3.36.99 \(\int \frac {e^{-\frac {3-x}{e^2-x}} (6-2 e^2+e^{\frac {3-x}{e^2-x}} (2 e^4-4 e^2 x+2 x^2))}{e^4-2 e^2 x+x^2} \, dx\)

Optimal. Leaf size=27 \[ \log \left (e^{-2 e^{-\frac {3-x}{e^2-x}}+2 x}\right ) \]

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Rubi [A]  time = 1.20, antiderivative size = 28, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, integrand size = 76, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.092, Rules used = {27, 6741, 6688, 12, 6742, 2230, 2209} \begin {gather*} 2 x-2 e^{-\frac {3-e^2}{e^2-x}-1} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

-2*E^(-1 - (3 - E^2)/(E^2 - x)) + 2*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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2230

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))*((g_.) + (h_.)*(x_))^(m_.), x_Symbol] :>
Int[(g + h*x)^m*F^((d*e + b*f)/d - (f*(b*c - a*d))/(d*(c + d*x))), x] /; FreeQ[{F, a, b, c, d, e, f, g, h, m},
x] && NeQ[b*c - a*d, 0] && EqQ[d*g - c*h, 0]

Rule 6688

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

Rule 6741

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

Rule 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\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 )}{\left (-e^2+x\right )^2} \, dx\\ &=\int \frac {e^{-\frac {3-x}{e^2-x}} \left (6 \left (1-\frac {e^2}{3}\right )+e^{\frac {3-x}{e^2-x}} \left (2 e^4-4 e^2 x+2 x^2\right )\right )}{\left (e^2-x\right )^2} \, dx\\ &=\int \frac {2 \left (e^4+e^{\frac {-3+x}{e^2-x}} \left (3-e^2\right )-2 e^2 x+x^2\right )}{\left (e^2-x\right )^2} \, dx\\ &=2 \int \frac {e^4+e^{\frac {-3+x}{e^2-x}} \left (3-e^2\right )-2 e^2 x+x^2}{\left (e^2-x\right )^2} \, dx\\ &=2 \int \left (1+\frac {e^{\frac {-3+x}{e^2-x}} \left (3-e^2\right )}{\left (e^2-x\right )^2}\right ) \, dx\\ &=2 x+\left (2 \left (3-e^2\right )\right ) \int \frac {e^{\frac {-3+x}{e^2-x}}}{\left (e^2-x\right )^2} \, dx\\ &=2 x+\left (2 \left (3-e^2\right )\right ) \int \frac {e^{-1-\frac {-3+e^2}{-e^2+x}}}{\left (e^2-x\right )^2} \, dx\\ &=-2 e^{-1-\frac {3-e^2}{e^2-x}}+2 x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.38, size = 21, normalized size = 0.78 \begin {gather*} -2 e^{\frac {-3+x}{e^2-x}}+2 x \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

-2*E^((-3 + x)/(E^2 - x)) + 2*x

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fricas [A]  time = 0.90, size = 20, normalized size = 0.74 \begin {gather*} 2 \, x - 2 \, e^{\left (-\frac {x - 3}{x - e^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

2*x - 2*e^(-(x - 3)/(x - e^2))

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giac [B]  time = 8.87, size = 81, normalized size = 3.00 \begin {gather*} -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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

-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

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maple [A]  time = 0.16, size = 20, normalized size = 0.74

method	result	size
risch	\(2 x -2 \,{\mathrm e}^{\frac {x -3}{{\mathrm e}^{2}-x}}\)	\(20\)
norman	\(\frac {\left (2 \,{\mathrm e}^{4} {\mathrm e}^{\frac {3-x}{{\mathrm e}^{2}-x}}+2 x -2 x^{2} {\mathrm e}^{\frac {3-x}{{\mathrm e}^{2}-x}}-2 \,{\mathrm e}^{2}\right ) {\mathrm e}^{-\frac {3-x}{{\mathrm e}^{2}-x}}}{{\mathrm e}^{2}-x}\)	\(76\)
derivativedivides	\(-\left ({\mathrm e}^{2}-3\right ) \left (-\frac {18 \left (x -{\mathrm e}^{2}\right )}{\left ({\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9\right ) \left ({\mathrm e}^{2}-3\right )}+\frac {12 \,{\mathrm e}^{2} \left (x -{\mathrm e}^{2}\right )}{\left ({\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9\right ) \left ({\mathrm e}^{2}-3\right )}-\frac {2 \,{\mathrm e}^{4} \left (x -{\mathrm e}^{2}\right )}{\left ({\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9\right ) \left ({\mathrm e}^{2}-3\right )}-\frac {6 \,{\mathrm e}^{-1-\frac {{\mathrm e}^{2}-3}{x -{\mathrm e}^{2}}}}{{\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9}+\frac {2 \,{\mathrm e}^{2} {\mathrm e}^{-1-\frac {{\mathrm e}^{2}-3}{x -{\mathrm e}^{2}}}}{{\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9}\right )\)	\(158\)
default	\(-\left ({\mathrm e}^{2}-3\right ) \left (-\frac {18 \left (x -{\mathrm e}^{2}\right )}{\left ({\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9\right ) \left ({\mathrm e}^{2}-3\right )}+\frac {12 \,{\mathrm e}^{2} \left (x -{\mathrm e}^{2}\right )}{\left ({\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9\right ) \left ({\mathrm e}^{2}-3\right )}-\frac {2 \,{\mathrm e}^{4} \left (x -{\mathrm e}^{2}\right )}{\left ({\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9\right ) \left ({\mathrm e}^{2}-3\right )}-\frac {6 \,{\mathrm e}^{-1-\frac {{\mathrm e}^{2}-3}{x -{\mathrm e}^{2}}}}{{\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9}+\frac {2 \,{\mathrm e}^{2} {\mathrm e}^{-1-\frac {{\mathrm e}^{2}-3}{x -{\mathrm e}^{2}}}}{{\mathrm e}^{4}-6 \,{\mathrm e}^{2}+9}\right )\)	\(158\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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,method=_RETURNVERBOSE)

[Out]

2*x-2*exp((x-3)/(exp(2)-x))

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maxima [B]  time = 0.59, size = 120, normalized size = 4.44 \begin {gather*} 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

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)

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mupad [B]  time = 2.53, size = 29, normalized size = 1.07 \begin {gather*} 2\,x-2\,{\mathrm {e}}^{-\frac {x}{x-{\mathrm {e}}^2}}\,{\mathrm {e}}^{\frac {3}{x-{\mathrm {e}}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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))/(e
xp(4) - 2*x*exp(2) + x^2),x)

[Out]

2*x - 2*exp(-x/(x - exp(2)))*exp(3/(x - exp(2)))

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sympy [A]  time = 0.22, size = 14, normalized size = 0.52 \begin {gather*} 2 x - 2 e^{- \frac {3 - x}{- x + e^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

2*x - 2*exp(-(3 - x)/(-x + exp(2)))

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