3.1925 \(\int \frac{1}{\sqrt{a+\frac{b}{x^2}} x^2} \, dx\)

Optimal. Leaf size=28 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )}{\sqrt{b}} \]

[Out]

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

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Rubi [A]  time = 0.0135937, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {335, 217, 206} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b/x^2]*x^2),x]

[Out]

-(ArcTanh[Sqrt[b]/(Sqrt[a + b/x^2]*x)]/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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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}{\sqrt{a+\frac{b}{x^2}} x^2} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{x}\right )\\ &=-\operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{1}{\sqrt{a+\frac{b}{x^2}} x}\right )\\ &=-\frac{\tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a+\frac{b}{x^2}} x}\right )}{\sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.0141918, size = 50, normalized size = 1.79 \[ -\frac{\sqrt{a x^2+b} \tanh ^{-1}\left (\frac{\sqrt{a x^2+b}}{\sqrt{b}}\right )}{\sqrt{b} x \sqrt{a+\frac{b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b/x^2]*x^2),x]

[Out]

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

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Maple [B]  time = 0.005, size = 52, normalized size = 1.9 \begin{align*} -{\frac{1}{x}\sqrt{a{x}^{2}+b}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a{x}^{2}+b}+b}{x}} \right ){\frac{1}{\sqrt{{\frac{a{x}^{2}+b}{{x}^{2}}}}}}{\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/((a*x^2+b)/x^2)^(1/2)/x*(a*x^2+b)^(1/2)/b^(1/2)*ln(2*(b^(1/2)*(a*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/(a+b/x^2)^(1/2)/x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.51497, size = 192, normalized size = 6.86 \begin{align*} \left [\frac{\log \left (-\frac{a x^{2} - 2 \, \sqrt{b} x \sqrt{\frac{a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right )}{2 \, \sqrt{b}}, \frac{\sqrt{-b} \arctan \left (\frac{\sqrt{-b} x \sqrt{\frac{a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right )}{b}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.53551, size = 19, normalized size = 0.68 \begin{align*} - \frac{\operatorname{asinh}{\left (\frac{\sqrt{b}}{\sqrt{a} x} \right )}}{\sqrt{b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(a + b/x^2)*x^2), x)