3.2.45 \(\int \frac {1}{\sqrt {b x^n}} \, dx\) [145]

Optimal. Leaf size=19 \[ \frac {2 x}{(2-n) \sqrt {b x^n}} \]

[Out]

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

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Rubi [A]
time = 0.00, antiderivative size = 19, 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}{(2-n) \sqrt {b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[b*x^n],x]

[Out]

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

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Mathematica [A]
time = 0.00, size = 17, normalized size = 0.89 \begin {gather*} -\frac {2 x}{(-2+n) \sqrt {b x^n}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[b*x^n],x]

[Out]

(-2*x)/((-2 + n)*Sqrt[b*x^n])

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Maple [A]
time = 0.01, size = 16, normalized size = 0.84

method	result	size
gosper	\(-\frac {2 x}{\left (-2+n \right ) \sqrt {b \,x^{n}}}\)	\(16\)
risch	\(-\frac {2 x}{\left (-2+n \right ) \sqrt {b \,x^{n}}}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*x/(-2+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(1/(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(-n/2>0)', see `assume?` for mo
re details)I

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-2*x/(n*sqrt(b*x**n) - 2*sqrt(b*x**n)), Ne(n, 2)), (Integral(1/sqrt(b*x**2), 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(1/(b*x^n)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(b*x^n), x)

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Mupad [B]
time = 0.97, size = 24, normalized size = 1.26 \begin {gather*} -\frac {2\,x^{1-n}\,\sqrt {b\,x^n}}{b\,\left (n-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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