\(\int \frac {x^2}{(a+\frac {b}{x^2})^{5/2}} \, dx\) [1951]

Optimal result

Integrand size = 15, antiderivative size = 82 \[ \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {2 b x}{3 a^2 \left (a+\frac {b}{x^2}\right )^{3/2}}+\frac {8 b x}{3 a^3 \sqrt {a+\frac {b}{x^2}}}-\frac {16 b \sqrt {a+\frac {b}{x^2}} x}{3 a^4}+\frac {x^3}{3 a \left (a+\frac {b}{x^2}\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {277, 198, 197} \[ \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=-\frac {16 b x \sqrt {a+\frac {b}{x^2}}}{3 a^4}+\frac {8 b x}{3 a^3 \sqrt {a+\frac {b}{x^2}}}+\frac {2 b x}{3 a^2 \left (a+\frac {b}{x^2}\right )^{3/2}}+\frac {x^3}{3 a \left (a+\frac {b}{x^2}\right )^{3/2}} \]

[In]

Int[x^2/(a + b/x^2)^(5/2),x]

[Out]

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

Rule 197

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

Rule 198

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

Rule 277

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.72 \[ \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {\left (b+a x^2\right ) \left (-16 b^3-24 a b^2 x^2-6 a^2 b x^4+a^3 x^6\right )}{3 a^4 \left (a+\frac {b}{x^2}\right )^{5/2} x^5} \]

[In]

Integrate[x^2/(a + b/x^2)^(5/2),x]

[Out]

((b + a*x^2)*(-16*b^3 - 24*a*b^2*x^2 - 6*a^2*b*x^4 + a^3*x^6))/(3*a^4*(a + b/x^2)^(5/2)*x^5)

Maple [A] (verified)

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	\(\frac {\left (a \,x^{2}-8 b \right ) \left (a \,x^{2}+b \right )}{3 a^{4} \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}-\frac {\left (a \,x^{2}+b \right ) \left (9 a \,x^{2}+8 b \right ) b^{2}}{3 a^{4} \left (a^{2} x^{4}+2 a b \,x^{2}+b^{2}\right ) \sqrt {\frac {a \,x^{2}+b}{x^{2}}}\, x}\)	\(100\)

[In]

int(x^2/(a+b/x^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3*(a*x^2+b)*(a^3*x^6-6*a^2*b*x^4-24*a*b^2*x^2-16*b^3)/a^4/x^5/((a*x^2+b)/x^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.48 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.89 \[ \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {{\left (a^{3} x^{7} - 6 \, a^{2} b x^{5} - 24 \, a b^{2} x^{3} - 16 \, b^{3} x\right )} \sqrt {\frac {a x^{2} + b}{x^{2}}}}{3 \, {\left (a^{6} x^{4} + 2 \, a^{5} b x^{2} + a^{4} b^{2}\right )}} \]

[In]

integrate(x^2/(a+b/x^2)^(5/2),x, algorithm="fricas")

[Out]

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)

Sympy [B] (verification not implemented)

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 \[ \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {a^{4} b^{\frac {19}{2}} x^{8} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{7} b^{9} x^{6} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{2} + 3 a^{4} b^{12}} - \frac {5 a^{3} b^{\frac {21}{2}} x^{6} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{7} b^{9} x^{6} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{2} + 3 a^{4} b^{12}} - \frac {30 a^{2} b^{\frac {23}{2}} x^{4} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{7} b^{9} x^{6} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{2} + 3 a^{4} b^{12}} - \frac {40 a b^{\frac {25}{2}} x^{2} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{7} b^{9} x^{6} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{2} + 3 a^{4} b^{12}} - \frac {16 b^{\frac {27}{2}} \sqrt {\frac {a x^{2}}{b} + 1}}{3 a^{7} b^{9} x^{6} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{2} + 3 a^{4} b^{12}} \]

[In]

integrate(x**2/(a+b/x**2)**(5/2),x)

[Out]

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**1
2) - 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**1
1*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)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.87 \[ \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {{\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} x^{3} - 9 \, \sqrt {a + \frac {b}{x^{2}}} b x}{3 \, a^{4}} - \frac {9 \, {\left (a + \frac {b}{x^{2}}\right )} b^{2} x^{2} - b^{3}}{3 \, {\left (a + \frac {b}{x^{2}}\right )}^{\frac {3}{2}} a^{4} x^{3}} \]

[In]

integrate(x^2/(a+b/x^2)^(5/2),x, algorithm="maxima")

[Out]

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)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.04 \[ \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {16 \, b^{\frac {3}{2}} \mathrm {sgn}\left (x\right )}{3 \, a^{4}} - \frac {9 \, {\left (a x^{2} + b\right )} b^{2} - b^{3}}{3 \, {\left (a x^{2} + b\right )}^{\frac {3}{2}} a^{4} \mathrm {sgn}\left (x\right )} + \frac {{\left (a x^{2} + b\right )}^{\frac {3}{2}} a^{8} - 9 \, \sqrt {a x^{2} + b} a^{8} b}{3 \, a^{12} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(x^2/(a+b/x^2)^(5/2),x, algorithm="giac")

[Out]

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))

Mupad [B] (verification not implemented)

Time = 6.34 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.99 \[ \int \frac {x^2}{\left (a+\frac {b}{x^2}\right )^{5/2}} \, dx=\frac {6\,a^2\,\left (a+\frac {b}{x^2}\right )-24\,a\,{\left (a+\frac {b}{x^2}\right )}^2+16\,{\left (a+\frac {b}{x^2}\right )}^3+a^3}{\left (\frac {3\,a^5}{b\,x}-\frac {3\,a^4\,\left (a+\frac {b}{x^2}\right )}{b\,x}\right )\,{\left (a+\frac {b}{x^2}\right )}^{3/2}} \]

[In]

int(x^2/(a + b/x^2)^(5/2),x)

[Out]

(6*a^2*(a + b/x^2) - 24*a*(a + b/x^2)^2 + 16*(a + b/x^2)^3 + a^3)/(((3*a^5)/(b*x) - (3*a^4*(a + b/x^2))/(b*x))
*(a + b/x^2)^(3/2))