### 3.138 $$\int \frac{1}{x \sqrt{3+x^2}} \, dx$$

Optimal. Leaf size=23 $-\frac{\tanh ^{-1}\left (\frac{\sqrt{x^2+3}}{\sqrt{3}}\right )}{\sqrt{3}}$

[Out]

-(ArcTanh[Sqrt[3 + x^2]/Sqrt]/Sqrt)

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Rubi [A]  time = 0.0110963, antiderivative size = 23, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.231, Rules used = {266, 63, 207} $-\frac{\tanh ^{-1}\left (\frac{\sqrt{x^2+3}}{\sqrt{3}}\right )}{\sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/(x*Sqrt[3 + x^2]),x]

[Out]

-(ArcTanh[Sqrt[3 + x^2]/Sqrt]/Sqrt)

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0042559, size = 23, normalized size = 1. $-\frac{\tanh ^{-1}\left (\frac{\sqrt{x^2+3}}{\sqrt{3}}\right )}{\sqrt{3}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/(x*Sqrt[3 + x^2]),x]

[Out]

-(ArcTanh[Sqrt[3 + x^2]/Sqrt]/Sqrt)

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Maple [A]  time = 0.004, size = 18, normalized size = 0.8 \begin{align*} -{\frac{\sqrt{3}}{3}{\it Artanh} \left ({\sqrt{3}{\frac{1}{\sqrt{{x}^{2}+3}}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(x^2+3)^(1/2),x)

[Out]

-1/3*3^(1/2)*arctanh(3^(1/2)/(x^2+3)^(1/2))

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Maxima [A]  time = 1.40106, size = 19, normalized size = 0.83 \begin{align*} -\frac{1}{3} \, \sqrt{3} \operatorname{arsinh}\left (\frac{\sqrt{3}}{{\left | x \right |}}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*sqrt(3)*arcsinh(sqrt(3)/abs(x))

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Fricas [A]  time = 2.2903, size = 63, normalized size = 2.74 \begin{align*} \frac{1}{3} \, \sqrt{3} \log \left (-\frac{\sqrt{3} - \sqrt{x^{2} + 3}}{x}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.987104, size = 15, normalized size = 0.65 \begin{align*} - \frac{\sqrt{3} \operatorname{asinh}{\left (\frac{\sqrt{3}}{x} \right )}}{3} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(3)*asinh(sqrt(3)/x)/3

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Giac [B]  time = 1.05048, size = 50, normalized size = 2.17 \begin{align*} -\frac{1}{6} \, \sqrt{3} \log \left (\sqrt{3} + \sqrt{x^{2} + 3}\right ) + \frac{1}{6} \, \sqrt{3} \log \left (-\sqrt{3} + \sqrt{x^{2} + 3}\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*sqrt(3)*log(sqrt(3) + sqrt(x^2 + 3)) + 1/6*sqrt(3)*log(-sqrt(3) + sqrt(x^2 + 3))