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

Optimal. Leaf size=17 \[ \tanh ^{-1}\left (\frac {x}{\sqrt {-x^2-4 x-3}}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1027, 206} \begin {gather*} \tanh ^{-1}\left (\frac {x}{\sqrt {-x^2-4 x-3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[x/Sqrt[-3 - 4*x - x^2]]

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])

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]

Rubi steps

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

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Mathematica [C]  time = 0.29, size = 165, normalized size = 9.71 \begin {gather*} \frac {1}{6} \left (\sqrt {1-2 i \sqrt {2}} \left (\sqrt {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 )+\sqrt {1+2 i \sqrt {2}} \left (\sqrt {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 ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - (2*I)*Sqrt[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]]*(-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

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IntegrateAlgebraic [A]  time = 0.30, size = 17, normalized size = 1.00 \begin {gather*} \tanh ^{-1}\left (\frac {x}{\sqrt {-x^2-4 x-3}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTanh[x/Sqrt[-3 - 4*x - x^2]]

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fricas [B]  time = 0.42, size = 56, normalized size = 3.29 \begin {gather*} -\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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)

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giac [B]  time = 0.23, size = 98, normalized size = 5.76 \begin {gather*} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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maple [B]  time = 0.02, size = 94, normalized size = 5.53 \begin {gather*} -\frac {\sqrt {3}\, \sqrt {4}\, \sqrt {\frac {3 x^{2}}{\left (-x -\frac {3}{2}\right )^{2}}-12}\, \arctanh \left (\frac {3 x}{\left (-x -\frac {3}{2}\right ) \sqrt {\frac {3 x^{2}}{\left (-x -\frac {3}{2}\right )^{2}}-12}}\right )}{6 \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((3+2*x)/(2*x^2+4*x+3)/(-x^2-4*x-3)^(1/2),x)

[Out]

-1/6*3^(1/2)*4^(1/2)/((x^2/(-3/2-x)^2-4)/(x/(-3/2-x)+1)^2)^(1/2)/(x/(-3/2-x)+1)*(3*x^2/(-3/2-x)^2-12)^(1/2)*ar
ctanh(3*x/(-3/2-x)/(3*x^2/(-3/2-x)^2-12)^(1/2))

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

[Out]

integrate((2*x + 3)/((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 = -0.06 \begin {gather*} \int \frac {2\,x+3}{\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((2*x + 3)/((- 4*x - x^2 - 3)^(1/2)*(4*x + 2*x^2 + 3)),x)

[Out]

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

[Out]

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

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