\(\int \frac {1}{(a+\frac {b}{x})^{5/2} x^5} \, dx\) [1749]

Optimal result

Integrand size = 15, antiderivative size = 76 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^5} \, dx=-\frac {2 a^3}{3 b^4 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {6 a^2}{b^4 \sqrt {a+\frac {b}{x}}}+\frac {6 a \sqrt {a+\frac {b}{x}}}{b^4}-\frac {2 \left (a+\frac {b}{x}\right )^{3/2}}{3 b^4} \]

[Out]

-2/3*a^3/b^4/(a+b/x)^(3/2)-2/3*(a+b/x)^(3/2)/b^4+6*a^2/b^4/(a+b/x)^(1/2)+6*a*(a+b/x)^(1/2)/b^4

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^5} \, dx=-\frac {2 a^3}{3 b^4 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {6 a^2}{b^4 \sqrt {a+\frac {b}{x}}}+\frac {6 a \sqrt {a+\frac {b}{x}}}{b^4}-\frac {2 \left (a+\frac {b}{x}\right )^{3/2}}{3 b^4} \]

[In]

Int[1/((a + b/x)^(5/2)*x^5),x]

[Out]

(-2*a^3)/(3*b^4*(a + b/x)^(3/2)) + (6*a^2)/(b^4*Sqrt[a + b/x]) + (6*a*Sqrt[a + b/x])/b^4 - (2*(a + b/x)^(3/2))
/(3*b^4)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 {x^3}{(a+b x)^{5/2}} \, dx,x,\frac {1}{x}\right ) \\ & = -\text {Subst}\left (\int \left (-\frac {a^3}{b^3 (a+b x)^{5/2}}+\frac {3 a^2}{b^3 (a+b x)^{3/2}}-\frac {3 a}{b^3 \sqrt {a+b x}}+\frac {\sqrt {a+b x}}{b^3}\right ) \, dx,x,\frac {1}{x}\right ) \\ & = -\frac {2 a^3}{3 b^4 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {6 a^2}{b^4 \sqrt {a+\frac {b}{x}}}+\frac {6 a \sqrt {a+\frac {b}{x}}}{b^4}-\frac {2 \left (a+\frac {b}{x}\right )^{3/2}}{3 b^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^5} \, dx=\frac {2 \sqrt {\frac {b+a x}{x}} \left (-b^3+6 a b^2 x+24 a^2 b x^2+16 a^3 x^3\right )}{3 b^4 x (b+a x)^2} \]

[In]

Integrate[1/((a + b/x)^(5/2)*x^5),x]

[Out]

(2*Sqrt[(b + a*x)/x]*(-b^3 + 6*a*b^2*x + 24*a^2*b*x^2 + 16*a^3*x^3))/(3*b^4*x*(b + a*x)^2)

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.72

method	result	size
gosper	\(\frac {2 \left (a x +b \right ) \left (16 a^{3} x^{3}+24 a^{2} b \,x^{2}+6 a \,b^{2} x -b^{3}\right )}{3 x^{4} b^{4} \left (\frac {a x +b}{x}\right )^{\frac {5}{2}}}\)	\(55\)
trager	\(\frac {2 \left (2 a x +b \right ) \left (8 a^{2} x^{2}+8 a b x -b^{2}\right ) \sqrt {-\frac {-a x -b}{x}}}{3 x \,b^{4} \left (a x +b \right )^{2}}\)	\(56\)
risch	\(\frac {2 \left (a x +b \right ) \left (8 a x -b \right )}{3 b^{4} x^{2} \sqrt {\frac {a x +b}{x}}}+\frac {2 a^{2} \left (8 a x +9 b \right )}{3 \left (a x +b \right ) b^{4} \sqrt {\frac {a x +b}{x}}}\)	\(68\)
default	\(\frac {2 \sqrt {\frac {a x +b}{x}}\, \left (9 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{4} x^{4}-9 \left (x \left (a x +b \right )\right )^{\frac {3}{2}} a^{4} x^{4}+26 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{3} b \,x^{3}-10 \left (x \left (a x +b \right )\right )^{\frac {3}{2}} a^{3} b \,x^{3}+24 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{2} b^{2} x^{2}+6 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a \,b^{3} x -\left (a \,x^{2}+b x \right )^{\frac {3}{2}} b^{4}\right )}{3 x^{2} \sqrt {x \left (a x +b \right )}\, b^{5} \left (a x +b \right )^{3}}\)	\(167\)

[In]

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

[Out]

2/3*(a*x+b)*(16*a^3*x^3+24*a^2*b*x^2+6*a*b^2*x-b^3)/x^4/b^4/((a*x+b)/x)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^5} \, dx=\frac {2 \, {\left (16 \, a^{3} x^{3} + 24 \, a^{2} b x^{2} + 6 \, a b^{2} x - b^{3}\right )} \sqrt {\frac {a x + b}{x}}}{3 \, {\left (a^{2} b^{4} x^{3} + 2 \, a b^{5} x^{2} + b^{6} x\right )}} \]

[In]

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

[Out]

2/3*(16*a^3*x^3 + 24*a^2*b*x^2 + 6*a*b^2*x - b^3)*sqrt((a*x + b)/x)/(a^2*b^4*x^3 + 2*a*b^5*x^2 + b^6*x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 187 vs. \(2 (65) = 130\).

Time = 0.67 (sec) , antiderivative size = 187, normalized size of antiderivative = 2.46 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^5} \, dx=\begin {cases} \frac {32 a^{3} x^{3}}{3 a b^{4} x^{3} \sqrt {a + \frac {b}{x}} + 3 b^{5} x^{2} \sqrt {a + \frac {b}{x}}} + \frac {48 a^{2} b x^{2}}{3 a b^{4} x^{3} \sqrt {a + \frac {b}{x}} + 3 b^{5} x^{2} \sqrt {a + \frac {b}{x}}} + \frac {12 a b^{2} x}{3 a b^{4} x^{3} \sqrt {a + \frac {b}{x}} + 3 b^{5} x^{2} \sqrt {a + \frac {b}{x}}} - \frac {2 b^{3}}{3 a b^{4} x^{3} \sqrt {a + \frac {b}{x}} + 3 b^{5} x^{2} \sqrt {a + \frac {b}{x}}} & \text {for}\: b \neq 0 \\- \frac {1}{4 a^{\frac {5}{2}} x^{4}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((32*a**3*x**3/(3*a*b**4*x**3*sqrt(a + b/x) + 3*b**5*x**2*sqrt(a + b/x)) + 48*a**2*b*x**2/(3*a*b**4*x
**3*sqrt(a + b/x) + 3*b**5*x**2*sqrt(a + b/x)) + 12*a*b**2*x/(3*a*b**4*x**3*sqrt(a + b/x) + 3*b**5*x**2*sqrt(a
+ b/x)) - 2*b**3/(3*a*b**4*x**3*sqrt(a + b/x) + 3*b**5*x**2*sqrt(a + b/x)), Ne(b, 0)), (-1/(4*a**(5/2)*x**4),
True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^5} \, dx=-\frac {2 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}}}{3 \, b^{4}} + \frac {6 \, \sqrt {a + \frac {b}{x}} a}{b^{4}} + \frac {6 \, a^{2}}{\sqrt {a + \frac {b}{x}} b^{4}} - \frac {2 \, a^{3}}{3 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b^{4}} \]

[In]

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

[Out]

-2/3*(a + b/x)^(3/2)/b^4 + 6*sqrt(a + b/x)*a/b^4 + 6*a^2/(sqrt(a + b/x)*b^4) - 2/3*a^3/((a + b/x)^(3/2)*b^4)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^5} \, dx=\frac {2 \, {\left (2 \, {\left (4 \, x {\left (\frac {2 \, a^{3} x}{b^{4} \mathrm {sgn}\left (x\right )} + \frac {3 \, a^{2}}{b^{3} \mathrm {sgn}\left (x\right )}\right )} + \frac {3 \, a}{b^{2} \mathrm {sgn}\left (x\right )}\right )} x - \frac {1}{b \mathrm {sgn}\left (x\right )}\right )}}{3 \, {\left (a x^{2} + b x\right )}^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.66 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.71 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^5} \, dx=\frac {2\,\sqrt {a+\frac {b}{x}}\,\left (16\,a^3\,x^3+24\,a^2\,b\,x^2+6\,a\,b^2\,x-b^3\right )}{3\,b^4\,x\,{\left (b+a\,x\right )}^2} \]

[In]

int(1/(x^5*(a + b/x)^(5/2)),x)

[Out]

(2*(a + b/x)^(1/2)*(16*a^3*x^3 - b^3 + 24*a^2*b*x^2 + 6*a*b^2*x))/(3*b^4*x*(b + a*x)^2)