\(\int \frac {1}{(a+\frac {b}{x}) \sqrt {x}} \, dx\) [1668]

Optimal result

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

[Out]

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

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, 52, 65, 211} \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) \sqrt {x}} \, dx=\frac {2 \sqrt {x}}{a}-\frac {2 \sqrt {b} \arctan \left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{3/2}} \]

[In]

Int[1/((a + b/x)*Sqrt[x]),x]

[Out]

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

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

Mathematica [A] (verified)

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

[In]

Integrate[1/((a + b/x)*Sqrt[x]),x]

[Out]

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

Maple [A] (verified)

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

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.12 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) \sqrt {x}} \, dx=\left [\frac {\sqrt {-\frac {b}{a}} \log \left (\frac {a x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - b}{a x + b}\right ) + 2 \, \sqrt {x}}{a}, -\frac {2 \, {\left (\sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {x} \sqrt {\frac {b}{a}}}{b}\right ) - \sqrt {x}\right )}}{a}\right ] \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (36) = 72\).

Time = 0.43 (sec) , antiderivative size = 88, normalized size of antiderivative = 2.20 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) \sqrt {x}} \, dx=\begin {cases} \tilde {\infty } x^{\frac {3}{2}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {2 x^{\frac {3}{2}}}{3 b} & \text {for}\: a = 0 \\\frac {2 \sqrt {x}}{a} & \text {for}\: b = 0 \\\frac {2 \sqrt {x}}{a} - \frac {b \log {\left (\sqrt {x} - \sqrt {- \frac {b}{a}} \right )}}{a^{2} \sqrt {- \frac {b}{a}}} + \frac {b \log {\left (\sqrt {x} + \sqrt {- \frac {b}{a}} \right )}}{a^{2} \sqrt {- \frac {b}{a}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) \sqrt {x}} \, dx=-\frac {2 \, b \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a} + \frac {2 \, \sqrt {x}}{a} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.70 \[ \int \frac {1}{\left (a+\frac {b}{x}\right ) \sqrt {x}} \, dx=\frac {2\,\sqrt {x}}{a}-\frac {2\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{a^{3/2}} \]

[In]

int(1/(x^(1/2)*(a + b/x)),x)

[Out]

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