3.2.31 \(\int \frac {1}{x \sqrt {-3-4 x-x^2} (3+4 x+2 x^2)} \, dx\) [131]

Optimal. Leaf size=130 \[ -\frac {\tan ^{-1}\left (\frac {3+2 x}{\sqrt {3} \sqrt {-3-4 x-x^2}}\right )}{3 \sqrt {3}}+\frac {1}{9} \sqrt {2} \tan ^{-1}\left (\frac {1-\frac {3+x}{\sqrt {-3-4 x-x^2}}}{\sqrt {2}}\right )-\frac {1}{9} \sqrt {2} \tan ^{-1}\left (\frac {1+\frac {3+x}{\sqrt {-3-4 x-x^2}}}{\sqrt {2}}\right )-\frac {4}{9} \tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right ) \]

[Out]

-4/9*arctanh(x/(-x^2-4*x-3)^(1/2))+1/9*arctan(1/2*(1+(-3-x)/(-x^2-4*x-3)^(1/2))*2^(1/2))*2^(1/2)-1/9*arctan(1/
2*(1+(3+x)/(-x^2-4*x-3)^(1/2))*2^(1/2))*2^(1/2)-1/9*arctan(1/3*(3+2*x)*3^(1/2)/(-x^2-4*x-3)^(1/2))*3^(1/2)

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Rubi [A]
time = 0.26, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 11, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {6860, 738, 210, 1042, 1000, 12, 1040, 1175, 632, 1041, 212} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {2 x+3}{\sqrt {3} \sqrt {-x^2-4 x-3}}\right )}{3 \sqrt {3}}+\frac {1}{9} \sqrt {2} \text {ArcTan}\left (\frac {1-\frac {x+3}{\sqrt {-x^2-4 x-3}}}{\sqrt {2}}\right )-\frac {1}{9} \sqrt {2} \text {ArcTan}\left (\frac {\frac {x+3}{\sqrt {-x^2-4 x-3}}+1}{\sqrt {2}}\right )-\frac {4}{9} \tanh ^{-1}\left (\frac {x}{\sqrt {-x^2-4 x-3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*ArcTan[(3 + 2*x)/(Sqrt[3]*Sqrt[-3 - 4*x - x^2])]/Sqrt[3] + (Sqrt[2]*ArcTan[(1 - (3 + x)/Sqrt[-3 - 4*x - x
^2])/Sqrt[2]])/9 - (Sqrt[2]*ArcTan[(1 + (3 + x)/Sqrt[-3 - 4*x - x^2])/Sqrt[2]])/9 - (4*ArcTanh[x/Sqrt[-3 - 4*x
- x^2]])/9

Rule 12

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

Rule 210

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

Rule 212

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

Rule 632

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 738

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 1000

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 1040

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 1041

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 1042

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 1175

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 6860

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 \sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx &=\int \left (\frac {1}{3 x \sqrt {-3-4 x-x^2}}-\frac {2 (2+x)}{3 \sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )}\right ) \, dx\\ &=\frac {1}{3} \int \frac {1}{x \sqrt {-3-4 x-x^2}} \, dx-\frac {2}{3} \int \frac {2+x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx\\ &=\frac {1}{6} \int \frac {-6-4 x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\frac {1}{3} \int \frac {1}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\frac {2}{3} \text {Subst}\left (\int \frac {1}{-12-x^2} \, dx,x,\frac {-6-4 x}{\sqrt {-3-4 x-x^2}}\right )\\ &=-\frac {\tan ^{-1}\left (\frac {3+2 x}{\sqrt {3} \sqrt {-3-4 x-x^2}}\right )}{3 \sqrt {3}}+\frac {1}{18} \int \frac {-6-4 x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\frac {1}{18} \int -\frac {4 x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\text {Subst}\left (\int \frac {1}{3-3 x^2} \, dx,x,\frac {x}{\sqrt {-3-4 x-x^2}}\right )\\ &=-\frac {\tan ^{-1}\left (\frac {3+2 x}{\sqrt {3} \sqrt {-3-4 x-x^2}}\right )}{3 \sqrt {3}}-\frac {1}{3} \tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )+\frac {2}{9} \int \frac {x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\frac {1}{3} \text {Subst}\left (\int \frac {1}{3-3 x^2} \, dx,x,\frac {x}{\sqrt {-3-4 x-x^2}}\right )\\ &=-\frac {\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 {16}{9} \text {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 {\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 {2}{27} \text {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 {2}{27} \text {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 {\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} \text {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 {4}{27} \text {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 {\tan ^{-1}\left (\frac {3+2 x}{\sqrt {3} \sqrt {-3-4 x-x^2}}\right )}{3 \sqrt {3}}+\frac {1}{9} \sqrt {2} \tan ^{-1}\left (\frac {1-\frac {3+x}{\sqrt {-3-4 x-x^2}}}{\sqrt {2}}\right )-\frac {1}{9} \sqrt {2} \tan ^{-1}\left (\frac {1+\frac {3+x}{\sqrt {-3-4 x-x^2}}}{\sqrt {2}}\right )-\frac {4}{9} \tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 90, normalized size = 0.69 \begin {gather*} \frac {1}{9} \left (-\sqrt {2} \tan ^{-1}\left (\frac {3+2 x}{\sqrt {2} \sqrt {-3-4 x-x^2}}\right )+2 \sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt {-3-4 x-x^2}}{3+x}\right )-4 \tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.37, size = 152, normalized size = 1.17

method	result	size
default	\(\frac {\sqrt {3}\, \sqrt {4}\, \sqrt {\frac {3 x^{2}}{\left (-\frac {3}{2}-x \right )^{2}}-12}\, \left (\sqrt {2}\, \arctan \left (\frac {\sqrt {\frac {3 x^{2}}{\left (-\frac {3}{2}-x \right )^{2}}-12}\, \sqrt {2}}{6}\right )+4 \arctanh \left (\frac {3 x}{\left (-\frac {3}{2}-x \right ) \sqrt {\frac {3 x^{2}}{\left (-\frac {3}{2}-x \right )^{2}}-12}}\right )\right )}{54 \sqrt {\frac {\frac {x^{2}}{\left (-\frac {3}{2}-x \right )^{2}}-4}{\left (1+\frac {x}{-\frac {3}{2}-x}\right )^{2}}}\, \left (1+\frac {x}{-\frac {3}{2}-x}\right )}+\frac {\sqrt {3}\, \arctan \left (\frac {\left (-6-4 x \right ) \sqrt {3}}{6 \sqrt {-x^{2}-4 x -3}}\right )}{9}\)	\(152\)
trager	\(\RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) \ln \left (-\frac {40500 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )^{2} x +4680 \sqrt {-x^{2}-4 x -3}\, \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )-40500 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )^{2}-3636 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) x +287 \sqrt {-x^{2}-4 x -3}+41886 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )-1731 x -8078}{18 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) x -18 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )-8 x -1}\right )+\frac {4 \ln \left (\frac {-4500 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )^{2} x +520 \sqrt {-x^{2}-4 x -3}\, \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )+4500 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )^{2}+3596 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) x -263 \sqrt {-x^{2}-4 x -3}+654 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )-517 x -282}{2 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) x -2 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )+1}\right )}{9}-\ln \left (\frac {-4500 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )^{2} x +520 \sqrt {-x^{2}-4 x -3}\, \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )+4500 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )^{2}+3596 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) x -263 \sqrt {-x^{2}-4 x -3}+654 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )-517 x -282}{2 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right ) x -2 \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )+1}\right ) \RootOf \left (18 \textit {\_Z}^{2}-8 \textit {\_Z} +1\right )-\frac {\RootOf \left (\textit {\_Z}^{2}+3\right ) \ln \left (\frac {-2 \RootOf \left (\textit {\_Z}^{2}+3\right ) x +3 \sqrt {-x^{2}-4 x -3}-3 \RootOf \left (\textit {\_Z}^{2}+3\right )}{x}\right )}{9}\)	\(483\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/54*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))+4*ar
ctanh(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/
9*3^(1/2)*arctan(1/6*(-6-4*x)*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*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(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), x)

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Fricas [A]
time = 0.38, size = 170, normalized size = 1.31 \begin {gather*} \frac {1}{9} \, \sqrt {3} \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 ) + \frac {1}{18} \, \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}{18} \, \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}{9} \, \log \left (-\frac {2 \, \sqrt {-x^{2} - 4 \, x - 3} x + 4 \, x + 3}{x^{2}}\right ) - \frac {1}{9} \, \log \left (\frac {2 \, \sqrt {-x^{2} - 4 \, x - 3} x - 4 \, x - 3}{x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*sqrt(3)*arctan(1/3*sqrt(3)*sqrt(-x^2 - 4*x - 3)*(2*x + 3)/(x^2 + 4*x + 3)) + 1/18*sqrt(2)*arctan(1/2*(sqrt
(2)*x + 3*sqrt(2)*sqrt(-x^2 - 4*x - 3))/(2*x + 3)) + 1/18*sqrt(2)*arctan(-1/2*(sqrt(2)*x - 3*sqrt(2)*sqrt(-x^2
- 4*x - 3))/(2*x + 3)) + 1/9*log(-(2*sqrt(-x^2 - 4*x - 3)*x + 4*x + 3)/x^2) - 1/9*log((2*sqrt(-x^2 - 4*x - 3)
*x - 4*x - 3)/x^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x \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*x**2+4*x+3)/(-x**2-4*x-3)**(1/2),x)

[Out]

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

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Giac [A]
time = 3.09, size = 199, normalized size = 1.53 \begin {gather*} \frac {1}{9} \, \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 {2}{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 {1}{9} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\frac {\sqrt {-x^{2} - 4 \, x - 3} - 1}{x + 2} + 1\right )}\right ) - \frac {2}{9} \, \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 {2}{9} \, \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*x^2+4*x+3)/(-x^2-4*x-3)^(1/2),x, algorithm="giac")

[Out]

1/9*sqrt(2)*arctan(1/2*sqrt(2)*(3*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1)) + 2/9*sqrt(3)*arctan(1/3*sqrt(3)*(2
*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1)) + 1/9*sqrt(2)*arctan(1/2*sqrt(2)*((sqrt(-x^2 - 4*x - 3) - 1)/(x + 2)
+ 1)) - 2/9*log(2*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 3*(sqrt(-x^2 - 4*x - 3) - 1)^2/(x + 2)^2 + 1) + 2/9*lo
g(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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x\,\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*(- 4*x - x^2 - 3)^(1/2)*(4*x + 2*x^2 + 3)),x)

[Out]

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

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