Integrand size = 15, antiderivative size = 85 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {16 b^2 \sqrt {a+\frac {b}{x^2}} x}{5 a^4}-\frac {8 b \sqrt {a+\frac {b}{x^2}} x^3}{5 a^3}-\frac {x^5}{a \sqrt {a+\frac {b}{x^2}}}+\frac {6 \sqrt {a+\frac {b}{x^2}} x^5}{5 a^2} \] Output:
16/5*b^2*(a+b/x^2)^(1/2)*x/a^4-8/5*b*(a+b/x^2)^(1/2)*x^3/a^3-x^5/a/(a+b/x^ 2)^(1/2)+6/5*(a+b/x^2)^(1/2)*x^5/a^2
Time = 0.03 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.61 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {16 b^3+8 a b^2 x^2-2 a^2 b x^4+a^3 x^6}{5 a^4 \sqrt {a+\frac {b}{x^2}} x} \] Input:
Integrate[x^4/(a + b/x^2)^(3/2),x]
Output:
(16*b^3 + 8*a*b^2*x^2 - 2*a^2*b*x^4 + a^3*x^6)/(5*a^4*Sqrt[a + b/x^2]*x)
Time = 0.36 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {803, 803, 773, 245, 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.
\(\displaystyle \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {x^5}{5 a \sqrt {a+\frac {b}{x^2}}}-\frac {6 b \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{3/2}}dx}{5 a}\) |
\(\Big \downarrow \) 803 |
\(\displaystyle \frac {x^5}{5 a \sqrt {a+\frac {b}{x^2}}}-\frac {6 b \left (\frac {x^3}{3 a \sqrt {a+\frac {b}{x^2}}}-\frac {4 b \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2}}dx}{3 a}\right )}{5 a}\) |
\(\Big \downarrow \) 773 |
\(\displaystyle \frac {x^5}{5 a \sqrt {a+\frac {b}{x^2}}}-\frac {6 b \left (\frac {4 b \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{3/2}}d\frac {1}{x}}{3 a}+\frac {x^3}{3 a \sqrt {a+\frac {b}{x^2}}}\right )}{5 a}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle \frac {x^5}{5 a \sqrt {a+\frac {b}{x^2}}}-\frac {6 b \left (\frac {4 b \left (-\frac {2 b \int \frac {1}{\left (a+\frac {b}{x^2}\right )^{3/2}}d\frac {1}{x}}{a}-\frac {x}{a \sqrt {a+\frac {b}{x^2}}}\right )}{3 a}+\frac {x^3}{3 a \sqrt {a+\frac {b}{x^2}}}\right )}{5 a}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {x^5}{5 a \sqrt {a+\frac {b}{x^2}}}-\frac {6 b \left (\frac {4 b \left (-\frac {2 b}{a^2 x \sqrt {a+\frac {b}{x^2}}}-\frac {x}{a \sqrt {a+\frac {b}{x^2}}}\right )}{3 a}+\frac {x^3}{3 a \sqrt {a+\frac {b}{x^2}}}\right )}{5 a}\) |
Input:
Int[x^4/(a + b/x^2)^(3/2),x]
Output:
x^5/(5*a*Sqrt[a + b/x^2]) - (6*b*(x^3/(3*a*Sqrt[a + b/x^2]) + (4*b*((-2*b) /(a^2*Sqrt[a + b/x^2]*x) - x/(a*Sqrt[a + b/x^2])))/(3*a)))/(5*a)
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
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 = 0.10 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.66
method | result | size |
orering | \(\frac {\left (a^{3} x^{6}-2 a^{2} b \,x^{4}+8 a \,b^{2} x^{2}+16 b^{3}\right ) \left (a \,x^{2}+b \right )}{5 a^{4} x^{3} \left (a +\frac {b}{x^{2}}\right )^{\frac {3}{2}}}\) | \(56\) |
gosper | \(\frac {\left (a \,x^{2}+b \right ) \left (a^{3} x^{6}-2 a^{2} b \,x^{4}+8 a \,b^{2} x^{2}+16 b^{3}\right )}{5 a^{4} x^{3} \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}}}\) | \(60\) |
default | \(\frac {\left (a \,x^{2}+b \right ) \left (a^{3} x^{6}-2 a^{2} b \,x^{4}+8 a \,b^{2} x^{2}+16 b^{3}\right )}{5 a^{4} x^{3} \left (\frac {a \,x^{2}+b}{x^{2}}\right )^{\frac {3}{2}}}\) | \(60\) |
trager | \(\frac {\left (a^{3} x^{6}-2 a^{2} b \,x^{4}+8 a \,b^{2} x^{2}+16 b^{3}\right ) x \sqrt {-\frac {-a \,x^{2}-b}{x^{2}}}}{5 a^{4} \left (a \,x^{2}+b \right )}\) | \(64\) |
risch | \(\frac {\left (a^{2} x^{4}-3 a b \,x^{2}+11 b^{2}\right ) \left (a \,x^{2}+b \right )}{5 a^{4} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}+\frac {b^{3}}{a^{4} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}\) | \(73\) |
Input:
int(x^4/(a+b/x^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/5*(a^3*x^6-2*a^2*b*x^4+8*a*b^2*x^2+16*b^3)/a^4*(a*x^2+b)/x^3/(a+b/x^2)^( 3/2)
Time = 0.08 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.73 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {{\left (a^{3} x^{7} - 2 \, a^{2} b x^{5} + 8 \, a b^{2} x^{3} + 16 \, b^{3} x\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{5 \, {\left (a^{5} x^{2} + a^{4} b\right )}} \] Input:
integrate(x^4/(a+b/x^2)^(3/2),x, algorithm="fricas")
Output:
1/5*(a^3*x^7 - 2*a^2*b*x^5 + 8*a*b^2*x^3 + 16*b^3*x)*sqrt((a*x^2 + b)/x^2) /(a^5*x^2 + a^4*b)
Leaf count of result is larger than twice the leaf count of optimal. 337 vs. \(2 (78) = 156\).
Time = 0.94 (sec) , antiderivative size = 337, normalized size of antiderivative = 3.96 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {a^{5} b^{\frac {19}{2}} x^{10} \sqrt {\frac {a x^{2}}{b} + 1}}{5 a^{7} b^{9} x^{6} + 15 a^{6} b^{10} x^{4} + 15 a^{5} b^{11} x^{2} + 5 a^{4} b^{12}} + \frac {5 a^{3} b^{\frac {23}{2}} x^{6} \sqrt {\frac {a x^{2}}{b} + 1}}{5 a^{7} b^{9} x^{6} + 15 a^{6} b^{10} x^{4} + 15 a^{5} b^{11} x^{2} + 5 a^{4} b^{12}} + \frac {30 a^{2} b^{\frac {25}{2}} x^{4} \sqrt {\frac {a x^{2}}{b} + 1}}{5 a^{7} b^{9} x^{6} + 15 a^{6} b^{10} x^{4} + 15 a^{5} b^{11} x^{2} + 5 a^{4} b^{12}} + \frac {40 a b^{\frac {27}{2}} x^{2} \sqrt {\frac {a x^{2}}{b} + 1}}{5 a^{7} b^{9} x^{6} + 15 a^{6} b^{10} x^{4} + 15 a^{5} b^{11} x^{2} + 5 a^{4} b^{12}} + \frac {16 b^{\frac {29}{2}} \sqrt {\frac {a x^{2}}{b} + 1}}{5 a^{7} b^{9} x^{6} + 15 a^{6} b^{10} x^{4} + 15 a^{5} b^{11} x^{2} + 5 a^{4} b^{12}} \] Input:
integrate(x**4/(a+b/x**2)**(3/2),x)
Output:
a**5*b**(19/2)*x**10*sqrt(a*x**2/b + 1)/(5*a**7*b**9*x**6 + 15*a**6*b**10* x**4 + 15*a**5*b**11*x**2 + 5*a**4*b**12) + 5*a**3*b**(23/2)*x**6*sqrt(a*x **2/b + 1)/(5*a**7*b**9*x**6 + 15*a**6*b**10*x**4 + 15*a**5*b**11*x**2 + 5 *a**4*b**12) + 30*a**2*b**(25/2)*x**4*sqrt(a*x**2/b + 1)/(5*a**7*b**9*x**6 + 15*a**6*b**10*x**4 + 15*a**5*b**11*x**2 + 5*a**4*b**12) + 40*a*b**(27/2 )*x**2*sqrt(a*x**2/b + 1)/(5*a**7*b**9*x**6 + 15*a**6*b**10*x**4 + 15*a**5 *b**11*x**2 + 5*a**4*b**12) + 16*b**(29/2)*sqrt(a*x**2/b + 1)/(5*a**7*b**9 *x**6 + 15*a**6*b**10*x**4 + 15*a**5*b**11*x**2 + 5*a**4*b**12)
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.81 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {b^{3}}{\sqrt {a + \frac {b}{x^{2}}} a^{4} x} + \frac {{\left (a + \frac {b}{x^{2}}\right )}^{\frac {5}{2}} x^{5} - 5 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} b x^{3} + 15 \, \sqrt {a + \frac {b}{x^{2}}} b^{2} x}{5 \, a^{4}} \] Input:
integrate(x^4/(a+b/x^2)^(3/2),x, algorithm="maxima")
Output:
b^3/(sqrt(a + b/x^2)*a^4*x) + 1/5*((a + b/x^2)^(5/2)*x^5 - 5*(a + b/x^2)^( 3/2)*b*x^3 + 15*sqrt(a + b/x^2)*b^2*x)/a^4
Time = 0.12 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.01 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=-\frac {16 \, b^{\frac {5}{2}} \mathrm {sgn}\left (x\right )}{5 \, a^{4}} + \frac {b^{3}}{\sqrt {a x^{2} + b} a^{4} \mathrm {sgn}\left (x\right )} + \frac {{\left (a x^{2} + b\right )}^{\frac {5}{2}} a^{16} - 5 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} a^{16} b + 15 \, \sqrt {a x^{2} + b} a^{16} b^{2}}{5 \, a^{20} \mathrm {sgn}\left (x\right )} \] Input:
integrate(x^4/(a+b/x^2)^(3/2),x, algorithm="giac")
Output:
-16/5*b^(5/2)*sgn(x)/a^4 + b^3/(sqrt(a*x^2 + b)*a^4*sgn(x)) + 1/5*((a*x^2 + b)^(5/2)*a^16 - 5*(a*x^2 + b)^(3/2)*a^16*b + 15*sqrt(a*x^2 + b)*a^16*b^2 )/(a^20*sgn(x))
Time = 1.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.56 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {a^3\,x^6-2\,a^2\,b\,x^4+8\,a\,b^2\,x^2+16\,b^3}{5\,a^4\,x\,\sqrt {a+\frac {b}{x^2}}} \] Input:
int(x^4/(a + b/x^2)^(3/2),x)
Output:
(16*b^3 + a^3*x^6 + 8*a*b^2*x^2 - 2*a^2*b*x^4)/(5*a^4*x*(a + b/x^2)^(1/2))
Time = 0.22 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.62 \[ \int \frac {x^4}{\left (a+\frac {b}{x^2}\right )^{3/2}} \, dx=\frac {\sqrt {a \,x^{2}+b}\, \left (a^{3} x^{6}-2 a^{2} b \,x^{4}+8 a \,b^{2} x^{2}+16 b^{3}\right )}{5 a^{4} \left (a \,x^{2}+b \right )} \] Input:
int(x^4/(a+b/x^2)^(3/2),x)
Output:
(sqrt(a*x**2 + b)*(a**3*x**6 - 2*a**2*b*x**4 + 8*a*b**2*x**2 + 16*b**3))/( 5*a**4*(a*x**2 + b))