Integrand size = 77, antiderivative size = 25 \[ \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-5+e^{e^2} \left (5 x-10 e^x x\right )\right )}{e^{2 e^{e^2} \left (100-2 e^x+x\right )}-2 e^{e^{e^2} \left (100-2 e^x+x\right )} x+x^2} \, dx=\frac {5 x}{-e^{e^{e^2} \left (100-2 e^x+x\right )}+x} \] Output:
x/(1/5*x-1/5*exp((-2*exp(x)+x+100)*exp(exp(2))))
Time = 0.57 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-5+e^{e^2} \left (5 x-10 e^x x\right )\right )}{e^{2 e^{e^2} \left (100-2 e^x+x\right )}-2 e^{e^{e^2} \left (100-2 e^x+x\right )} x+x^2} \, dx=-\frac {5 x}{e^{e^{e^2} \left (100-2 e^x+x\right )}-x} \] Input:
Integrate[(E^(E^E^2*(100 - 2*E^x + x))*(-5 + E^E^2*(5*x - 10*E^x*x)))/(E^( 2*E^E^2*(100 - 2*E^x + x)) - 2*E^(E^E^2*(100 - 2*E^x + x))*x + x^2),x]
Output:
(-5*x)/(E^(E^E^2*(100 - 2*E^x + x)) - x)
Time = 2.39 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {7292, 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.
\(\displaystyle \int \frac {e^{e^{e^2} \left (x-2 e^x+100\right )} \left (e^{e^2} \left (5 x-10 e^x x\right )-5\right )}{x^2-2 e^{e^{e^2} \left (x-2 e^x+100\right )} x+e^{2 e^{e^2} \left (x-2 e^x+100\right )}} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{e^{e^2} \left (x-2 e^x+100\right )} \left (e^{e^2} \left (5 x-10 e^x x\right )-5\right )}{\left (e^{e^{e^2} \left (x-2 e^x+100\right )}-x\right )^2}dx\) |
\(\Big \downarrow \) 7262 |
\(\displaystyle 5 \int \frac {1}{\left (\frac {e^{e^{e^2} \left (x-2 e^x+100\right )}}{x}-1\right )^2}d\frac {e^{e^{e^2} \left (x-2 e^x+100\right )}}{x}\) |
\(\Big \downarrow \) 17 |
\(\displaystyle \frac {5}{1-\frac {e^{e^{e^2} \left (x-2 e^x+100\right )}}{x}}\) |
Input:
Int[(E^(E^E^2*(100 - 2*E^x + x))*(-5 + E^E^2*(5*x - 10*E^x*x)))/(E^(2*E^E^ 2*(100 - 2*E^x + x)) - 2*E^(E^E^2*(100 - 2*E^x + x))*x + x^2),x]
Output:
5/(1 - E^(E^E^2*(100 - 2*E^x + x))/x)
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[(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 = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(\frac {5 x}{x -{\mathrm e}^{\left (-2 \,{\mathrm e}^{x}+x +100\right ) {\mathrm e}^{{\mathrm e}^{2}}}}\) | \(22\) |
risch | \(\frac {5 x}{x -{\mathrm e}^{-\left (2 \,{\mathrm e}^{x}-x -100\right ) {\mathrm e}^{{\mathrm e}^{2}}}}\) | \(25\) |
norman | \(\frac {5 \,{\mathrm e}^{\left (-2 \,{\mathrm e}^{x}+x +100\right ) {\mathrm e}^{{\mathrm e}^{2}}}}{x -{\mathrm e}^{\left (-2 \,{\mathrm e}^{x}+x +100\right ) {\mathrm e}^{{\mathrm e}^{2}}}}\) | \(33\) |
Input:
int(((-10*exp(x)*x+5*x)*exp(exp(2))-5)*exp((-2*exp(x)+x+100)*exp(exp(2)))/ (exp((-2*exp(x)+x+100)*exp(exp(2)))^2-2*x*exp((-2*exp(x)+x+100)*exp(exp(2) ))+x^2),x,method=_RETURNVERBOSE)
Output:
5*x/(x-exp((-2*exp(x)+x+100)*exp(exp(2))))
Time = 0.08 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-5+e^{e^2} \left (5 x-10 e^x x\right )\right )}{e^{2 e^{e^2} \left (100-2 e^x+x\right )}-2 e^{e^{e^2} \left (100-2 e^x+x\right )} x+x^2} \, dx=\frac {5 \, x}{x - e^{\left ({\left (x - 2 \, e^{x} + 100\right )} e^{\left (e^{2}\right )}\right )}} \] Input:
integrate(((-10*exp(x)*x+5*x)*exp(exp(2))-5)*exp((-2*exp(x)+x+100)*exp(exp (2)))/(exp((-2*exp(x)+x+100)*exp(exp(2)))^2-2*x*exp((-2*exp(x)+x+100)*exp( exp(2)))+x^2),x, algorithm="fricas")
Output:
5*x/(x - e^((x - 2*e^x + 100)*e^(e^2)))
Time = 0.10 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-5+e^{e^2} \left (5 x-10 e^x x\right )\right )}{e^{2 e^{e^2} \left (100-2 e^x+x\right )}-2 e^{e^{e^2} \left (100-2 e^x+x\right )} x+x^2} \, dx=- \frac {5 x}{- x + e^{\left (x - 2 e^{x} + 100\right ) e^{e^{2}}}} \] Input:
integrate(((-10*exp(x)*x+5*x)*exp(exp(2))-5)*exp((-2*exp(x)+x+100)*exp(exp (2)))/(exp((-2*exp(x)+x+100)*exp(exp(2)))**2-2*x*exp((-2*exp(x)+x+100)*exp (exp(2)))+x**2),x)
Output:
-5*x/(-x + exp((x - 2*exp(x) + 100)*exp(exp(2))))
Time = 0.15 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-5+e^{e^2} \left (5 x-10 e^x x\right )\right )}{e^{2 e^{e^2} \left (100-2 e^x+x\right )}-2 e^{e^{e^2} \left (100-2 e^x+x\right )} x+x^2} \, dx=\frac {5 \, e^{\left (x e^{\left (e^{2}\right )} + 100 \, e^{\left (e^{2}\right )}\right )}}{x e^{\left (2 \, e^{\left (x + e^{2}\right )}\right )} - e^{\left (x e^{\left (e^{2}\right )} + 100 \, e^{\left (e^{2}\right )}\right )}} \] Input:
integrate(((-10*exp(x)*x+5*x)*exp(exp(2))-5)*exp((-2*exp(x)+x+100)*exp(exp (2)))/(exp((-2*exp(x)+x+100)*exp(exp(2)))^2-2*x*exp((-2*exp(x)+x+100)*exp( exp(2)))+x^2),x, algorithm="maxima")
Output:
5*e^(x*e^(e^2) + 100*e^(e^2))/(x*e^(2*e^(x + e^2)) - e^(x*e^(e^2) + 100*e^ (e^2)))
\[ \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-5+e^{e^2} \left (5 x-10 e^x x\right )\right )}{e^{2 e^{e^2} \left (100-2 e^x+x\right )}-2 e^{e^{e^2} \left (100-2 e^x+x\right )} x+x^2} \, dx=\int { -\frac {5 \, {\left ({\left (2 \, x e^{x} - x\right )} e^{\left (e^{2}\right )} + 1\right )} e^{\left ({\left (x - 2 \, e^{x} + 100\right )} e^{\left (e^{2}\right )}\right )}}{x^{2} - 2 \, x e^{\left ({\left (x - 2 \, e^{x} + 100\right )} e^{\left (e^{2}\right )}\right )} + e^{\left (2 \, {\left (x - 2 \, e^{x} + 100\right )} e^{\left (e^{2}\right )}\right )}} \,d x } \] Input:
integrate(((-10*exp(x)*x+5*x)*exp(exp(2))-5)*exp((-2*exp(x)+x+100)*exp(exp (2)))/(exp((-2*exp(x)+x+100)*exp(exp(2)))^2-2*x*exp((-2*exp(x)+x+100)*exp( exp(2)))+x^2),x, algorithm="giac")
Output:
integrate(-5*((2*x*e^x - x)*e^(e^2) + 1)*e^((x - 2*e^x + 100)*e^(e^2))/(x^ 2 - 2*x*e^((x - 2*e^x + 100)*e^(e^2)) + e^(2*(x - 2*e^x + 100)*e^(e^2))), x)
Time = 6.64 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-5+e^{e^2} \left (5 x-10 e^x x\right )\right )}{e^{2 e^{e^2} \left (100-2 e^x+x\right )}-2 e^{e^{e^2} \left (100-2 e^x+x\right )} x+x^2} \, dx=\frac {5\,x}{x-{\mathrm {e}}^{x\,{\mathrm {e}}^{{\mathrm {e}}^2}}\,{\mathrm {e}}^{-2\,{\mathrm {e}}^{{\mathrm {e}}^2}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{100\,{\mathrm {e}}^{{\mathrm {e}}^2}}} \] Input:
int((exp(exp(exp(2))*(x - 2*exp(x) + 100))*(exp(exp(2))*(5*x - 10*x*exp(x) ) - 5))/(exp(2*exp(exp(2))*(x - 2*exp(x) + 100)) - 2*x*exp(exp(exp(2))*(x - 2*exp(x) + 100)) + x^2),x)
Output:
(5*x)/(x - exp(x*exp(exp(2)))*exp(-2*exp(exp(2))*exp(x))*exp(100*exp(exp(2 ))))
Time = 0.17 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92 \[ \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-5+e^{e^2} \left (5 x-10 e^x x\right )\right )}{e^{2 e^{e^2} \left (100-2 e^x+x\right )}-2 e^{e^{e^2} \left (100-2 e^x+x\right )} x+x^2} \, dx=-\frac {5 e^{2 e^{e^{2}+x}} x}{e^{e^{e^{2}} x +100 e^{e^{2}}}-e^{2 e^{e^{2}+x}} x} \] Input:
int(((-10*exp(x)*x+5*x)*exp(exp(2))-5)*exp((-2*exp(x)+x+100)*exp(exp(2)))/ (exp((-2*exp(x)+x+100)*exp(exp(2)))^2-2*x*exp((-2*exp(x)+x+100)*exp(exp(2) ))+x^2),x)
Output:
( - 5*e**(2*e**(e**2 + x))*x)/(e**(e**(e**2)*x + 100*e**(e**2)) - e**(2*e* *(e**2 + x))*x)