3.1928 \(\int \frac{1}{\sqrt{2+\frac{b}{x^2}} x^2} \, dx\)

Optimal. Leaf size=20 \[ -\frac{\text{csch}^{-1}\left (\frac{\sqrt{2} x}{\sqrt{b}}\right )}{\sqrt{b}} \]

[Out]

-(ArcCsch[(Sqrt[2]*x)/Sqrt[b]]/Sqrt[b])

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Rubi [A]  time = 0.0075717, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {335, 215} \[ -\frac{\text{csch}^{-1}\left (\frac{\sqrt{2} x}{\sqrt{b}}\right )}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[2 + b/x^2]*x^2),x]

[Out]

-(ArcCsch[(Sqrt[2]*x)/Sqrt[b]]/Sqrt[b])

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

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

Mathematica [B]  time = 0.0150686, size = 50, normalized size = 2.5 \[ -\frac{\sqrt{b+2 x^2} \tanh ^{-1}\left (\frac{\sqrt{b+2 x^2}}{\sqrt{b}}\right )}{\sqrt{b} x \sqrt{\frac{b}{x^2}+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[2 + b/x^2]*x^2),x]

[Out]

-((Sqrt[b + 2*x^2]*ArcTanh[Sqrt[b + 2*x^2]/Sqrt[b]])/(Sqrt[b]*Sqrt[2 + b/x^2]*x))

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Maple [B]  time = 0.009, size = 52, normalized size = 2.6 \begin{align*} -{\frac{1}{x}\sqrt{2\,{x}^{2}+b}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{2\,{x}^{2}+b}+b}{x}} \right ){\frac{1}{\sqrt{{\frac{2\,{x}^{2}+b}{{x}^{2}}}}}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(2+1/x^2*b)^(1/2),x)

[Out]

-1/((2*x^2+b)/x^2)^(1/2)/x*(2*x^2+b)^(1/2)/b^(1/2)*ln(2*(b^(1/2)*(2*x^2+b)^(1/2)+b)/x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.46846, size = 184, normalized size = 9.2 \begin{align*} \left [\frac{\log \left (-\frac{x^{2} - \sqrt{b} x \sqrt{\frac{2 \, x^{2} + b}{x^{2}}} + b}{x^{2}}\right )}{2 \, \sqrt{b}}, \frac{\sqrt{-b} \arctan \left (\frac{\sqrt{-b} x \sqrt{\frac{2 \, x^{2} + b}{x^{2}}}}{2 \, x^{2} + b}\right )}{b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*log(-(x^2 - sqrt(b)*x*sqrt((2*x^2 + b)/x^2) + b)/x^2)/sqrt(b), sqrt(-b)*arctan(sqrt(-b)*x*sqrt((2*x^2 + b
)/x^2)/(2*x^2 + b))/b]

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Sympy [A]  time = 1.19456, size = 20, normalized size = 1. \begin{align*} - \frac{\operatorname{asinh}{\left (\frac{\sqrt{2} \sqrt{b}}{2 x} \right )}}{\sqrt{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-asinh(sqrt(2)*sqrt(b)/(2*x))/sqrt(b)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \sqrt{\frac{b}{x^{2}} + 2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(x^2*sqrt(b/x^2 + 2)), x)