3.1783 \(\int \frac {1}{\sqrt {a+\frac {b}{x}} x^{5/2}} \, dx\)

Optimal. Leaf size=52 \[ \frac {a \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{b^{3/2}}-\frac {\sqrt {a+\frac {b}{x}}}{b \sqrt {x}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {337, 321, 217, 206} \[ \frac {a \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{b^{3/2}}-\frac {\sqrt {a+\frac {b}{x}}}{b \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b/x]*x^(5/2)),x]

[Out]

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

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])

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 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 337

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, -Dist[k/c, Subst[
Int[(a + b/(c^n*x^(k*n)))^p/x^(k*(m + 1) + 1), x], x, 1/(c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && ILtQ[n,
0] && FractionQ[m]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a+\frac {b}{x}} x^{5/2}} \, dx &=-\left (2 \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )\right )\\ &=-\frac {\sqrt {a+\frac {b}{x}}}{b \sqrt {x}}+\frac {a \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{b}\\ &=-\frac {\sqrt {a+\frac {b}{x}}}{b \sqrt {x}}+\frac {a \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{b}\\ &=-\frac {\sqrt {a+\frac {b}{x}}}{b \sqrt {x}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 77, normalized size = 1.48 \[ \frac {a^{3/2} x^{3/2} \sqrt {\frac {b}{a x}+1} \sinh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}}\right )-\sqrt {b} (a x+b)}{b^{3/2} x^{3/2} \sqrt {a+\frac {b}{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b/x]*x^(5/2)),x]

[Out]

(-(Sqrt[b]*(b + a*x)) + a^(3/2)*Sqrt[1 + b/(a*x)]*x^(3/2)*ArcSinh[Sqrt[b]/(Sqrt[a]*Sqrt[x])])/(b^(3/2)*Sqrt[a
+ b/x]*x^(3/2))

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fricas [A]  time = 0.95, size = 121, normalized size = 2.33 \[ \left [\frac {a \sqrt {b} x \log \left (\frac {a x + 2 \, \sqrt {b} \sqrt {x} \sqrt {\frac {a x + b}{x}} + 2 \, b}{x}\right ) - 2 \, b \sqrt {x} \sqrt {\frac {a x + b}{x}}}{2 \, b^{2} x}, -\frac {a \sqrt {-b} x \arctan \left (\frac {\sqrt {-b} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{b}\right ) + b \sqrt {x} \sqrt {\frac {a x + b}{x}}}{b^{2} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(1/2)/x^(5/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.21, size = 47, normalized size = 0.90 \[ -\frac {\frac {a^{2} \arctan \left (\frac {\sqrt {a x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} + \frac {\sqrt {a x + b} a}{b x}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(1/2)/x^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 55, normalized size = 1.06 \[ -\frac {\sqrt {\frac {a x +b}{x}}\, \left (-a x \arctanh \left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right )+\sqrt {a x +b}\, \sqrt {b}\right )}{\sqrt {a x +b}\, b^{\frac {3}{2}} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^(1/2)/x^(5/2),x)

[Out]

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

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maxima [A]  time = 2.42, size = 80, normalized size = 1.54 \[ -\frac {\sqrt {a + \frac {b}{x}} a \sqrt {x}}{{\left (a + \frac {b}{x}\right )} b x - b^{2}} - \frac {a \log \left (\frac {\sqrt {a + \frac {b}{x}} \sqrt {x} - \sqrt {b}}{\sqrt {a + \frac {b}{x}} \sqrt {x} + \sqrt {b}}\right )}{2 \, b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^(1/2)/x^(5/2),x, algorithm="maxima")

[Out]

-sqrt(a + b/x)*a*sqrt(x)/((a + b/x)*b*x - b^2) - 1/2*a*log((sqrt(a + b/x)*sqrt(x) - sqrt(b))/(sqrt(a + b/x)*sq
rt(x) + sqrt(b)))/b^(3/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{x^{5/2}\,\sqrt {a+\frac {b}{x}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(a + b/x)^(1/2)),x)

[Out]

int(1/(x^(5/2)*(a + b/x)^(1/2)), x)

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sympy [A]  time = 8.69, size = 44, normalized size = 0.85 \[ - \frac {\sqrt {a} \sqrt {1 + \frac {b}{a x}}}{b \sqrt {x}} + \frac {a \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}} \right )}}{b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)**(1/2)/x**(5/2),x)

[Out]

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

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