Integrand size = 17, antiderivative size = 126 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx=\frac {256 b^4}{15 a^5 \left (a+\frac {b}{x}\right )^{3/2} x^{3/2}}+\frac {128 b^3}{5 a^4 \left (a+\frac {b}{x}\right )^{3/2} \sqrt {x}}+\frac {32 b^2 \sqrt {x}}{5 a^3 \left (a+\frac {b}{x}\right )^{3/2}}-\frac {16 b x^{3/2}}{15 a^2 \left (a+\frac {b}{x}\right )^{3/2}}+\frac {2 x^{5/2}}{5 a \left (a+\frac {b}{x}\right )^{3/2}} \]
256/15*b^4/a^5/(a+b/x)^(3/2)/x^(3/2)-16/15*b*x^(3/2)/a^2/(a+b/x)^(3/2)+2/5 *x^(5/2)/a/(a+b/x)^(3/2)+128/5*b^3/a^4/(a+b/x)^(3/2)/x^(1/2)+32/5*b^2*x^(1 /2)/a^3/(a+b/x)^(3/2)
Time = 6.34 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.56 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx=\frac {2 \sqrt {a+\frac {b}{x}} \sqrt {x} \left (128 b^4+192 a b^3 x+48 a^2 b^2 x^2-8 a^3 b x^3+3 a^4 x^4\right )}{15 a^5 (b+a x)^2} \]
(2*Sqrt[a + b/x]*Sqrt[x]*(128*b^4 + 192*a*b^3*x + 48*a^2*b^2*x^2 - 8*a^3*b *x^3 + 3*a^4*x^4))/(15*a^5*(b + a*x)^2)
Time = 0.25 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {803, 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.
\(\displaystyle \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {2 x^{5/2}}{5 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {8 b \int \frac {\sqrt {x}}{\left (a+\frac {b}{x}\right )^{5/2}}dx}{5 a}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {2 x^{5/2}}{5 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {8 b \left (\frac {2 x^{3/2}}{3 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {2 b \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} \sqrt {x}}dx}{a}\right )}{5 a}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {2 x^{5/2}}{5 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {8 b \left (\frac {2 x^{3/2}}{3 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {2 b \left (\frac {2 \sqrt {x}}{a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {4 b \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{3/2}}dx}{a}\right )}{a}\right )}{5 a}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {2 x^{5/2}}{5 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {8 b \left (\frac {2 x^{3/2}}{3 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {2 b \left (\frac {2 \sqrt {x}}{a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {4 b \left (\frac {2 b \int \frac {1}{\left (a+\frac {b}{x}\right )^{5/2} x^{5/2}}dx}{a}-\frac {2}{a \sqrt {x} \left (a+\frac {b}{x}\right )^{3/2}}\right )}{a}\right )}{a}\right )}{5 a}\) |
\(\Big \downarrow \) 796 |
\(\displaystyle \frac {2 x^{5/2}}{5 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {8 b \left (\frac {2 x^{3/2}}{3 a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {2 b \left (\frac {2 \sqrt {x}}{a \left (a+\frac {b}{x}\right )^{3/2}}-\frac {4 b \left (-\frac {4 b}{3 a^2 x^{3/2} \left (a+\frac {b}{x}\right )^{3/2}}-\frac {2}{a \sqrt {x} \left (a+\frac {b}{x}\right )^{3/2}}\right )}{a}\right )}{a}\right )}{5 a}\) |
(2*x^(5/2))/(5*a*(a + b/x)^(3/2)) - (8*b*((-2*b*((-4*b*((-4*b)/(3*a^2*(a + b/x)^(3/2)*x^(3/2)) - 2/(a*(a + b/x)^(3/2)*Sqrt[x])))/a + (2*Sqrt[x])/(a* (a + b/x)^(3/2))))/a + (2*x^(3/2))/(3*a*(a + b/x)^(3/2))))/(5*a)
3.18.96.3.1 Defintions of rubi rules used
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 = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.52
method | result | size |
gosper | \(\frac {2 \left (a x +b \right ) \left (3 a^{4} x^{4}-8 a^{3} b \,x^{3}+48 a^{2} b^{2} x^{2}+192 a \,b^{3} x +128 b^{4}\right )}{15 a^{5} x^{\frac {5}{2}} \left (\frac {a x +b}{x}\right )^{\frac {5}{2}}}\) | \(66\) |
default | \(\frac {2 \sqrt {\frac {a x +b}{x}}\, \sqrt {x}\, \left (3 a^{4} x^{4}-8 a^{3} b \,x^{3}+48 a^{2} b^{2} x^{2}+192 a \,b^{3} x +128 b^{4}\right )}{15 \left (a x +b \right )^{2} a^{5}}\) | \(68\) |
risch | \(\frac {2 \left (3 a^{2} x^{2}-14 a b x +73 b^{2}\right ) \left (a x +b \right )}{15 a^{5} \sqrt {x}\, \sqrt {\frac {a x +b}{x}}}+\frac {2 b^{3} \left (12 a x +11 b \right )}{3 a^{5} \left (a x +b \right ) \sqrt {x}\, \sqrt {\frac {a x +b}{x}}}\) | \(82\) |
2/15*(a*x+b)*(3*a^4*x^4-8*a^3*b*x^3+48*a^2*b^2*x^2+192*a*b^3*x+128*b^4)/a^ 5/x^(5/2)/((a*x+b)/x)^(5/2)
Time = 0.26 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.64 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx=\frac {2 \, {\left (3 \, a^{4} x^{4} - 8 \, a^{3} b x^{3} + 48 \, a^{2} b^{2} x^{2} + 192 \, a b^{3} x + 128 \, b^{4}\right )} \sqrt {x} \sqrt {\frac {a x + b}{x}}}{15 \, {\left (a^{7} x^{2} + 2 \, a^{6} b x + a^{5} b^{2}\right )}} \]
2/15*(3*a^4*x^4 - 8*a^3*b*x^3 + 48*a^2*b^2*x^2 + 192*a*b^3*x + 128*b^4)*sq rt(x)*sqrt((a*x + b)/x)/(a^7*x^2 + 2*a^6*b*x + a^5*b^2)
Leaf count of result is larger than twice the leaf count of optimal. 536 vs. \(2 (110) = 220\).
Time = 3.49 (sec) , antiderivative size = 536, normalized size of antiderivative = 4.25 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx=\frac {6 a^{6} b^{\frac {33}{2}} x^{6} \sqrt {\frac {a x}{b} + 1}}{15 a^{9} b^{16} x^{4} + 60 a^{8} b^{17} x^{3} + 90 a^{7} b^{18} x^{2} + 60 a^{6} b^{19} x + 15 a^{5} b^{20}} - \frac {4 a^{5} b^{\frac {35}{2}} x^{5} \sqrt {\frac {a x}{b} + 1}}{15 a^{9} b^{16} x^{4} + 60 a^{8} b^{17} x^{3} + 90 a^{7} b^{18} x^{2} + 60 a^{6} b^{19} x + 15 a^{5} b^{20}} + \frac {70 a^{4} b^{\frac {37}{2}} x^{4} \sqrt {\frac {a x}{b} + 1}}{15 a^{9} b^{16} x^{4} + 60 a^{8} b^{17} x^{3} + 90 a^{7} b^{18} x^{2} + 60 a^{6} b^{19} x + 15 a^{5} b^{20}} + \frac {560 a^{3} b^{\frac {39}{2}} x^{3} \sqrt {\frac {a x}{b} + 1}}{15 a^{9} b^{16} x^{4} + 60 a^{8} b^{17} x^{3} + 90 a^{7} b^{18} x^{2} + 60 a^{6} b^{19} x + 15 a^{5} b^{20}} + \frac {1120 a^{2} b^{\frac {41}{2}} x^{2} \sqrt {\frac {a x}{b} + 1}}{15 a^{9} b^{16} x^{4} + 60 a^{8} b^{17} x^{3} + 90 a^{7} b^{18} x^{2} + 60 a^{6} b^{19} x + 15 a^{5} b^{20}} + \frac {896 a b^{\frac {43}{2}} x \sqrt {\frac {a x}{b} + 1}}{15 a^{9} b^{16} x^{4} + 60 a^{8} b^{17} x^{3} + 90 a^{7} b^{18} x^{2} + 60 a^{6} b^{19} x + 15 a^{5} b^{20}} + \frac {256 b^{\frac {45}{2}} \sqrt {\frac {a x}{b} + 1}}{15 a^{9} b^{16} x^{4} + 60 a^{8} b^{17} x^{3} + 90 a^{7} b^{18} x^{2} + 60 a^{6} b^{19} x + 15 a^{5} b^{20}} \]
6*a**6*b**(33/2)*x**6*sqrt(a*x/b + 1)/(15*a**9*b**16*x**4 + 60*a**8*b**17* x**3 + 90*a**7*b**18*x**2 + 60*a**6*b**19*x + 15*a**5*b**20) - 4*a**5*b**( 35/2)*x**5*sqrt(a*x/b + 1)/(15*a**9*b**16*x**4 + 60*a**8*b**17*x**3 + 90*a **7*b**18*x**2 + 60*a**6*b**19*x + 15*a**5*b**20) + 70*a**4*b**(37/2)*x**4 *sqrt(a*x/b + 1)/(15*a**9*b**16*x**4 + 60*a**8*b**17*x**3 + 90*a**7*b**18* x**2 + 60*a**6*b**19*x + 15*a**5*b**20) + 560*a**3*b**(39/2)*x**3*sqrt(a*x /b + 1)/(15*a**9*b**16*x**4 + 60*a**8*b**17*x**3 + 90*a**7*b**18*x**2 + 60 *a**6*b**19*x + 15*a**5*b**20) + 1120*a**2*b**(41/2)*x**2*sqrt(a*x/b + 1)/ (15*a**9*b**16*x**4 + 60*a**8*b**17*x**3 + 90*a**7*b**18*x**2 + 60*a**6*b* *19*x + 15*a**5*b**20) + 896*a*b**(43/2)*x*sqrt(a*x/b + 1)/(15*a**9*b**16* x**4 + 60*a**8*b**17*x**3 + 90*a**7*b**18*x**2 + 60*a**6*b**19*x + 15*a**5 *b**20) + 256*b**(45/2)*sqrt(a*x/b + 1)/(15*a**9*b**16*x**4 + 60*a**8*b**1 7*x**3 + 90*a**7*b**18*x**2 + 60*a**6*b**19*x + 15*a**5*b**20)
Time = 0.19 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.71 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx=\frac {2 \, {\left (3 \, {\left (a + \frac {b}{x}\right )}^{\frac {5}{2}} x^{\frac {5}{2}} - 20 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} b x^{\frac {3}{2}} + 90 \, \sqrt {a + \frac {b}{x}} b^{2} \sqrt {x}\right )}}{15 \, a^{5}} + \frac {2 \, {\left (12 \, {\left (a + \frac {b}{x}\right )} b^{3} x - b^{4}\right )}}{3 \, {\left (a + \frac {b}{x}\right )}^{\frac {3}{2}} a^{5} x^{\frac {3}{2}}} \]
2/15*(3*(a + b/x)^(5/2)*x^(5/2) - 20*(a + b/x)^(3/2)*b*x^(3/2) + 90*sqrt(a + b/x)*b^2*sqrt(x))/a^5 + 2/3*(12*(a + b/x)*b^3*x - b^4)/((a + b/x)^(3/2) *a^5*x^(3/2))
Time = 0.29 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.60 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx=\frac {2 \, {\left (12 \, {\left (a x + b\right )} b^{3} - b^{4}\right )}}{3 \, {\left (a x + b\right )}^{\frac {3}{2}} a^{5}} + \frac {2 \, {\left (3 \, {\left (a x + b\right )}^{\frac {5}{2}} a^{20} - 20 \, {\left (a x + b\right )}^{\frac {3}{2}} a^{20} b + 90 \, \sqrt {a x + b} a^{20} b^{2}\right )}}{15 \, a^{25}} \]
2/3*(12*(a*x + b)*b^3 - b^4)/((a*x + b)^(3/2)*a^5) + 2/15*(3*(a*x + b)^(5/ 2)*a^20 - 20*(a*x + b)^(3/2)*a^20*b + 90*sqrt(a*x + b)*a^20*b^2)/a^25
Time = 6.48 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.64 \[ \int \frac {x^{3/2}}{\left (a+\frac {b}{x}\right )^{5/2}} \, dx=\frac {\sqrt {a+\frac {b}{x}}\,\left (\frac {2\,x^{9/2}}{5\,a^3}-\frac {16\,b\,x^{7/2}}{15\,a^4}+\frac {32\,b^2\,x^{5/2}}{5\,a^5}+\frac {128\,b^3\,x^{3/2}}{5\,a^6}+\frac {256\,b^4\,\sqrt {x}}{15\,a^7}\right )}{x^2+\frac {b^2}{a^2}+\frac {2\,b\,x}{a}} \]