3.2.49 \(\int \frac {x^2}{(b x^n)^{3/2}} \, dx\) [149]

Optimal. Leaf size=30 \[ \frac {2 x^{3-n}}{3 b (2-n) \sqrt {b x^n}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {15, 30} \begin {gather*} \frac {2 x^{3-n}}{3 b (2-n) \sqrt {b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]
time = 0.01, size = 22, normalized size = 0.73 \begin {gather*} \frac {x^3}{\left (3-\frac {3 n}{2}\right ) \left (b x^n\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x^3/((3 - (3*n)/2)*(b*x^n)^(3/2))

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Maple [A]
time = 0.02, size = 18, normalized size = 0.60

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*x^3/(-2+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(x^2/(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(2-(3*n)/2>0)', see `assume?` f
or more deta

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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(x^2/(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}}{3 n \left (b x^{n}\right )^{\frac {3}{2}} - 6 \left (b x^{n}\right )^{\frac {3}{2}}} & \text {for}\: n \neq 2 \\\int \frac {x^{2}}{\left (b x^{2}\right )^{\frac {3}{2}}}\, dx & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*x**3/(3*n*(b*x**n)**(3/2) - 6*(b*x**n)**(3/2)), Ne(n, 2)), (Integral(x**2/(b*x**2)**(3/2), x), T
rue))

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

[Out]

integrate(x^2/(b*x^n)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x^2}{{\left (b\,x^n\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^2/(b*x^n)^(3/2), x)

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