}^{2}-x}} x -2 \,{\mathrm e}^{-\frac {-3+x}{{\mathrm e}^{2}-x}} x^{2}-2 \,{\mathrm e}^{2}+2 x \right ) {\mathrm e}^{\frac {-3+x}{{\mathrm e}^{2}-x}}}{{\mathrm e}^{2}-x}\) | \(92\) |
derivativedivides | \(-\left ({\mathrm e}^{2}-3\right ) \left (-\frac {18 \left (-{\mathrm e}^{2}+x \right )}{\left ({\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9\right ) \left ({\mathrm e}^{2}-3\right )}+\frac {12 \,{\mathrm e}^{2} \left (-{\mathrm e}^{2}+x \right )}{\left ({\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9\right ) \left ({\mathrm e}^{2}-3\right )}-\frac {2 \,{\mathrm e}^{4} \left (-{\mathrm e}^{2}+x \right )}{\left ({\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9\right ) \left ({\mathrm e}^{2}-3\right )}-\frac {6 \,{\mathrm e}^{-1-\frac {{\mathrm e}^{2}-3}{-{\mathrm e}^{2}+x}}}{{\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9}+\frac {2 \,{\mathrm e}^{2} {\mathrm e}^{-1-\frac {{\mathrm e}^{2}-3}{-{\mathrm e}^{2}+x}}}{{\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9}\right )\) | \(158\) |
default | \(-\left ({\mathrm e}^{2}-3\right ) \left (-\frac {18 \left (-{\mathrm e}^{2}+x \right )}{\left ({\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9\right ) \left ({\mathrm e}^{2}-3\right )}+\frac {12 \,{\mathrm e}^{2} \left (-{\mathrm e}^{2}+x \right )}{\left ({\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9\right ) \left ({\mathrm e}^{2}-3\right )}-\frac {2 \,{\mathrm e}^{4} \left (-{\mathrm e}^{2}+x \right )}{\left ({\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9\right ) \left ({\mathrm e}^{2}-3\right )}-\frac {6 \,{\mathrm e}^{-1-\frac {{\mathrm e}^{2}-3}{-{\mathrm e}^{2}+x}}}{{\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9}+\frac {2 \,{\mathrm e}^{2} {\mathrm e}^{-1-\frac {{\mathrm e}^{2}-3}{-{\mathrm e}^{2}+x}}}{{\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9}\right )\) | \(158\) |
Input:
int(((2*exp(2)^2-4*exp(2)*x+2*x^2)*exp((-x+3)/(exp(2)-x))-2*exp(2)+6)/(exp (2)^2-2*exp(2)*x+x^2)/exp((-x+3)/(exp(2)-x)),x,method=_RETURNVERBOSE)
Output:
2*x-2*exp((-3+x)/(exp(2)-x))
Time = 0.08 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.74 \[ \int \frac {e^{-\frac {3-x}{e^2-x}} \left (6-2 e^2+e^{\frac {3-x}{e^2-x}} \left (2 e^4-4 e^2 x+2 x^2\right )\right )}{e^4-2 e^2 x+x^2} \, dx=2 \, x - 2 \, e^{\left (-\frac {x - 3}{x - e^{2}}\right )} \] Input:
integrate(((2*exp(2)^2-4*exp(2)*x+2*x^2)*exp((3-x)/(exp(2)-x))-2*exp(2)+6) /(exp(2)^2-2*exp(2)*x+x^2)/exp((3-x)/(exp(2)-x)),x, algorithm="fricas")
Output:
2*x - 2*e^(-(x - 3)/(x - e^2))
Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.52 \[ \int \frac {e^{-\frac {3-x}{e^2-x}} \left (6-2 e^2+e^{\frac {3-x}{e^2-x}} \left (2 e^4-4 e^2 x+2 x^2\right )\right )}{e^4-2 e^2 x+x^2} \, dx=2 x - 2 e^{- \frac {3 - x}{- x + e^{2}}} \] Input:
integrate(((2*exp(2)**2-4*exp(2)*x+2*x**2)*exp((3-x)/(exp(2)-x))-2*exp(2)+ 6)/(exp(2)**2-2*exp(2)*x+x**2)/exp((3-x)/(exp(2)-x)),x)
Output:
2*x - 2*exp(-(3 - x)/(-x + exp(2)))
Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (22) = 44\).
Time = 0.04 (sec) , antiderivative size = 120, normalized size of antiderivative = 4.44 \[ \int \frac {e^{-\frac {3-x}{e^2-x}} \left (6-2 e^2+e^{\frac {3-x}{e^2-x}} \left (2 e^4-4 e^2 x+2 x^2\right )\right )}{e^4-2 e^2 x+x^2} \, dx=4 \, {\left (\frac {e^{2}}{x - e^{2}} - \log \left (x - e^{2}\right )\right )} e^{2} + 4 \, e^{2} \log \left (x - e^{2}\right ) + 2 \, x - \frac {4 \, e^{4}}{x - e^{2}} - \frac {2 \, e^{\left (-\frac {e^{2}}{x - e^{2}} + \frac {3}{x - e^{2}} + 1\right )}}{e^{2} - 3} + \frac {6 \, e^{\left (-\frac {e^{2}}{x - e^{2}} + \frac {3}{x - e^{2}}\right )}}{e^{3} - 3 \, e} \] Input:
integrate(((2*exp(2)^2-4*exp(2)*x+2*x^2)*exp((3-x)/(exp(2)-x))-2*exp(2)+6) /(exp(2)^2-2*exp(2)*x+x^2)/exp((3-x)/(exp(2)-x)),x, algorithm="maxima")
Output:
4*(e^2/(x - e^2) - log(x - e^2))*e^2 + 4*e^2*log(x - e^2) + 2*x - 4*e^4/(x - e^2) - 2*e^(-e^2/(x - e^2) + 3/(x - e^2) + 1)/(e^2 - 3) + 6*e^(-e^2/(x - e^2) + 3/(x - e^2))/(e^3 - 3*e)
Leaf count of result is larger than twice the leaf count of optimal. 81 vs. \(2 (22) = 44\).
Time = 1.60 (sec) , antiderivative size = 81, normalized size of antiderivative = 3.00 \[ \int \frac {e^{-\frac {3-x}{e^2-x}} \left (6-2 e^2+e^{\frac {3-x}{e^2-x}} \left (2 e^4-4 e^2 x+2 x^2\right )\right )}{e^4-2 e^2 x+x^2} \, dx=-2 \, {\left (\frac {2 \, e^{2} - 3}{{\left (e^{2} - 3\right )}^{2}} - \frac {e^{2}}{{\left (e^{2} - 3\right )}^{2}}\right )} e^{\left (\frac {x - 2 \, e^{2} + 3}{x - e^{2}}\right )} + 6 \, {\left (\frac {e^{2}}{{\left (e^{2} - 3\right )}^{2}} - \frac {3}{{\left (e^{2} - 3\right )}^{2}}\right )} e^{\left (-\frac {x - 3}{x - e^{2}}\right )} + 2 \, x \] Input:
integrate(((2*exp(2)^2-4*exp(2)*x+2*x^2)*exp((3-x)/(exp(2)-x))-2*exp(2)+6) /(exp(2)^2-2*exp(2)*x+x^2)/exp((3-x)/(exp(2)-x)),x, algorithm="giac")
Output:
-2*((2*e^2 - 3)/(e^2 - 3)^2 - e^2/(e^2 - 3)^2)*e^((x - 2*e^2 + 3)/(x - e^2 )) + 6*(e^2/(e^2 - 3)^2 - 3/(e^2 - 3)^2)*e^(-(x - 3)/(x - e^2)) + 2*x
Time = 2.72 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-\frac {3-x}{e^2-x}} \left (6-2 e^2+e^{\frac {3-x}{e^2-x}} \left (2 e^4-4 e^2 x+2 x^2\right )\right )}{e^4-2 e^2 x+x^2} \, dx=2\,x-2\,{\mathrm {e}}^{-\frac {x}{x-{\mathrm {e}}^2}}\,{\mathrm {e}}^{\frac {3}{x-{\mathrm {e}}^2}} \] Input:
int((exp(-(x - 3)/(x - exp(2)))*(exp((x - 3)/(x - exp(2)))*(2*exp(4) - 4*x *exp(2) + 2*x^2) - 2*exp(2) + 6))/(exp(4) - 2*x*exp(2) + x^2),x)
Output:
2*x - 2*exp(-x/(x - exp(2)))*exp(3/(x - exp(2)))
Time = 0.21 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.78 \[ \int \frac {e^{-\frac {3-x}{e^2-x}} \left (6-2 e^2+e^{\frac {3-x}{e^2-x}} \left (2 e^4-4 e^2 x+2 x^2\right )\right )}{e^4-2 e^2 x+x^2} \, dx=\frac {-2 e^{\frac {x}{e^{2}-x}}+2 e^{\frac {3}{e^{2}-x}} x}{e^{\frac {3}{e^{2}-x}}} \] Input:
int(((2*exp(2)^2-4*exp(2)*x+2*x^2)*exp((3-x)/(exp(2)-x))-2*exp(2)+6)/(exp( 2)^2-2*exp(2)*x+x^2)/exp((3-x)/(exp(2)-x)),x)
Output:
(2*( - e**(x/(e**2 - x)) + e**(3/(e**2 - x))*x))/e**(3/(e**2 - x))