3.1.46 \(\int \frac {1}{x^3 (b x^2)^{3/2}} \, dx\) [46]

Optimal. Leaf size=19 \[ -\frac {1}{5 b x^4 \sqrt {b x^2}} \]

[Out]

-1/5/b/x^4/(b*x^2)^(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 = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {15, 30} \begin {gather*} -\frac {1}{5 b x^4 \sqrt {b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(b*x^2)^(3/2)),x]

[Out]

-1/5*1/(b*x^4*Sqrt[b*x^2])

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}{x^3 \left (b x^2\right )^{3/2}} \, dx &=\frac {x \int \frac {1}{x^6} \, dx}{b \sqrt {b x^2}}\\ &=-\frac {1}{5 b x^4 \sqrt {b x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 16, normalized size = 0.84 \begin {gather*} -\frac {1}{5 x^2 \left (b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(b*x^2)^(3/2)),x]

[Out]

-1/5*1/(x^2*(b*x^2)^(3/2))

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Maple [A]
time = 0.03, size = 13, normalized size = 0.68

method	result	size
gosper	\(-\frac {1}{5 x^{2} \left (b \,x^{2}\right )^{\frac {3}{2}}}\)	\(13\)
default	\(-\frac {1}{5 x^{2} \left (b \,x^{2}\right )^{\frac {3}{2}}}\)	\(13\)
risch	\(-\frac {1}{5 b \,x^{4} \sqrt {b \,x^{2}}}\)	\(16\)
trager	\(\frac {\left (x -1\right ) \left (x^{4}+x^{3}+x^{2}+x +1\right ) \sqrt {b \,x^{2}}}{5 b^{2} x^{6}}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/x^2/(b*x^2)^(3/2)

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Maxima [A]
time = 0.26, size = 8, normalized size = 0.42 \begin {gather*} -\frac {1}{5 \, b^{\frac {3}{2}} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/(b^(3/2)*x^5)

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Fricas [A]
time = 0.34, size = 15, normalized size = 0.79 \begin {gather*} -\frac {\sqrt {b x^{2}}}{5 \, b^{2} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*sqrt(b*x^2)/(b^2*x^6)

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Sympy [A]
time = 0.27, size = 15, normalized size = 0.79 \begin {gather*} - \frac {1}{5 x^{2} \left (b x^{2}\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(5*x**2*(b*x**2)**(3/2))

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Giac [A]
time = 0.98, size = 12, normalized size = 0.63 \begin {gather*} -\frac {1}{5 \, b^{\frac {3}{2}} x^{5} \mathrm {sgn}\left (x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/(b^(3/2)*x^5*sgn(x))

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Mupad [B]
time = 0.95, size = 10, normalized size = 0.53 \begin {gather*} -\frac {1}{5\,b^{3/2}\,{\left (x^2\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(5*b^(3/2)*(x^2)^(5/2))

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