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]

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

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Rubi [A]  time = 0.02, antiderivative size = 73, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 0.01, 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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fricas [A]  time = 0.83, size = 186, normalized size = 2.55 \[ \left [-\frac {{\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {-a b} \log \left (\frac {a x - b - 2 \, \sqrt {-a b} \sqrt {x}}{a x + b}\right ) - 2 \, {\left (a^{2} b x - a b^{2}\right )} \sqrt {x}}{8 \, {\left (a^{4} b^{2} x^{2} + 2 \, a^{3} b^{3} x + a^{2} b^{4}\right )}}, -\frac {{\left (a^{2} x^{2} + 2 \, a b x + b^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{a \sqrt {x}}\right ) - {\left (a^{2} b x - a b^{2}\right )} \sqrt {x}}{4 \, {\left (a^{4} b^{2} x^{2} + 2 \, a^{3} b^{3} x + a^{2} b^{4}\right )}}\right ] \]

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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giac [A]  time = 0.16, size = 52, normalized size = 0.71 \[ \frac {\arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b} + \frac {a x^{\frac {3}{2}} - b \sqrt {x}}{4 \, {\left (a x + b\right )}^{2} a b} \]

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)

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maple [A]  time = 0.01, size = 52, normalized size = 0.71 \[ \frac {\arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, a b}+\frac {\frac {x^{\frac {3}{2}}}{4 b}-\frac {\sqrt {x}}{4 a}}{\left (a x +b \right )^{2}} \]

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

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maxima [A]  time = 2.20, size = 66, normalized size = 0.90 \[ \frac {\frac {a}{\sqrt {x}} - \frac {b}{x^{\frac {3}{2}}}}{4 \, {\left (a^{3} b + \frac {2 \, a^{2} b^{2}}{x} + \frac {a b^{3}}{x^{2}}\right )}} - \frac {\arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{4 \, \sqrt {a b} a b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.13, size = 57, normalized size = 0.78 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{4\,a^{3/2}\,b^{3/2}}-\frac {\frac {\sqrt {x}}{4\,a}-\frac {x^{3/2}}{4\,b}}{a^2\,x^2+2\,a\,b\,x+b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(a + b/x)^3),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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