3.2.44 \(\int \frac {x}{\sqrt {b x^n}} \, dx\) [144]

Optimal. Leaf size=21 \[ \frac {2 x^2}{(4-n) \sqrt {b x^n}} \]

[Out]

2*x^2/(4-n)/(b*x^n)^(1/2)

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Rubi [A]
time = 0.00, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {15, 30} \begin {gather*} \frac {2 x^2}{(4-n) \sqrt {b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[b*x^n],x]

[Out]

(2*x^2)/((4 - 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}{\sqrt {b x^n}} \, dx &=\frac {x^{n/2} \int x^{1-\frac {n}{2}} \, dx}{\sqrt {b x^n}}\\ &=\frac {2 x^2}{(4-n) \sqrt {b x^n}}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 19, normalized size = 0.90 \begin {gather*} -\frac {2 x^2}{(-4+n) \sqrt {b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[b*x^n],x]

[Out]

(-2*x^2)/((-4 + n)*Sqrt[b*x^n])

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Maple [A]
time = 0.01, size = 18, normalized size = 0.86

method	result	size
gosper	\(-\frac {2 x^{2}}{\left (-4+n \right ) \sqrt {b \,x^{n}}}\)	\(18\)
risch	\(-\frac {2 x^{2}}{\left (-4+n \right ) \sqrt {b \,x^{n}}}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x^2/(-4+n)/(b*x^n)^(1/2)

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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/(b*x^n)^(1/2),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(1-n/2>0)', see `assume?` for m
ore details)

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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/(b*x^n)^(1/2),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^{2}}{n \sqrt {b x^{n}} - 4 \sqrt {b x^{n}}} & \text {for}\: n \neq 4 \\\int \frac {x}{\sqrt {b x^{4}}}\, dx & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*x**2/(n*sqrt(b*x**n) - 4*sqrt(b*x**n)), Ne(n, 4)), (Integral(x/sqrt(b*x**4), x), True))

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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/(b*x^n)^(1/2),x, algorithm="giac")

[Out]

integrate(x/sqrt(b*x^n), x)

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Mupad [B]
time = 1.04, size = 24, normalized size = 1.14 \begin {gather*} -\frac {2\,x^{2-n}\,\sqrt {b\,x^n}}{b\,\left (n-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*x^(2 - n)*(b*x^n)^(1/2))/(b*(n - 4))

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