3.20.47 \(\int \frac {1}{(a+\frac {b}{x^2})^{5/2} x^3} \, dx\) [1947]

Optimal. Leaf size=18 \[ \frac {1}{3 b \left (a+\frac {b}{x^2}\right )^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {267} \begin {gather*} \frac {1}{3 b \left (a+\frac {b}{x^2}\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^(5/2)*x^3),x]

[Out]

1/(3*b*(a + b/x^2)^(3/2))

Rule 267

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

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 28, normalized size = 1.56 \begin {gather*} \frac {b+a x^2}{3 b \left (a+\frac {b}{x^2}\right )^{5/2} x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^(5/2)*x^3),x]

[Out]

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

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Maple [A]
time = 0.03, size = 29, normalized size = 1.61

method	result	size
derivativedivides	\(\frac {1}{3 b \left (\frac {b}{x^{2}}+a \right )^{\frac {3}{2}}}\)	\(15\)
gosper	\(\frac {a \,x^{2}+b}{3 x^{2} b \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}}}\)	\(29\)
default	\(\frac {a \,x^{2}+b}{3 x^{2} b \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}}}\)	\(29\)
trager	\(\frac {x^{4} \sqrt {-\frac {-a \,x^{2}-b}{x^{2}}}}{3 \left (a \,x^{2}+b \right )^{2} b}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b/x^2+a)^(5/2)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 14, normalized size = 0.78 \begin {gather*} \frac {1}{3 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x^3,x, algorithm="maxima")

[Out]

1/3/((a + b/x^2)^(3/2)*b)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (14) = 28\).
time = 0.36, size = 41, normalized size = 2.28 \begin {gather*} \frac {x^{4} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{3 \, {\left (a^{2} b x^{4} + 2 \, a b^{2} x^{2} + b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (14) = 28\).
time = 0.70, size = 48, normalized size = 2.67 \begin {gather*} \begin {cases} \frac {1}{3 a b \sqrt {a + \frac {b}{x^{2}}} + \frac {3 b^{2} \sqrt {a + \frac {b}{x^{2}}}}{x^{2}}} & \text {for}\: b \neq 0 \\- \frac {1}{2 a^{\frac {5}{2}} x^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**2)**(5/2)/x**3,x)

[Out]

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

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Giac [A]
time = 1.06, size = 21, normalized size = 1.17 \begin {gather*} \frac {x^{3}}{3 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} b \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^(5/2)/x^3,x, algorithm="giac")

[Out]

1/3*x^3/((a*x^2 + b)^(3/2)*b*sgn(x))

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Mupad [B]
time = 1.25, size = 14, normalized size = 0.78 \begin {gather*} \frac {1}{3\,b\,{\left (a+\frac {b}{x^2}\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b/x^2)^(5/2)),x)

[Out]

1/(3*b*(a + b/x^2)^(3/2))

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