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

Optimal. Leaf size=25 \[ \frac {5 x}{-e^{e^{e^2} \left (100-2 e^x+x\right )}+x} \]

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Rubi [A]  time = 2.25, antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.052, Rules used = {6688, 12, 6711, 32} \begin {gather*} \frac {5}{1-\frac {e^{e^{e^2} \left (x-2 e^x+100\right )}}{x}} \end {gather*}

Antiderivative was successfully verified.

[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 6688

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

Rule 6711

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 {gather*} \begin {aligned} \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 \operatorname {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 {aligned} \end {gather*}

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Mathematica [A]  time = 1.22, size = 25, normalized size = 1.00 \begin {gather*} -\frac {5 x}{e^{e^{e^2} \left (100-2 e^x+x\right )}-x} \end {gather*}

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.77, size = 21, normalized size = 0.84 \begin {gather*} \frac {5 \, x}{x - e^{\left ({\left (x - 2 \, e^{x} + 100\right )} e^{\left (e^{2}\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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giac [B]  time = 0.26, size = 149, normalized size = 5.96 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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maple [A]  time = 0.14, size = 25, normalized size = 1.00

method	result	size
risch	\(\frac {5 x}{x -{\mathrm e}^{-\left (2 \,{\mathrm e}^{x}-x -100\right ) {\mathrm e}^{{\mathrm e}^{2}}}}\)	\(25\)
norman	\(\frac {5 \,{\mathrm e}^{\left (-2 \,{\mathrm e}^{x}+x +100\right ) {\mathrm e}^{{\mathrm e}^{2}}}}{x -{\mathrm e}^{\left (-2 \,{\mathrm e}^{x}+x +100\right ) {\mathrm e}^{{\mathrm e}^{2}}}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[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))))

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maxima [A]  time = 0.47, size = 41, normalized size = 1.64 \begin {gather*} \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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)))

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mupad [B]  time = 5.86, size = 29, normalized size = 1.16 \begin {gather*} \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}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))))

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sympy [A]  time = 0.18, size = 20, normalized size = 0.80 \begin {gather*} - \frac {5 x}{- x + e^{\left (x - 2 e^{x} + 100\right ) e^{e^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))))

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