3.2.32 \(\int \frac {1}{x^2 \sqrt {-3-4 x-x^2} (3+4 x+2 x^2)} \, dx\)

Optimal. Leaf size=151 \[ \frac {\sqrt {-x^2-4 x-3}}{9 x}+\frac {2 \tan ^{-1}\left (\frac {2 x+3}{\sqrt {3} \sqrt {-x^2-4 x-3}}\right )}{3 \sqrt {3}}+\frac {2}{27} \sqrt {2} \tan ^{-1}\left (\frac {1-\frac {x+3}{\sqrt {-x^2-4 x-3}}}{\sqrt {2}}\right )-\frac {2}{27} \sqrt {2} \tan ^{-1}\left (\frac {\frac {x+3}{\sqrt {-x^2-4 x-3}}+1}{\sqrt {2}}\right )+\frac {10}{27} \tanh ^{-1}\left (\frac {x}{\sqrt {-x^2-4 x-3}}\right ) \]

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Rubi [A]  time = 0.45, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 12, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6728, 730, 724, 204, 1028, 986, 12, 1026, 1161, 618, 1027, 206} \begin {gather*} \frac {\sqrt {-x^2-4 x-3}}{9 x}+\frac {2 \tan ^{-1}\left (\frac {2 x+3}{\sqrt {3} \sqrt {-x^2-4 x-3}}\right )}{3 \sqrt {3}}+\frac {2}{27} \sqrt {2} \tan ^{-1}\left (\frac {1-\frac {x+3}{\sqrt {-x^2-4 x-3}}}{\sqrt {2}}\right )-\frac {2}{27} \sqrt {2} \tan ^{-1}\left (\frac {\frac {x+3}{\sqrt {-x^2-4 x-3}}+1}{\sqrt {2}}\right )+\frac {10}{27} \tanh ^{-1}\left (\frac {x}{\sqrt {-x^2-4 x-3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[-3 - 4*x - x^2]*(3 + 4*x + 2*x^2)),x]

[Out]

Sqrt[-3 - 4*x - x^2]/(9*x) + (2*ArcTan[(3 + 2*x)/(Sqrt[3]*Sqrt[-3 - 4*x - x^2])])/(3*Sqrt[3]) + (2*Sqrt[2]*Arc
Tan[(1 - (3 + x)/Sqrt[-3 - 4*x - x^2])/Sqrt[2]])/27 - (2*Sqrt[2]*ArcTan[(1 + (3 + x)/Sqrt[-3 - 4*x - x^2])/Sqr
t[2]])/27 + (10*ArcTanh[x/Sqrt[-3 - 4*x - x^2]])/27

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 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 724

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

Rule 730

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)
*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e
^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c,
0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m + 2*p + 3, 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 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]))

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {1}{x^2 \sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx &=\int \left (\frac {1}{3 x^2 \sqrt {-3-4 x-x^2}}-\frac {4}{9 x \sqrt {-3-4 x-x^2}}+\frac {2 (5+4 x)}{9 \sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )}\right ) \, dx\\ &=\frac {2}{9} \int \frac {5+4 x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx+\frac {1}{3} \int \frac {1}{x^2 \sqrt {-3-4 x-x^2}} \, dx-\frac {4}{9} \int \frac {1}{x \sqrt {-3-4 x-x^2}} \, dx\\ &=\frac {\sqrt {-3-4 x-x^2}}{9 x}-\frac {2}{9} \int \frac {1}{x \sqrt {-3-4 x-x^2}} \, dx-\frac {2}{9} \int \frac {1}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\frac {2}{9} \int \frac {-6-4 x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx+\frac {8}{9} \operatorname {Subst}\left (\int \frac {1}{-12-x^2} \, dx,x,\frac {-6-4 x}{\sqrt {-3-4 x-x^2}}\right )\\ &=\frac {\sqrt {-3-4 x-x^2}}{9 x}+\frac {4 \tan ^{-1}\left (\frac {3+2 x}{\sqrt {3} \sqrt {-3-4 x-x^2}}\right )}{9 \sqrt {3}}+\frac {1}{27} \int \frac {-6-4 x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\frac {1}{27} \int -\frac {4 x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx+\frac {4}{9} \operatorname {Subst}\left (\int \frac {1}{-12-x^2} \, dx,x,\frac {-6-4 x}{\sqrt {-3-4 x-x^2}}\right )+\frac {4}{3} \operatorname {Subst}\left (\int \frac {1}{3-3 x^2} \, dx,x,\frac {x}{\sqrt {-3-4 x-x^2}}\right )\\ &=\frac {\sqrt {-3-4 x-x^2}}{9 x}+\frac {2 \tan ^{-1}\left (\frac {3+2 x}{\sqrt {3} \sqrt {-3-4 x-x^2}}\right )}{3 \sqrt {3}}+\frac {4}{9} \tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )+\frac {4}{27} \int \frac {x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\frac {2}{9} \operatorname {Subst}\left (\int \frac {1}{3-3 x^2} \, dx,x,\frac {x}{\sqrt {-3-4 x-x^2}}\right )\\ &=\frac {\sqrt {-3-4 x-x^2}}{9 x}+\frac {2 \tan ^{-1}\left (\frac {3+2 x}{\sqrt {3} \sqrt {-3-4 x-x^2}}\right )}{3 \sqrt {3}}+\frac {10}{27} \tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )+\frac {32}{27} \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 {\sqrt {-3-4 x-x^2}}{9 x}+\frac {2 \tan ^{-1}\left (\frac {3+2 x}{\sqrt {3} \sqrt {-3-4 x-x^2}}\right )}{3 \sqrt {3}}+\frac {10}{27} \tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )-\frac {4}{81} \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 {4}{81} \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 {\sqrt {-3-4 x-x^2}}{9 x}+\frac {2 \tan ^{-1}\left (\frac {3+2 x}{\sqrt {3} \sqrt {-3-4 x-x^2}}\right )}{3 \sqrt {3}}+\frac {10}{27} \tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )+\frac {8}{81} \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 {8}{81} \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 {\sqrt {-3-4 x-x^2}}{9 x}+\frac {2 \tan ^{-1}\left (\frac {3+2 x}{\sqrt {3} \sqrt {-3-4 x-x^2}}\right )}{3 \sqrt {3}}+\frac {2}{27} \sqrt {2} \tan ^{-1}\left (\frac {1-\frac {3+x}{\sqrt {-3-4 x-x^2}}}{\sqrt {2}}\right )-\frac {2}{27} \sqrt {2} \tan ^{-1}\left (\frac {1+\frac {3+x}{\sqrt {-3-4 x-x^2}}}{\sqrt {2}}\right )+\frac {10}{27} \tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.41, size = 225, normalized size = 1.49 \begin {gather*} \frac {3 \left (\sqrt {-x^2-4 x-3}+2 \sqrt {3} x \tan ^{-1}\left (\frac {2 x+3}{\sqrt {3} \sqrt {-x^2-4 x-3}}\right )\right )+\sqrt {1-2 i \sqrt {2}} \left (2 \sqrt {2}+i\right ) x \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 )+\sqrt {1+2 i \sqrt {2}} \left (2 \sqrt {2}-i\right ) x \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 )}{27 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*Sqrt[-3 - 4*x - x^2]*(3 + 4*x + 2*x^2)),x]

[Out]

(3*(Sqrt[-3 - 4*x - x^2] + 2*Sqrt[3]*x*ArcTan[(3 + 2*x)/(Sqrt[3]*Sqrt[-3 - 4*x - x^2])]) + Sqrt[1 - (2*I)*Sqrt
[2]]*(I + 2*Sqrt[2])*x*ArcTanh[(2 - (2*I)*Sqrt[2] + 2*x - I*Sqrt[2]*x)/(Sqrt[2 + (4*I)*Sqrt[2]]*Sqrt[-3 - 4*x
- x^2])] + Sqrt[1 + (2*I)*Sqrt[2]]*(-I + 2*Sqrt[2])*x*ArcTanh[(2 + (2*I)*Sqrt[2] + (2 + I*Sqrt[2])*x)/(Sqrt[2
- (4*I)*Sqrt[2]]*Sqrt[-3 - 4*x - x^2])])/(27*x)

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IntegrateAlgebraic [A]  time = 0.43, size = 118, normalized size = 0.78 \begin {gather*} \frac {\sqrt {-x^2-4 x-3}}{9 x}-\frac {2}{27} \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x+\frac {3}{\sqrt {2}}}{\sqrt {-x^2-4 x-3}}\right )-\frac {4 \tan ^{-1}\left (\frac {\sqrt {3} \sqrt {-x^2-4 x-3}}{x+3}\right )}{3 \sqrt {3}}+\frac {10}{27} \tanh ^{-1}\left (\frac {x}{\sqrt {-x^2-4 x-3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^2*Sqrt[-3 - 4*x - x^2]*(3 + 4*x + 2*x^2)),x]

[Out]

Sqrt[-3 - 4*x - x^2]/(9*x) - (2*Sqrt[2]*ArcTan[(3/Sqrt[2] + Sqrt[2]*x)/Sqrt[-3 - 4*x - x^2]])/27 - (4*ArcTan[(
Sqrt[3]*Sqrt[-3 - 4*x - x^2])/(3 + x)])/(3*Sqrt[3]) + (10*ArcTanh[x/Sqrt[-3 - 4*x - x^2]])/27

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fricas [A]  time = 1.09, size = 194, normalized size = 1.28 \begin {gather*} -\frac {12 \, \sqrt {3} x \arctan \left (\frac {\sqrt {3} \sqrt {-x^{2} - 4 \, x - 3} {\left (2 \, x + 3\right )}}{3 \, {\left (x^{2} + 4 \, x + 3\right )}}\right ) - 2 \, \sqrt {2} x \arctan \left (\frac {\sqrt {2} x + 3 \, \sqrt {2} \sqrt {-x^{2} - 4 \, x - 3}}{2 \, {\left (2 \, x + 3\right )}}\right ) - 2 \, \sqrt {2} x \arctan \left (-\frac {\sqrt {2} x - 3 \, \sqrt {2} \sqrt {-x^{2} - 4 \, x - 3}}{2 \, {\left (2 \, x + 3\right )}}\right ) + 5 \, x \log \left (-\frac {2 \, \sqrt {-x^{2} - 4 \, x - 3} x + 4 \, x + 3}{x^{2}}\right ) - 5 \, x \log \left (\frac {2 \, \sqrt {-x^{2} - 4 \, x - 3} x - 4 \, x - 3}{x^{2}}\right ) - 6 \, \sqrt {-x^{2} - 4 \, x - 3}}{54 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/54*(12*sqrt(3)*x*arctan(1/3*sqrt(3)*sqrt(-x^2 - 4*x - 3)*(2*x + 3)/(x^2 + 4*x + 3)) - 2*sqrt(2)*x*arctan(1/
2*(sqrt(2)*x + 3*sqrt(2)*sqrt(-x^2 - 4*x - 3))/(2*x + 3)) - 2*sqrt(2)*x*arctan(-1/2*(sqrt(2)*x - 3*sqrt(2)*sqr
t(-x^2 - 4*x - 3))/(2*x + 3)) + 5*x*log(-(2*sqrt(-x^2 - 4*x - 3)*x + 4*x + 3)/x^2) - 5*x*log((2*sqrt(-x^2 - 4*
x - 3)*x - 4*x - 3)/x^2) - 6*sqrt(-x^2 - 4*x - 3))/x

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giac [B]  time = 0.48, size = 269, normalized size = 1.78 \begin {gather*} \frac {2}{27} \, \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 {4}{9} \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (\frac {2 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + 1\right )}\right ) + \frac {2}{27} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\frac {\sqrt {-x^{2} - 4 \, x - 3} - 1}{x + 2} + 1\right )}\right ) - \frac {\frac {\sqrt {-x^{2} - 4 \, x - 3} - 1}{x + 2} + 2}{18 \, {\left (\frac {\sqrt {-x^{2} - 4 \, x - 3} - 1}{x + 2} + \frac {{\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}^{2}}{{\left (x + 2\right )}^{2}} + 1\right )}} + \frac {5}{27} \, \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 {5}{27} \, \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/27*sqrt(2)*arctan(1/2*sqrt(2)*(3*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1)) - 4/9*sqrt(3)*arctan(1/3*sqrt(3)*(
2*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1)) + 2/27*sqrt(2)*arctan(1/2*sqrt(2)*((sqrt(-x^2 - 4*x - 3) - 1)/(x +
2) + 1)) - 1/18*((sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 2)/((sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + (sqrt(-x^2 - 4*
x - 3) - 1)^2/(x + 2)^2 + 1) + 5/27*log(2*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 3*(sqrt(-x^2 - 4*x - 3) - 1)^2/
(x + 2)^2 + 1) - 5/27*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 [A]  time = 0.02, size = 169, normalized size = 1.12 \begin {gather*} -\frac {2 \sqrt {3}\, \arctan \left (\frac {\left (-4 x -6\right ) \sqrt {3}}{6 \sqrt {-x^{2}-4 x -3}}\right )}{9}+\frac {\sqrt {-x^{2}-4 x -3}}{9 x}+\frac {\sqrt {3}\, \sqrt {4}\, \sqrt {\frac {3 x^{2}}{\left (-x -\frac {3}{2}\right )^{2}}-12}\, \left (-5 \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 )}{81 \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/81*3^(1/2)*4^(1/2)*(3/(-x-3/2)^2*x^2-12)^(1/2)*(2^(1/2)*arctan(1/6*(3/(-x-3/2)^2*x^2-12)^(1/2)*2^(1/2))-5*ar
ctanh(3/(-x-3/2)/(3/(-x-3/2)^2*x^2-12)^(1/2)*x))/((1/(-x-3/2)^2*x^2-4)/(1/(-x-3/2)*x+1)^2)^(1/2)/(1/(-x-3/2)*x
+1)+1/9*(-x^2-4*x-3)^(1/2)/x-2/9*3^(1/2)*arctan(1/6*(-4*x-6)*3^(1/2)/(-x^2-4*x-3)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^2\,\sqrt {-x^2-4\,x-3}\,\left (2\,x^2+4\,x+3\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/(x^2*(- 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 \begin {gather*} \int \frac {1}{x^{2} \sqrt {- \left (x + 1\right ) \left (x + 3\right )} \left (2 x^{2} + 4 x + 3\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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