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]

-1/4*b^2*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(3/2)+1/4*b*x*(a+b/x)^(1/2)/a+1/2*x^2*(a+b/x)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 69, normalized size of antiderivative = 1.00, 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 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]

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]]

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*}

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Mathematica [C]  time = 0.01, 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)

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fricas [A]  time = 1.06, size = 125, normalized size = 1.81 \[ \left [\frac {\sqrt {a} b^{2} \log \left (2 \, a x - 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (2 \, a^{2} x^{2} + a b x\right )} \sqrt {\frac {a x + b}{x}}}{8 \, a^{2}}, \frac {\sqrt {-a} b^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) + {\left (2 \, a^{2} x^{2} + a b x\right )} \sqrt {\frac {a x + b}{x}}}{4 \, a^{2}}\right ] \]

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]

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giac [A]  time = 0.21, size = 78, normalized size = 1.13 \[ -\frac {b^{2} \log \left ({\left | b \right |}\right ) \mathrm {sgn}\relax (x)}{8 \, a^{\frac {3}{2}}} + \frac {1}{8} \, {\left (2 \, \sqrt {a x^{2} + b x} {\left (2 \, x + \frac {b}{a}\right )} + \frac {b^{2} \log \left ({\left | -2 \, {\left (\sqrt {a} x - \sqrt {a x^{2} + b x}\right )} \sqrt {a} - b \right |}\right )}{a^{\frac {3}{2}}}\right )} \mathrm {sgn}\relax (x) \]

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)

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maple [A]  time = 0.01, size = 96, normalized size = 1.39 \[ \frac {\sqrt {\frac {a x +b}{x}}\, \left (-a \,b^{2} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )+4 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} x +2 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b \right ) x}{8 \sqrt {\left (a x +b \right ) x}\, a^{\frac {5}{2}}} \]

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+b+2*(a*
x^2+b*x)^(1/2)*a^(1/2))/a^(1/2))*a)/((a*x+b)*x)^(1/2)/a^(5/2)

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maxima [A]  time = 2.39, size = 100, normalized size = 1.45 \[ \frac {b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{8 \, a^{\frac {3}{2}}} + \frac {{\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b^{2} + \sqrt {a + \frac {b}{x}} a b^{2}}{4 \, {\left ({\left (a + \frac {b}{x}\right )}^{2} a - 2 \, {\left (a + \frac {b}{x}\right )} a^{2} + a^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*b^2*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^(3/2) + 1/4*((a + b/x)^(3/2)*b^2 + sqrt(a +
b/x)*a*b^2)/((a + b/x)^2*a - 2*(a + b/x)*a^2 + a^3)

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mupad [B]  time = 1.20, size = 54, normalized size = 0.78 \[ \frac {x^2\,\sqrt {a+\frac {b}{x}}}{4}-\frac {b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4\,a^{3/2}}+\frac {x^2\,{\left (a+\frac {b}{x}\right )}^{3/2}}{4\,a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b/x)^(1/2),x)

[Out]

(x^2*(a + b/x)^(1/2))/4 - (b^2*atanh((a + b/x)^(1/2)/a^(1/2)))/(4*a^(3/2)) + (x^2*(a + b/x)^(3/2))/(4*a)

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sympy [A]  time = 4.18, size = 97, normalized size = 1.41 \[ \frac {a x^{\frac {5}{2}}}{2 \sqrt {b} \sqrt {\frac {a x}{b} + 1}} + \frac {3 \sqrt {b} x^{\frac {3}{2}}}{4 \sqrt {\frac {a x}{b} + 1}} + \frac {b^{\frac {3}{2}} \sqrt {x}}{4 a \sqrt {\frac {a x}{b} + 1}} - \frac {b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{4 a^{\frac {3}{2}}} \]

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))

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