3.1787 \(\int \frac {x^{3/2}}{(a+\frac {b}{x})^{3/2}} \, dx\)

Optimal. Leaf size=100 \[ \frac {32 b^3}{5 a^4 \sqrt {x} \sqrt {a+\frac {b}{x}}}+\frac {16 b^2 \sqrt {x}}{5 a^3 \sqrt {a+\frac {b}{x}}}-\frac {4 b x^{3/2}}{5 a^2 \sqrt {a+\frac {b}{x}}}+\frac {2 x^{5/2}}{5 a \sqrt {a+\frac {b}{x}}} \]

[Out]

-4/5*b*x^(3/2)/a^2/(a+b/x)^(1/2)+2/5*x^(5/2)/a/(a+b/x)^(1/2)+32/5*b^3/a^4/(a+b/x)^(1/2)/x^(1/2)+16/5*b^2*x^(1/
2)/a^3/(a+b/x)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {271, 264} \[ \frac {32 b^3}{5 a^4 \sqrt {x} \sqrt {a+\frac {b}{x}}}+\frac {16 b^2 \sqrt {x}}{5 a^3 \sqrt {a+\frac {b}{x}}}-\frac {4 b x^{3/2}}{5 a^2 \sqrt {a+\frac {b}{x}}}+\frac {2 x^{5/2}}{5 a \sqrt {a+\frac {b}{x}}} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(a + b/x)^(3/2),x]

[Out]

(32*b^3)/(5*a^4*Sqrt[a + b/x]*Sqrt[x]) + (16*b^2*Sqrt[x])/(5*a^3*Sqrt[a + b/x]) - (4*b*x^(3/2))/(5*a^2*Sqrt[a
+ b/x]) + (2*x^(5/2))/(5*a*Sqrt[a + b/x])

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx &=\frac {2 x^{5/2}}{5 a \sqrt {a+\frac {b}{x}}}-\frac {(6 b) \int \frac {\sqrt {x}}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx}{5 a}\\ &=-\frac {4 b x^{3/2}}{5 a^2 \sqrt {a+\frac {b}{x}}}+\frac {2 x^{5/2}}{5 a \sqrt {a+\frac {b}{x}}}+\frac {\left (8 b^2\right ) \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} \sqrt {x}} \, dx}{5 a^2}\\ &=\frac {16 b^2 \sqrt {x}}{5 a^3 \sqrt {a+\frac {b}{x}}}-\frac {4 b x^{3/2}}{5 a^2 \sqrt {a+\frac {b}{x}}}+\frac {2 x^{5/2}}{5 a \sqrt {a+\frac {b}{x}}}-\frac {\left (16 b^3\right ) \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{3/2}} \, dx}{5 a^3}\\ &=\frac {32 b^3}{5 a^4 \sqrt {a+\frac {b}{x}} \sqrt {x}}+\frac {16 b^2 \sqrt {x}}{5 a^3 \sqrt {a+\frac {b}{x}}}-\frac {4 b x^{3/2}}{5 a^2 \sqrt {a+\frac {b}{x}}}+\frac {2 x^{5/2}}{5 a \sqrt {a+\frac {b}{x}}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 52, normalized size = 0.52 \[ \frac {2 \left (a^3 x^3-2 a^2 b x^2+8 a b^2 x+16 b^3\right )}{5 a^4 \sqrt {x} \sqrt {a+\frac {b}{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(a + b/x)^(3/2),x]

[Out]

(2*(16*b^3 + 8*a*b^2*x - 2*a^2*b*x^2 + a^3*x^3))/(5*a^4*Sqrt[a + b/x]*Sqrt[x])

________________________________________________________________________________________

fricas [A]  time = 0.83, size = 58, normalized size = 0.58 \[ \frac {2 \, {\left (a^{3} x^{3} - 2 \, a^{2} b x^{2} + 8 \, a b^{2} x + 16 \, b^{3}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{5 \, {\left (a^{5} x + a^{4} b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(a^3*x^3 - 2*a^2*b*x^2 + 8*a*b^2*x + 16*b^3)*sqrt(x)*sqrt((a*x + b)/x)/(a^5*x + a^4*b)

________________________________________________________________________________________

giac [A]  time = 0.21, size = 69, normalized size = 0.69 \[ -\frac {32 \, b^{\frac {5}{2}}}{5 \, a^{4}} + \frac {2 \, b^{3}}{\sqrt {a x + b} a^{4}} + \frac {2 \, {\left ({\left (a x + b\right )}^{\frac {5}{2}} a^{16} - 5 \, {\left (a x + b\right )}^{\frac {3}{2}} a^{16} b + 15 \, \sqrt {a x + b} a^{16} b^{2}\right )}}{5 \, a^{20}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-32/5*b^(5/2)/a^4 + 2*b^3/(sqrt(a*x + b)*a^4) + 2/5*((a*x + b)^(5/2)*a^16 - 5*(a*x + b)^(3/2)*a^16*b + 15*sqrt
(a*x + b)*a^16*b^2)/a^20

________________________________________________________________________________________

maple [A]  time = 0.01, size = 54, normalized size = 0.54 \[ \frac {2 \left (a x +b \right ) \left (a^{3} x^{3}-2 a^{2} b \,x^{2}+8 a \,b^{2} x +16 b^{3}\right )}{5 \left (\frac {a x +b}{x}\right )^{\frac {3}{2}} a^{4} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(a*x+b)*(a^3*x^3-2*a^2*b*x^2+8*a*b^2*x+16*b^3)/a^4/x^(3/2)/((a*x+b)/x)^(3/2)

________________________________________________________________________________________

maxima [A]  time = 1.04, size = 72, normalized size = 0.72 \[ \frac {2 \, b^{3}}{\sqrt {a + \frac {b}{x}} a^{4} \sqrt {x}} + \frac {2 \, {\left ({\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} x^{\frac {5}{2}} - 5 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b x^{\frac {3}{2}} + 15 \, \sqrt {a + \frac {b}{x}} b^{2} \sqrt {x}\right )}}{5 \, a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^3/(sqrt(a + b/x)*a^4*sqrt(x)) + 2/5*((a + b/x)^(5/2)*x^(5/2) - 5*(a + b/x)^(3/2)*b*x^(3/2) + 15*sqrt(a + b
/x)*b^2*sqrt(x))/a^4

________________________________________________________________________________________

mupad [B]  time = 1.52, size = 59, normalized size = 0.59 \[ \frac {\sqrt {a+\frac {b}{x}}\,\left (\frac {2\,x^{7/2}}{5\,a^2}-\frac {4\,b\,x^{5/2}}{5\,a^3}+\frac {16\,b^2\,x^{3/2}}{5\,a^4}+\frac {32\,b^3\,\sqrt {x}}{5\,a^5}\right )}{x+\frac {b}{a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(a + b/x)^(3/2),x)

[Out]

((a + b/x)^(1/2)*((2*x^(7/2))/(5*a^2) - (4*b*x^(5/2))/(5*a^3) + (16*b^2*x^(3/2))/(5*a^4) + (32*b^3*x^(1/2))/(5
*a^5)))/(x + b/a)

________________________________________________________________________________________

sympy [B]  time = 5.10, size = 320, normalized size = 3.20 \[ \frac {2 a^{5} b^{\frac {19}{2}} x^{5} \sqrt {\frac {a x}{b} + 1}}{5 a^{7} b^{9} x^{3} + 15 a^{6} b^{10} x^{2} + 15 a^{5} b^{11} x + 5 a^{4} b^{12}} + \frac {10 a^{3} b^{\frac {23}{2}} x^{3} \sqrt {\frac {a x}{b} + 1}}{5 a^{7} b^{9} x^{3} + 15 a^{6} b^{10} x^{2} + 15 a^{5} b^{11} x + 5 a^{4} b^{12}} + \frac {60 a^{2} b^{\frac {25}{2}} x^{2} \sqrt {\frac {a x}{b} + 1}}{5 a^{7} b^{9} x^{3} + 15 a^{6} b^{10} x^{2} + 15 a^{5} b^{11} x + 5 a^{4} b^{12}} + \frac {80 a b^{\frac {27}{2}} x \sqrt {\frac {a x}{b} + 1}}{5 a^{7} b^{9} x^{3} + 15 a^{6} b^{10} x^{2} + 15 a^{5} b^{11} x + 5 a^{4} b^{12}} + \frac {32 b^{\frac {29}{2}} \sqrt {\frac {a x}{b} + 1}}{5 a^{7} b^{9} x^{3} + 15 a^{6} b^{10} x^{2} + 15 a^{5} b^{11} x + 5 a^{4} b^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a**5*b**(19/2)*x**5*sqrt(a*x/b + 1)/(5*a**7*b**9*x**3 + 15*a**6*b**10*x**2 + 15*a**5*b**11*x + 5*a**4*b**12)
+ 10*a**3*b**(23/2)*x**3*sqrt(a*x/b + 1)/(5*a**7*b**9*x**3 + 15*a**6*b**10*x**2 + 15*a**5*b**11*x + 5*a**4*b*
*12) + 60*a**2*b**(25/2)*x**2*sqrt(a*x/b + 1)/(5*a**7*b**9*x**3 + 15*a**6*b**10*x**2 + 15*a**5*b**11*x + 5*a**
4*b**12) + 80*a*b**(27/2)*x*sqrt(a*x/b + 1)/(5*a**7*b**9*x**3 + 15*a**6*b**10*x**2 + 15*a**5*b**11*x + 5*a**4*
b**12) + 32*b**(29/2)*sqrt(a*x/b + 1)/(5*a**7*b**9*x**3 + 15*a**6*b**10*x**2 + 15*a**5*b**11*x + 5*a**4*b**12)

________________________________________________________________________________________