3.1733 \(\int \frac {x}{(a+\frac {b}{x})^{3/2}} \, dx\)

Optimal. Leaf size=93 \[ \frac {15 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4 a^{7/2}}-\frac {15 b^2}{4 a^3 \sqrt {a+\frac {b}{x}}}-\frac {5 b x}{4 a^2 \sqrt {a+\frac {b}{x}}}+\frac {x^2}{2 a \sqrt {a+\frac {b}{x}}} \]

[Out]

15/4*b^2*arctanh((a+b/x)^(1/2)/a^(1/2))/a^(7/2)-15/4*b^2/a^3/(a+b/x)^(1/2)-5/4*b*x/a^2/(a+b/x)^(1/2)+1/2*x^2/a
/(a+b/x)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 91, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {266, 51, 63, 208} \[ \frac {15 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4 a^{7/2}}+\frac {5 x^2 \sqrt {a+\frac {b}{x}}}{2 a^2}-\frac {15 b x \sqrt {a+\frac {b}{x}}}{4 a^3}-\frac {2 x^2}{a \sqrt {a+\frac {b}{x}}} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b/x)^(3/2),x]

[Out]

(-15*b*Sqrt[a + b/x]*x)/(4*a^3) - (2*x^2)/(a*Sqrt[a + b/x]) + (5*Sqrt[a + b/x]*x^2)/(2*a^2) + (15*b^2*ArcTanh[
Sqrt[a + b/x]/Sqrt[a]])/(4*a^(7/2))

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}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{x^3 (a+b x)^{3/2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {2 x^2}{a \sqrt {a+\frac {b}{x}}}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{a}\\ &=-\frac {2 x^2}{a \sqrt {a+\frac {b}{x}}}+\frac {5 \sqrt {a+\frac {b}{x}} x^2}{2 a^2}+\frac {(15 b) \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{4 a^2}\\ &=-\frac {15 b \sqrt {a+\frac {b}{x}} x}{4 a^3}-\frac {2 x^2}{a \sqrt {a+\frac {b}{x}}}+\frac {5 \sqrt {a+\frac {b}{x}} x^2}{2 a^2}-\frac {\left (15 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\frac {1}{x}\right )}{8 a^3}\\ &=-\frac {15 b \sqrt {a+\frac {b}{x}} x}{4 a^3}-\frac {2 x^2}{a \sqrt {a+\frac {b}{x}}}+\frac {5 \sqrt {a+\frac {b}{x}} x^2}{2 a^2}-\frac {(15 b) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+\frac {b}{x}}\right )}{4 a^3}\\ &=-\frac {15 b \sqrt {a+\frac {b}{x}} x}{4 a^3}-\frac {2 x^2}{a \sqrt {a+\frac {b}{x}}}+\frac {5 \sqrt {a+\frac {b}{x}} x^2}{2 a^2}+\frac {15 b^2 \tanh ^{-1}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4 a^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 37, normalized size = 0.40 \[ -\frac {2 b^2 \, _2F_1\left (-\frac {1}{2},3;\frac {1}{2};\frac {b}{a x}+1\right )}{a^3 \sqrt {a+\frac {b}{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b/x)^(3/2),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.96, size = 186, normalized size = 2.00 \[ \left [\frac {15 \, {\left (a b^{2} x + b^{3}\right )} \sqrt {a} \log \left (2 \, a x + 2 \, \sqrt {a} x \sqrt {\frac {a x + b}{x}} + b\right ) + 2 \, {\left (2 \, a^{3} x^{3} - 5 \, a^{2} b x^{2} - 15 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{8 \, {\left (a^{5} x + a^{4} b\right )}}, -\frac {15 \, {\left (a b^{2} x + b^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {\frac {a x + b}{x}}}{a}\right ) - {\left (2 \, a^{3} x^{3} - 5 \, a^{2} b x^{2} - 15 \, a b^{2} x\right )} \sqrt {\frac {a x + b}{x}}}{4 \, {\left (a^{5} x + a^{4} b\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b/x)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(15*(a*b^2*x + b^3)*sqrt(a)*log(2*a*x + 2*sqrt(a)*x*sqrt((a*x + b)/x) + b) + 2*(2*a^3*x^3 - 5*a^2*b*x^2 -
15*a*b^2*x)*sqrt((a*x + b)/x))/(a^5*x + a^4*b), -1/4*(15*(a*b^2*x + b^3)*sqrt(-a)*arctan(sqrt(-a)*sqrt((a*x +
b)/x)/a) - (2*a^3*x^3 - 5*a^2*b*x^2 - 15*a*b^2*x)*sqrt((a*x + b)/x))/(a^5*x + a^4*b)]

________________________________________________________________________________________

giac [A]  time = 0.17, size = 105, normalized size = 1.13 \[ -\frac {1}{4} \, b^{2} {\left (\frac {15 \, \arctan \left (\frac {\sqrt {\frac {a x + b}{x}}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {8}{a^{3} \sqrt {\frac {a x + b}{x}}} - \frac {9 \, a \sqrt {\frac {a x + b}{x}} - \frac {7 \, {\left (a x + b\right )} \sqrt {\frac {a x + b}{x}}}{x}}{{\left (a - \frac {a x + b}{x}\right )}^{2} a^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b/x)^(3/2),x, algorithm="giac")

[Out]

-1/4*b^2*(15*arctan(sqrt((a*x + b)/x)/sqrt(-a))/(sqrt(-a)*a^3) + 8/(a^3*sqrt((a*x + b)/x)) - (9*a*sqrt((a*x +
b)/x) - 7*(a*x + b)*sqrt((a*x + b)/x)/x)/((a - (a*x + b)/x)^2*a^3))

________________________________________________________________________________________

maple [B]  time = 0.01, size = 395, normalized size = 4.25 \[ \frac {\sqrt {\frac {a x +b}{x}}\, \left (16 a^{3} b^{2} x^{2} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-a^{3} b^{2} x^{2} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )+4 \sqrt {a \,x^{2}+b x}\, a^{\frac {9}{2}} x^{3}+32 a^{2} b^{3} x \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-2 a^{2} b^{3} x \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )+10 \sqrt {a \,x^{2}+b x}\, a^{\frac {7}{2}} b \,x^{2}-32 \sqrt {\left (a x +b \right ) x}\, a^{\frac {7}{2}} b \,x^{2}+16 a \,b^{4} \ln \left (\frac {2 a x +b +2 \sqrt {\left (a x +b \right ) x}\, \sqrt {a}}{2 \sqrt {a}}\right )-a \,b^{4} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x}\, \sqrt {a}}{2 \sqrt {a}}\right )+8 \sqrt {a \,x^{2}+b x}\, a^{\frac {5}{2}} b^{2} x -64 \sqrt {\left (a x +b \right ) x}\, a^{\frac {5}{2}} b^{2} x +2 \sqrt {a \,x^{2}+b x}\, a^{\frac {3}{2}} b^{3}-32 \sqrt {\left (a x +b \right ) x}\, a^{\frac {3}{2}} b^{3}+16 \left (\left (a x +b \right ) x \right )^{\frac {3}{2}} a^{\frac {5}{2}} b \right ) x}{8 \sqrt {\left (a x +b \right ) x}\, \left (a x +b \right )^{2} a^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a+b/x)^(3/2),x)

[Out]

1/8*((a*x+b)/x)^(1/2)*x/a^(9/2)*(4*(a*x^2+b*x)^(1/2)*a^(9/2)*x^3+10*(a*x^2+b*x)^(1/2)*a^(7/2)*x^2*b-32*a^(7/2)
*((a*x+b)*x)^(1/2)*x^2*b+16*a^3*ln(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*x^2*b^2+8*(a*x^2+b*x)^(1
/2)*a^(5/2)*b^2*x+16*a^(5/2)*((a*x+b)*x)^(3/2)*b-64*a^(5/2)*((a*x+b)*x)^(1/2)*x*b^2+32*a^2*ln(1/2*(2*a*x+b+2*(
(a*x+b)*x)^(1/2)*a^(1/2))/a^(1/2))*x*b^3-ln(1/2*(2*a*x+b+2*(a*x^2+b*x)^(1/2)*a^(1/2))/a^(1/2))*x^2*a^3*b^2+2*(
a*x^2+b*x)^(1/2)*a^(3/2)*b^3-32*((a*x+b)*x)^(1/2)*a^(3/2)*b^3+16*a*b^4*ln(1/2*(2*a*x+b+2*((a*x+b)*x)^(1/2)*a^(
1/2))/a^(1/2))-2*ln(1/2*(2*a*x+b+2*(a*x^2+b*x)^(1/2)*a^(1/2))/a^(1/2))*x*a^2*b^3-a*b^4*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*x+b)^2

________________________________________________________________________________________

maxima [A]  time = 2.37, size = 122, normalized size = 1.31 \[ -\frac {15 \, {\left (a + \frac {b}{x}\right )}^{2} b^{2} - 25 \, {\left (a + \frac {b}{x}\right )} a b^{2} + 8 \, a^{2} b^{2}}{4 \, {\left ({\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} a^{3} - 2 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{4} + \sqrt {a + \frac {b}{x}} a^{5}\right )}} - \frac {15 \, b^{2} \log \left (\frac {\sqrt {a + \frac {b}{x}} - \sqrt {a}}{\sqrt {a + \frac {b}{x}} + \sqrt {a}}\right )}{8 \, a^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b/x)^(3/2),x, algorithm="maxima")

[Out]

-1/4*(15*(a + b/x)^2*b^2 - 25*(a + b/x)*a*b^2 + 8*a^2*b^2)/((a + b/x)^(5/2)*a^3 - 2*(a + b/x)^(3/2)*a^4 + sqrt
(a + b/x)*a^5) - 15/8*b^2*log((sqrt(a + b/x) - sqrt(a))/(sqrt(a + b/x) + sqrt(a)))/a^(7/2)

________________________________________________________________________________________

mupad [B]  time = 1.35, size = 73, normalized size = 0.78 \[ \frac {15\,b^2\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{x}}}{\sqrt {a}}\right )}{4\,a^{7/2}}-\frac {15\,b^2}{4\,a^3\,\sqrt {a+\frac {b}{x}}}+\frac {x^2}{2\,a\,\sqrt {a+\frac {b}{x}}}-\frac {5\,b\,x}{4\,a^2\,\sqrt {a+\frac {b}{x}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b/x)^(3/2),x)

[Out]

(15*b^2*atanh((a + b/x)^(1/2)/a^(1/2)))/(4*a^(7/2)) - (15*b^2)/(4*a^3*(a + b/x)^(1/2)) + x^2/(2*a*(a + b/x)^(1
/2)) - (5*b*x)/(4*a^2*(a + b/x)^(1/2))

________________________________________________________________________________________

sympy [A]  time = 5.97, size = 105, normalized size = 1.13 \[ \frac {x^{\frac {5}{2}}}{2 a \sqrt {b} \sqrt {\frac {a x}{b} + 1}} - \frac {5 \sqrt {b} x^{\frac {3}{2}}}{4 a^{2} \sqrt {\frac {a x}{b} + 1}} - \frac {15 b^{\frac {3}{2}} \sqrt {x}}{4 a^{3} \sqrt {\frac {a x}{b} + 1}} + \frac {15 b^{2} \operatorname {asinh}{\left (\frac {\sqrt {a} \sqrt {x}}{\sqrt {b}} \right )}}{4 a^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b/x)**(3/2),x)

[Out]

x**(5/2)/(2*a*sqrt(b)*sqrt(a*x/b + 1)) - 5*sqrt(b)*x**(3/2)/(4*a**2*sqrt(a*x/b + 1)) - 15*b**(3/2)*sqrt(x)/(4*
a**3*sqrt(a*x/b + 1)) + 15*b**2*asinh(sqrt(a)*sqrt(x)/sqrt(b))/(4*a**(7/2))

________________________________________________________________________________________