\(\int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^5} \, dx\) [1919]

Optimal result

Integrand size = 15, antiderivative size = 35 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^5} \, dx=\frac {a \sqrt {a+\frac {b}{x^2}}}{b^2}-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{3 b^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, 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} \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^5} \, dx=\frac {a \sqrt {a+\frac {b}{x^2}}}{b^2}-\frac {\left (a+\frac {b}{x^2}\right )^{3/2}}{3 b^2} \]

[In]

Int[1/(Sqrt[a + b/x^2]*x^5),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^5} \, dx=\frac {\sqrt {a+\frac {b}{x^2}} \left (-b+2 a x^2\right )}{3 b^2 x^2} \]

[In]

Integrate[1/(Sqrt[a + b/x^2]*x^5),x]

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.03

method	result	size
trager	\(\frac {\left (2 a \,x^{2}-b \right ) \sqrt {-\frac {-a \,x^{2}-b}{x^{2}}}}{3 x^{2} b^{2}}\)	\(36\)
gosper	\(\frac {\left (a \,x^{2}+b \right ) \left (2 a \,x^{2}-b \right )}{3 x^{4} b^{2} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}\)	\(39\)
default	\(\frac {\left (a \,x^{2}+b \right ) \left (2 a \,x^{2}-b \right )}{3 x^{4} b^{2} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}\)	\(39\)
risch	\(\frac {\left (a \,x^{2}+b \right ) \left (2 a \,x^{2}-b \right )}{3 x^{4} b^{2} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}}\)	\(39\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^5} \, dx=\frac {{\left (2 \, a x^{2} - b\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{3 \, b^{2} x^{2}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 231 vs. \(2 (29) = 58\).

Time = 0.77 (sec) , antiderivative size = 231, normalized size of antiderivative = 6.60 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^5} \, dx=\frac {2 a^{\frac {7}{2}} b^{\frac {3}{2}} x^{4} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{\frac {5}{2}} b^{3} x^{5} + 3 a^{\frac {3}{2}} b^{4} x^{3}} + \frac {a^{\frac {5}{2}} b^{\frac {5}{2}} x^{2} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{\frac {5}{2}} b^{3} x^{5} + 3 a^{\frac {3}{2}} b^{4} x^{3}} - \frac {a^{\frac {3}{2}} b^{\frac {7}{2}} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{\frac {5}{2}} b^{3} x^{5} + 3 a^{\frac {3}{2}} b^{4} x^{3}} - \frac {2 a^{4} b x^{5}}{3 a^{\frac {5}{2}} b^{3} x^{5} + 3 a^{\frac {3}{2}} b^{4} x^{3}} - \frac {2 a^{3} b^{2} x^{3}}{3 a^{\frac {5}{2}} b^{3} x^{5} + 3 a^{\frac {3}{2}} b^{4} x^{3}} \]

[In]

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

[Out]

2*a**(7/2)*b**(3/2)*x**4*sqrt(a*x**2/b + 1)/(3*a**(5/2)*b**3*x**5 + 3*a**(3/2)*b**4*x**3) + a**(5/2)*b**(5/2)*
x**2*sqrt(a*x**2/b + 1)/(3*a**(5/2)*b**3*x**5 + 3*a**(3/2)*b**4*x**3) - a**(3/2)*b**(7/2)*sqrt(a*x**2/b + 1)/(
3*a**(5/2)*b**3*x**5 + 3*a**(3/2)*b**4*x**3) - 2*a**4*b*x**5/(3*a**(5/2)*b**3*x**5 + 3*a**(3/2)*b**4*x**3) - 2
*a**3*b**2*x**3/(3*a**(5/2)*b**3*x**5 + 3*a**(3/2)*b**4*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^5} \, dx=-\frac {{\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}}}{3 \, b^{2}} + \frac {\sqrt {a + \frac {b}{x^{2}}} a}{b^{2}} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (29) = 58\).

Time = 0.31 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.69 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^5} \, dx=\frac {4 \, {\left (3 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b}\right )}^{2} - b\right )} a^{\frac {3}{2}}}{3 \, {\left ({\left (\sqrt {a} x - \sqrt {a x^{2} + b}\right )}^{2} - b\right )}^{3} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.71 \[ \int \frac {1}{\sqrt {a+\frac {b}{x^2}} x^5} \, dx=-\frac {\sqrt {a+\frac {b}{x^2}}\,\left (b-2\,a\,x^2\right )}{3\,b^2\,x^2} \]

[In]

int(1/(x^5*(a + b/x^2)^(1/2)),x)

[Out]

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