3.14 \(\int \frac {\sqrt {-3-4 x-x^2}}{3+4 x+2 x^2} \, dx\)

Optimal. Leaf size=98 \[ -\frac {\tan ^{-1}\left (\frac {1-\frac {x+3}{\sqrt {-x^2-4 x-3}}}{\sqrt {2}}\right )}{\sqrt {2}}+\frac {\tan ^{-1}\left (\frac {\frac {x+3}{\sqrt {-x^2-4 x-3}}+1}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt {-x^2-4 x-3}}\right )-\frac {1}{2} \sin ^{-1}(x+2) \]

[Out]

-1/2*arcsin(2+x)-1/2*arctanh(x/(-x^2-4*x-3)^(1/2))-1/2*arctan(1/2*(1+(-3-x)/(-x^2-4*x-3)^(1/2))*2^(1/2))*2^(1/
2)+1/2*arctan(1/2*(1+(3+x)/(-x^2-4*x-3)^(1/2))*2^(1/2))*2^(1/2)

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Rubi [A]  time = 0.21, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {989, 619, 216, 1028, 986, 12, 1026, 1161, 618, 204, 1027, 206} \[ -\frac {\tan ^{-1}\left (\frac {1-\frac {x+3}{\sqrt {-x^2-4 x-3}}}{\sqrt {2}}\right )}{\sqrt {2}}+\frac {\tan ^{-1}\left (\frac {\frac {x+3}{\sqrt {-x^2-4 x-3}}+1}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt {-x^2-4 x-3}}\right )-\frac {1}{2} \sin ^{-1}(x+2) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[-3 - 4*x - x^2]/(3 + 4*x + 2*x^2),x]

[Out]

-ArcSin[2 + x]/2 - ArcTan[(1 - (3 + x)/Sqrt[-3 - 4*x - x^2])/Sqrt[2]]/Sqrt[2] + ArcTan[(1 + (3 + x)/Sqrt[-3 -
4*x - x^2])/Sqrt[2]]/Sqrt[2] - ArcTanh[x/Sqrt[-3 - 4*x - x^2]]/2

Rule 12

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

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 986

Int[1/(((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_.) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q = Rt
[(c*d - a*f)^2 - (b*d - a*e)*(c*e - b*f), 2]}, Dist[1/(2*q), Int[(c*d - a*f + q + (c*e - b*f)*x)/((a + b*x + c
*x^2)*Sqrt[d + e*x + f*x^2]), x], x] - Dist[1/(2*q), Int[(c*d - a*f - q + (c*e - b*f)*x)/((a + b*x + c*x^2)*Sq
rt[d + e*x + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] &&
NeQ[c*e - b*f, 0] && NegQ[b^2 - 4*a*c]

Rule 989

Int[Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]/((d_) + (e_.)*(x_) + (f_.)*(x_)^2), x_Symbol] :> Dist[c/f, Int[1/Sq
rt[a + b*x + c*x^2], x], x] - Dist[1/f, Int[(c*d - a*f + (c*e - b*f)*x)/(Sqrt[a + b*x + c*x^2]*(d + e*x + f*x^
2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0]

Rule 1026

Int[(x_)/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol] :> Dist[-2*e,
Subst[Int[(1 - d*x^2)/(c*e - b*f - e*(2*c*d - b*e + 2*a*f)*x^2 + d^2*(c*e - b*f)*x^4), x], x, (1 + ((e + Sqrt[
e^2 - 4*d*f])*x)/(2*d))/Sqrt[d + e*x + f*x^2]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[e^2 - 4*d*f, 0] && EqQ[b*d - a*e, 0]

Rule 1027

Int[((g_) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol]
:> Dist[g, Subst[Int[1/(a + (c*d - a*f)*x^2), x], x, x/Sqrt[d + e*x + f*x^2]], x] /; FreeQ[{a, b, c, d, e, f,
g, h}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0] && EqQ[b*d - a*e, 0] && EqQ[2*h*d - g*e, 0]

Rule 1028

Int[((g_) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (e_.)*(x_) + (f_.)*(x_)^2]), x_Symbol]
:> -Dist[(2*h*d - g*e)/e, Int[1/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] + Dist[h/e, Int[(2*d + e*x)
/((a + b*x + c*x^2)*Sqrt[d + e*x + f*x^2]), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0
] && NeQ[e^2 - 4*d*f, 0] && EqQ[b*d - a*e, 0] && NeQ[2*h*d - g*e, 0]

Rule 1161

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e - b/c, 2]},
Dist[e/(2*c), Int[1/Simp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /
; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[(2*d)/e - b/c, 0] || ( !Lt
Q[(2*d)/e - b/c, 0] && EqQ[d - e*Rt[a/c, 2], 0]))

Rubi steps

\begin {align*} \int \frac {\sqrt {-3-4 x-x^2}}{3+4 x+2 x^2} \, dx &=-\left (\frac {1}{2} \int \frac {1}{\sqrt {-3-4 x-x^2}} \, dx\right )-\frac {1}{2} \int \frac {3+4 x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx\\ &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{4}}} \, dx,x,-4-2 x\right )+\frac {1}{2} \int \frac {-6-4 x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx+\frac {3}{2} \int \frac {1}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx\\ &=-\frac {1}{2} \sin ^{-1}(2+x)-\frac {1}{4} \int \frac {-6-4 x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx+\frac {1}{4} \int -\frac {4 x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-3 \operatorname {Subst}\left (\int \frac {1}{3-3 x^2} \, dx,x,\frac {x}{\sqrt {-3-4 x-x^2}}\right )\\ &=-\frac {1}{2} \sin ^{-1}(2+x)-\tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )+\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{3-3 x^2} \, dx,x,\frac {x}{\sqrt {-3-4 x-x^2}}\right )-\int \frac {x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx\\ &=-\frac {1}{2} \sin ^{-1}(2+x)-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )-8 \operatorname {Subst}\left (\int \frac {1+3 x^2}{-4-8 x^2-36 x^4} \, dx,x,\frac {1+\frac {x}{3}}{\sqrt {-3-4 x-x^2}}\right )\\ &=-\frac {1}{2} \sin ^{-1}(2+x)-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{\frac {1}{3}-\frac {2 x}{3}+x^2} \, dx,x,\frac {1+\frac {x}{3}}{\sqrt {-3-4 x-x^2}}\right )+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{\frac {1}{3}+\frac {2 x}{3}+x^2} \, dx,x,\frac {1+\frac {x}{3}}{\sqrt {-3-4 x-x^2}}\right )\\ &=-\frac {1}{2} \sin ^{-1}(2+x)-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )-\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{-\frac {8}{9}-x^2} \, dx,x,\frac {2}{3} \left (-1+\frac {3+x}{\sqrt {-3-4 x-x^2}}\right )\right )-\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{-\frac {8}{9}-x^2} \, dx,x,\frac {2}{3} \left (1+\frac {3+x}{\sqrt {-3-4 x-x^2}}\right )\right )\\ &=-\frac {1}{2} \sin ^{-1}(2+x)-\frac {\tan ^{-1}\left (\frac {1-\frac {3+x}{\sqrt {-3-4 x-x^2}}}{\sqrt {2}}\right )}{\sqrt {2}}+\frac {\tan ^{-1}\left (\frac {1+\frac {3+x}{\sqrt {-3-4 x-x^2}}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.39, size = 159, normalized size = 1.62 \[ \frac {1}{4} \left (-i \sqrt {1-2 i \sqrt {2}} \tanh ^{-1}\left (\frac {i \sqrt {2} x+2 x+2 i \sqrt {2}+2}{\sqrt {2-4 i \sqrt {2}} \sqrt {-x^2-4 x-3}}\right )+i \sqrt {1+2 i \sqrt {2}} \tanh ^{-1}\left (\frac {\left (2-i \sqrt {2}\right ) x-2 i \sqrt {2}+2}{\sqrt {2+4 i \sqrt {2}} \sqrt {-x^2-4 x-3}}\right )-2 \sin ^{-1}(x+2)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-3 - 4*x - x^2]/(3 + 4*x + 2*x^2),x]

[Out]

(-2*ArcSin[2 + x] - I*Sqrt[1 - (2*I)*Sqrt[2]]*ArcTanh[(2 + (2*I)*Sqrt[2] + 2*x + I*Sqrt[2]*x)/(Sqrt[2 - (4*I)*
Sqrt[2]]*Sqrt[-3 - 4*x - x^2])] + I*Sqrt[1 + (2*I)*Sqrt[2]]*ArcTanh[(2 - (2*I)*Sqrt[2] + (2 - I*Sqrt[2])*x)/(S
qrt[2 + (4*I)*Sqrt[2]]*Sqrt[-3 - 4*x - x^2])])/4

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fricas [A]  time = 0.96, size = 161, normalized size = 1.64 \[ -\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} x + 3 \, \sqrt {2} \sqrt {-x^{2} - 4 \, x - 3}}{2 \, {\left (2 \, x + 3\right )}}\right ) - \frac {1}{4} \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} x - 3 \, \sqrt {2} \sqrt {-x^{2} - 4 \, x - 3}}{2 \, {\left (2 \, x + 3\right )}}\right ) + \frac {1}{2} \, \arctan \left (\frac {\sqrt {-x^{2} - 4 \, x - 3} {\left (x + 2\right )}}{x^{2} + 4 \, x + 3}\right ) + \frac {1}{8} \, \log \left (-\frac {2 \, \sqrt {-x^{2} - 4 \, x - 3} x + 4 \, x + 3}{x^{2}}\right ) - \frac {1}{8} \, \log \left (\frac {2 \, \sqrt {-x^{2} - 4 \, x - 3} x - 4 \, x - 3}{x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(2)*arctan(1/2*(sqrt(2)*x + 3*sqrt(2)*sqrt(-x^2 - 4*x - 3))/(2*x + 3)) - 1/4*sqrt(2)*arctan(-1/2*(sqr
t(2)*x - 3*sqrt(2)*sqrt(-x^2 - 4*x - 3))/(2*x + 3)) + 1/2*arctan(sqrt(-x^2 - 4*x - 3)*(x + 2)/(x^2 + 4*x + 3))
+ 1/8*log(-(2*sqrt(-x^2 - 4*x - 3)*x + 4*x + 3)/x^2) - 1/8*log((2*sqrt(-x^2 - 4*x - 3)*x - 4*x - 3)/x^2)

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giac [B]  time = 0.21, size = 171, normalized size = 1.74 \[ -\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\frac {3 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + 1\right )}\right ) - \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\frac {\sqrt {-x^{2} - 4 \, x - 3} - 1}{x + 2} + 1\right )}\right ) - \frac {1}{2} \, \arcsin \left (x + 2\right ) - \frac {1}{4} \, \log \left (\frac {2 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + \frac {3 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}^{2}}{{\left (x + 2\right )}^{2}} + 1\right ) + \frac {1}{4} \, \log \left (\frac {2 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + \frac {{\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}^{2}}{{\left (x + 2\right )}^{2}} + 3\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(2)*arctan(1/2*sqrt(2)*(3*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1)) - 1/2*sqrt(2)*arctan(1/2*sqrt(2)*(
(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1)) - 1/2*arcsin(x + 2) - 1/4*log(2*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) +
3*(sqrt(-x^2 - 4*x - 3) - 1)^2/(x + 2)^2 + 1) + 1/4*log(2*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + (sqrt(-x^2 - 4*
x - 3) - 1)^2/(x + 2)^2 + 3)

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maple [B]  time = 0.03, size = 341, normalized size = 3.48 \[ -\frac {\arcsin \left (x +2\right )}{2}-\frac {\sqrt {3}\, \sqrt {4}\, \sqrt {\frac {3 x^{2}}{\left (-x -\frac {3}{2}\right )^{2}}-12}\, \sqrt {2}\, \arctan \left (\frac {\sqrt {\frac {3 x^{2}}{\left (-x -\frac {3}{2}\right )^{2}}-12}\, \sqrt {2}}{6}\right )}{3 \sqrt {\frac {\frac {x^{2}}{\left (-x -\frac {3}{2}\right )^{2}}-4}{\left (\frac {x}{-x -\frac {3}{2}}+1\right )^{2}}}\, \left (\frac {x}{-x -\frac {3}{2}}+1\right )}+\frac {\sqrt {3}\, \sqrt {4}\, \sqrt {\frac {3 x^{2}}{\left (-x -\frac {3}{2}\right )^{2}}-12}\, \left (-\arctanh \left (\frac {3 x}{\left (-x -\frac {3}{2}\right ) \sqrt {\frac {3 x^{2}}{\left (-x -\frac {3}{2}\right )^{2}}-12}}\right )+\sqrt {2}\, \arctan \left (\frac {\sqrt {\frac {3 x^{2}}{\left (-x -\frac {3}{2}\right )^{2}}-12}\, \sqrt {2}}{6}\right )\right )}{12 \sqrt {\frac {\frac {x^{2}}{\left (-x -\frac {3}{2}\right )^{2}}-4}{\left (\frac {x}{-x -\frac {3}{2}}+1\right )^{2}}}\, \left (\frac {x}{-x -\frac {3}{2}}+1\right )}+\frac {\sqrt {3}\, \sqrt {4}\, \sqrt {\frac {3 x^{2}}{\left (-x -\frac {3}{2}\right )^{2}}-12}\, \left (\arctanh \left (\frac {3 x}{\left (-x -\frac {3}{2}\right ) \sqrt {\frac {3 x^{2}}{\left (-x -\frac {3}{2}\right )^{2}}-12}}\right )+\sqrt {2}\, \arctan \left (\frac {\sqrt {\frac {3 x^{2}}{\left (-x -\frac {3}{2}\right )^{2}}-12}\, \sqrt {2}}{6}\right )\right )}{6 \sqrt {\frac {\frac {x^{2}}{\left (-x -\frac {3}{2}\right )^{2}}-4}{\left (\frac {x}{-x -\frac {3}{2}}+1\right )^{2}}}\, \left (\frac {x}{-x -\frac {3}{2}}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2-4*x-3)^(1/2)/(2*x^2+4*x+3),x)

[Out]

-1/2*arcsin(2+x)+1/12*3^(1/2)*4^(1/2)*(3*x^2/(-3/2-x)^2-12)^(1/2)*(2^(1/2)*arctan(1/6*(3*x^2/(-3/2-x)^2-12)^(1
/2)*2^(1/2))-arctanh(3*x/(-3/2-x)/(3*x^2/(-3/2-x)^2-12)^(1/2)))/((x^2/(-3/2-x)^2-4)/(1+x/(-3/2-x))^2)^(1/2)/(1
+x/(-3/2-x))-1/3*3^(1/2)*4^(1/2)/((x^2/(-3/2-x)^2-4)/(1+x/(-3/2-x))^2)^(1/2)/(1+x/(-3/2-x))*(3*x^2/(-3/2-x)^2-
12)^(1/2)*2^(1/2)*arctan(1/6*(3*x^2/(-3/2-x)^2-12)^(1/2)*2^(1/2))+1/6*3^(1/2)*4^(1/2)*(3*x^2/(-3/2-x)^2-12)^(1
/2)*(2^(1/2)*arctan(1/6*(3*x^2/(-3/2-x)^2-12)^(1/2)*2^(1/2))+arctanh(3*x/(-3/2-x)/(3*x^2/(-3/2-x)^2-12)^(1/2))
)/((x^2/(-3/2-x)^2-4)/(1+x/(-3/2-x))^2)^(1/2)/(1+x/(-3/2-x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-x^{2} - 4 \, x - 3}}{2 \, x^{2} + 4 \, x + 3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(-x^2 - 4*x - 3)/(2*x^2 + 4*x + 3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {-x^2-4\,x-3}}{2\,x^2+4\,x+3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((- 4*x - x^2 - 3)^(1/2)/(4*x + 2*x^2 + 3),x)

[Out]

int((- 4*x - x^2 - 3)^(1/2)/(4*x + 2*x^2 + 3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {- \left (x + 1\right ) \left (x + 3\right )}}{2 x^{2} + 4 x + 3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**2-4*x-3)**(1/2)/(2*x**2+4*x+3),x)

[Out]

Integral(sqrt(-(x + 1)*(x + 3))/(2*x**2 + 4*x + 3), x)

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