3.6.11 \(\int \frac {x}{(a+b x^2)^{5/2}} \, dx\) [511]

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

[Out]

-1/3/b/(b*x^2+a)^(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 = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {267} \begin {gather*} -\frac {1}{3 b \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a + b*x^2)^(5/2),x]

[Out]

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

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Mathematica [A]
time = 0.00, size = 18, normalized size = 1.00 \begin {gather*} -\frac {1}{3 b \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*x^2)^(5/2),x]

[Out]

-1/3*1/(b*(a + b*x^2)^(3/2))

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Maple [A]
time = 0.03, size = 15, normalized size = 0.83

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/((b*x^2 + a)^(3/2)*b)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 35 vs. \(2 (14) = 28\).
time = 0.90, size = 35, normalized size = 1.94 \begin {gather*} -\frac {\sqrt {b x^{2} + a}}{3 \, {\left (b^{3} x^{4} + 2 \, a b^{2} x^{2} + a^{2} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (15) = 30\).
time = 0.33, size = 46, normalized size = 2.56 \begin {gather*} \begin {cases} - \frac {1}{3 a b \sqrt {a + b x^{2}} + 3 b^{2} x^{2} \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.63, size = 14, normalized size = 0.78 \begin {gather*} -\frac {1}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/((b*x^2 + a)^(3/2)*b)

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Mupad [B]
time = 4.94, size = 14, normalized size = 0.78 \begin {gather*} -\frac {1}{3\,b\,{\left (b\,x^2+a\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*x^2)^(5/2),x)

[Out]

-1/(3*b*(a + b*x^2)^(3/2))

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