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]

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

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Rubi [A]  time = 0.01, antiderivative size = 33, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.71, size = 25, normalized size = 0.76 \[ \frac {1}{2} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (x - 1\right )}}{\sqrt {3 \, x^{2} + 4 \, x - 2}}\right ) \]

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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giac [A]  time = 0.20, size = 30, normalized size = 0.91 \[ \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 4 \, x - 2}\right )}\right ) \]

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)))

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maple [A]  time = 0.05, size = 29, normalized size = 0.88 \[ \frac {\sqrt {2}\, \arctan \left (\frac {\left (4 x -4\right ) \sqrt {2}}{4 \sqrt {3 x^{2}+4 x -2}}\right )}{2} \]

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 = 1.99, size = 26, normalized size = 0.79 \[ \frac {1}{2} \, \sqrt {2} \arcsin \left (\frac {\sqrt {10} x}{5 \, {\left | x \right |}} - \frac {\sqrt {10}}{5 \, {\left | x \right |}}\right ) \]

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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mupad [B]  time = 0.37, size = 34, normalized size = 1.03 \[ \frac {\sqrt {2}\,\ln \left (\frac {2\,x-2+\sqrt {2}\,\sqrt {3\,x^2+4\,x-2}\,1{}\mathrm {i}}{x}\right )\,1{}\mathrm {i}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(4*x + 3*x^2 - 2)^(1/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \sqrt {3 x^{2} + 4 x - 2}}\, dx \]

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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