3.1895 \(\int \frac{\sqrt{a+\frac{b}{x^2}}}{x} \, dx\)

Optimal. Leaf size=38 \[ \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )-\sqrt{a+\frac{b}{x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0227823, antiderivative size = 38, 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 = {266, 50, 63, 208} \[ \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )-\sqrt{a+\frac{b}{x^2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x^2]/x,x]

[Out]

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

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 208

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

Rubi steps

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

Mathematica [A]  time = 0.0620198, size = 69, normalized size = 1.82 \[ -\frac{\sqrt{a+\frac{b}{x^2}} \left (-\sqrt{a} \sqrt{b} x \sqrt{\frac{a x^2}{b}+1} \sinh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )+a x^2+b\right )}{a x^2+b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/x^2]/x,x]

[Out]

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

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Maple [B]  time = 0.005, size = 81, normalized size = 2.1 \begin{align*}{\frac{1}{b}\sqrt{{\frac{a{x}^{2}+b}{{x}^{2}}}} \left ({a}^{{\frac{3}{2}}}\sqrt{a{x}^{2}+b}{x}^{2}- \left ( a{x}^{2}+b \right ) ^{{\frac{3}{2}}}\sqrt{a}+\ln \left ( x\sqrt{a}+\sqrt{a{x}^{2}+b} \right ) xab \right ){\frac{1}{\sqrt{a{x}^{2}+b}}}{\frac{1}{\sqrt{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54115, size = 252, normalized size = 6.63 \begin{align*} \left [\frac{1}{2} \, \sqrt{a} \log \left (-2 \, a x^{2} - 2 \, \sqrt{a} x^{2} \sqrt{\frac{a x^{2} + b}{x^{2}}} - b\right ) - \sqrt{\frac{a x^{2} + b}{x^{2}}}, -\sqrt{-a} \arctan \left (\frac{\sqrt{-a} x^{2} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) - \sqrt{\frac{a x^{2} + b}{x^{2}}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.42436, size = 56, normalized size = 1.47 \begin{align*} \sqrt{a} \operatorname{asinh}{\left (\frac{\sqrt{a} x}{\sqrt{b}} \right )} - \frac{a x}{\sqrt{b} \sqrt{\frac{a x^{2}}{b} + 1}} - \frac{\sqrt{b}}{x \sqrt{\frac{a x^{2}}{b} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.18872, size = 82, normalized size = 2.16 \begin{align*} -\frac{1}{2} \, \sqrt{a} \log \left ({\left (\sqrt{a} x - \sqrt{a x^{2} + b}\right )}^{2}\right ) \mathrm{sgn}\left (x\right ) + \frac{2 \, \sqrt{a} b \mathrm{sgn}\left (x\right )}{{\left (\sqrt{a} x - \sqrt{a x^{2} + b}\right )}^{2} - b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*sqrt(a)*log((sqrt(a)*x - sqrt(a*x^2 + b))^2)*sgn(x) + 2*sqrt(a)*b*sgn(x)/((sqrt(a)*x - sqrt(a*x^2 + b))^2
- b)