Integrand size = 17, antiderivative size = 95 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {32 b^2 \sqrt {a+\frac {b}{x}} \sqrt {x}}{5 a^4}-\frac {16 b \sqrt {a+\frac {b}{x}} x^{3/2}}{5 a^3}-\frac {2 x^{5/2}}{a \sqrt {a+\frac {b}{x}}}+\frac {12 \sqrt {a+\frac {b}{x}} x^{5/2}}{5 a^2} \] Output:
32/5*b^2*(a+b/x)^(1/2)*x^(1/2)/a^4-16/5*b*(a+b/x)^(1/2)*x^(3/2)/a^3-2*x^(5 /2)/a/(a+b/x)^(1/2)+12/5*(a+b/x)^(1/2)*x^(5/2)/a^2
Time = 3.98 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.62 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {2 \sqrt {a+\frac {b}{x}} \sqrt {x} \left (16 b^3+8 a b^2 x-2 a^2 b x^2+a^3 x^3\right )}{5 a^4 (b+a x)} \] Input:
Integrate[x^(3/2)/(a + b/x)^(3/2),x]
Output:
(2*Sqrt[a + b/x]*Sqrt[x]*(16*b^3 + 8*a*b^2*x - 2*a^2*b*x^2 + a^3*x^3))/(5* a^4*(b + a*x))
Time = 0.36 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.14, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {803, 803, 803, 796}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {2 x^{5/2}}{5 a \sqrt {a+\frac {b}{x}}}-\frac {6 b \int \frac {\sqrt {x}}{\left (a+\frac {b}{x}\right )^{3/2}}dx}{5 a}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {2 x^{5/2}}{5 a \sqrt {a+\frac {b}{x}}}-\frac {6 b \left (\frac {2 x^{3/2}}{3 a \sqrt {a+\frac {b}{x}}}-\frac {4 b \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} \sqrt {x}}dx}{3 a}\right )}{5 a}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {2 x^{5/2}}{5 a \sqrt {a+\frac {b}{x}}}-\frac {6 b \left (\frac {2 x^{3/2}}{3 a \sqrt {a+\frac {b}{x}}}-\frac {4 b \left (\frac {2 \sqrt {x}}{a \sqrt {a+\frac {b}{x}}}-\frac {2 b \int \frac {1}{\left (a+\frac {b}{x}\right )^{3/2} x^{3/2}}dx}{a}\right )}{3 a}\right )}{5 a}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle \frac {2 x^{5/2}}{5 a \sqrt {a+\frac {b}{x}}}-\frac {6 b \left (\frac {2 x^{3/2}}{3 a \sqrt {a+\frac {b}{x}}}-\frac {4 b \left (\frac {4 b}{a^2 \sqrt {x} \sqrt {a+\frac {b}{x}}}+\frac {2 \sqrt {x}}{a \sqrt {a+\frac {b}{x}}}\right )}{3 a}\right )}{5 a}\) |
Input:
Int[x^(3/2)/(a + b/x)^(3/2),x]
Output:
(2*x^(5/2))/(5*a*Sqrt[a + b/x]) - (6*b*((-4*b*((4*b)/(a^2*Sqrt[a + b/x]*Sq rt[x]) + (2*Sqrt[x])/(a*Sqrt[a + b/x])))/(3*a) + (2*x^(3/2))/(3*a*Sqrt[a + b/x])))/(5*a)
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]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x^(m + 1)*(( a + b*x^n)^(p + 1)/(a*(m + 1))), x] - Simp[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] && I LtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]
Time = 0.15 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.55
method | result | size |
orering | \(\frac {2 \left (a^{3} x^{3}-2 a^{2} b \,x^{2}+8 a \,b^{2} x +16 b^{3}\right ) \left (a x +b \right )}{5 a^{4} x^{\frac {3}{2}} \left (a +\frac {b}{x}\right )^{\frac {3}{2}}}\) | \(52\) |
gosper | \(\frac {2 \left (a x +b \right ) \left (a^{3} x^{3}-2 a^{2} b \,x^{2}+8 a \,b^{2} x +16 b^{3}\right )}{5 a^{4} x^{\frac {3}{2}} \left (\frac {a x +b}{x}\right )^{\frac {3}{2}}}\) | \(54\) |
default | \(\frac {2 \sqrt {\frac {a x +b}{x}}\, \sqrt {x}\, \left (a^{3} x^{3}-2 a^{2} b \,x^{2}+8 a \,b^{2} x +16 b^{3}\right )}{5 \left (a x +b \right ) a^{4}}\) | \(56\) |
risch | \(\frac {2 \left (a^{2} x^{2}-3 a b x +11 b^{2}\right ) \left (a x +b \right )}{5 a^{4} \sqrt {\frac {a x +b}{x}}\, \sqrt {x}}+\frac {2 b^{3}}{a^{4} \sqrt {\frac {a x +b}{x}}\, \sqrt {x}}\) | \(66\) |
Input:
int(x^(3/2)/(a+b/x)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/5*(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)/(a+b/x)^(3/ 2)
Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.61 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {2 \, {\left (a^{3} x^{3} - 2 \, a^{2} b x^{2} + 8 \, a b^{2} x + 16 \, b^{3}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{5 \, {\left (a^{5} x + a^{4} b\right )}} \] Input:
integrate(x^(3/2)/(a+b/x)^(3/2),x, algorithm="fricas")
Output:
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)
Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (82) = 164\).
Time = 2.08 (sec) , antiderivative size = 320, normalized size of antiderivative = 3.37 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {2 a^{5} b^{\frac {19}{2}} x^{5} \sqrt {\frac {a x}{b} + 1}}{5 a^{7} b^{9} x^{3} + 15 a^{6} b^{10} x^{2} + 15 a^{5} b^{11} x + 5 a^{4} b^{12}} + \frac {10 a^{3} b^{\frac {23}{2}} x^{3} \sqrt {\frac {a x}{b} + 1}}{5 a^{7} b^{9} x^{3} + 15 a^{6} b^{10} x^{2} + 15 a^{5} b^{11} x + 5 a^{4} b^{12}} + \frac {60 a^{2} b^{\frac {25}{2}} x^{2} \sqrt {\frac {a x}{b} + 1}}{5 a^{7} b^{9} x^{3} + 15 a^{6} b^{10} x^{2} + 15 a^{5} b^{11} x + 5 a^{4} b^{12}} + \frac {80 a b^{\frac {27}{2}} x \sqrt {\frac {a x}{b} + 1}}{5 a^{7} b^{9} x^{3} + 15 a^{6} b^{10} x^{2} + 15 a^{5} b^{11} x + 5 a^{4} b^{12}} + \frac {32 b^{\frac {29}{2}} \sqrt {\frac {a x}{b} + 1}}{5 a^{7} b^{9} x^{3} + 15 a^{6} b^{10} x^{2} + 15 a^{5} b^{11} x + 5 a^{4} b^{12}} \] Input:
integrate(x**(3/2)/(a+b/x)**(3/2),x)
Output:
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**1 2) + 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**1 2) + 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)
Time = 0.03 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.76 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {2 \, b^{3}}{\sqrt {a + \frac {b}{x}} a^{4} \sqrt {x}} + \frac {2 \, {\left ({\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} x^{\frac {5}{2}} - 5 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b x^{\frac {3}{2}} + 15 \, \sqrt {a + \frac {b}{x}} b^{2} \sqrt {x}\right )}}{5 \, a^{4}} \] Input:
integrate(x^(3/2)/(a+b/x)^(3/2),x, algorithm="maxima")
Output:
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
Time = 0.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.83 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=-\frac {32 \, b^{\frac {5}{2}} \mathrm {sgn}\left (x\right )}{5 \, a^{4}} + \frac {2 \, b^{3}}{\sqrt {a x + b} a^{4} \mathrm {sgn}\left (x\right )} + \frac {2 \, {\left ({\left (a x + b\right )}^{\frac {5}{2}} a^{16} - 5 \, {\left (a x + b\right )}^{\frac {3}{2}} a^{16} b + 15 \, \sqrt {a x + b} a^{16} b^{2}\right )}}{5 \, a^{20} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^(3/2)/(a+b/x)^(3/2),x, algorithm="giac")
Output:
-32/5*b^(5/2)*sgn(x)/a^4 + 2*b^3/(sqrt(a*x + b)*a^4*sgn(x)) + 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^2 0*sgn(x))
Time = 0.76 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.62 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {\sqrt {a+\frac {b}{x}}\,\left (\frac {2\,x^{7/2}}{5\,a^2}-\frac {4\,b\,x^{5/2}}{5\,a^3}+\frac {16\,b^2\,x^{3/2}}{5\,a^4}+\frac {32\,b^3\,\sqrt {x}}{5\,a^5}\right )}{x+\frac {b}{a}} \] Input:
int(x^(3/2)/(a + b/x)^(3/2),x)
Output:
((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)
Time = 0.21 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.44 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{3/2}} \, dx=\frac {\frac {2}{5} a^{3} x^{3}-\frac {4}{5} a^{2} b \,x^{2}+\frac {16}{5} a \,b^{2} x +\frac {32}{5} b^{3}}{\sqrt {a x +b}\, a^{4}} \] Input:
int(x^(3/2)/(a+b/x)^(3/2),x)
Output:
(2*(a**3*x**3 - 2*a**2*b*x**2 + 8*a*b**2*x + 16*b**3))/(5*sqrt(a*x + b)*a* *4)