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

Optimal. Leaf size=95 \[ -\frac{1}{3} \sqrt{2} \tan ^{-1}\left (\frac{1-\frac{x+3}{\sqrt{-x^2-4 x-3}}}{\sqrt{2}}\right )+\frac{1}{3} \sqrt{2} \tan ^{-1}\left (\frac{\frac{x+3}{\sqrt{-x^2-4 x-3}}+1}{\sqrt{2}}\right )+\frac{1}{3} \tanh ^{-1}\left (\frac{x}{\sqrt{-x^2-4 x-3}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.111173, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {986, 12, 1026, 1161, 618, 204, 1027, 206} \[ -\frac{1}{3} \sqrt{2} \tan ^{-1}\left (\frac{1-\frac{x+3}{\sqrt{-x^2-4 x-3}}}{\sqrt{2}}\right )+\frac{1}{3} \sqrt{2} \tan ^{-1}\left (\frac{\frac{x+3}{\sqrt{-x^2-4 x-3}}+1}{\sqrt{2}}\right )+\frac{1}{3} \tanh ^{-1}\left (\frac{x}{\sqrt{-x^2-4 x-3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.109496, size = 150, normalized size = 1.58 \[ \frac{1}{6} i \left (\sqrt{1-2 i \sqrt{2}} \tanh ^{-1}\left (\frac{\left (2-i \sqrt{2}\right ) x-2 i \sqrt{2}+2}{\sqrt{2+4 i \sqrt{2}} \sqrt{-x^2-4 x-3}}\right )-\sqrt{1+2 i \sqrt{2}} \tanh ^{-1}\left (\frac{\left (2+i \sqrt{2}\right ) x+2 i \sqrt{2}+2}{\sqrt{2-4 i \sqrt{2}} \sqrt{-x^2-4 x-3}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I/6)*(Sqrt[1 - (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])] - Sqrt[1 + (2*I)*Sqrt[2]]*ArcTanh[(2 + (2*I)*Sqrt[2] + (2 + I*Sqrt[2])*x)/(Sqrt[2 - (4*I)*Sqr
t[2]]*Sqrt[-3 - 4*x - x^2])])

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Maple [A]  time = 0.1, size = 121, normalized size = 1.3 \begin{align*} -{\frac{\sqrt{4}\sqrt{3}}{18}\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 ) +{\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/(2*x^2+4*x+3)/(-x^2-4*x-3)^(1/2),x)

[Out]

-1/18*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))+arc
tanh(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}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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)

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Fricas [A]  time = 1.60765, size = 365, normalized size = 3.84 \begin{align*} -\frac{1}{6} \, \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}{6} \, \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}{12} \, \log \left (-\frac{2 \, \sqrt{-x^{2} - 4 \, x - 3} x + 4 \, x + 3}{x^{2}}\right ) + \frac{1}{12} \, \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/(2*x^2+4*x+3)/(-x^2-4*x-3)^(1/2),x, algorithm="fricas")

[Out]

-1/6*sqrt(2)*arctan(1/2*(sqrt(2)*x + 3*sqrt(2)*sqrt(-x^2 - 4*x - 3))/(2*x + 3)) - 1/6*sqrt(2)*arctan(-1/2*(sqr
t(2)*x - 3*sqrt(2)*sqrt(-x^2 - 4*x - 3))/(2*x + 3)) - 1/12*log(-(2*sqrt(-x^2 - 4*x - 3)*x + 4*x + 3)/x^2) + 1/
12*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}{\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/(2*x**2+4*x+3)/(-x**2-4*x-3)**(1/2),x)

[Out]

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

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Giac [B]  time = 1.25614, size = 223, normalized size = 2.35 \begin{align*} -\frac{1}{3} \, \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}{3} \, \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}{6} \, \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}{6} \, \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/(2*x^2+4*x+3)/(-x^2-4*x-3)^(1/2),x, algorithm="giac")

[Out]

-1/3*sqrt(2)*arctan(1/2*sqrt(2)*(3*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1)) - 1/3*sqrt(2)*arctan(1/2*sqrt(2)*(
(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 1)) + 1/6*log(2*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + 3*(sqrt(-x^2 - 4*x -
3) - 1)^2/(x + 2)^2 + 1) - 1/6*log(2*(sqrt(-x^2 - 4*x - 3) - 1)/(x + 2) + (sqrt(-x^2 - 4*x - 3) - 1)^2/(x + 2
)^2 + 3)