\(\int \frac {x^3}{\sqrt {-3-4 x-x^2} (3+4 x+2 x^2)} \, dx\) [127]

Optimal result

Integrand size = 30, antiderivative size = 115 \[ \int \frac {x^3}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx=-\frac {1}{2} \sqrt {-3-4 x-x^2}-2 \arcsin (2+x)+\frac {\arctan \left (\frac {1-\frac {3+x}{\sqrt {-3-4 x-x^2}}}{\sqrt {2}}\right )}{2 \sqrt {2}}-\frac {\arctan \left (\frac {1+\frac {3+x}{\sqrt {-3-4 x-x^2}}}{\sqrt {2}}\right )}{2 \sqrt {2}}+\text {arctanh}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.433, Rules used = {6860, 633, 222, 654, 1042, 1000, 12, 1040, 1175, 632, 210, 1041, 212} \[ \int \frac {x^3}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx=-2 \arcsin (x+2)+\frac {\arctan \left (\frac {1-\frac {x+3}{\sqrt {-x^2-4 x-3}}}{\sqrt {2}}\right )}{2 \sqrt {2}}-\frac {\arctan \left (\frac {\frac {x+3}{\sqrt {-x^2-4 x-3}}+1}{\sqrt {2}}\right )}{2 \sqrt {2}}+\text {arctanh}\left (\frac {x}{\sqrt {-x^2-4 x-3}}\right )-\frac {1}{2} \sqrt {-x^2-4 x-3} \]

[In]

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

[Out]

-1/2*Sqrt[-3 - 4*x - x^2] - 2*ArcSin[2 + x] + ArcTan[(1 - (3 + x)/Sqrt[-3 - 4*x - x^2])/Sqrt[2]]/(2*Sqrt[2]) -
ArcTan[(1 + (3 + x)/Sqrt[-3 - 4*x - x^2])/Sqrt[2]]/(2*Sqrt[2]) + ArcTanh[x/Sqrt[-3 - 4*x - x^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 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 222

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 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 633

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 654

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

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*} \text {integral}& = \int \left (-\frac {1}{\sqrt {-3-4 x-x^2}}+\frac {x}{2 \sqrt {-3-4 x-x^2}}+\frac {6+5 x}{2 \sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )}\right ) \, dx \\ & = \frac {1}{2} \int \frac {x}{\sqrt {-3-4 x-x^2}} \, dx+\frac {1}{2} \int \frac {6+5 x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\int \frac {1}{\sqrt {-3-4 x-x^2}} \, dx \\ & = -\frac {1}{2} \sqrt {-3-4 x-x^2}+\frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{4}}} \, dx,x,-4-2 x\right )-\frac {5}{8} \int \frac {-6-4 x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\frac {3}{4} \int \frac {1}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\int \frac {1}{\sqrt {-3-4 x-x^2}} \, dx \\ & = -\frac {1}{2} \sqrt {-3-4 x-x^2}-\sin ^{-1}(2+x)+\frac {1}{8} \int \frac {-6-4 x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\frac {1}{8} \int -\frac {4 x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx+\frac {1}{2} \text {Subst}\left (\int \frac {1}{\sqrt {1-\frac {x^2}{4}}} \, dx,x,-4-2 x\right )+\frac {15}{4} \text {Subst}\left (\int \frac {1}{3-3 x^2} \, dx,x,\frac {x}{\sqrt {-3-4 x-x^2}}\right ) \\ & = -\frac {1}{2} \sqrt {-3-4 x-x^2}-2 \sin ^{-1}(2+x)+\frac {5}{4} \tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )+\frac {1}{2} \int \frac {x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx-\frac {3}{4} \text {Subst}\left (\int \frac {1}{3-3 x^2} \, dx,x,\frac {x}{\sqrt {-3-4 x-x^2}}\right ) \\ & = -\frac {1}{2} \sqrt {-3-4 x-x^2}-2 \sin ^{-1}(2+x)+\tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )+4 \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 {1}{2} \sqrt {-3-4 x-x^2}-2 \sin ^{-1}(2+x)+\tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )-\frac {1}{6} \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 {1}{6} \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 {1}{2} \sqrt {-3-4 x-x^2}-2 \sin ^{-1}(2+x)+\tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right )+\frac {1}{3} \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 {1}{3} \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 {1}{2} \sqrt {-3-4 x-x^2}-2 \sin ^{-1}(2+x)+\frac {\tan ^{-1}\left (\frac {1-\frac {3+x}{\sqrt {-3-4 x-x^2}}}{\sqrt {2}}\right )}{2 \sqrt {2}}-\frac {\tan ^{-1}\left (\frac {1+\frac {3+x}{\sqrt {-3-4 x-x^2}}}{\sqrt {2}}\right )}{2 \sqrt {2}}+\tanh ^{-1}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.82 \[ \int \frac {x^3}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx=-\frac {1}{2} \sqrt {-3-4 x-x^2}-\frac {\arctan \left (\frac {3+2 x}{\sqrt {2} \sqrt {-3-4 x-x^2}}\right )}{2 \sqrt {2}}+4 \arctan \left (\frac {\sqrt {-3-4 x-x^2}}{3+x}\right )+\text {arctanh}\left (\frac {x}{\sqrt {-3-4 x-x^2}}\right ) \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.25

method	result	size
default	\(-2 \arcsin \left (2+x \right )-\frac {\sqrt {-x^{2}-4 x -3}}{2}+\frac {\sqrt {3}\, \sqrt {4}\, \sqrt {\frac {3 x^{2}}{\left (-\frac {3}{2}-x \right )^{2}}-12}\, \left (\sqrt {2}\, \arctan \left (\frac {\sqrt {\frac {3 x^{2}}{\left (-\frac {3}{2}-x \right )^{2}}-12}\, \sqrt {2}}{6}\right )-4 \,\operatorname {arctanh}\left (\frac {3 x}{\left (-\frac {3}{2}-x \right ) \sqrt {\frac {3 x^{2}}{\left (-\frac {3}{2}-x \right )^{2}}-12}}\right )\right )}{24 \sqrt {\frac {\frac {x^{2}}{\left (-\frac {3}{2}-x \right )^{2}}-4}{\left (1+\frac {x}{-\frac {3}{2}-x}\right )^{2}}}\, \left (1+\frac {x}{-\frac {3}{2}-x}\right )}\)	\(144\)
risch	\(\frac {x^{2}+4 x +3}{2 \sqrt {-x^{2}-4 x -3}}-2 \arcsin \left (2+x \right )+\frac {\sqrt {3}\, \sqrt {4}\, \sqrt {\frac {3 x^{2}}{\left (-\frac {3}{2}-x \right )^{2}}-12}\, \left (\sqrt {2}\, \arctan \left (\frac {\sqrt {\frac {3 x^{2}}{\left (-\frac {3}{2}-x \right )^{2}}-12}\, \sqrt {2}}{6}\right )-4 \,\operatorname {arctanh}\left (\frac {3 x}{\left (-\frac {3}{2}-x \right ) \sqrt {\frac {3 x^{2}}{\left (-\frac {3}{2}-x \right )^{2}}-12}}\right )\right )}{24 \sqrt {\frac {\frac {x^{2}}{\left (-\frac {3}{2}-x \right )^{2}}-4}{\left (1+\frac {x}{-\frac {3}{2}-x}\right )^{2}}}\, \left (1+\frac {x}{-\frac {3}{2}-x}\right )}\)	\(152\)
trager	\(-\frac {\sqrt {-x^{2}-4 x -3}}{2}+\frac {3 \operatorname {RootOf}\left (24 \textit {\_Z}^{2}-16 \textit {\_Z} +3\right ) \ln \left (\frac {72 \operatorname {RootOf}\left (24 \textit {\_Z}^{2}-16 \textit {\_Z} +3\right )^{2} x -72 \operatorname {RootOf}\left (24 \textit {\_Z}^{2}-16 \textit {\_Z} +3\right ) x -36 \operatorname {RootOf}\left (24 \textit {\_Z}^{2}-16 \textit {\_Z} +3\right )-3 \sqrt {-x^{2}-4 x -3}+16 x +12}{12 \operatorname {RootOf}\left (24 \textit {\_Z}^{2}-16 \textit {\_Z} +3\right ) x -2 x +3}\right )}{2}+\ln \left (\frac {24 \operatorname {RootOf}\left (24 \textit {\_Z}^{2}-16 \textit {\_Z} +3\right )^{2} x -8 \operatorname {RootOf}\left (24 \textit {\_Z}^{2}-16 \textit {\_Z} +3\right ) x +12 \operatorname {RootOf}\left (24 \textit {\_Z}^{2}-16 \textit {\_Z} +3\right )-\sqrt {-x^{2}-4 x -3}-4}{4 \operatorname {RootOf}\left (24 \textit {\_Z}^{2}-16 \textit {\_Z} +3\right ) x -2 x -1}\right )-\frac {3 \ln \left (\frac {24 \operatorname {RootOf}\left (24 \textit {\_Z}^{2}-16 \textit {\_Z} +3\right )^{2} x -8 \operatorname {RootOf}\left (24 \textit {\_Z}^{2}-16 \textit {\_Z} +3\right ) x +12 \operatorname {RootOf}\left (24 \textit {\_Z}^{2}-16 \textit {\_Z} +3\right )-\sqrt {-x^{2}-4 x -3}-4}{4 \operatorname {RootOf}\left (24 \textit {\_Z}^{2}-16 \textit {\_Z} +3\right ) x -2 x -1}\right ) \operatorname {RootOf}\left (24 \textit {\_Z}^{2}-16 \textit {\_Z} +3\right )}{2}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+\sqrt {-x^{2}-4 x -3}\right )\)	\(330\)

[In]

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

[Out]

-2*arcsin(2+x)-1/2*(-x^2-4*x-3)^(1/2)+1/24*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))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.52 \[ \int \frac {x^3}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx=\frac {1}{8} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} x + 3 \, \sqrt {2} \sqrt {-x^{2} - 4 \, x - 3}}{2 \, {\left (2 \, x + 3\right )}}\right ) + \frac {1}{8} \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} x - 3 \, \sqrt {2} \sqrt {-x^{2} - 4 \, x - 3}}{2 \, {\left (2 \, x + 3\right )}}\right ) - \frac {1}{2} \, \sqrt {-x^{2} - 4 \, x - 3} + 2 \, \arctan \left (\frac {\sqrt {-x^{2} - 4 \, x - 3} {\left (x + 2\right )}}{x^{2} + 4 \, x + 3}\right ) - \frac {1}{4} \, \log \left (-\frac {2 \, \sqrt {-x^{2} - 4 \, x - 3} x + 4 \, x + 3}{x^{2}}\right ) + \frac {1}{4} \, \log \left (\frac {2 \, \sqrt {-x^{2} - 4 \, x - 3} x - 4 \, x - 3}{x^{2}}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {x^3}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx=\int { \frac {x^{3}}{{\left (2 \, x^{2} + 4 \, x + 3\right )} \sqrt {-x^{2} - 4 \, x - 3}} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.61 \[ \int \frac {x^3}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx=\frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\frac {3 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + 1\right )}\right ) + \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\frac {\sqrt {-x^{2} - 4 \, x - 3} - 1}{x + 2} + 1\right )}\right ) - \frac {1}{2} \, \sqrt {-x^{2} - 4 \, x - 3} - 2 \, \arcsin \left (x + 2\right ) + \frac {1}{2} \, \log \left (\frac {2 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + \frac {3 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}^{2}}{{\left (x + 2\right )}^{2}} + 1\right ) - \frac {1}{2} \, \log \left (\frac {2 \, {\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}}{x + 2} + \frac {{\left (\sqrt {-x^{2} - 4 \, x - 3} - 1\right )}^{2}}{{\left (x + 2\right )}^{2}} + 3\right ) \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^3}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx=\int \frac {x^3}{\sqrt {-x^2-4\,x-3}\,\left (2\,x^2+4\,x+3\right )} \,d x \]

[In]

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

[Out]

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