3.1697 \(\int \frac{\sqrt{a+\frac{b}{x}}}{x^3} \, dx\)

Optimal. Leaf size=38 \[ \frac{2 a \left (a+\frac{b}{x}\right )^{3/2}}{3 b^2}-\frac{2 \left (a+\frac{b}{x}\right )^{5/2}}{5 b^2} \]

[Out]

(2*a*(a + b/x)^(3/2))/(3*b^2) - (2*(a + b/x)^(5/2))/(5*b^2)

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Rubi [A]  time = 0.0174295, antiderivative size = 38, 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} \[ \frac{2 a \left (a+\frac{b}{x}\right )^{3/2}}{3 b^2}-\frac{2 \left (a+\frac{b}{x}\right )^{5/2}}{5 b^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x]/x^3,x]

[Out]

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

Mathematica [A]  time = 0.0146367, size = 34, normalized size = 0.89 \[ \frac{2 \sqrt{a+\frac{b}{x}} (a x+b) (2 a x-3 b)}{15 b^2 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/x]/x^3,x]

[Out]

(2*Sqrt[a + b/x]*(b + a*x)*(-3*b + 2*a*x))/(15*b^2*x^2)

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Maple [A]  time = 0.004, size = 33, normalized size = 0.9 \begin{align*}{\frac{ \left ( 2\,ax+2\,b \right ) \left ( 2\,ax-3\,b \right ) }{15\,{b}^{2}{x}^{2}}\sqrt{{\frac{ax+b}{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(a*x+b)*(2*a*x-3*b)*((a*x+b)/x)^(1/2)/b^2/x^2

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Maxima [A]  time = 0.9994, size = 41, normalized size = 1.08 \begin{align*} -\frac{2 \,{\left (a + \frac{b}{x}\right )}^{\frac{5}{2}}}{5 \, b^{2}} + \frac{2 \,{\left (a + \frac{b}{x}\right )}^{\frac{3}{2}} a}{3 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(a + b/x)^(5/2)/b^2 + 2/3*(a + b/x)^(3/2)*a/b^2

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Fricas [A]  time = 1.73122, size = 84, normalized size = 2.21 \begin{align*} \frac{2 \,{\left (2 \, a^{2} x^{2} - a b x - 3 \, b^{2}\right )} \sqrt{\frac{a x + b}{x}}}{15 \, b^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(2*a^2*x^2 - a*b*x - 3*b^2)*sqrt((a*x + b)/x)/(b^2*x^2)

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Sympy [B]  time = 1.54957, size = 304, normalized size = 8. \begin{align*} \frac{4 a^{\frac{11}{2}} b^{\frac{3}{2}} x^{3} \sqrt{\frac{a x}{b} + 1}}{15 a^{\frac{7}{2}} b^{3} x^{\frac{7}{2}} + 15 a^{\frac{5}{2}} b^{4} x^{\frac{5}{2}}} + \frac{2 a^{\frac{9}{2}} b^{\frac{5}{2}} x^{2} \sqrt{\frac{a x}{b} + 1}}{15 a^{\frac{7}{2}} b^{3} x^{\frac{7}{2}} + 15 a^{\frac{5}{2}} b^{4} x^{\frac{5}{2}}} - \frac{8 a^{\frac{7}{2}} b^{\frac{7}{2}} x \sqrt{\frac{a x}{b} + 1}}{15 a^{\frac{7}{2}} b^{3} x^{\frac{7}{2}} + 15 a^{\frac{5}{2}} b^{4} x^{\frac{5}{2}}} - \frac{6 a^{\frac{5}{2}} b^{\frac{9}{2}} \sqrt{\frac{a x}{b} + 1}}{15 a^{\frac{7}{2}} b^{3} x^{\frac{7}{2}} + 15 a^{\frac{5}{2}} b^{4} x^{\frac{5}{2}}} - \frac{4 a^{6} b x^{\frac{7}{2}}}{15 a^{\frac{7}{2}} b^{3} x^{\frac{7}{2}} + 15 a^{\frac{5}{2}} b^{4} x^{\frac{5}{2}}} - \frac{4 a^{5} b^{2} x^{\frac{5}{2}}}{15 a^{\frac{7}{2}} b^{3} x^{\frac{7}{2}} + 15 a^{\frac{5}{2}} b^{4} x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*a**(11/2)*b**(3/2)*x**3*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) + 2*a**(9/2)
*b**(5/2)*x**2*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 8*a**(7/2)*b**(7/2)*x
*sqrt(a*x/b + 1)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 6*a**(5/2)*b**(9/2)*sqrt(a*x/b + 1)
/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2)) - 4*a**6*b*x**(7/2)/(15*a**(7/2)*b**3*x**(7/2) + 15*a
**(5/2)*b**4*x**(5/2)) - 4*a**5*b**2*x**(5/2)/(15*a**(7/2)*b**3*x**(7/2) + 15*a**(5/2)*b**4*x**(5/2))

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Giac [B]  time = 1.15617, size = 155, normalized size = 4.08 \begin{align*} \frac{2 \,{\left (15 \,{\left (\sqrt{a} x - \sqrt{a x^{2} + b x}\right )}^{3} a^{\frac{3}{2}} \mathrm{sgn}\left (x\right ) + 25 \,{\left (\sqrt{a} x - \sqrt{a x^{2} + b x}\right )}^{2} a b \mathrm{sgn}\left (x\right ) + 15 \,{\left (\sqrt{a} x - \sqrt{a x^{2} + b x}\right )} \sqrt{a} b^{2} \mathrm{sgn}\left (x\right ) + 3 \, b^{3} \mathrm{sgn}\left (x\right )\right )}}{15 \,{\left (\sqrt{a} x - \sqrt{a x^{2} + b x}\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(15*(sqrt(a)*x - sqrt(a*x^2 + b*x))^3*a^(3/2)*sgn(x) + 25*(sqrt(a)*x - sqrt(a*x^2 + b*x))^2*a*b*sgn(x) +
15*(sqrt(a)*x - sqrt(a*x^2 + b*x))*sqrt(a)*b^2*sgn(x) + 3*b^3*sgn(x))/(sqrt(a)*x - sqrt(a*x^2 + b*x))^5