3.1737 \(\int \frac{1}{(a+\frac{b}{x})^{3/2} x^3} \, dx\)

Optimal. Leaf size=34 \[ -\frac{2 a}{b^2 \sqrt{a+\frac{b}{x}}}-\frac{2 \sqrt{a+\frac{b}{x}}}{b^2} \]

[Out]

(-2*a)/(b^2*Sqrt[a + b/x]) - (2*Sqrt[a + b/x])/b^2

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Rubi [A]  time = 0.0169184, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {266, 43} \[ -\frac{2 a}{b^2 \sqrt{a+\frac{b}{x}}}-\frac{2 \sqrt{a+\frac{b}{x}}}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^(3/2)*x^3),x]

[Out]

(-2*a)/(b^2*Sqrt[a + b/x]) - (2*Sqrt[a + b/x])/b^2

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0173901, size = 25, normalized size = 0.74 \[ -\frac{2 (2 a x+b)}{b^2 x \sqrt{a+\frac{b}{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^(3/2)*x^3),x]

[Out]

(-2*(b + 2*a*x))/(b^2*Sqrt[a + b/x]*x)

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Maple [A]  time = 0.004, size = 31, normalized size = 0.9 \begin{align*} -2\,{\frac{ \left ( ax+b \right ) \left ( 2\,ax+b \right ) }{{b}^{2}{x}^{2}} \left ({\frac{ax+b}{x}} \right ) ^{-3/2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^(3/2)/x^3,x)

[Out]

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

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Maxima [A]  time = 0.986999, size = 41, normalized size = 1.21 \begin{align*} -\frac{2 \, \sqrt{a + \frac{b}{x}}}{b^{2}} - \frac{2 \, a}{\sqrt{a + \frac{b}{x}} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(a + b/x)/b^2 - 2*a/(sqrt(a + b/x)*b^2)

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Fricas [A]  time = 1.47794, size = 68, normalized size = 2. \begin{align*} -\frac{2 \,{\left (2 \, a x + b\right )} \sqrt{\frac{a x + b}{x}}}{a b^{2} x + b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(2*a*x + b)*sqrt((a*x + b)/x)/(a*b^2*x + b^3)

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Sympy [A]  time = 2.25887, size = 42, normalized size = 1.24 \begin{align*} \begin{cases} - \frac{4 a}{b^{2} \sqrt{a + \frac{b}{x}}} - \frac{2}{b x \sqrt{a + \frac{b}{x}}} & \text{for}\: b \neq 0 \\- \frac{1}{2 a^{\frac{3}{2}} x^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-4*a/(b**2*sqrt(a + b/x)) - 2/(b*x*sqrt(a + b/x)), Ne(b, 0)), (-1/(2*a**(3/2)*x**2), True))

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Giac [A]  time = 1.24599, size = 47, normalized size = 1.38 \begin{align*} -2 \, b{\left (\frac{a}{b^{3} \sqrt{\frac{a x + b}{x}}} + \frac{\sqrt{\frac{a x + b}{x}}}{b^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*b*(a/(b^3*sqrt((a*x + b)/x)) + sqrt((a*x + b)/x)/b^3)