3.1686 \(\int \frac{1}{(a+\frac{b}{x})^3 x^{5/2}} \, dx\)

Optimal. Leaf size=73 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 a^{3/2} b^{3/2}}+\frac{\sqrt{x}}{4 a b (a x+b)}-\frac{\sqrt{x}}{2 a (a x+b)^2} \]

[Out]

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

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Rubi [A]  time = 0.0222098, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {263, 47, 51, 63, 205} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 a^{3/2} b^{3/2}}+\frac{\sqrt{x}}{4 a b (a x+b)}-\frac{\sqrt{x}}{2 a (a x+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^3*x^(5/2)),x]

[Out]

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

Rule 263

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+\frac{b}{x}\right )^3 x^{5/2}} \, dx &=\int \frac{\sqrt{x}}{(b+a x)^3} \, dx\\ &=-\frac{\sqrt{x}}{2 a (b+a x)^2}+\frac{\int \frac{1}{\sqrt{x} (b+a x)^2} \, dx}{4 a}\\ &=-\frac{\sqrt{x}}{2 a (b+a x)^2}+\frac{\sqrt{x}}{4 a b (b+a x)}+\frac{\int \frac{1}{\sqrt{x} (b+a x)} \, dx}{8 a b}\\ &=-\frac{\sqrt{x}}{2 a (b+a x)^2}+\frac{\sqrt{x}}{4 a b (b+a x)}+\frac{\operatorname{Subst}\left (\int \frac{1}{b+a x^2} \, dx,x,\sqrt{x}\right )}{4 a b}\\ &=-\frac{\sqrt{x}}{2 a (b+a x)^2}+\frac{\sqrt{x}}{4 a b (b+a x)}+\frac{\tan ^{-1}\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}}\right )}{4 a^{3/2} b^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0047959, size = 27, normalized size = 0.37 \[ \frac{2 x^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};-\frac{a x}{b}\right )}{3 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^3*x^(5/2)),x]

[Out]

(2*x^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, -((a*x)/b)])/(3*b^3)

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Maple [A]  time = 0.01, size = 52, normalized size = 0.7 \begin{align*} 2\,{\frac{1}{ \left ( ax+b \right ) ^{2}} \left ( 1/8\,{\frac{{x}^{3/2}}{b}}-1/8\,{\frac{\sqrt{x}}{a}} \right ) }+{\frac{1}{4\,ab}\arctan \left ({a\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x)^3/x^(5/2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^3/x^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.74412, size = 412, normalized size = 5.64 \begin{align*} \left [-\frac{{\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt{-a b} \log \left (\frac{a x - b - 2 \, \sqrt{-a b} \sqrt{x}}{a x + b}\right ) - 2 \,{\left (a^{2} b x - a b^{2}\right )} \sqrt{x}}{8 \,{\left (a^{4} b^{2} x^{2} + 2 \, a^{3} b^{3} x + a^{2} b^{4}\right )}}, -\frac{{\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b}}{a \sqrt{x}}\right ) -{\left (a^{2} b x - a b^{2}\right )} \sqrt{x}}{4 \,{\left (a^{4} b^{2} x^{2} + 2 \, a^{3} b^{3} x + a^{2} b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^3/x^(5/2),x, algorithm="fricas")

[Out]

[-1/8*((a^2*x^2 + 2*a*b*x + b^2)*sqrt(-a*b)*log((a*x - b - 2*sqrt(-a*b)*sqrt(x))/(a*x + b)) - 2*(a^2*b*x - a*b
^2)*sqrt(x))/(a^4*b^2*x^2 + 2*a^3*b^3*x + a^2*b^4), -1/4*((a^2*x^2 + 2*a*b*x + b^2)*sqrt(a*b)*arctan(sqrt(a*b)
/(a*sqrt(x))) - (a^2*b*x - a*b^2)*sqrt(x))/(a^4*b^2*x^2 + 2*a^3*b^3*x + a^2*b^4)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)**3/x**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.09483, size = 70, normalized size = 0.96 \begin{align*} \frac{\arctan \left (\frac{a \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} a b} + \frac{a x^{\frac{3}{2}} - b \sqrt{x}}{4 \,{\left (a x + b\right )}^{2} a b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x)^3/x^(5/2),x, algorithm="giac")

[Out]

1/4*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a*b) + 1/4*(a*x^(3/2) - b*sqrt(x))/((a*x + b)^2*a*b)