3.1729 \(\int \frac{1}{\sqrt{a+\frac{b}{x}} x^4} \, dx\)

Optimal. Leaf size=57 \[ -\frac{2 a^2 \sqrt{a+\frac{b}{x}}}{b^3}-\frac{2 \left (a+\frac{b}{x}\right )^{5/2}}{5 b^3}+\frac{4 a \left (a+\frac{b}{x}\right )^{3/2}}{3 b^3} \]

[Out]

(-2*a^2*Sqrt[a + b/x])/b^3 + (4*a*(a + b/x)^(3/2))/(3*b^3) - (2*(a + b/x)^(5/2))/(5*b^3)

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Rubi [A]  time = 0.023212, antiderivative size = 57, 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} \[ -\frac{2 a^2 \sqrt{a+\frac{b}{x}}}{b^3}-\frac{2 \left (a+\frac{b}{x}\right )^{5/2}}{5 b^3}+\frac{4 a \left (a+\frac{b}{x}\right )^{3/2}}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[a + b/x]*x^4),x]

[Out]

(-2*a^2*Sqrt[a + b/x])/b^3 + (4*a*(a + b/x)^(3/2))/(3*b^3) - (2*(a + b/x)^(5/2))/(5*b^3)

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

Mathematica [A]  time = 0.0220955, size = 40, normalized size = 0.7 \[ -\frac{2 \sqrt{a+\frac{b}{x}} \left (8 a^2 x^2-4 a b x+3 b^2\right )}{15 b^3 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[a + b/x]*x^4),x]

[Out]

(-2*Sqrt[a + b/x]*(3*b^2 - 4*a*b*x + 8*a^2*x^2))/(15*b^3*x^2)

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Maple [A]  time = 0.004, size = 44, normalized size = 0.8 \begin{align*} -{\frac{ \left ( 2\,ax+2\,b \right ) \left ( 8\,{a}^{2}{x}^{2}-4\,xab+3\,{b}^{2} \right ) }{15\,{b}^{3}{x}^{3}}{\frac{1}{\sqrt{{\frac{ax+b}{x}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(a+b/x)^(1/2),x)

[Out]

-2/15*(a*x+b)*(8*a^2*x^2-4*a*b*x+3*b^2)/x^3/b^3/((a*x+b)/x)^(1/2)

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Maxima [A]  time = 1.14922, size = 63, normalized size = 1.11 \begin{align*} -\frac{2 \,{\left (a + \frac{b}{x}\right )}^{\frac{5}{2}}}{5 \, b^{3}} + \frac{4 \,{\left (a + \frac{b}{x}\right )}^{\frac{3}{2}} a}{3 \, b^{3}} - \frac{2 \, \sqrt{a + \frac{b}{x}} a^{2}}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b/x)^(1/2),x, algorithm="maxima")

[Out]

-2/5*(a + b/x)^(5/2)/b^3 + 4/3*(a + b/x)^(3/2)*a/b^3 - 2*sqrt(a + b/x)*a^2/b^3

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Fricas [A]  time = 1.4491, size = 88, normalized size = 1.54 \begin{align*} -\frac{2 \,{\left (8 \, a^{2} x^{2} - 4 \, a b x + 3 \, b^{2}\right )} \sqrt{\frac{a x + b}{x}}}{15 \, b^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b/x)^(1/2),x, algorithm="fricas")

[Out]

-2/15*(8*a^2*x^2 - 4*a*b*x + 3*b^2)*sqrt((a*x + b)/x)/(b^3*x^2)

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Sympy [B]  time = 1.78961, size = 813, normalized size = 14.26 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(a+b/x)**(1/2),x)

[Out]

-16*a**(15/2)*b**(9/2)*x**5*sqrt(a*x/b + 1)/(15*a**(11/2)*b**7*x**(11/2) + 45*a**(9/2)*b**8*x**(9/2) + 45*a**(
7/2)*b**9*x**(7/2) + 15*a**(5/2)*b**10*x**(5/2)) - 40*a**(13/2)*b**(11/2)*x**4*sqrt(a*x/b + 1)/(15*a**(11/2)*b
**7*x**(11/2) + 45*a**(9/2)*b**8*x**(9/2) + 45*a**(7/2)*b**9*x**(7/2) + 15*a**(5/2)*b**10*x**(5/2)) - 30*a**(1
1/2)*b**(13/2)*x**3*sqrt(a*x/b + 1)/(15*a**(11/2)*b**7*x**(11/2) + 45*a**(9/2)*b**8*x**(9/2) + 45*a**(7/2)*b**
9*x**(7/2) + 15*a**(5/2)*b**10*x**(5/2)) - 10*a**(9/2)*b**(15/2)*x**2*sqrt(a*x/b + 1)/(15*a**(11/2)*b**7*x**(1
1/2) + 45*a**(9/2)*b**8*x**(9/2) + 45*a**(7/2)*b**9*x**(7/2) + 15*a**(5/2)*b**10*x**(5/2)) - 10*a**(7/2)*b**(1
7/2)*x*sqrt(a*x/b + 1)/(15*a**(11/2)*b**7*x**(11/2) + 45*a**(9/2)*b**8*x**(9/2) + 45*a**(7/2)*b**9*x**(7/2) +
15*a**(5/2)*b**10*x**(5/2)) - 6*a**(5/2)*b**(19/2)*sqrt(a*x/b + 1)/(15*a**(11/2)*b**7*x**(11/2) + 45*a**(9/2)*
b**8*x**(9/2) + 45*a**(7/2)*b**9*x**(7/2) + 15*a**(5/2)*b**10*x**(5/2)) + 16*a**8*b**4*x**(11/2)/(15*a**(11/2)
*b**7*x**(11/2) + 45*a**(9/2)*b**8*x**(9/2) + 45*a**(7/2)*b**9*x**(7/2) + 15*a**(5/2)*b**10*x**(5/2)) + 48*a**
7*b**5*x**(9/2)/(15*a**(11/2)*b**7*x**(11/2) + 45*a**(9/2)*b**8*x**(9/2) + 45*a**(7/2)*b**9*x**(7/2) + 15*a**(
5/2)*b**10*x**(5/2)) + 48*a**6*b**6*x**(7/2)/(15*a**(11/2)*b**7*x**(11/2) + 45*a**(9/2)*b**8*x**(9/2) + 45*a**
(7/2)*b**9*x**(7/2) + 15*a**(5/2)*b**10*x**(5/2)) + 16*a**5*b**7*x**(5/2)/(15*a**(11/2)*b**7*x**(11/2) + 45*a*
*(9/2)*b**8*x**(9/2) + 45*a**(7/2)*b**9*x**(7/2) + 15*a**(5/2)*b**10*x**(5/2))

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Giac [A]  time = 1.15703, size = 90, normalized size = 1.58 \begin{align*} -\frac{2 \,{\left (15 \, a^{2} \sqrt{\frac{a x + b}{x}} - \frac{10 \,{\left (a x + b\right )} a \sqrt{\frac{a x + b}{x}}}{x} + \frac{3 \,{\left (a x + b\right )}^{2} \sqrt{\frac{a x + b}{x}}}{x^{2}}\right )}}{15 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b/x)^(1/2),x, algorithm="giac")

[Out]

-2/15*(15*a^2*sqrt((a*x + b)/x) - 10*(a*x + b)*a*sqrt((a*x + b)/x)/x + 3*(a*x + b)^2*sqrt((a*x + b)/x)/x^2)/b^
3