3.1668 \(\int \frac {1}{(a+\frac {b}{x}) \sqrt {x}} \, dx\)

Optimal. Leaf size=40 \[ \frac {2 \sqrt {x}}{a}-\frac {2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 40, 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, 50, 63, 205} \[ \frac {2 \sqrt {x}}{a}-\frac {2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{3/2}} \]

Antiderivative was successfully verified.

[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 50

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 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 ) \sqrt {x}} \, dx &=\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) \operatorname {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*}

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Mathematica [A]  time = 0.01, size = 40, normalized size = 1.00 \[ \frac {2 \sqrt {x}}{a}-\frac {2 \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{3/2}} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.94, size = 85, normalized size = 2.12 \[ \left [\frac {\sqrt {-\frac {b}{a}} \log \left (\frac {a x - 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - b}{a x + b}\right ) + 2 \, \sqrt {x}}{a}, -\frac {2 \, {\left (\sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {x} \sqrt {\frac {b}{a}}}{b}\right ) - \sqrt {x}\right )}}{a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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giac [A]  time = 0.15, size = 31, normalized size = 0.78 \[ -\frac {2 \, b \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a} + \frac {2 \, \sqrt {x}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.01, size = 32, normalized size = 0.80 \[ -\frac {2 b \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a}+\frac {2 \sqrt {x}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.65, size = 31, normalized size = 0.78 \[ \frac {2 \, b \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} a} + \frac {2 \, \sqrt {x}}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 0.04, size = 28, normalized size = 0.70 \[ \frac {2\,\sqrt {x}}{a}-\frac {2\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{a^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 1.35, size = 92, normalized size = 2.30 \[ \begin {cases} \frac {2 \sqrt {x}}{a} + \frac {i \sqrt {b} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{a^{2} \sqrt {\frac {1}{a}}} - \frac {i \sqrt {b} \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{a^{2} \sqrt {\frac {1}{a}}} & \text {for}\: a \neq 0 \\\frac {2 x^{\frac {3}{2}}}{3 b} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*sqrt(x)/a + I*sqrt(b)*log(-I*sqrt(b)*sqrt(1/a) + sqrt(x))/(a**2*sqrt(1/a)) - I*sqrt(b)*log(I*sqrt
(b)*sqrt(1/a) + sqrt(x))/(a**2*sqrt(1/a)), Ne(a, 0)), (2*x**(3/2)/(3*b), True))

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