3.1758 \(\int \frac {\sqrt {a+\frac {b}{x}}}{x^{5/2}} \, dx\)

Optimal. Leaf size=80 \[ \frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{4 b^{3/2}}-\frac {\sqrt {a+\frac {b}{x}}}{2 x^{3/2}}-\frac {a \sqrt {a+\frac {b}{x}}}{4 b \sqrt {x}} \]

[Out]

1/4*a^2*arctanh(b^(1/2)/(a+b/x)^(1/2)/x^(1/2))/b^(3/2)-1/2*(a+b/x)^(1/2)/x^(3/2)-1/4*a*(a+b/x)^(1/2)/b/x^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {337, 279, 321, 217, 206} \[ \frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {x} \sqrt {a+\frac {b}{x}}}\right )}{4 b^{3/2}}-\frac {\sqrt {a+\frac {b}{x}}}{2 x^{3/2}}-\frac {a \sqrt {a+\frac {b}{x}}}{4 b \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x]/x^(5/2),x]

[Out]

-Sqrt[a + b/x]/(2*x^(3/2)) - (a*Sqrt[a + b/x])/(4*b*Sqrt[x]) + (a^2*ArcTanh[Sqrt[b]/(Sqrt[a + b/x]*Sqrt[x])])/
(4*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 279

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

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 {\sqrt {a+\frac {b}{x}}}{x^{5/2}} \, dx &=-\left (2 \operatorname {Subst}\left (\int x^2 \sqrt {a+b x^2} \, dx,x,\frac {1}{\sqrt {x}}\right )\right )\\ &=-\frac {\sqrt {a+\frac {b}{x}}}{2 x^{3/2}}-\frac {1}{2} a \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )\\ &=-\frac {\sqrt {a+\frac {b}{x}}}{2 x^{3/2}}-\frac {a \sqrt {a+\frac {b}{x}}}{4 b \sqrt {x}}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{4 b}\\ &=-\frac {\sqrt {a+\frac {b}{x}}}{2 x^{3/2}}-\frac {a \sqrt {a+\frac {b}{x}}}{4 b \sqrt {x}}+\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{4 b}\\ &=-\frac {\sqrt {a+\frac {b}{x}}}{2 x^{3/2}}-\frac {a \sqrt {a+\frac {b}{x}}}{4 b \sqrt {x}}+\frac {a^2 \tanh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a+\frac {b}{x}} \sqrt {x}}\right )}{4 b^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.17, size = 77, normalized size = 0.96 \[ \frac {\sqrt {a+\frac {b}{x}} \left (\frac {a^{3/2} \sinh ^{-1}\left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}}\right )}{\sqrt {\frac {b}{a x}+1}}-\frac {\sqrt {b} (a x+2 b)}{x^{3/2}}\right )}{4 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/x]/x^(5/2),x]

[Out]

(Sqrt[a + b/x]*(-((Sqrt[b]*(2*b + a*x))/x^(3/2)) + (a^(3/2)*ArcSinh[Sqrt[b]/(Sqrt[a]*Sqrt[x])])/Sqrt[1 + b/(a*
x)]))/(4*b^(3/2))

________________________________________________________________________________________

fricas [A]  time = 0.88, size = 147, normalized size = 1.84 \[ \left [\frac {a^{2} \sqrt {b} x^{2} \log \left (\frac {a x + 2 \, \sqrt {b} \sqrt {x} \sqrt {\frac {a x + b}{x}} + 2 \, b}{x}\right ) - 2 \, {\left (a b x + 2 \, b^{2}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{8 \, b^{2} x^{2}}, -\frac {a^{2} \sqrt {-b} x^{2} \arctan \left (\frac {\sqrt {-b} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{b}\right ) + {\left (a b x + 2 \, b^{2}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{4 \, b^{2} x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(a^2*sqrt(b)*x^2*log((a*x + 2*sqrt(b)*sqrt(x)*sqrt((a*x + b)/x) + 2*b)/x) - 2*(a*b*x + 2*b^2)*sqrt(x)*sqr
t((a*x + b)/x))/(b^2*x^2), -1/4*(a^2*sqrt(-b)*x^2*arctan(sqrt(-b)*sqrt(x)*sqrt((a*x + b)/x)/b) + (a*b*x + 2*b^
2)*sqrt(x)*sqrt((a*x + b)/x))/(b^2*x^2)]

________________________________________________________________________________________

giac [A]  time = 0.29, size = 68, normalized size = 0.85 \[ -\frac {{\left (\frac {a^{3} \arctan \left (\frac {\sqrt {a x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b} + \frac {{\left (a x + b\right )}^{\frac {3}{2}} a^{3} + \sqrt {a x + b} a^{3} b}{a^{2} b x^{2}}\right )} \mathrm {sgn}\relax (x)}{4 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(a^3*arctan(sqrt(a*x + b)/sqrt(-b))/(sqrt(-b)*b) + ((a*x + b)^(3/2)*a^3 + sqrt(a*x + b)*a^3*b)/(a^2*b*x^2
))*sgn(x)/a

________________________________________________________________________________________

maple [A]  time = 0.02, size = 73, normalized size = 0.91 \[ -\frac {\sqrt {\frac {a x +b}{x}}\, \left (-a^{2} x^{2} \arctanh \left (\frac {\sqrt {a x +b}}{\sqrt {b}}\right )+\sqrt {a x +b}\, a \sqrt {b}\, x +2 \sqrt {a x +b}\, b^{\frac {3}{2}}\right )}{4 \sqrt {a x +b}\, b^{\frac {3}{2}} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((a*x+b)/x)^(1/2)*(-arctanh((a*x+b)^(1/2)/b^(1/2))*a^2*x^2+2*b^(3/2)*(a*x+b)^(1/2)+x*a*(a*x+b)^(1/2)*b^(1
/2))/x^(3/2)/b^(3/2)/(a*x+b)^(1/2)

________________________________________________________________________________________

maxima [B]  time = 2.30, size = 118, normalized size = 1.48 \[ -\frac {a^{2} \log \left (\frac {\sqrt {a + \frac {b}{x}} \sqrt {x} - \sqrt {b}}{\sqrt {a + \frac {b}{x}} \sqrt {x} + \sqrt {b}}\right )}{8 \, b^{\frac {3}{2}}} - \frac {{\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{2} x^{\frac {3}{2}} + \sqrt {a + \frac {b}{x}} a^{2} b \sqrt {x}}{4 \, {\left ({\left (a + \frac {b}{x}\right )}^{2} b x^{2} - 2 \, {\left (a + \frac {b}{x}\right )} b^{2} x + b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*a^2*log((sqrt(a + b/x)*sqrt(x) - sqrt(b))/(sqrt(a + b/x)*sqrt(x) + sqrt(b)))/b^(3/2) - 1/4*((a + b/x)^(3/
2)*a^2*x^(3/2) + sqrt(a + b/x)*a^2*b*sqrt(x))/((a + b/x)^2*b*x^2 - 2*(a + b/x)*b^2*x + b^3)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+\frac {b}{x}}}{x^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^(1/2)/x^(5/2),x)

[Out]

int((a + b/x)^(1/2)/x^(5/2), x)

________________________________________________________________________________________

sympy [A]  time = 8.94, size = 97, normalized size = 1.21 \[ - \frac {a^{\frac {3}{2}}}{4 b \sqrt {x} \sqrt {1 + \frac {b}{a x}}} - \frac {3 \sqrt {a}}{4 x^{\frac {3}{2}} \sqrt {1 + \frac {b}{a x}}} + \frac {a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b}}{\sqrt {a} \sqrt {x}} \right )}}{4 b^{\frac {3}{2}}} - \frac {b}{2 \sqrt {a} x^{\frac {5}{2}} \sqrt {1 + \frac {b}{a x}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(3/2)/(4*b*sqrt(x)*sqrt(1 + b/(a*x))) - 3*sqrt(a)/(4*x**(3/2)*sqrt(1 + b/(a*x))) + a**2*asinh(sqrt(b)/(sqr
t(a)*sqrt(x)))/(4*b**(3/2)) - b/(2*sqrt(a)*x**(5/2)*sqrt(1 + b/(a*x)))

________________________________________________________________________________________