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]

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

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Rubi [A]  time = 0.02, antiderivative size = 38, normalized size of antiderivative = 1.00, 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 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]

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]]

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*}

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Mathematica [A]  time = 0.07, 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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fricas [A]  time = 0.82, size = 109, normalized size = 2.87 \[ \left [\frac {1}{2} \, \sqrt {a} \log \left (-2 \, a x^{2} - 2 \, \sqrt {a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}} - b\right ) - \sqrt {\frac {a x^{2} + b}{x^{2}}}, -\sqrt {-a} \arctan \left (\frac {\sqrt {-a} x^{2} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) - \sqrt {\frac {a x^{2} + b}{x^{2}}}\right ] \]

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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giac [B]  time = 0.22, size = 61, normalized size = 1.61 \[ -\frac {1}{2} \, \sqrt {a} \log \left ({\left (\sqrt {a} x - \sqrt {a x^{2} + b}\right )}^{2}\right ) \mathrm {sgn}\relax (x) + \frac {2 \, \sqrt {a} b \mathrm {sgn}\relax (x)}{{\left (\sqrt {a} x - \sqrt {a x^{2} + b}\right )}^{2} - b} \]

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)

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maple [B]  time = 0.01, size = 83, normalized size = 2.18 \[ -\frac {\sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, \left (-a b x \ln \left (\sqrt {a}\, x +\sqrt {a \,x^{2}+b}\right )-\sqrt {a \,x^{2}+b}\, a^{\frac {3}{2}} x^{2}+\left (a \,x^{2}+b \right )^{\frac {3}{2}} \sqrt {a}\right )}{\sqrt {a \,x^{2}+b}\, \sqrt {a}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.93, size = 49, normalized size = 1.29 \[ -\frac {1}{2} \, \sqrt {a} \log \left (\frac {\sqrt {a + \frac {b}{x^{2}}} - \sqrt {a}}{\sqrt {a + \frac {b}{x^{2}}} + \sqrt {a}}\right ) - \sqrt {a + \frac {b}{x^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.30, size = 30, normalized size = 0.79 \[ \sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x^2}}}{\sqrt {a}}\right )-\sqrt {a+\frac {b}{x^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^2)^(1/2)/x,x)

[Out]

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

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sympy [A]  time = 1.59, size = 56, normalized size = 1.47 \[ \sqrt {a} \operatorname {asinh}{\left (\frac {\sqrt {a} x}{\sqrt {b}} \right )} - \frac {a x}{\sqrt {b} \sqrt {\frac {a x^{2}}{b} + 1}} - \frac {\sqrt {b}}{x \sqrt {\frac {a x^{2}}{b} + 1}} \]

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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