3.1667 \(\int \frac {\sqrt {x}}{a+\frac {b}{x}} \, dx\)

Optimal. Leaf size=53 \[ \frac {2 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{5/2}}-\frac {2 b \sqrt {x}}{a^2}+\frac {2 x^{3/2}}{3 a} \]

[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

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Rubi [A]  time = 0.02, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {263, 50, 63, 205} \[ \frac {2 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{5/2}}-\frac {2 b \sqrt {x}}{a^2}+\frac {2 x^{3/2}}{3 a} \]

Antiderivative was successfully verified.

[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 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 {\sqrt {x}}{a+\frac {b}{x}} \, dx &=\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 ) \operatorname {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*}

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Mathematica [A]  time = 0.02, size = 49, normalized size = 0.92 \[ \frac {2 b^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}}\right )}{a^{5/2}}+\frac {2 \sqrt {x} (a x-3 b)}{3 a^2} \]

Antiderivative was successfully verified.

[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)

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fricas [A]  time = 0.86, size = 103, normalized size = 1.94 \[ \left [\frac {3 \, b \sqrt {-\frac {b}{a}} \log \left (\frac {a x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - b}{a x + b}\right ) + 2 \, {\left (a x - 3 \, b\right )} \sqrt {x}}{3 \, a^{2}}, \frac {2 \, {\left (3 \, b \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {x} \sqrt {\frac {b}{a}}}{b}\right ) + {\left (a x - 3 \, b\right )} \sqrt {x}\right )}}{3 \, a^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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giac [A]  time = 0.17, size = 45, normalized size = 0.85 \[ \frac {2 \, b^{2} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{2}} + \frac {2 \, {\left (a^{2} x^{\frac {3}{2}} - 3 \, a b \sqrt {x}\right )}}{3 \, a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.01, size = 43, normalized size = 0.81 \[ \frac {2 b^{2} \arctan \left (\frac {a \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b}\, a^{2}}+\frac {2 x^{\frac {3}{2}}}{3 a}-\frac {2 b \sqrt {x}}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.33, size = 41, normalized size = 0.77 \[ \frac {2 \, {\left (a - \frac {3 \, b}{x}\right )} x^{\frac {3}{2}}}{3 \, a^{2}} - \frac {2 \, b^{2} \arctan \left (\frac {b}{\sqrt {a b} \sqrt {x}}\right )}{\sqrt {a b} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.05, size = 37, normalized size = 0.70 \[ \frac {2\,x^{3/2}}{3\,a}-\frac {2\,b\,\sqrt {x}}{a^2}+\frac {2\,b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {x}}{\sqrt {b}}\right )}{a^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 1.37, size = 105, normalized size = 1.98 \[ \begin {cases} \frac {2 x^{\frac {3}{2}}}{3 a} - \frac {2 b \sqrt {x}}{a^{2}} - \frac {i b^{\frac {3}{2}} \log {\left (- i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{a^{3} \sqrt {\frac {1}{a}}} + \frac {i b^{\frac {3}{2}} \log {\left (i \sqrt {b} \sqrt {\frac {1}{a}} + \sqrt {x} \right )}}{a^{3} \sqrt {\frac {1}{a}}} & \text {for}\: a \neq 0 \\\frac {2 x^{\frac {5}{2}}}{5 b} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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