\(\int \frac {1}{(a+\frac {b}{x})^{5/2}} \, dx\) [262]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [A] (verification not implemented)
   Sympy [B] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 11, antiderivative size = 79 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx=\frac {5 b}{3 a^2 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {5 b}{a^3 \sqrt {a+\frac {b}{x}}}+\frac {x}{a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {5 b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{7/2}} \]

[Out]

5/3*b/a^2/(a+b/x)^(3/2)+x/a/(a+b/x)^(3/2)-5*b*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(7/2)+5*b/a^3/(a+b/x)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {248, 44, 53, 65, 214} \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx=-\frac {5 b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{7/2}}+\frac {5 b}{a^3 \sqrt {a+\frac {b}{x}}}+\frac {5 b}{3 a^2 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {x}{a \left (a+\frac {b}{x}\right )^{3/2}} \]

[In]

Int[(a + b/x)^(-5/2),x]

[Out]

(5*b)/(3*a^2*(a + b/x)^(3/2)) + (5*b)/(a^3*Sqrt[a + b/x]) + x/(a*(a + b/x)^(3/2)) - (5*b*ArcTanh[Sqrt[a + b/x]
/Sqrt[a]])/a^(7/2)

Rule 44

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

Rule 53

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

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 248

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^2, x], x, 1/x] /; FreeQ[{a, b, p},
x] && ILtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {1}{x^2 (a+b x)^{5/2}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {x}{a \left (a+\frac {b}{x}\right )^{3/2}}+\frac {(5 b) \text {Subst}\left (\int \frac {1}{x (a+b x)^{5/2}} \, dx,x,\frac {1}{x}\right )}{2 a} \\ & = \frac {5 b}{3 a^2 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {x}{a \left (a+\frac {b}{x}\right )^{3/2}}+\frac {(5 b) \text {Subst}\left (\int \frac {1}{x (a+b x)^{3/2}} \, dx,x,\frac {1}{x}\right )}{2 a^2} \\ & = \frac {5 b}{3 a^2 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {5 b}{a^3 \sqrt {a+\frac {b}{x}}}+\frac {x}{a \left (a+\frac {b}{x}\right )^{3/2}}+\frac {(5 b) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{2 a^3} \\ & = \frac {5 b}{3 a^2 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {5 b}{a^3 \sqrt {a+\frac {b}{x}}}+\frac {x}{a \left (a+\frac {b}{x}\right )^{3/2}}+\frac {5 \text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )}{a^3} \\ & = \frac {5 b}{3 a^2 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {5 b}{a^3 \sqrt {a+\frac {b}{x}}}+\frac {x}{a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {5 b \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx=\frac {\sqrt {a+\frac {b}{x}} x \left (15 b^2+20 a b x+3 a^2 x^2\right )}{3 a^3 (b+a x)^2}-\frac {5 b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{a^{7/2}} \]

[In]

Integrate[(a + b/x)^(-5/2),x]

[Out]

(Sqrt[a + b/x]*x*(15*b^2 + 20*a*b*x + 3*a^2*x^2))/(3*a^3*(b + a*x)^2) - (5*b*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/a
^(7/2)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(156\) vs. \(2(65)=130\).

Time = 0.16 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.99

method result size
risch \(\frac {a x +b}{a^{3} \sqrt {\frac {a x +b}{x}}}+\frac {\left (-\frac {5 b \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x}\right )}{2 a^{\frac {7}{2}}}-\frac {2 b^{2} \sqrt {a \left (x +\frac {b}{a}\right )^{2}-b \left (x +\frac {b}{a}\right )}}{3 a^{5} \left (x +\frac {b}{a}\right )^{2}}+\frac {14 b \sqrt {a \left (x +\frac {b}{a}\right )^{2}-b \left (x +\frac {b}{a}\right )}}{3 a^{4} \left (x +\frac {b}{a}\right )}\right ) \sqrt {x \left (a x +b \right )}}{x \sqrt {\frac {a x +b}{x}}}\) \(157\)
default \(\frac {\sqrt {\frac {a x +b}{x}}\, x \left (30 \sqrt {x \left (a x +b \right )}\, a^{\frac {7}{2}} x^{3}-24 \left (x \left (a x +b \right )\right )^{\frac {3}{2}} a^{\frac {5}{2}} x +90 \sqrt {x \left (a x +b \right )}\, a^{\frac {5}{2}} b \,x^{2}-15 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{3} b \,x^{3}-20 b \,a^{\frac {3}{2}} \left (x \left (a x +b \right )\right )^{\frac {3}{2}}+90 \sqrt {x \left (a x +b \right )}\, a^{\frac {3}{2}} b^{2} x -45 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{2} b^{2} x^{2}-45 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{3} x +30 \sqrt {x \left (a x +b \right )}\, \sqrt {a}\, b^{3}-15 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) b^{4}\right )}{6 a^{\frac {7}{2}} \sqrt {x \left (a x +b \right )}\, \left (a x +b \right )^{3}}\) \(271\)

[In]

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

[Out]

1/a^3*(a*x+b)/((a*x+b)/x)^(1/2)+(-5/2/a^(7/2)*b*ln((1/2*b+a*x)/a^(1/2)+(a*x^2+b*x)^(1/2))-2/3/a^5*b^2/(x+b/a)^
2*(a*(x+b/a)^2-b*(x+b/a))^(1/2)+14/3/a^4*b/(x+b/a)*(a*(x+b/a)^2-b*(x+b/a))^(1/2))/x/((a*x+b)/x)^(1/2)*(x*(a*x+
b))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 225, normalized size of antiderivative = 2.85 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx=\left [\frac {15 \, {\left (a^{2} b x^{2} + 2 \, a b^{2} x + b^{3}\right )} \sqrt {a} \log \left (2 \, a x - 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (3 \, a^{3} x^{3} + 20 \, a^{2} b x^{2} + 15 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{6 \, {\left (a^{6} x^{2} + 2 \, a^{5} b x + a^{4} b^{2}\right )}}, \frac {15 \, {\left (a^{2} b x^{2} + 2 \, a b^{2} x + b^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (3 \, a^{3} x^{3} + 20 \, a^{2} b x^{2} + 15 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{3 \, {\left (a^{6} x^{2} + 2 \, a^{5} b x + a^{4} b^{2}\right )}}\right ] \]

[In]

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

[Out]

[1/6*(15*(a^2*b*x^2 + 2*a*b^2*x + b^3)*sqrt(a)*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(3*a^3*x^3 +
 20*a^2*b*x^2 + 15*a*b^2*x)*sqrt((a*x + b)/x))/(a^6*x^2 + 2*a^5*b*x + a^4*b^2), 1/3*(15*(a^2*b*x^2 + 2*a*b^2*x
 + b^3)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (3*a^3*x^3 + 20*a^2*b*x^2 + 15*a*b^2*x)*sqrt((a*x + b)
/x))/(a^6*x^2 + 2*a^5*b*x + a^4*b^2)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 774 vs. \(2 (66) = 132\).

Time = 2.71 (sec) , antiderivative size = 774, normalized size of antiderivative = 9.80 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx=\frac {6 a^{17} x^{4} \sqrt {1 + \frac {b}{a x}}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} + \frac {46 a^{16} b x^{3} \sqrt {1 + \frac {b}{a x}}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} + \frac {15 a^{16} b x^{3} \log {\left (\frac {b}{a x} \right )}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} - \frac {30 a^{16} b x^{3} \log {\left (\sqrt {1 + \frac {b}{a x}} + 1 \right )}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} + \frac {70 a^{15} b^{2} x^{2} \sqrt {1 + \frac {b}{a x}}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} + \frac {45 a^{15} b^{2} x^{2} \log {\left (\frac {b}{a x} \right )}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} - \frac {90 a^{15} b^{2} x^{2} \log {\left (\sqrt {1 + \frac {b}{a x}} + 1 \right )}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} + \frac {30 a^{14} b^{3} x \sqrt {1 + \frac {b}{a x}}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} + \frac {45 a^{14} b^{3} x \log {\left (\frac {b}{a x} \right )}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} - \frac {90 a^{14} b^{3} x \log {\left (\sqrt {1 + \frac {b}{a x}} + 1 \right )}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} + \frac {15 a^{13} b^{4} \log {\left (\frac {b}{a x} \right )}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} - \frac {30 a^{13} b^{4} \log {\left (\sqrt {1 + \frac {b}{a x}} + 1 \right )}}{6 a^{\frac {39}{2}} x^{3} + 18 a^{\frac {37}{2}} b x^{2} + 18 a^{\frac {35}{2}} b^{2} x + 6 a^{\frac {33}{2}} b^{3}} \]

[In]

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

[Out]

6*a**17*x**4*sqrt(1 + b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**
3) + 46*a**16*b*x**3*sqrt(1 + b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(3
3/2)*b**3) + 15*a**16*b*x**3*log(b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a*
*(33/2)*b**3) - 30*a**16*b*x**3*log(sqrt(1 + b/(a*x)) + 1)/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35
/2)*b**2*x + 6*a**(33/2)*b**3) + 70*a**15*b**2*x**2*sqrt(1 + b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2
+ 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) + 45*a**15*b**2*x**2*log(b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b
*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) - 90*a**15*b**2*x**2*log(sqrt(1 + b/(a*x)) + 1)/(6*a**(39/2)*x
**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) + 30*a**14*b**3*x*sqrt(1 + b/(a*x))/(6*a**
(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) + 45*a**14*b**3*x*log(b/(a*x))/(6*
a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) - 90*a**14*b**3*x*log(sqrt(1 +
b/(a*x)) + 1)/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) + 15*a**13*b**
4*log(b/(a*x))/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3) - 30*a**13*b*
*4*log(sqrt(1 + b/(a*x)) + 1)/(6*a**(39/2)*x**3 + 18*a**(37/2)*b*x**2 + 18*a**(35/2)*b**2*x + 6*a**(33/2)*b**3
)

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.28 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx=\frac {15 \, {\left (a + \frac {b}{x}\right )}^{2} b - 10 \, {\left (a + \frac {b}{x}\right )} a b - 2 \, a^{2} b}{3 \, {\left ({\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} a^{3} - {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{4}\right )}} + \frac {5 \, b \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{2 \, a^{\frac {7}{2}}} \]

[In]

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

[Out]

1/3*(15*(a + b/x)^2*b - 10*(a + b/x)*a*b - 2*a^2*b)/((a + b/x)^(5/2)*a^3 - (a + b/x)^(3/2)*a^4) + 5/2*b*log((s
qrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^(7/2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (65) = 130\).

Time = 0.30 (sec) , antiderivative size = 171, normalized size of antiderivative = 2.16 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx=-\frac {{\left (15 \, b \log \left ({\left | b \right |}\right ) + 28 \, b\right )} \mathrm {sgn}\left (x\right )}{6 \, a^{\frac {7}{2}}} + \frac {5 \, b \log \left ({\left | 2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} + b \right |}\right )}{2 \, a^{\frac {7}{2}} \mathrm {sgn}\left (x\right )} + \frac {\sqrt {a x^{2} + b x}}{a^{3} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left (9 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )}^{2} a b^{2} + 15 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} b^{3} + 7 \, b^{4}\right )}}{3 \, {\left ({\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} + b\right )}^{3} a^{\frac {7}{2}} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

-1/6*(15*b*log(abs(b)) + 28*b)*sgn(x)/a^(7/2) + 5/2*b*log(abs(2*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) + b))/
(a^(7/2)*sgn(x)) + sqrt(a*x^2 + b*x)/(a^3*sgn(x)) + 2/3*(9*(sqrt(a)*x - sqrt(a*x^2 + b*x))^2*a*b^2 + 15*(sqrt(
a)*x - sqrt(a*x^2 + b*x))*sqrt(a)*b^3 + 7*b^4)/(((sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a) + b)^3*a^(7/2)*sgn(x)
)

Mupad [B] (verification not implemented)

Time = 5.87 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.43 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx=\frac {2\,x\,{\left (\frac {a\,x}{b}+1\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (\frac {5}{2},\frac {7}{2};\ \frac {9}{2};\ -\frac {a\,x}{b}\right )}{7\,{\left (a+\frac {b}{x}\right )}^{5/2}} \]

[In]

int(1/(a + b/x)^(5/2),x)

[Out]

(2*x*((a*x)/b + 1)^(5/2)*hypergeom([5/2, 7/2], 9/2, -(a*x)/b))/(7*(a + b/x)^(5/2))