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*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)

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Rubi [A]  time = 0.0181793, antiderivative size = 53, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [A]  time = 0.017519, 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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Maple [A]  time = 0.006, size = 43, normalized size = 0.8 \begin{align*}{\frac{2}{3\,a}{x}^{{\frac{3}{2}}}}-2\,{\frac{b\sqrt{x}}{{a}^{2}}}+2\,{\frac{{b}^{2}}{{a}^{2}\sqrt{ab}}\arctan \left ({\frac{a\sqrt{x}}{\sqrt{ab}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.72931, size = 244, normalized size = 4.6 \begin{align*} \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 ] \end{align*}

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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Sympy [A]  time = 1.32869, size = 105, normalized size = 1.98 \begin{align*} \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} \end{align*}

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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Giac [A]  time = 1.10151, size = 61, normalized size = 1.15 \begin{align*} \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}} \end{align*}

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