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

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

[Out]

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

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Rubi [A]  time = 0.417422, antiderivative size = 130, normalized size of antiderivative = 1., 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 = {6728, 724, 204, 1028, 986, 12, 1026, 1161, 618, 1027, 206} \[ -\frac{\tan ^{-1}\left (\frac{2 x+3}{\sqrt{3} \sqrt{-x^2-4 x-3}}\right )}{3 \sqrt{3}}+\frac{1}{9} \sqrt{2} \tan ^{-1}\left (\frac{1-\frac{x+3}{\sqrt{-x^2-4 x-3}}}{\sqrt{2}}\right )-\frac{1}{9} \sqrt{2} \tan ^{-1}\left (\frac{\frac{x+3}{\sqrt{-x^2-4 x-3}}+1}{\sqrt{2}}\right )-\frac{4}{9} \tanh ^{-1}\left (\frac{x}{\sqrt{-x^2-4 x-3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 12

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

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 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 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 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 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])

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} \operatorname{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-\operatorname{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} \operatorname{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} \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{\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} \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{2}{27} \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{\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} \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{4}{27} \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{\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*}

Mathematica [C]  time = 0.453093, size = 200, normalized size = 1.54 \[ \frac{1}{54} \left (-6 \sqrt{3} \tan ^{-1}\left (\frac{2 x+3}{\sqrt{3} \sqrt{-x^2-4 x-3}}\right )-3 \sqrt{1-2 i \sqrt{2}} \left (\sqrt{2}+2 i\right ) \tanh ^{-1}\left (\frac{\left (2-i \sqrt{2}\right ) x-2 i \sqrt{2}+2}{\sqrt{2+4 i \sqrt{2}} \sqrt{-x^2-4 x-3}}\right )-3 \sqrt{1+2 i \sqrt{2}} \left (\sqrt{2}-2 i\right ) \tanh ^{-1}\left (\frac{\left (2+i \sqrt{2}\right ) x+2 i \sqrt{2}+2}{\sqrt{2-4 i \sqrt{2}} \sqrt{-x^2-4 x-3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*Sqrt[3]*ArcTan[(3 + 2*x)/(Sqrt[3]*Sqrt[-3 - 4*x - x^2])] - 3*Sqrt[1 - (2*I)*Sqrt[2]]*(2*I + Sqrt[2])*ArcTa
nh[(2 - (2*I)*Sqrt[2] + (2 - I*Sqrt[2])*x)/(Sqrt[2 + (4*I)*Sqrt[2]]*Sqrt[-3 - 4*x - x^2])] - 3*Sqrt[1 + (2*I)*
Sqrt[2]]*(-2*I + Sqrt[2])*ArcTanh[(2 + (2*I)*Sqrt[2] + (2 + I*Sqrt[2])*x)/(Sqrt[2 - (4*I)*Sqrt[2]]*Sqrt[-3 - 4
*x - x^2])])/54

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Maple [A]  time = 0.102, size = 152, normalized size = 1.2 \begin{align*}{\frac{\sqrt{3}}{9}\arctan \left ({\frac{ \left ( -6-4\,x \right ) \sqrt{3}}{6}{\frac{1}{\sqrt{-{x}^{2}-4\,x-3}}}} \right ) }+{\frac{\sqrt{4}\sqrt{3}}{54}\sqrt{3\,{\frac{{x}^{2}}{ \left ( -3/2-x \right ) ^{2}}}-12} \left ( \sqrt{2}\arctan \left ({\frac{\sqrt{2}}{6}\sqrt{3\,{\frac{{x}^{2}}{ \left ( -3/2-x \right ) ^{2}}}-12}} \right ) +4\,{\it Artanh} \left ( 3\,{\frac{x}{-3/2-x}{\frac{1}{\sqrt{3\,{\frac{{x}^{2}}{ \left ( -3/2-x \right ) ^{2}}}-12}}}} \right ) \right ){\frac{1}{\sqrt{{ \left ({{x}^{2} \left ( -{\frac{3}{2}}-x \right ) ^{-2}}-4 \right ) \left ( 1+{x \left ( -{\frac{3}{2}}-x \right ) ^{-1}} \right ) ^{-2}}}}} \left ( 1+{x \left ( -{\frac{3}{2}}-x \right ) ^{-1}} \right ) ^{-1}} \end{align*}

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)

[Out]

1/9*3^(1/2)*arctan(1/6*(-6-4*x)*3^(1/2)/(-x^2-4*x-3)^(1/2))+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*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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (2 \, x^{2} + 4 \, x + 3\right )} \sqrt{-x^{2} - 4 \, x - 3} x}\,{d x} \end{align*}

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 = 1.63283, size = 473, normalized size = 3.64 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \sqrt{- \left (x + 1\right ) \left (x + 3\right )} \left (2 x^{2} + 4 x + 3\right )}\, dx \end{align*}

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 = 1.30737, size = 269, normalized size = 2.07 \begin{align*} \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{align*}

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)