\(\int \frac {1}{(a+\frac {b}{x}) x^{5/2}} \, dx\) [1670]

Optimal result

Integrand size = 15, antiderivative size = 40 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^{5/2}} \, dx=-\frac {2}{b \sqrt {x}}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {269, 53, 65, 211} \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^{5/2}} \, dx=-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2}}-\frac {2}{b \sqrt {x}} \]

[In]

Int[1/((a + b/x)*x^(5/2)),x]

[Out]

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

Rule 53

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 65

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 211

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

Rule 269

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^{5/2}} \, dx=-\frac {2}{b \sqrt {x}}-\frac {2 \sqrt {a} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{b^{3/2}} \]

[In]

Integrate[1/((a + b/x)*x^(5/2)),x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.80

method	result	size
derivativedivides	\(-\frac {2 a \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{b \sqrt {a b}}-\frac {2}{\sqrt {x}\, b}\)	\(32\)
default	\(-\frac {2 a \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{b \sqrt {a b}}-\frac {2}{\sqrt {x}\, b}\)	\(32\)
risch	\(-\frac {2 a \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{b \sqrt {a b}}-\frac {2}{\sqrt {x}\, b}\)	\(32\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 93, normalized size of antiderivative = 2.32 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^{5/2}} \, dx=\left [\frac {x \sqrt {-\frac {a}{b}} \log \left (\frac {a x - 2 \, b \sqrt {x} \sqrt {-\frac {a}{b}} - b}{a x + b}\right ) - 2 \, \sqrt {x}}{b x}, \frac {2 \, {\left (x \sqrt {\frac {a}{b}} \arctan \left (\frac {b \sqrt {\frac {a}{b}}}{a \sqrt {x}}\right ) - \sqrt {x}\right )}}{b x}\right ] \]

[In]

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

[Out]

[(x*sqrt(-a/b)*log((a*x - 2*b*sqrt(x)*sqrt(-a/b) - b)/(a*x + b)) - 2*sqrt(x))/(b*x), 2*(x*sqrt(a/b)*arctan(b*s
qrt(a/b)/(a*sqrt(x))) - sqrt(x))/(b*x)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (37) = 74\).

Time = 1.67 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.12 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^{5/2}} \, dx=\begin {cases} \frac {\tilde {\infty }}{\sqrt {x}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2}{b \sqrt {x}} & \text {for}\: a = 0 \\- \frac {2}{3 a x^{\frac {3}{2}}} & \text {for}\: b = 0 \\- \frac {\log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{b \sqrt {- \frac {b}{a}}} + \frac {\log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{b \sqrt {- \frac {b}{a}}} - \frac {2}{b \sqrt {x}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^{5/2}} \, dx=\frac {2 \, a \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} b} - \frac {2}{b \sqrt {x}} \]

[In]

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

[Out]

2*a*arctan(b/(sqrt(a*b)*sqrt(x)))/(sqrt(a*b)*b) - 2/(b*sqrt(x))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^{5/2}} \, dx=-\frac {2 \, a \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} b} - \frac {2}{b \sqrt {x}} \]

[In]

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

[Out]

-2*a*arctan(a*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b) - 2/(b*sqrt(x))

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) x^{5/2}} \, dx=-\frac {2}{b\,\sqrt {x}}-\frac {2\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{b^{3/2}} \]

[In]

int(1/(x^(5/2)*(a + b/x)),x)

[Out]

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