3.1926 \(\int \frac{1}{\sqrt{a+\frac{b}{x^2}} x^4} \, dx\)

Optimal. Leaf size=53 \[ \frac{a \tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )}{2 b^{3/2}}-\frac{\sqrt{a+\frac{b}{x^2}}}{2 b x} \]

[Out]

-Sqrt[a + b/x^2]/(2*b*x) + (a*ArcTanh[Sqrt[b]/(Sqrt[a + b/x^2]*x)])/(2*b^(3/2))

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Rubi [A]  time = 0.0231338, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {335, 321, 217, 206} \[ \frac{a \tanh ^{-1}\left (\frac{\sqrt{b}}{x \sqrt{a+\frac{b}{x^2}}}\right )}{2 b^{3/2}}-\frac{\sqrt{a+\frac{b}{x^2}}}{2 b x} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b/x^2]*x^4),x]

[Out]

-Sqrt[a + b/x^2]/(2*b*x) + (a*ArcTanh[Sqrt[b]/(Sqrt[a + b/x^2]*x)])/(2*b^(3/2))

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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^4} \, dx &=-\operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{\sqrt{a+\frac{b}{x^2}}}{2 b x}+\frac{a \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{x}\right )}{2 b}\\ &=-\frac{\sqrt{a+\frac{b}{x^2}}}{2 b x}+\frac{a \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{1}{\sqrt{a+\frac{b}{x^2}} x}\right )}{2 b}\\ &=-\frac{\sqrt{a+\frac{b}{x^2}}}{2 b x}+\frac{a \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a+\frac{b}{x^2}} x}\right )}{2 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0467315, size = 71, normalized size = 1.34 \[ \frac{a \left (a x^2+b\right ) \left (\frac{\tanh ^{-1}\left (\sqrt{\frac{a x^2}{b}+1}\right )}{2 \sqrt{\frac{a x^2}{b}+1}}-\frac{b}{2 a x^2}\right )}{b^2 x \sqrt{a+\frac{b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b/x^2]*x^4),x]

[Out]

(a*(b + a*x^2)*(-b/(2*a*x^2) + ArcTanh[Sqrt[1 + (a*x^2)/b]]/(2*Sqrt[1 + (a*x^2)/b])))/(b^2*Sqrt[a + b/x^2]*x)

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Maple [A]  time = 0.006, size = 73, normalized size = 1.4 \begin{align*} -{\frac{1}{2\,{x}^{3}}\sqrt{a{x}^{2}+b} \left ( -a\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a{x}^{2}+b}+b}{x}} \right ){x}^{2}b+\sqrt{a{x}^{2}+b}{b}^{{\frac{3}{2}}} \right ){\frac{1}{\sqrt{{\frac{a{x}^{2}+b}{{x}^{2}}}}}}{b}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+1/x^2*b)^(1/2)/x^4,x)

[Out]

-1/2*(a*x^2+b)^(1/2)*(-a*ln(2*(b^(1/2)*(a*x^2+b)^(1/2)+b)/x)*x^2*b+(a*x^2+b)^(1/2)*b^(3/2))/((a*x^2+b)/x^2)^(1
/2)/x^3/b^(5/2)

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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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.58837, size = 306, normalized size = 5.77 \begin{align*} \left [\frac{a \sqrt{b} x \log \left (-\frac{a x^{2} + 2 \, \sqrt{b} x \sqrt{\frac{a x^{2} + b}{x^{2}}} + 2 \, b}{x^{2}}\right ) - 2 \, b \sqrt{\frac{a x^{2} + b}{x^{2}}}}{4 \, b^{2} x}, -\frac{a \sqrt{-b} x \arctan \left (\frac{\sqrt{-b} x \sqrt{\frac{a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + b \sqrt{\frac{a x^{2} + b}{x^{2}}}}{2 \, b^{2} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(1/2)/x^4,x, algorithm="fricas")

[Out]

[1/4*(a*sqrt(b)*x*log(-(a*x^2 + 2*sqrt(b)*x*sqrt((a*x^2 + b)/x^2) + 2*b)/x^2) - 2*b*sqrt((a*x^2 + b)/x^2))/(b^
2*x), -1/2*(a*sqrt(-b)*x*arctan(sqrt(-b)*x*sqrt((a*x^2 + b)/x^2)/(a*x^2 + b)) + b*sqrt((a*x^2 + b)/x^2))/(b^2*
x)]

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Sympy [A]  time = 2.51181, size = 42, normalized size = 0.79 \begin{align*} - \frac{\sqrt{a} \sqrt{1 + \frac{b}{a x^{2}}}}{2 b x} + \frac{a \operatorname{asinh}{\left (\frac{\sqrt{b}}{\sqrt{a} x} \right )}}{2 b^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**2)**(1/2)/x**4,x)

[Out]

-sqrt(a)*sqrt(1 + b/(a*x**2))/(2*b*x) + a*asinh(sqrt(b)/(sqrt(a)*x))/(2*b**(3/2))

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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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(1/2)/x^4,x, algorithm="giac")

[Out]

integrate(1/(sqrt(a + b/x^2)*x^4), x)