3.1929 \(\int \frac{1}{\sqrt{2-\frac{b}{x^2}} x^2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0079047, antiderivative size = 20, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {335, 216} \[ -\frac{\csc ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{b}}\right )}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[2 - b/x^2]*x^2),x]

[Out]

-(ArcCsc[(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 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[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{\csc ^{-1}\left (\frac{\sqrt{2} x}{\sqrt{b}}\right )}{\sqrt{b}}\\ \end{align*}

Mathematica [B]  time = 0.0174186, size = 54, normalized size = 2.7 \[ \frac{\sqrt{2 x^2-b} \tan ^{-1}\left (\frac{\sqrt{2 x^2-b}}{\sqrt{b}}\right )}{\sqrt{b} x \sqrt{2-\frac{b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[2 - b/x^2]*x^2),x]

[Out]

(Sqrt[-b + 2*x^2]*ArcTan[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 = 64, normalized size = 3.2 \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 = 1.52683, size = 186, normalized size = 9.3 \begin{align*} \left [-\frac{\sqrt{-b} \log \left (-\frac{x^{2} - \sqrt{-b} x \sqrt{\frac{2 \, x^{2} - b}{x^{2}}} - b}{x^{2}}\right )}{2 \, b}, -\frac{\arctan \left (\frac{\sqrt{b} x \sqrt{\frac{2 \, x^{2} - b}{x^{2}}}}{2 \, x^{2} - b}\right )}{\sqrt{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*sqrt(-b)*log(-(x^2 - sqrt(-b)*x*sqrt((2*x^2 - b)/x^2) - b)/x^2)/b, -arctan(sqrt(b)*x*sqrt((2*x^2 - b)/x^
2)/(2*x^2 - b))/sqrt(b)]

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Sympy [A]  time = 1.25341, size = 51, normalized size = 2.55 \begin{align*} \begin{cases} \frac{i \operatorname{acosh}{\left (\frac{\sqrt{2} \sqrt{b}}{2 x} \right )}}{\sqrt{b}} & \text{for}\: \frac{\left |{b}\right |}{2 \left |{x^{2}}\right |} > 1 \\- \frac{\operatorname{asin}{\left (\frac{\sqrt{2} \sqrt{b}}{2 x} \right )}}{\sqrt{b}} & \text{otherwise} \end{cases} \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]

Piecewise((I*acosh(sqrt(2)*sqrt(b)/(2*x))/sqrt(b), Abs(b)/(2*Abs(x**2)) > 1), (-asin(sqrt(2)*sqrt(b)/(2*x))/sq
rt(b), True))

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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)