3.2430 \(\int \frac{1}{x \sqrt{2+4 x+3 x^2}} \, dx\)

Optimal. Leaf size=31 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{2} (x+1)}{\sqrt{3 x^2+4 x+2}}\right )}{\sqrt{2}} \]

[Out]

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

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Rubi [A]  time = 0.0117503, antiderivative size = 31, 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, 206} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{2} (x+1)}{\sqrt{3 x^2+4 x+2}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*Sqrt[2 + 4*x + 3*x^2]),x]

[Out]

-(ArcTanh[(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 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])

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{\tanh ^{-1}\left (\frac{\sqrt{2} (1+x)}{\sqrt{2+4 x+3 x^2}}\right )}{\sqrt{2}}\\ \end{align*}

Mathematica [A]  time = 0.0082974, size = 28, normalized size = 0.9 \[ -\frac{\tanh ^{-1}\left (\frac{x+1}{\sqrt{\frac{3 x^2}{2}+2 x+1}}\right )}{\sqrt{2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[(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}{\it Artanh} \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)*arctanh(1/4*(4+4*x)*2^(1/2)/(3*x^2+4*x+2)^(1/2))

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Maxima [A]  time = 1.56766, size = 32, normalized size = 1.03 \begin{align*} -\frac{1}{2} \, \sqrt{2} \operatorname{arsinh}\left (\frac{\sqrt{2} x}{{\left | x \right |}} + \frac{\sqrt{2}}{{\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)*arcsinh(sqrt(2)*x/abs(x) + sqrt(2)/abs(x))

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Fricas [A]  time = 2.00629, size = 111, normalized size = 3.58 \begin{align*} \frac{1}{4} \, \sqrt{2} \log \left (\frac{2 \, \sqrt{2} \sqrt{3 \, x^{2} + 4 \, x + 2}{\left (x + 1\right )} - 5 \, x^{2} - 8 \, x - 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]

1/4*sqrt(2)*log((2*sqrt(2)*sqrt(3*x^2 + 4*x + 2)*(x + 1) - 5*x^2 - 8*x - 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 [B]  time = 1.12745, size = 81, normalized size = 2.61 \begin{align*} -\frac{1}{2} \, \sqrt{2} \log \left (-\sqrt{3} x + \sqrt{2} + \sqrt{3 \, x^{2} + 4 \, x + 2}\right ) + \frac{1}{2} \, \sqrt{2} \log \left ({\left | -\sqrt{3} x - \sqrt{2} + \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]

-1/2*sqrt(2)*log(-sqrt(3)*x + sqrt(2) + sqrt(3*x^2 + 4*x + 2)) + 1/2*sqrt(2)*log(abs(-sqrt(3)*x - sqrt(2) + sq
rt(3*x^2 + 4*x + 2)))