3.1678 \(\int \frac {1}{(a+\frac {b}{x})^2 x^{5/2}} \, dx\)

Optimal. Leaf size=45 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {a} b^{3/2}}+\frac {\sqrt {x}}{b (a x+b)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {263, 51, 63, 205} \[ \frac {\tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {a} b^{3/2}}+\frac {\sqrt {x}}{b (a x+b)} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x)^2*x^(5/2)),x]

[Out]

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 45, normalized size = 1.00 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{\sqrt {a} b^{3/2}}+\frac {\sqrt {x}}{a b x+b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x)^2*x^(5/2)),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.91, size = 116, normalized size = 2.58 \[ \left [\frac {2 \, a b \sqrt {x} - \sqrt {-a b} {\left (a x + b\right )} \log \left (\frac {a x - b - 2 \, \sqrt {-a b} \sqrt {x}}{a x + b}\right )}{2 \, {\left (a^{2} b^{2} x + a b^{3}\right )}}, \frac {a b \sqrt {x} - \sqrt {a b} {\left (a x + b\right )} \arctan \left (\frac {\sqrt {a b}}{a \sqrt {x}}\right )}{a^{2} b^{2} x + a b^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.16, size = 35, normalized size = 0.78 \[ \frac {\arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b} + \frac {\sqrt {x}}{{\left (a x + b\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b) + sqrt(x)/((a*x + b)*b)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 36, normalized size = 0.80 \[ \frac {\arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, b}+\frac {\sqrt {x}}{\left (a x +b \right ) b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 2.33, size = 39, normalized size = 0.87 \[ -\frac {\arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} b} + \frac {1}{{\left (a b + \frac {b^{2}}{x}\right )} \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.10, size = 33, normalized size = 0.73 \[ \frac {\sqrt {x}}{b\,\left (b+a\,x\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{\sqrt {a}\,b^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^(5/2)*(a + b/x)^2),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 102.66, size = 328, normalized size = 7.29 \[ \begin {cases} \tilde {\infty } \sqrt {x} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 \sqrt {x}}{b^{2}} & \text {for}\: a = 0 \\- \frac {2}{3 a^{2} x^{\frac {3}{2}}} & \text {for}\: b = 0 \\\frac {2 i a \sqrt {b} \sqrt {x} \sqrt {\frac {1}{a}}}{2 i a^{2} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 2 i a b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} + \frac {a x \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{2 i a^{2} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 2 i a b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} - \frac {a x \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{2 i a^{2} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 2 i a b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} + \frac {b \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{2 i a^{2} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 2 i a b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} - \frac {b \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{2 i a^{2} b^{\frac {3}{2}} x \sqrt {\frac {1}{a}} + 2 i a b^{\frac {5}{2}} \sqrt {\frac {1}{a}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*sqrt(x), Eq(a, 0) & Eq(b, 0)), (2*sqrt(x)/b**2, Eq(a, 0)), (-2/(3*a**2*x**(3/2)), Eq(b, 0)), (2
*I*a*sqrt(b)*sqrt(x)*sqrt(1/a)/(2*I*a**2*b**(3/2)*x*sqrt(1/a) + 2*I*a*b**(5/2)*sqrt(1/a)) + a*x*log(-I*sqrt(b)
*sqrt(1/a) + sqrt(x))/(2*I*a**2*b**(3/2)*x*sqrt(1/a) + 2*I*a*b**(5/2)*sqrt(1/a)) - a*x*log(I*sqrt(b)*sqrt(1/a)
+ sqrt(x))/(2*I*a**2*b**(3/2)*x*sqrt(1/a) + 2*I*a*b**(5/2)*sqrt(1/a)) + b*log(-I*sqrt(b)*sqrt(1/a) + sqrt(x))
/(2*I*a**2*b**(3/2)*x*sqrt(1/a) + 2*I*a*b**(5/2)*sqrt(1/a)) - b*log(I*sqrt(b)*sqrt(1/a) + sqrt(x))/(2*I*a**2*b
**(3/2)*x*sqrt(1/a) + 2*I*a*b**(5/2)*sqrt(1/a)), True))

________________________________________________________________________________________