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

Optimal. Leaf size=68 \[ \frac {\tan ^{-1}\left (\frac {\frac {3 \sqrt {-x-1}}{\sqrt {x+3}}+1}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {\tan ^{-1}\left (\frac {1-\frac {3 \sqrt {-x-1}}{\sqrt {x+3}}}{\sqrt {2}}\right )}{\sqrt {2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 69, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1026, 1161, 618, 204} \begin {gather*} \frac {\tan ^{-1}\left (\frac {1-\frac {x+3}{\sqrt {-x^2-4 x-3}}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {\tan ^{-1}\left (\frac {\frac {x+3}{\sqrt {-x^2-4 x-3}}+1}{\sqrt {2}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]))

Rubi steps

\begin {align*} \int \frac {x}{\sqrt {-3-4 x-x^2} \left (3+4 x+2 x^2\right )} \, dx &=8 \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 )\\ &=-\left (\frac {1}{3} \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 )\right )-\frac {1}{3} \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}{3} \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 {2}{3} \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 {1-\frac {3+x}{\sqrt {-3-4 x-x^2}}}{\sqrt {2}}\right )}{\sqrt {2}}-\frac {\tan ^{-1}\left (\frac {1+\frac {3+x}{\sqrt {-3-4 x-x^2}}}{\sqrt {2}}\right )}{\sqrt {2}}\\ \end {align*}

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Mathematica [C]  time = 0.17, size = 174, normalized size = 2.56 \begin {gather*} \frac {\left (1-i \sqrt {2}\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 )+\left (1+i \sqrt {2}\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 )}{6 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - I*Sqrt[2])*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])] + (1 + I*Sqrt[2])*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])])/(6*Sqrt[2])

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IntegrateAlgebraic [A]  time = 0.25, size = 38, normalized size = 0.56 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {2} x+\frac {3}{\sqrt {2}}}{\sqrt {-x^2-4 x-3}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.87, size = 50, normalized size = 0.74 \begin {gather*} \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (6 \, x^{2} + 20 \, x + 15\right )} \sqrt {-x^{2} - 4 \, x - 3}}{4 \, {\left (2 \, x^{3} + 11 \, x^{2} + 18 \, x + 9\right )}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*arctan(1/4*sqrt(2)*(6*x^2 + 20*x + 15)*sqrt(-x^2 - 4*x - 3)/(2*x^3 + 11*x^2 + 18*x + 9))

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giac [A]  time = 0.18, size = 68, normalized size = 1.00 \begin {gather*} \frac {1}{2} \, \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}{2} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\frac {\sqrt {-x^{2} - 4 \, x - 3} - 1}{x + 2} + 1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 92, normalized size = 1.35 \begin {gather*} \frac {\sqrt {3}\, \sqrt {4}\, \sqrt {\frac {3 x^{2}}{\left (-x -\frac {3}{2}\right )^{2}}-12}\, \sqrt {2}\, \arctan \left (\frac {\sqrt {\frac {3 x^{2}}{\left (-x -\frac {3}{2}\right )^{2}}-12}\, \sqrt {2}}{6}\right )}{12 \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(x/(2*x^2+4*x+3)/(-x^2-4*x-3)^(1/2),x)

[Out]

1/12*3^(1/2)*4^(1/2)/((1/(-x-3/2)^2*x^2-4)/(1/(-x-3/2)*x+1)^2)^(1/2)/(1/(-x-3/2)*x+1)*(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))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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

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