3.2434 \(\int \frac{1}{x \sqrt{-2+4 x+3 x^2}} \, dx\)

Optimal. Leaf size=33 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{2} (1-x)}{\sqrt{3 x^2+4 x-2}}\right )}{\sqrt{2}} \]

[Out]

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

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Rubi [A]  time = 0.0106804, antiderivative size = 33, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {724, 204} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{2} (1-x)}{\sqrt{3 x^2+4 x-2}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[-2 + 4*x + 3*x^2]),x]

[Out]

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

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])

Rubi steps

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

Mathematica [A]  time = 0.0087675, size = 27, normalized size = 0.82 \[ \frac{\tan ^{-1}\left (\frac{x-1}{\sqrt{\frac{3 x^2}{2}+2 x-1}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.042, size = 29, normalized size = 0.9 \begin{align*}{\frac{\sqrt{2}}{2}\arctan \left ({\frac{ \left ( -4+4\,x \right ) \sqrt{2}}{4}{\frac{1}{\sqrt{3\,{x}^{2}+4\,x-2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.54439, size = 35, normalized size = 1.06 \begin{align*} \frac{1}{2} \, \sqrt{2} \arcsin \left (\frac{\sqrt{10} x}{5 \,{\left | x \right |}} - \frac{\sqrt{10}}{5 \,{\left | x \right |}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*arcsin(1/5*sqrt(10)*x/abs(x) - 1/5*sqrt(10)/abs(x))

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Fricas [A]  time = 2.06923, size = 80, normalized size = 2.42 \begin{align*} \frac{1}{2} \, \sqrt{2} \arctan \left (\frac{\sqrt{2}{\left (x - 1\right )}}{\sqrt{3 \, x^{2} + 4 \, x - 2}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \sqrt{3 x^{2} + 4 x - 2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13588, size = 41, normalized size = 1.24 \begin{align*} \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{3} x - \sqrt{3 \, x^{2} + 4 \, x - 2}\right )}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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