3.31.82 \(\int \frac {-x^2-4 x^3-4 x^4+e^2 (-1-4 x-4 x^2)+e^{\frac {2-x}{-x-2 x^2+e (1+2 x)}} (-4+10 e-16 x+4 x^2)+e^{\frac {2 (2-x)}{-x-2 x^2+e (1+2 x)}} (-4+10 e-16 x+4 x^2)+e (2 x+8 x^2+8 x^3)}{x^2+4 x^3+4 x^4+e^2 (1+4 x+4 x^2)+e (-2 x-8 x^2-8 x^3)} \, dx\)

Optimal. Leaf size=31 \[ 3-\left (1+e^{\frac {-2+x}{(-e+x) (1+2 x)}}\right )^2-x \]

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Rubi [B]  time = 3.94, antiderivative size = 245, normalized size of antiderivative = 7.90, number of steps used = 13, number of rules used = 5, integrand size = 174, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {6688, 6742, 43, 88, 6706} \begin {gather*} -x-2 e^{\frac {2-x}{(e-x) (2 x+1)}}-e^{\frac {2 (2-x)}{(e-x) (2 x+1)}}-\frac {4 e^4}{(1+2 e)^2 (e-x)}-\frac {4 e^3}{(1+2 e)^2 (e-x)}-\frac {e^2}{(1+2 e)^2 (e-x)}+\frac {e^2}{e-x}-\frac {4 e^2 (3+2 e) \log (e-x)}{(1+2 e)^3}-\frac {16 e^3 (1+e) \log (e-x)}{(1+2 e)^3}-\frac {2 e \log (e-x)}{(1+2 e)^3}+2 e \log (e-x)-\frac {(1+6 e) \log (2 x+1)}{(1+2 e)^3}+\frac {(1+4 e) \log (2 x+1)}{(1+2 e)^3}+\frac {2 e \log (2 x+1)}{(1+2 e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-x^2 - 4*x^3 - 4*x^4 + E^2*(-1 - 4*x - 4*x^2) + E^((2 - x)/(-x - 2*x^2 + E*(1 + 2*x)))*(-4 + 10*E - 16*x
+ 4*x^2) + E^((2*(2 - x))/(-x - 2*x^2 + E*(1 + 2*x)))*(-4 + 10*E - 16*x + 4*x^2) + E*(2*x + 8*x^2 + 8*x^3))/(x
^2 + 4*x^3 + 4*x^4 + E^2*(1 + 4*x + 4*x^2) + E*(-2*x - 8*x^2 - 8*x^3)),x]

[Out]

-2*E^((2 - x)/((E - x)*(1 + 2*x))) - E^((2*(2 - x))/((E - x)*(1 + 2*x))) + E^2/(E - x) - E^2/((1 + 2*E)^2*(E -
x)) - (4*E^3)/((1 + 2*E)^2*(E - x)) - (4*E^4)/((1 + 2*E)^2*(E - x)) - x + 2*E*Log[E - x] - (2*E*Log[E - x])/(
1 + 2*E)^3 - (16*E^3*(1 + E)*Log[E - x])/(1 + 2*E)^3 - (4*E^2*(3 + 2*E)*Log[E - x])/(1 + 2*E)^3 + (2*E*Log[1 +
2*x])/(1 + 2*E)^3 + ((1 + 4*E)*Log[1 + 2*x])/(1 + 2*E)^3 - ((1 + 6*E)*Log[1 + 2*x])/(1 + 2*E)^3

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 6688

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

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

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 {-x^2-4 x^3-4 x^4+2 e x (1+2 x)^2-(e+2 e x)^2+2 e^{\frac {2-x}{(e-x) (1+2 x)}} \left (-2+5 e-8 x+2 x^2\right )+2 e^{-\frac {2 (-2+x)}{(e-x) (1+2 x)}} \left (-2+5 e-8 x+2 x^2\right )}{(e-x)^2 (1+2 x)^2} \, dx\\ &=\int \left (-\frac {e^2}{(e-x)^2}+\frac {2 e x}{(e-x)^2}-\frac {x^2}{(e-x)^2 (1+2 x)^2}-\frac {4 x^3}{(e-x)^2 (1+2 x)^2}-\frac {4 x^4}{(e-x)^2 (1+2 x)^2}+\frac {2 e^{-\frac {2 (-2+x)}{(e-x) (1+2 x)}} \left (-2+5 e-8 x+2 x^2\right )}{(e-x)^2 (1+2 x)^2}+\frac {2 e^{-\frac {-2+x}{(e-x) (1+2 x)}} \left (-2+5 e-8 x+2 x^2\right )}{(e-x)^2 (1+2 x)^2}\right ) \, dx\\ &=-\frac {e^2}{e-x}+2 \int \frac {e^{-\frac {2 (-2+x)}{(e-x) (1+2 x)}} \left (-2+5 e-8 x+2 x^2\right )}{(e-x)^2 (1+2 x)^2} \, dx+2 \int \frac {e^{-\frac {-2+x}{(e-x) (1+2 x)}} \left (-2+5 e-8 x+2 x^2\right )}{(e-x)^2 (1+2 x)^2} \, dx-4 \int \frac {x^3}{(e-x)^2 (1+2 x)^2} \, dx-4 \int \frac {x^4}{(e-x)^2 (1+2 x)^2} \, dx+(2 e) \int \frac {x}{(e-x)^2} \, dx-\int \frac {x^2}{(e-x)^2 (1+2 x)^2} \, dx\\ &=-2 e^{\frac {2-x}{(e-x) (1+2 x)}}-e^{\frac {2 (2-x)}{(e-x) (1+2 x)}}-\frac {e^2}{e-x}-4 \int \left (\frac {1}{4}+\frac {e^4}{(1+2 e)^2 (e-x)^2}-\frac {4 e^3 (1+e)}{(1+2 e)^3 (e-x)}+\frac {1}{4 (1+2 e)^2 (1+2 x)^2}+\frac {-1-4 e}{2 (1+2 e)^3 (1+2 x)}\right ) \, dx-4 \int \left (\frac {e^3}{(1+2 e)^2 (e-x)^2}-\frac {e^2 (3+2 e)}{(1+2 e)^3 (e-x)}-\frac {1}{2 (1+2 e)^2 (1+2 x)^2}+\frac {1+6 e}{2 (1+2 e)^3 (1+2 x)}\right ) \, dx+(2 e) \int \left (\frac {e}{(e-x)^2}+\frac {1}{-e+x}\right ) \, dx-\int \left (\frac {e^2}{(1+2 e)^2 (e-x)^2}-\frac {2 e}{(1+2 e)^3 (e-x)}+\frac {1}{(1+2 e)^2 (1+2 x)^2}-\frac {4 e}{(1+2 e)^3 (1+2 x)}\right ) \, dx\\ &=-2 e^{\frac {2-x}{(e-x) (1+2 x)}}-e^{\frac {2 (2-x)}{(e-x) (1+2 x)}}+\frac {e^2}{e-x}-\frac {e^2}{(1+2 e)^2 (e-x)}-\frac {4 e^3}{(1+2 e)^2 (e-x)}-\frac {4 e^4}{(1+2 e)^2 (e-x)}-x+2 e \log (e-x)-\frac {2 e \log (e-x)}{(1+2 e)^3}-\frac {16 e^3 (1+e) \log (e-x)}{(1+2 e)^3}-\frac {4 e^2 (3+2 e) \log (e-x)}{(1+2 e)^3}+\frac {2 e \log (1+2 x)}{(1+2 e)^3}+\frac {(1+4 e) \log (1+2 x)}{(1+2 e)^3}-\frac {(1+6 e) \log (1+2 x)}{(1+2 e)^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.23, size = 50, normalized size = 1.61 \begin {gather*} -e^{-\frac {2 (-2+x)}{(e-x) (1+2 x)}}-2 e^{-\frac {-2+x}{(e-x) (1+2 x)}}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-x^2 - 4*x^3 - 4*x^4 + E^2*(-1 - 4*x - 4*x^2) + E^((2 - x)/(-x - 2*x^2 + E*(1 + 2*x)))*(-4 + 10*E -
16*x + 4*x^2) + E^((2*(2 - x))/(-x - 2*x^2 + E*(1 + 2*x)))*(-4 + 10*E - 16*x + 4*x^2) + E*(2*x + 8*x^2 + 8*x^
3))/(x^2 + 4*x^3 + 4*x^4 + E^2*(1 + 4*x + 4*x^2) + E*(-2*x - 8*x^2 - 8*x^3)),x]

[Out]

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

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fricas [A]  time = 0.53, size = 55, normalized size = 1.77 \begin {gather*} -x - e^{\left (\frac {2 \, {\left (x - 2\right )}}{2 \, x^{2} - {\left (2 \, x + 1\right )} e + x}\right )} - 2 \, e^{\left (\frac {x - 2}{2 \, x^{2} - {\left (2 \, x + 1\right )} e + x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*exp(1)+4*x^2-16*x-4)*exp((2-x)/((2*x+1)*exp(1)-2*x^2-x))^2+(10*exp(1)+4*x^2-16*x-4)*exp((2-x)/(
(2*x+1)*exp(1)-2*x^2-x))+(-4*x^2-4*x-1)*exp(1)^2+(8*x^3+8*x^2+2*x)*exp(1)-4*x^4-4*x^3-x^2)/((4*x^2+4*x+1)*exp(
1)^2+(-8*x^3-8*x^2-2*x)*exp(1)+4*x^4+4*x^3+x^2),x, algorithm="fricas")

[Out]

-x - e^(2*(x - 2)/(2*x^2 - (2*x + 1)*e + x)) - 2*e^((x - 2)/(2*x^2 - (2*x + 1)*e + x))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {4 \, x^{4} + 4 \, x^{3} + x^{2} + {\left (4 \, x^{2} + 4 \, x + 1\right )} e^{2} - 2 \, {\left (4 \, x^{3} + 4 \, x^{2} + x\right )} e - 2 \, {\left (2 \, x^{2} - 8 \, x + 5 \, e - 2\right )} e^{\left (\frac {2 \, {\left (x - 2\right )}}{2 \, x^{2} - {\left (2 \, x + 1\right )} e + x}\right )} - 2 \, {\left (2 \, x^{2} - 8 \, x + 5 \, e - 2\right )} e^{\left (\frac {x - 2}{2 \, x^{2} - {\left (2 \, x + 1\right )} e + x}\right )}}{4 \, x^{4} + 4 \, x^{3} + x^{2} + {\left (4 \, x^{2} + 4 \, x + 1\right )} e^{2} - 2 \, {\left (4 \, x^{3} + 4 \, x^{2} + x\right )} e}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*exp(1)+4*x^2-16*x-4)*exp((2-x)/((2*x+1)*exp(1)-2*x^2-x))^2+(10*exp(1)+4*x^2-16*x-4)*exp((2-x)/(
(2*x+1)*exp(1)-2*x^2-x))+(-4*x^2-4*x-1)*exp(1)^2+(8*x^3+8*x^2+2*x)*exp(1)-4*x^4-4*x^3-x^2)/((4*x^2+4*x+1)*exp(
1)^2+(-8*x^3-8*x^2-2*x)*exp(1)+4*x^4+4*x^3+x^2),x, algorithm="giac")

[Out]

integrate(-(4*x^4 + 4*x^3 + x^2 + (4*x^2 + 4*x + 1)*e^2 - 2*(4*x^3 + 4*x^2 + x)*e - 2*(2*x^2 - 8*x + 5*e - 2)*
e^(2*(x - 2)/(2*x^2 - (2*x + 1)*e + x)) - 2*(2*x^2 - 8*x + 5*e - 2)*e^((x - 2)/(2*x^2 - (2*x + 1)*e + x)))/(4*
x^4 + 4*x^3 + x^2 + (4*x^2 + 4*x + 1)*e^2 - 2*(4*x^3 + 4*x^2 + x)*e), x)

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maple [A]  time = 1.06, size = 51, normalized size = 1.65

method	result	size
risch	\(-{\mathrm e}^{-\frac {2 \left (x -2\right )}{\left (2 x +1\right ) \left ({\mathrm e}-x \right )}}-x -2 \,{\mathrm e}^{-\frac {x -2}{\left (2 x +1\right ) \left ({\mathrm e}-x \right )}}\)	\(51\)
norman	\(\frac {\left (-\frac {1}{2}-2 \,{\mathrm e}^{2}+{\mathrm e}\right ) x +\left (2-4 \,{\mathrm e}\right ) x \,{\mathrm e}^{\frac {2-x}{\left (2 x +1\right ) {\mathrm e}-2 x^{2}-x}}+\left (-2 \,{\mathrm e}+1\right ) x \,{\mathrm e}^{\frac {4-2 x}{\left (2 x +1\right ) {\mathrm e}-2 x^{2}-x}}+2 x^{3}+4 x^{2} {\mathrm e}^{\frac {2-x}{\left (2 x +1\right ) {\mathrm e}-2 x^{2}-x}}+2 x^{2} {\mathrm e}^{\frac {4-2 x}{\left (2 x +1\right ) {\mathrm e}-2 x^{2}-x}}-2 \,{\mathrm e} \,{\mathrm e}^{\frac {2-x}{\left (2 x +1\right ) {\mathrm e}-2 x^{2}-x}}-{\mathrm e} \,{\mathrm e}^{\frac {4-2 x}{\left (2 x +1\right ) {\mathrm e}-2 x^{2}-x}}-{\mathrm e}^{2}+\frac {{\mathrm e}}{2}}{\left (2 x +1\right ) \left ({\mathrm e}-x \right )}\)	\(241\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*exp(1)+4*x^2-16*x-4)*exp((2-x)/((2*x+1)*exp(1)-2*x^2-x))^2+(10*exp(1)+4*x^2-16*x-4)*exp((2-x)/((2*x+1
)*exp(1)-2*x^2-x))+(-4*x^2-4*x-1)*exp(1)^2+(8*x^3+8*x^2+2*x)*exp(1)-4*x^4-4*x^3-x^2)/((4*x^2+4*x+1)*exp(1)^2+(
-8*x^3-8*x^2-2*x)*exp(1)+4*x^4+4*x^3+x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.90, size = 1252, normalized size = 40.39 result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*exp(1)+4*x^2-16*x-4)*exp((2-x)/((2*x+1)*exp(1)-2*x^2-x))^2+(10*exp(1)+4*x^2-16*x-4)*exp((2-x)/(
(2*x+1)*exp(1)-2*x^2-x))+(-4*x^2-4*x-1)*exp(1)^2+(8*x^3+8*x^2+2*x)*exp(1)-4*x^4-4*x^3-x^2)/((4*x^2+4*x+1)*exp(
1)^2+(-8*x^3-8*x^2-2*x)*exp(1)+4*x^4+4*x^3+x^2),x, algorithm="maxima")

[Out]

-4*((2*e - 1)*log(2*x + 1)/(8*e^3 + 12*e^2 + 6*e + 1) - (2*e - 1)*log(x - e)/(8*e^3 + 12*e^2 + 6*e + 1) - (x*(
2*e - 1) + 2*e)/(2*x^2*(4*e^2 + 4*e + 1) - x*(8*e^3 + 4*e^2 - 2*e - 1) - 4*e^3 - 4*e^2 - e))*e^2 + 2*(4*e*log(
2*x + 1)/(8*e^3 + 12*e^2 + 6*e + 1) - 4*e*log(x - e)/(8*e^3 + 12*e^2 + 6*e + 1) + (x*(4*e^2 + 1) + 2*e^2 - e)/
(2*x^2*(4*e^2 + 4*e + 1) - x*(8*e^3 + 4*e^2 - 2*e - 1) - 4*e^3 - 4*e^2 - e))*e^2 + ((4*x - 2*e + 1)/(2*x^2*(4*
e^2 + 4*e + 1) - x*(8*e^3 + 4*e^2 - 2*e - 1) - 4*e^3 - 4*e^2 - e) - 4*log(2*x + 1)/(8*e^3 + 12*e^2 + 6*e + 1)
+ 4*log(x - e)/(8*e^3 + 12*e^2 + 6*e + 1))*e^2 + 2*((6*e + 1)*log(2*x + 1)/(8*e^3 + 12*e^2 + 6*e + 1) + 4*(2*e
^3 + 3*e^2)*log(x - e)/(8*e^3 + 12*e^2 + 6*e + 1) - (x*(8*e^3 - 1) + 4*e^3 + e)/(2*x^2*(4*e^2 + 4*e + 1) - x*(
8*e^3 + 4*e^2 - 2*e - 1) - 4*e^3 - 4*e^2 - e))*e + 2*((2*e - 1)*log(2*x + 1)/(8*e^3 + 12*e^2 + 6*e + 1) - (2*e
- 1)*log(x - e)/(8*e^3 + 12*e^2 + 6*e + 1) - (x*(2*e - 1) + 2*e)/(2*x^2*(4*e^2 + 4*e + 1) - x*(8*e^3 + 4*e^2
- 2*e - 1) - 4*e^3 - 4*e^2 - e))*e - 4*(4*e*log(2*x + 1)/(8*e^3 + 12*e^2 + 6*e + 1) - 4*e*log(x - e)/(8*e^3 +
12*e^2 + 6*e + 1) + (x*(4*e^2 + 1) + 2*e^2 - e)/(2*x^2*(4*e^2 + 4*e + 1) - x*(8*e^3 + 4*e^2 - 2*e - 1) - 4*e^3
- 4*e^2 - e))*e - (e^(2*e/(x*(2*e + 1) - 2*e^2 - e) + 10/(2*x*(2*e + 1) + 2*e + 1)) + 2*e^(e/(x*(2*e + 1) - 2
*e^2 - e) + 5/(2*x*(2*e + 1) + 2*e + 1) + 2/(x*(2*e + 1) - 2*e^2 - e)))*e^(-4/(x*(2*e + 1) - 2*e^2 - e)) - x -
(6*e + 1)*log(2*x + 1)/(8*e^3 + 12*e^2 + 6*e + 1) + (4*e + 1)*log(2*x + 1)/(8*e^3 + 12*e^2 + 6*e + 1) + 2*e*l
og(2*x + 1)/(8*e^3 + 12*e^2 + 6*e + 1) - 16*(e^4 + e^3)*log(x - e)/(8*e^3 + 12*e^2 + 6*e + 1) - 4*(2*e^3 + 3*e
^2)*log(x - e)/(8*e^3 + 12*e^2 + 6*e + 1) - 2*e*log(x - e)/(8*e^3 + 12*e^2 + 6*e + 1) + 1/2*(x*(16*e^4 + 1) +
8*e^4 - e)/(2*x^2*(4*e^2 + 4*e + 1) - x*(8*e^3 + 4*e^2 - 2*e - 1) - 4*e^3 - 4*e^2 - e) + (x*(8*e^3 - 1) + 4*e^
3 + e)/(2*x^2*(4*e^2 + 4*e + 1) - x*(8*e^3 + 4*e^2 - 2*e - 1) - 4*e^3 - 4*e^2 - e) + 1/2*(x*(4*e^2 + 1) + 2*e^
2 - e)/(2*x^2*(4*e^2 + 4*e + 1) - x*(8*e^3 + 4*e^2 - 2*e - 1) - 4*e^3 - 4*e^2 - e)

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mupad [B]  time = 2.79, size = 93, normalized size = 3.00 \begin {gather*} -x-2\,{\mathrm {e}}^{-\frac {2}{x-\mathrm {e}-2\,x\,\mathrm {e}+2\,x^2}}\,{\mathrm {e}}^{\frac {x}{x-\mathrm {e}-2\,x\,\mathrm {e}+2\,x^2}}-{\mathrm {e}}^{-\frac {4}{x-\mathrm {e}-2\,x\,\mathrm {e}+2\,x^2}}\,{\mathrm {e}}^{\frac {2\,x}{x-\mathrm {e}-2\,x\,\mathrm {e}+2\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2)*(4*x + 4*x^2 + 1) - exp(1)*(2*x + 8*x^2 + 8*x^3) + exp((x - 2)/(x + 2*x^2 - exp(1)*(2*x + 1)))*(1
6*x - 10*exp(1) - 4*x^2 + 4) + exp((2*(x - 2))/(x + 2*x^2 - exp(1)*(2*x + 1)))*(16*x - 10*exp(1) - 4*x^2 + 4)
+ x^2 + 4*x^3 + 4*x^4)/(exp(2)*(4*x + 4*x^2 + 1) - exp(1)*(2*x + 8*x^2 + 8*x^3) + x^2 + 4*x^3 + 4*x^4),x)

[Out]

- x - 2*exp(-2/(x - exp(1) - 2*x*exp(1) + 2*x^2))*exp(x/(x - exp(1) - 2*x*exp(1) + 2*x^2)) - exp(-4/(x - exp(1
) - 2*x*exp(1) + 2*x^2))*exp((2*x)/(x - exp(1) - 2*x*exp(1) + 2*x^2))

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sympy [B]  time = 0.82, size = 46, normalized size = 1.48 \begin {gather*} - x - e^{\frac {2 \left (2 - x\right )}{- 2 x^{2} - x + e \left (2 x + 1\right )}} - 2 e^{\frac {2 - x}{- 2 x^{2} - x + e \left (2 x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*exp(1)+4*x**2-16*x-4)*exp((2-x)/((2*x+1)*exp(1)-2*x**2-x))**2+(10*exp(1)+4*x**2-16*x-4)*exp((2-
x)/((2*x+1)*exp(1)-2*x**2-x))+(-4*x**2-4*x-1)*exp(1)**2+(8*x**3+8*x**2+2*x)*exp(1)-4*x**4-4*x**3-x**2)/((4*x**
2+4*x+1)*exp(1)**2+(-8*x**3-8*x**2-2*x)*exp(1)+4*x**4+4*x**3+x**2),x)

[Out]

-x - exp(2*(2 - x)/(-2*x**2 - x + E*(2*x + 1))) - 2*exp((2 - x)/(-2*x**2 - x + E*(2*x + 1)))

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