3.1724 \(\int \frac {x}{\sqrt {a+\frac {b}{x}}} \, dx\)

Optimal. Leaf size=72 \[ \frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4 a^{5/2}}-\frac {3 b x \sqrt {a+\frac {b}{x}}}{4 a^2}+\frac {x^2 \sqrt {a+\frac {b}{x}}}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {266, 51, 63, 208} \[ \frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4 a^{5/2}}-\frac {3 b x \sqrt {a+\frac {b}{x}}}{4 a^2}+\frac {x^2 \sqrt {a+\frac {b}{x}}}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a + b/x],x]

[Out]

(-3*b*Sqrt[a + b/x]*x)/(4*a^2) + (Sqrt[a + b/x]*x^2)/(2*a) + (3*b^2*ArcTanh[Sqrt[a + b/x]/Sqrt[a]])/(4*a^(5/2)
)

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 \frac {x}{\sqrt {a+\frac {b}{x}}} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )\\ &=\frac {\sqrt {a+\frac {b}{x}} x^2}{2 a}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{4 a}\\ &=-\frac {3 b \sqrt {a+\frac {b}{x}} x}{4 a^2}+\frac {\sqrt {a+\frac {b}{x}} x^2}{2 a}-\frac {\left (3 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{8 a^2}\\ &=-\frac {3 b \sqrt {a+\frac {b}{x}} x}{4 a^2}+\frac {\sqrt {a+\frac {b}{x}} x^2}{2 a}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )}{4 a^2}\\ &=-\frac {3 b \sqrt {a+\frac {b}{x}} x}{4 a^2}+\frac {\sqrt {a+\frac {b}{x}} x^2}{2 a}+\frac {3 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4 a^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 37, normalized size = 0.51 \[ \frac {2 b^2 \sqrt {a+\frac {b}{x}} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};\frac {b}{a x}+1\right )}{a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[a + b/x],x]

[Out]

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

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fricas [A]  time = 1.05, size = 130, normalized size = 1.81 \[ \left [\frac {3 \, \sqrt {a} b^{2} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (2 \, a^{2} x^{2} - 3 \, a b x\right )} \sqrt {\frac {a x + b}{x}}}{8 \, a^{3}}, -\frac {3 \, \sqrt {-a} b^{2} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) - {\left (2 \, a^{2} x^{2} - 3 \, a b x\right )} \sqrt {\frac {a x + b}{x}}}{4 \, a^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 89, normalized size = 1.24 \[ -\frac {1}{4} \, b^{2} {\left (\frac {3 \, \arctan \left (\frac {\sqrt {\frac {a x + b}{x}}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} - \frac {5 \, a \sqrt {\frac {a x + b}{x}} - \frac {3 \, {\left (a x + b\right )} \sqrt {\frac {a x + b}{x}}}{x}}{{\left (a - \frac {a x + b}{x}\right )}^{2} a^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.01, size = 142, normalized size = 1.97 \[ -\frac {\sqrt {\frac {a x +b}{x}}\, \left (-4 a \,b^{2} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )+a \,b^{2} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )-4 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} x +8 \sqrt {\left (a x +b \right ) x}\, a^{\frac {3}{2}} b -2 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b \right ) x}{8 \sqrt {\left (a x +b \right ) x}\, a^{\frac {7}{2}}} \]

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+8*((a*x+b)*x)^(1/2)*a^(3/2)*b-2*(a*x^2+b*x)^(1/2)*a^(
3/2)*b-4*a*ln(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*b^2+a*b^2*ln(1/2*(2*a*x+b+2*(a*x^2+b*x)^(1/2)
*a^(1/2))/a^(1/2)))/((a*x+b)*x)^(1/2)/a^(7/2)

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maxima [A]  time = 2.34, size = 104, normalized size = 1.44 \[ -\frac {3 \, b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{8 \, a^{\frac {5}{2}}} - \frac {3 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b^{2} - 5 \, \sqrt {a + \frac {b}{x}} a b^{2}}{4 \, {\left ({\left (a + \frac {b}{x}\right )}^{2} a^{2} - 2 \, {\left (a + \frac {b}{x}\right )} a^{3} + a^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*b^2*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^(5/2) - 1/4*(3*(a + b/x)^(3/2)*b^2 - 5*sqr
t(a + b/x)*a*b^2)/((a + b/x)^2*a^2 - 2*(a + b/x)*a^3 + a^4)

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mupad [B]  time = 1.21, size = 57, normalized size = 0.79 \[ \frac {3\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4\,a^{5/2}}+\frac {5\,x^2\,\sqrt {a+\frac {b}{x}}}{4\,a}-\frac {3\,x^2\,{\left (a+\frac {b}{x}\right )}^{3/2}}{4\,a^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b/x)^(1/2),x)

[Out]

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

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sympy [A]  time = 4.48, size = 100, normalized size = 1.39 \[ \frac {x^{\frac {5}{2}}}{2 \sqrt {b} \sqrt {\frac {a x}{b} + 1}} - \frac {\sqrt {b} x^{\frac {3}{2}}}{4 a \sqrt {\frac {a x}{b} + 1}} - \frac {3 b^{\frac {3}{2}} \sqrt {x}}{4 a^{2} \sqrt {\frac {a x}{b} + 1}} + \frac {3 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{4 a^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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