\(\int \frac {e^{e^{e^2} (100-2 e^x+x)} (-5+e^{e^2} (5 x-10 e^x x))}{e^{2 e^{e^2} (100-2 e^x+x)}-2 e^{e^{e^2} (100-2 e^x+x)} x+x^2} \, dx\) [9845]

Optimal result

Integrand size = 77, antiderivative size = 25 \[ \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-5+e^{e^2} \left (5 x-10 e^x x\right )\right )}{e^{2 e^{e^2} \left (100-2 e^x+x\right )}-2 e^{e^{e^2} \left (100-2 e^x+x\right )} x+x^2} \, dx=\frac {5 x}{-e^{e^{e^2} \left (100-2 e^x+x\right )}+x} \]

[Out]

x/(1/5*x-1/5*exp((-2*exp(x)+x+100)*exp(exp(2))))

Rubi [A] (verified)

Time = 1.48 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {6820, 12, 6843, 32} \[ \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-5+e^{e^2} \left (5 x-10 e^x x\right )\right )}{e^{2 e^{e^2} \left (100-2 e^x+x\right )}-2 e^{e^{e^2} \left (100-2 e^x+x\right )} x+x^2} \, dx=\frac {5}{1-\frac {e^{e^{e^2} \left (x-2 e^x+100\right )}}{x}} \]

[In]

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*(1
00 - 2*E^x + x))*x + x^2),x]

[Out]

5/(1 - E^(E^E^2*(100 - 2*E^x + x))/x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6843

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])]}, Dist[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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {5 e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-1+e^{e^2} x-2 e^{e^2+x} x\right )}{\left (e^{e^{e^2} \left (100-2 e^x+x\right )}-x\right )^2} \, dx \\ & = 5 \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-1+e^{e^2} x-2 e^{e^2+x} x\right )}{\left (e^{e^{e^2} \left (100-2 e^x+x\right )}-x\right )^2} \, dx \\ & = 5 \text {Subst}\left (\int \frac {1}{(-1+x)^2} \, dx,x,\frac {e^{e^{e^2} \left (100-2 e^x+x\right )}}{x}\right ) \\ & = \frac {5}{1-\frac {e^{e^{e^2} \left (100-2 e^x+x\right )}}{x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.90 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-5+e^{e^2} \left (5 x-10 e^x x\right )\right )}{e^{2 e^{e^2} \left (100-2 e^x+x\right )}-2 e^{e^{e^2} \left (100-2 e^x+x\right )} x+x^2} \, dx=-\frac {5 x}{e^{e^{e^2} \left (100-2 e^x+x\right )}-x} \]

[In]

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]

[Out]

(-5*x)/(E^(E^E^2*(100 - 2*E^x + x)) - x)

Maple [A] (verified)

Time = 0.50 (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	\(\frac {5 \,{\mathrm e}^{\left (-2 \,{\mathrm e}^{x}+x +100\right ) {\mathrm e}^{{\mathrm e}^{2}}}}{x -{\mathrm e}^{\left (-2 \,{\mathrm e}^{x}+x +100\right ) {\mathrm e}^{{\mathrm e}^{2}}}}\)	\(33\)

[In]

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)

[Out]

5*x/(x-exp((-2*exp(x)+x+100)*exp(exp(2))))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-5+e^{e^2} \left (5 x-10 e^x x\right )\right )}{e^{2 e^{e^2} \left (100-2 e^x+x\right )}-2 e^{e^{e^2} \left (100-2 e^x+x\right )} x+x^2} \, dx=\frac {5 \, x}{x - e^{\left ({\left (x - 2 \, e^{x} + 100\right )} e^{\left (e^{2}\right )}\right )}} \]

[In]

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")

[Out]

5*x/(x - e^((x - 2*e^x + 100)*e^(e^2)))

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-5+e^{e^2} \left (5 x-10 e^x x\right )\right )}{e^{2 e^{e^2} \left (100-2 e^x+x\right )}-2 e^{e^{e^2} \left (100-2 e^x+x\right )} x+x^2} \, dx=- \frac {5 x}{- x + e^{\left (x - 2 e^{x} + 100\right ) e^{e^{2}}}} \]

[In]

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)

[Out]

-5*x/(-x + exp((x - 2*exp(x) + 100)*exp(exp(2))))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.64 \[ \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-5+e^{e^2} \left (5 x-10 e^x x\right )\right )}{e^{2 e^{e^2} \left (100-2 e^x+x\right )}-2 e^{e^{e^2} \left (100-2 e^x+x\right )} x+x^2} \, dx=\frac {5 \, e^{\left (x e^{\left (e^{2}\right )} + 100 \, e^{\left (e^{2}\right )}\right )}}{x e^{\left (2 \, e^{\left (x + e^{2}\right )}\right )} - e^{\left (x e^{\left (e^{2}\right )} + 100 \, e^{\left (e^{2}\right )}\right )}} \]

[In]

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")

[Out]

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)))

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (21) = 42\).

Time = 0.33 (sec) , antiderivative size = 149, normalized size of antiderivative = 5.96 \[ \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-5+e^{e^2} \left (5 x-10 e^x x\right )\right )}{e^{2 e^{e^2} \left (100-2 e^x+x\right )}-2 e^{e^{e^2} \left (100-2 e^x+x\right )} x+x^2} \, dx=\frac {5 \, {\left (2 \, x^{2} e^{\left (x + e^{2} + 100 \, e^{\left (e^{2}\right )}\right )} - x^{2} e^{\left (e^{2} + 100 \, e^{\left (e^{2}\right )}\right )} + x e^{\left (100 \, e^{\left (e^{2}\right )}\right )}\right )}}{2 \, x^{2} e^{\left (x + e^{2} + 100 \, e^{\left (e^{2}\right )}\right )} - x^{2} e^{\left (e^{2} + 100 \, e^{\left (e^{2}\right )}\right )} - 2 \, x e^{\left (x e^{\left (e^{2}\right )} + x + e^{2} - 2 \, e^{\left (x + e^{2}\right )} + 200 \, e^{\left (e^{2}\right )}\right )} + x e^{\left (x e^{\left (e^{2}\right )} + e^{2} - 2 \, e^{\left (x + e^{2}\right )} + 200 \, e^{\left (e^{2}\right )}\right )} + x e^{\left (100 \, e^{\left (e^{2}\right )}\right )} - e^{\left (x e^{\left (e^{2}\right )} - 2 \, e^{\left (x + e^{2}\right )} + 200 \, e^{\left (e^{2}\right )}\right )}} \]

[In]

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")

[Out]

5*(2*x^2*e^(x + e^2 + 100*e^(e^2)) - x^2*e^(e^2 + 100*e^(e^2)) + x*e^(100*e^(e^2)))/(2*x^2*e^(x + e^2 + 100*e^
(e^2)) - x^2*e^(e^2 + 100*e^(e^2)) - 2*x*e^(x*e^(e^2) + x + e^2 - 2*e^(x + e^2) + 200*e^(e^2)) + x*e^(x*e^(e^2
) + e^2 - 2*e^(x + e^2) + 200*e^(e^2)) + x*e^(100*e^(e^2)) - e^(x*e^(e^2) - 2*e^(x + e^2) + 200*e^(e^2)))

Mupad [B] (verification not implemented)

Time = 15.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {e^{e^{e^2} \left (100-2 e^x+x\right )} \left (-5+e^{e^2} \left (5 x-10 e^x x\right )\right )}{e^{2 e^{e^2} \left (100-2 e^x+x\right )}-2 e^{e^{e^2} \left (100-2 e^x+x\right )} x+x^2} \, dx=\frac {5\,x}{x-{\mathrm {e}}^{x\,{\mathrm {e}}^{{\mathrm {e}}^2}}\,{\mathrm {e}}^{-2\,{\mathrm {e}}^{{\mathrm {e}}^2}\,{\mathrm {e}}^x}\,{\mathrm {e}}^{100\,{\mathrm {e}}^{{\mathrm {e}}^2}}} \]

[In]

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*ex
p(x) + 100)) - 2*x*exp(exp(exp(2))*(x - 2*exp(x) + 100)) + x^2),x)

[Out]

(5*x)/(x - exp(x*exp(exp(2)))*exp(-2*exp(exp(2))*exp(x))*exp(100*exp(exp(2))))