Integrand size = 15, antiderivative size = 82 \[ \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {2 b x}{3 a^2 \left (a+\frac {b}{x^2}\right )^{3/2}}+\frac {8 b x}{3 a^3 \sqrt {a+\frac {b}{x^2}}}-\frac {16 b \sqrt {a+\frac {b}{x^2}} x}{3 a^4}+\frac {x^3}{3 a \left (a+\frac {b}{x^2}\right )^{3/2}} \]
2/3*b*x/a^2/(a+b/x^2)^(3/2)+1/3*x^3/a/(a+b/x^2)^(3/2)+8/3*b*x/a^3/(a+b/x^2 )^(1/2)-16/3*b*x*(a+b/x^2)^(1/2)/a^4
Time = 0.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.72 \[ \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {\left (b+a x^2\right ) \left (-16 b^3-24 a b^2 x^2-6 a^2 b x^4+a^3 x^6\right )}{3 a^4 \left (a+\frac {b}{x^2}\right )^{5/2} x^5} \]
Time = 0.20 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.16, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {803, 773, 245, 209, 208}
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.
\(\displaystyle \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {x^3}{3 a \left (a+\frac {b}{x^2}\right )^{3/2}}-\frac {2 b \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2}}dx}{a}\) |
\(\Big \downarrow \) 773 |
\(\displaystyle \frac {2 b \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}}d\frac {1}{x}}{a}+\frac {x^3}{3 a \left (a+\frac {b}{x^2}\right )^{3/2}}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle \frac {2 b \left (-\frac {4 b \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{5/2}}d\frac {1}{x}}{a}-\frac {x}{a \left (a+\frac {b}{x^2}\right )^{3/2}}\right )}{a}+\frac {x^3}{3 a \left (a+\frac {b}{x^2}\right )^{3/2}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {2 b \left (-\frac {4 b \left (\frac {2 \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2}}d\frac {1}{x}}{3 a}+\frac {1}{3 a x \left (a+\frac {b}{x^2}\right )^{3/2}}\right )}{a}-\frac {x}{a \left (a+\frac {b}{x^2}\right )^{3/2}}\right )}{a}+\frac {x^3}{3 a \left (a+\frac {b}{x^2}\right )^{3/2}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {2 b \left (-\frac {4 b \left (\frac {2}{3 a^2 x \sqrt {a+\frac {b}{x^2}}}+\frac {1}{3 a x \left (a+\frac {b}{x^2}\right )^{3/2}}\right )}{a}-\frac {x}{a \left (a+\frac {b}{x^2}\right )^{3/2}}\right )}{a}+\frac {x^3}{3 a \left (a+\frac {b}{x^2}\right )^{3/2}}\) |
x^3/(3*a*(a + b/x^2)^(3/2)) + (2*b*((-4*b*(1/(3*a*(a + b/x^2)^(3/2)*x) + 2 /(3*a^2*Sqrt[a + b/x^2]*x)))/a - x/(a*(a + b/x^2)^(3/2))))/a
3.20.51.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^ 2, x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && !IntegerQ[p]
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 = 0.06 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.73
method | result | size |
gosper | \(\frac {\left (a \,x^{2}+b \right ) \left (x^{6} a^{3}-6 a^{2} b \,x^{4}-24 a \,b^{2} x^{2}-16 b^{3}\right )}{3 a^{4} x^{5} \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}}}\) | \(60\) |
default | \(\frac {\left (a \,x^{2}+b \right ) \left (x^{6} a^{3}-6 a^{2} b \,x^{4}-24 a \,b^{2} x^{2}-16 b^{3}\right )}{3 a^{4} x^{5} \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {5}{2}}}\) | \(60\) |
trager | \(\frac {x \left (x^{6} a^{3}-6 a^{2} b \,x^{4}-24 a \,b^{2} x^{2}-16 b^{3}\right ) \sqrt {-\frac {-a \,x^{2}-b}{x^{2}}}}{3 a^{4} \left (a \,x^{2}+b \right )^{2}}\) | \(64\) |
risch | \(\frac {\left (a \,x^{2}-8 b \right ) \left (a \,x^{2}+b \right )}{3 a^{4} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}-\frac {\left (a \,x^{2}+b \right ) \left (9 a \,x^{2}+8 b \right ) b^{2}}{3 a^{4} \left (a^{2} x^{4}+2 a b \,x^{2}+b^{2}\right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}\) | \(100\) |
Time = 0.48 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89 \[ \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {{\left (a^{3} x^{7} - 6 \, a^{2} b x^{5} - 24 \, a b^{2} x^{3} - 16 \, b^{3} x\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{3 \, {\left (a^{6} x^{4} + 2 \, a^{5} b x^{2} + a^{4} b^{2}\right )}} \]
1/3*(a^3*x^7 - 6*a^2*b*x^5 - 24*a*b^2*x^3 - 16*b^3*x)*sqrt((a*x^2 + b)/x^2 )/(a^6*x^4 + 2*a^5*b*x^2 + a^4*b^2)
Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (76) = 152\).
Time = 0.90 (sec) , antiderivative size = 337, normalized size of antiderivative = 4.11 \[ \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {a^{4} b^{\frac {19}{2}} x^{8} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{7} b^{9} x^{6} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{2} + 3 a^{4} b^{12}} - \frac {5 a^{3} b^{\frac {21}{2}} x^{6} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{7} b^{9} x^{6} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{2} + 3 a^{4} b^{12}} - \frac {30 a^{2} b^{\frac {23}{2}} x^{4} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{7} b^{9} x^{6} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{2} + 3 a^{4} b^{12}} - \frac {40 a b^{\frac {25}{2}} x^{2} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{7} b^{9} x^{6} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{2} + 3 a^{4} b^{12}} - \frac {16 b^{\frac {27}{2}} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{7} b^{9} x^{6} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{2} + 3 a^{4} b^{12}} \]
a**4*b**(19/2)*x**8*sqrt(a*x**2/b + 1)/(3*a**7*b**9*x**6 + 9*a**6*b**10*x* *4 + 9*a**5*b**11*x**2 + 3*a**4*b**12) - 5*a**3*b**(21/2)*x**6*sqrt(a*x**2 /b + 1)/(3*a**7*b**9*x**6 + 9*a**6*b**10*x**4 + 9*a**5*b**11*x**2 + 3*a**4 *b**12) - 30*a**2*b**(23/2)*x**4*sqrt(a*x**2/b + 1)/(3*a**7*b**9*x**6 + 9* a**6*b**10*x**4 + 9*a**5*b**11*x**2 + 3*a**4*b**12) - 40*a*b**(25/2)*x**2* sqrt(a*x**2/b + 1)/(3*a**7*b**9*x**6 + 9*a**6*b**10*x**4 + 9*a**5*b**11*x* *2 + 3*a**4*b**12) - 16*b**(27/2)*sqrt(a*x**2/b + 1)/(3*a**7*b**9*x**6 + 9 *a**6*b**10*x**4 + 9*a**5*b**11*x**2 + 3*a**4*b**12)
Time = 0.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.87 \[ \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {{\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} x^{3} - 9 \, \sqrt {a + \frac {b}{x^{2}}} b x}{3 \, a^{4}} - \frac {9 \, {\left (a + \frac {b}{x^{2}}\right )} b^{2} x^{2} - b^{3}}{3 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{4} x^{3}} \]
1/3*((a + b/x^2)^(3/2)*x^3 - 9*sqrt(a + b/x^2)*b*x)/a^4 - 1/3*(9*(a + b/x^ 2)*b^2*x^2 - b^3)/((a + b/x^2)^(3/2)*a^4*x^3)
Time = 0.29 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.04 \[ \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {16 \, b^{\frac {3}{2}} \mathrm {sgn}\left (x\right )}{3 \, a^{4}} - \frac {9 \, {\left (a x^{2} + b\right )} b^{2} - b^{3}}{3 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} a^{4} \mathrm {sgn}\left (x\right )} + \frac {{\left (a x^{2} + b\right )}^{\frac {3}{2}} a^{8} - 9 \, \sqrt {a x^{2} + b} a^{8} b}{3 \, a^{12} \mathrm {sgn}\left (x\right )} \]
16/3*b^(3/2)*sgn(x)/a^4 - 1/3*(9*(a*x^2 + b)*b^2 - b^3)/((a*x^2 + b)^(3/2) *a^4*sgn(x)) + 1/3*((a*x^2 + b)^(3/2)*a^8 - 9*sqrt(a*x^2 + b)*a^8*b)/(a^12 *sgn(x))
Time = 6.34 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.99 \[ \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {6\,a^2\,\left (a+\frac {b}{x^2}\right )-24\,a\,{\left (a+\frac {b}{x^2}\right )}^2+16\,{\left (a+\frac {b}{x^2}\right )}^3+a^3}{\left (\frac {3\,a^5}{b\,x}-\frac {3\,a^4\,\left (a+\frac {b}{x^2}\right )}{b\,x}\right )\,{\left (a+\frac {b}{x^2}\right )}^{3/2}} \]