\(\int \frac {1}{(a+\frac {b}{x})^{5/2} x^4} \, dx\) [1748]

Optimal result

Integrand size = 15, antiderivative size = 55 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4} \, dx=\frac {2 a^2}{3 b^3 \left (a+\frac {b}{x}\right )^{3/2}}-\frac {4 a}{b^3 \sqrt {a+\frac {b}{x}}}-\frac {2 \sqrt {a+\frac {b}{x}}}{b^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, 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} \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4} \, dx=\frac {2 a^2}{3 b^3 \left (a+\frac {b}{x}\right )^{3/2}}-\frac {4 a}{b^3 \sqrt {a+\frac {b}{x}}}-\frac {2 \sqrt {a+\frac {b}{x}}}{b^3} \]

[In]

Int[1/((a + b/x)^(5/2)*x^4),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4} \, dx=-\frac {2 \sqrt {\frac {b+a x}{x}} \left (3 b^2+12 a b x+8 a^2 x^2\right )}{3 b^3 (b+a x)^2} \]

[In]

Integrate[1/((a + b/x)^(5/2)*x^4),x]

[Out]

(-2*Sqrt[(b + a*x)/x]*(3*b^2 + 12*a*b*x + 8*a^2*x^2))/(3*b^3*(b + a*x)^2)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.80

method	result	size
gosper	\(-\frac {2 \left (a x +b \right ) \left (8 a^{2} x^{2}+12 a b x +3 b^{2}\right )}{3 x^{3} b^{3} \left (\frac {a x +b}{x}\right )^{\frac {5}{2}}}\)	\(44\)
trager	\(-\frac {2 \left (8 a^{2} x^{2}+12 a b x +3 b^{2}\right ) \sqrt {-\frac {-a x -b}{x}}}{3 b^{3} \left (a x +b \right )^{2}}\)	\(47\)
risch	\(-\frac {2 \left (a x +b \right )}{b^{3} x \sqrt {\frac {a x +b}{x}}}-\frac {2 a \left (5 a x +6 b \right )}{3 \left (a x +b \right ) b^{3} \sqrt {\frac {a x +b}{x}}}\)	\(58\)
default	\(-\frac {\sqrt {\frac {a x +b}{x}}\, \left (6 \sqrt {a \,x^{2}+b x}\, a^{\frac {9}{2}} x^{5}+3 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{4} b \,x^{5}+6 a^{\frac {9}{2}} \sqrt {x \left (a x +b \right )}\, x^{5}-3 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{4} b \,x^{5}+12 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {7}{2}} x^{3}+18 \sqrt {a \,x^{2}+b x}\, a^{\frac {7}{2}} b \,x^{4}+9 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{3} b^{2} x^{4}-24 a^{\frac {7}{2}} \left (x \left (a x +b \right )\right )^{\frac {3}{2}} x^{3}+18 a^{\frac {7}{2}} \sqrt {x \left (a x +b \right )}\, b \,x^{4}-9 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{3} b^{2} x^{4}+36 a^{\frac {5}{2}} \left (a \,x^{2}+b x \right )^{\frac {3}{2}} b \,x^{2}+18 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} b^{2} x^{3}+9 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{2} b^{3} x^{3}-28 a^{\frac {5}{2}} \left (x \left (a x +b \right )\right )^{\frac {3}{2}} b \,x^{2}+18 a^{\frac {5}{2}} \sqrt {x \left (a x +b \right )}\, b^{2} x^{3}-9 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a^{2} b^{3} x^{3}+36 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} a^{\frac {3}{2}} b^{2} x +6 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b^{3} x^{2}+3 \ln \left (\frac {2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{4} x^{2}+6 a^{\frac {3}{2}} \sqrt {x \left (a x +b \right )}\, b^{3} x^{2}-3 \ln \left (\frac {2 \sqrt {x \left (a x +b \right )}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) a \,b^{4} x^{2}+12 \left (a \,x^{2}+b x \right )^{\frac {3}{2}} \sqrt {a}\, b^{3}\right )}{6 x \sqrt {x \left (a x +b \right )}\, b^{4} \sqrt {a}\, \left (a x +b \right )^{3}}\)	\(607\)

[In]

int(1/(a+b/x)^(5/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

-2/3*(a*x+b)*(8*a^2*x^2+12*a*b*x+3*b^2)/x^3/b^3/((a*x+b)/x)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4} \, dx=-\frac {2 \, {\left (8 \, a^{2} x^{2} + 12 \, a b x + 3 \, b^{2}\right )} \sqrt {\frac {a x + b}{x}}}{3 \, {\left (a^{2} b^{3} x^{2} + 2 \, a b^{4} x + b^{5}\right )}} \]

[In]

integrate(1/(a+b/x)^(5/2)/x^4,x, algorithm="fricas")

[Out]

-2/3*(8*a^2*x^2 + 12*a*b*x + 3*b^2)*sqrt((a*x + b)/x)/(a^2*b^3*x^2 + 2*a*b^4*x + b^5)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (46) = 92\).

Time = 0.61 (sec) , antiderivative size = 136, normalized size of antiderivative = 2.47 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4} \, dx=\begin {cases} - \frac {16 a^{2} x^{2}}{3 a b^{3} x^{2} \sqrt {a + \frac {b}{x}} + 3 b^{4} x \sqrt {a + \frac {b}{x}}} - \frac {24 a b x}{3 a b^{3} x^{2} \sqrt {a + \frac {b}{x}} + 3 b^{4} x \sqrt {a + \frac {b}{x}}} - \frac {6 b^{2}}{3 a b^{3} x^{2} \sqrt {a + \frac {b}{x}} + 3 b^{4} x \sqrt {a + \frac {b}{x}}} & \text {for}\: b \neq 0 \\- \frac {1}{3 a^{\frac {5}{2}} x^{3}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(a+b/x)**(5/2)/x**4,x)

[Out]

Piecewise((-16*a**2*x**2/(3*a*b**3*x**2*sqrt(a + b/x) + 3*b**4*x*sqrt(a + b/x)) - 24*a*b*x/(3*a*b**3*x**2*sqrt
(a + b/x) + 3*b**4*x*sqrt(a + b/x)) - 6*b**2/(3*a*b**3*x**2*sqrt(a + b/x) + 3*b**4*x*sqrt(a + b/x)), Ne(b, 0))
, (-1/(3*a**(5/2)*x**3), True))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4} \, dx=-\frac {2 \, \sqrt {a + \frac {b}{x}}}{b^{3}} - \frac {4 \, a}{\sqrt {a + \frac {b}{x}} b^{3}} + \frac {2 \, a^{2}}{3 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b^{3}} \]

[In]

integrate(1/(a+b/x)^(5/2)/x^4,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4} \, dx=\int { \frac {1}{{\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} x^{4}} \,d x } \]

[In]

integrate(1/(a+b/x)^(5/2)/x^4,x, algorithm="giac")

[Out]

integrate(1/((a + b/x)^(5/2)*x^4), x)

Mupad [B] (verification not implemented)

Time = 5.76 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.73 \[ \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^4} \, dx=-\frac {2\,\sqrt {a+\frac {b}{x}}\,\left (8\,a^2\,x^2+12\,a\,b\,x+3\,b^2\right )}{3\,b^3\,{\left (b+a\,x\right )}^2} \]

[In]

int(1/(x^4*(a + b/x)^(5/2)),x)

[Out]

-(2*(a + b/x)^(1/2)*(3*b^2 + 8*a^2*x^2 + 12*a*b*x))/(3*b^3*(b + a*x)^2)