3.2.51 \(\int \frac {1}{(b x^n)^{3/2}} \, dx\) [151]

Optimal. Leaf size=28 \[ \frac {2 x^{1-n}}{b (2-3 n) \sqrt {b x^n}} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {15, 30} \begin {gather*} \frac {2 x^{1-n}}{b (2-3 n) \sqrt {b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^n)^(-3/2),x]

[Out]

(2*x^(1 - n))/(b*(2 - 3*n)*Sqrt[b*x^n])

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 20, normalized size = 0.71 \begin {gather*} \frac {x}{\left (1-\frac {3 n}{2}\right ) \left (b x^n\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^n)^(-3/2),x]

[Out]

x/((1 - (3*n)/2)*(b*x^n)^(3/2))

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Maple [A]
time = 0.01, size = 18, normalized size = 0.64

method	result	size
gosper	\(-\frac {2 x}{\left (-2+3 n \right ) \left (b \,x^{n}\right )^{\frac {3}{2}}}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^n)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2*x/(-2+3*n)/(b*x^n)^(3/2)

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^n)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(-(3*n)/2>0)', see `assume?` fo
r more detai

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^n)^(3/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (ha
s polynomial part)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \begin {cases} - \frac {2 x}{3 n \left (b x^{n}\right )^{\frac {3}{2}} - 2 \left (b x^{n}\right )^{\frac {3}{2}}} & \text {for}\: n \neq \frac {2}{3} \\\int \frac {1}{\left (b x^{\frac {2}{3}}\right )^{\frac {3}{2}}}\, dx & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**n)**(3/2),x)

[Out]

Piecewise((-2*x/(3*n*(b*x**n)**(3/2) - 2*(b*x**n)**(3/2)), Ne(n, 2/3)), (Integral((b*x**(2/3))**(-3/2), x), Tr
ue))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^n)^(3/2),x, algorithm="giac")

[Out]

integrate((b*x^n)^(-3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {1}{{\left (b\,x^n\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^n)^(3/2),x)

[Out]

int(1/(b*x^n)^(3/2), x)

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