\(\int \frac {\sqrt {x}}{a+\frac {b}{x}} \, dx\) [1667]

Optimal result

Integrand size = 15, antiderivative size = 53 \[ \int \frac {\sqrt {x}}{a+\frac {b}{x}} \, dx=-\frac {2 b \sqrt {x}}{a^2}+\frac {2 x^{3/2}}{3 a}+\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{5/2}} \]

[Out]

2/3*x^(3/2)/a+2*b^(3/2)*arctan(a^(1/2)*x^(1/2)/b^(1/2))/a^(5/2)-2*b*x^(1/2)/a^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {269, 52, 65, 211} \[ \int \frac {\sqrt {x}}{a+\frac {b}{x}} \, dx=\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{5/2}}-\frac {2 b \sqrt {x}}{a^2}+\frac {2 x^{3/2}}{3 a} \]

[In]

Int[Sqrt[x]/(a + b/x),x]

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 {x^{3/2}}{b+a x} \, dx \\ & = \frac {2 x^{3/2}}{3 a}-\frac {b \int \frac {\sqrt {x}}{b+a x} \, dx}{a} \\ & = -\frac {2 b \sqrt {x}}{a^2}+\frac {2 x^{3/2}}{3 a}+\frac {b^2 \int \frac {1}{\sqrt {x} (b+a x)} \, dx}{a^2} \\ & = -\frac {2 b \sqrt {x}}{a^2}+\frac {2 x^{3/2}}{3 a}+\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\sqrt {x}\right )}{a^2} \\ & = -\frac {2 b \sqrt {x}}{a^2}+\frac {2 x^{3/2}}{3 a}+\frac {2 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {x}}{a+\frac {b}{x}} \, dx=\frac {2 \sqrt {x} (-3 b+a x)}{3 a^2}+\frac {2 b^{3/2} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{5/2}} \]

[In]

Integrate[Sqrt[x]/(a + b/x),x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.77

method	result	size
risch	\(\frac {2 \left (a x -3 b \right ) \sqrt {x}}{3 a^{2}}+\frac {2 b^{2} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{a^{2} \sqrt {a b}}\)	\(41\)
derivativedivides	\(\frac {\frac {2 a \,x^{\frac {3}{2}}}{3}-2 b \sqrt {x}}{a^{2}}+\frac {2 b^{2} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{a^{2} \sqrt {a b}}\)	\(44\)
default	\(\frac {\frac {2 a \,x^{\frac {3}{2}}}{3}-2 b \sqrt {x}}{a^{2}}+\frac {2 b^{2} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{a^{2} \sqrt {a b}}\)	\(44\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.94 \[ \int \frac {\sqrt {x}}{a+\frac {b}{x}} \, dx=\left [\frac {3 \, b \sqrt {-\frac {b}{a}} \log \left (\frac {a x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - b}{a x + b}\right ) + 2 \, {\left (a x - 3 \, b\right )} \sqrt {x}}{3 \, a^{2}}, \frac {2 \, {\left (3 \, b \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {x} \sqrt {\frac {b}{a}}}{b}\right ) + {\left (a x - 3 \, b\right )} \sqrt {x}\right )}}{3 \, a^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 107 vs. \(2 (49) = 98\).

Time = 0.39 (sec) , antiderivative size = 107, normalized size of antiderivative = 2.02 \[ \int \frac {\sqrt {x}}{a+\frac {b}{x}} \, dx=\begin {cases} \tilde {\infty } x^{\frac {5}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {5}{2}}}{5 b} & \text {for}\: a = 0 \\\frac {2 x^{\frac {3}{2}}}{3 a} & \text {for}\: b = 0 \\\frac {2 x^{\frac {3}{2}}}{3 a} - \frac {2 b \sqrt {x}}{a^{2}} + \frac {b^{2} \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{a^{3} \sqrt {- \frac {b}{a}}} - \frac {b^{2} \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{a^{3} \sqrt {- \frac {b}{a}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {x}}{a+\frac {b}{x}} \, dx=\frac {2 \, {\left (a - \frac {3 \, b}{x}\right )} x^{\frac {3}{2}}}{3 \, a^{2}} - \frac {2 \, b^{2} \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} a^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {x}}{a+\frac {b}{x}} \, dx=\frac {2 \, b^{2} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {2 \, {\left (a^{2} x^{\frac {3}{2}} - 3 \, a b \sqrt {x}\right )}}{3 \, a^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {x}}{a+\frac {b}{x}} \, dx=\frac {2\,x^{3/2}}{3\,a}-\frac {2\,b\,\sqrt {x}}{a^2}+\frac {2\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{a^{5/2}} \]

[In]

int(x^(1/2)/(a + b/x),x)

[Out]

(2*x^(3/2))/(3*a) - (2*b*x^(1/2))/a^2 + (2*b^(3/2)*atan((a^(1/2)*x^(1/2))/b^(1/2)))/a^(5/2)