3.1952 \(\int \frac {1}{(a+\frac {b}{x^2})^{5/2}} \, dx\)

Optimal. Leaf size=58 \[ \frac {8 x \sqrt {a+\frac {b}{x^2}}}{3 a^3}-\frac {4 x}{3 a^2 \sqrt {a+\frac {b}{x^2}}}-\frac {x}{3 a \left (a+\frac {b}{x^2}\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {192, 191} \[ \frac {8 x \sqrt {a+\frac {b}{x^2}}}{3 a^3}-\frac {4 x}{3 a^2 \sqrt {a+\frac {b}{x^2}}}-\frac {x}{3 a \left (a+\frac {b}{x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^(-5/2),x]

[Out]

-x/(3*a*(a + b/x^2)^(3/2)) - (4*x)/(3*a^2*Sqrt[a + b/x^2]) + (8*Sqrt[a + b/x^2]*x)/(3*a^3)

Rule 191

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

Rule 192

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

Rubi steps

\begin {align*} \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx &=-\frac {x}{3 a \left (a+\frac {b}{x^2}\right )^{3/2}}+\frac {4 \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx}{3 a}\\ &=-\frac {x}{3 a \left (a+\frac {b}{x^2}\right )^{3/2}}-\frac {4 x}{3 a^2 \sqrt {a+\frac {b}{x^2}}}+\frac {8 \int \frac {1}{\sqrt {a+\frac {b}{x^2}}} \, dx}{3 a^2}\\ &=-\frac {x}{3 a \left (a+\frac {b}{x^2}\right )^{3/2}}-\frac {4 x}{3 a^2 \sqrt {a+\frac {b}{x^2}}}+\frac {8 \sqrt {a+\frac {b}{x^2}} x}{3 a^3}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 51, normalized size = 0.88 \[ \frac {3 a^2 x^4+12 a b x^2+8 b^2}{3 a^3 x \sqrt {a+\frac {b}{x^2}} \left (a x^2+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^(-5/2),x]

[Out]

(8*b^2 + 12*a*b*x^2 + 3*a^2*x^4)/(3*a^3*Sqrt[a + b/x^2]*x*(b + a*x^2))

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fricas [A]  time = 0.64, size = 63, normalized size = 1.09 \[ \frac {{\left (3 \, a^{2} x^{5} + 12 \, a b x^{3} + 8 \, b^{2} x\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{3 \, {\left (a^{5} x^{4} + 2 \, a^{4} b x^{2} + a^{3} b^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*a^2*x^5 + 12*a*b*x^3 + 8*b^2*x)*sqrt((a*x^2 + b)/x^2)/(a^5*x^4 + 2*a^4*b*x^2 + a^3*b^2)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is
real):Check [sign(t_nostep),sign(t_nostep+sqrt(b)/a*sign(t_nostep))]sym2poly/r2sym(const gen & e,const index_
m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.01, size = 50, normalized size = 0.86 \[ \frac {\left (a \,x^{2}+b \right ) \left (3 a^{2} x^{4}+12 a b \,x^{2}+8 b^{2}\right )}{3 \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}} a^{3} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(a*x^2+b)*(3*a^2*x^4+12*a*b*x^2+8*b^2)/a^3/x^5/((a*x^2+b)/x^2)^(5/2)

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maxima [A]  time = 0.81, size = 51, normalized size = 0.88 \[ \frac {\sqrt {a + \frac {b}{x^{2}}} x}{a^{3}} + \frac {6 \, {\left (a + \frac {b}{x^{2}}\right )} b x^{2} - b^{2}}{3 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.33, size = 43, normalized size = 0.74 \[ \frac {x\,{\left (\frac {a\,x^2}{b}+1\right )}^{5/2}\,\sqrt {x^{10}}\,{{}}_2{\mathrm {F}}_1\left (\frac {5}{2},3;\ 4;\ -\frac {a\,x^2}{b}\right )}{6\,{\left (a\,x^2+b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/x^2)^(5/2),x)

[Out]

(x*((a*x^2)/b + 1)^(5/2)*(x^10)^(1/2)*hypergeom([5/2, 3], 4, -(a*x^2)/b))/(6*(b + a*x^2)^(5/2))

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sympy [B]  time = 1.74, size = 163, normalized size = 2.81 \[ \frac {3 a^{2} b^{\frac {9}{2}} x^{4} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{5} b^{4} x^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6}} + \frac {12 a b^{\frac {11}{2}} x^{2} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{5} b^{4} x^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6}} + \frac {8 b^{\frac {13}{2}} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{5} b^{4} x^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*a**2*b**(9/2)*x**4*sqrt(a*x**2/b + 1)/(3*a**5*b**4*x**4 + 6*a**4*b**5*x**2 + 3*a**3*b**6) + 12*a*b**(11/2)*x
**2*sqrt(a*x**2/b + 1)/(3*a**5*b**4*x**4 + 6*a**4*b**5*x**2 + 3*a**3*b**6) + 8*b**(13/2)*sqrt(a*x**2/b + 1)/(3
*a**5*b**4*x**4 + 6*a**4*b**5*x**2 + 3*a**3*b**6)

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