3.1693 \(\int \sqrt{a+\frac{b}{x}} x \, dx\)

Optimal. Leaf size=69 \[ -\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{4 a^{3/2}}+\frac{1}{2} x^2 \sqrt{a+\frac{b}{x}}+\frac{b x \sqrt{a+\frac{b}{x}}}{4 a} \]

[Out]

(b*Sqrt[a + b/x]*x)/(4*a) + (Sqrt[a + b/x]*x^2)/2 - (b^2*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(4*a^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.0274541, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {266, 47, 51, 63, 208} \[ -\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{4 a^{3/2}}+\frac{1}{2} x^2 \sqrt{a+\frac{b}{x}}+\frac{b x \sqrt{a+\frac{b}{x}}}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x]*x,x]

[Out]

(b*Sqrt[a + b/x]*x)/(4*a) + (Sqrt[a + b/x]*x^2)/2 - (b^2*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(4*a^(3/2))

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \sqrt{a+\frac{b}{x}} x \, dx &=-\operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x^3} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{2} \sqrt{a+\frac{b}{x}} x^2-\frac{1}{4} b \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )\\ &=\frac{b \sqrt{a+\frac{b}{x}} x}{4 a}+\frac{1}{2} \sqrt{a+\frac{b}{x}} x^2+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x}\right )}{8 a}\\ &=\frac{b \sqrt{a+\frac{b}{x}} x}{4 a}+\frac{1}{2} \sqrt{a+\frac{b}{x}} x^2+\frac{b \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x}}\right )}{4 a}\\ &=\frac{b \sqrt{a+\frac{b}{x}} x}{4 a}+\frac{1}{2} \sqrt{a+\frac{b}{x}} x^2-\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x}}}{\sqrt{a}}\right )}{4 a^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0115659, size = 39, normalized size = 0.57 \[ \frac{2 b^2 \left (a+\frac{b}{x}\right )^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{b}{a x}+1\right )}{3 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/x]*x,x]

[Out]

(2*b^2*(a + b/x)^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, 1 + b/(a*x)])/(3*a^3)

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 96, normalized size = 1.4 \begin{align*}{\frac{x}{8}\sqrt{{\frac{ax+b}{x}}} \left ( 4\,\sqrt{a{x}^{2}+bx}{a}^{5/2}x+2\,\sqrt{a{x}^{2}+bx}{a}^{3/2}b-{b}^{2}\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{a{x}^{2}+bx}\sqrt{a}+2\,ax+b \right ){\frac{1}{\sqrt{a}}}} \right ) a \right ){\frac{1}{\sqrt{ \left ( ax+b \right ) x}}}{a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((a*x+b)/x)^(1/2)*x*(4*(a*x^2+b*x)^(1/2)*a^(5/2)*x+2*(a*x^2+b*x)^(1/2)*a^(3/2)*b-b^2*ln(1/2*(2*(a*x^2+b*x)
^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*a)/((a*x+b)*x)^(1/2)/a^(5/2)

________________________________________________________________________________________

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*(a+b/x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.79178, size = 293, normalized size = 4.25 \begin{align*} \left [\frac{\sqrt{a} b^{2} \log \left (2 \, a x - 2 \, \sqrt{a} x \sqrt{\frac{a x + b}{x}} + b\right ) + 2 \,{\left (2 \, a^{2} x^{2} + a b x\right )} \sqrt{\frac{a x + b}{x}}}{8 \, a^{2}}, \frac{\sqrt{-a} b^{2} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a x + b}{x}}}{a}\right ) +{\left (2 \, a^{2} x^{2} + a b x\right )} \sqrt{\frac{a x + b}{x}}}{4 \, a^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(sqrt(a)*b^2*log(2*a*x - 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(2*a^2*x^2 + a*b*x)*sqrt((a*x + b)/x))/a^
2, 1/4*(sqrt(-a)*b^2*arctan(sqrt(-a)*sqrt((a*x + b)/x)/a) + (2*a^2*x^2 + a*b*x)*sqrt((a*x + b)/x))/a^2]

________________________________________________________________________________________

Sympy [A]  time = 4.76507, size = 97, normalized size = 1.41 \begin{align*} \frac{a x^{\frac{5}{2}}}{2 \sqrt{b} \sqrt{\frac{a x}{b} + 1}} + \frac{3 \sqrt{b} x^{\frac{3}{2}}}{4 \sqrt{\frac{a x}{b} + 1}} + \frac{b^{\frac{3}{2}} \sqrt{x}}{4 a \sqrt{\frac{a x}{b} + 1}} - \frac{b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a} \sqrt{x}}{\sqrt{b}} \right )}}{4 a^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**(5/2)/(2*sqrt(b)*sqrt(a*x/b + 1)) + 3*sqrt(b)*x**(3/2)/(4*sqrt(a*x/b + 1)) + b**(3/2)*sqrt(x)/(4*a*sqrt(a
*x/b + 1)) - b**2*asinh(sqrt(a)*sqrt(x)/sqrt(b))/(4*a**(3/2))

________________________________________________________________________________________

Giac [A]  time = 1.16925, size = 105, normalized size = 1.52 \begin{align*} -\frac{b^{2} \log \left ({\left | b \right |}\right ) \mathrm{sgn}\left (x\right )}{8 \, a^{\frac{3}{2}}} + \frac{1}{8} \,{\left (2 \, \sqrt{a x^{2} + b x}{\left (2 \, x + \frac{b}{a}\right )} + \frac{b^{2} \log \left ({\left | -2 \,{\left (\sqrt{a} x - \sqrt{a x^{2} + b x}\right )} \sqrt{a} - b \right |}\right )}{a^{\frac{3}{2}}}\right )} \mathrm{sgn}\left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*b^2*log(abs(b))*sgn(x)/a^(3/2) + 1/8*(2*sqrt(a*x^2 + b*x)*(2*x + b/a) + b^2*log(abs(-2*(sqrt(a)*x - sqrt(
a*x^2 + b*x))*sqrt(a) - b))/a^(3/2))*sgn(x)