\(\int \frac {1}{\sqrt {a+\frac {b}{x}} x} \, dx\) [1726]

Optimal result

Integrand size = 15, antiderivative size = 25 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {272, 65, 214} \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[In]

Int[1/(Sqrt[a + b/x]*x),x]

[Out]

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

Rule 65

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 214

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

Rule 272

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {\frac {b+a x}{x}}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[In]

Integrate[1/(Sqrt[a + b/x]*x),x]

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(118\) vs. \(2(19)=38\).

Time = 0.04 (sec) , antiderivative size = 119, normalized size of antiderivative = 4.76

method	result	size
default	\(\frac {\sqrt {\frac {a x +b}{x}}\, x \left (2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+b \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right )-2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+b \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right )\right )}{2 \sqrt {x \left (a x +b \right )}\, b \sqrt {a}}\)	\(119\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 60, normalized size of antiderivative = 2.40 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x} \, dx=\left [\frac {\log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right )}{\sqrt {a}}, -\frac {2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right )}{a}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.55 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x} \, dx=\frac {2 \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{\sqrt {a}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x} \, dx=-\frac {\log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{\sqrt {a}} \]

[In]

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

[Out]

-log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/sqrt(a)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (19) = 38\).

Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.88 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x} \, dx=\frac {\log \left ({\left | b \right |}\right ) \mathrm {sgn}\left (x\right )}{\sqrt {a}} - \frac {\log \left ({\left | 2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} + b \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.06 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\sqrt {a+\frac {b}{x}} x} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[In]

int(1/(x*(a + b/x)^(1/2)),x)

[Out]

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