Integrand size = 9, antiderivative size = 71 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3} \, dx=\frac {x}{a^3}-\frac {b^2 x}{4 a^3 \left (b+a x^2\right )^2}+\frac {9 b x}{8 a^3 \left (b+a x^2\right )}-\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 a^{7/2}} \] Output:
x/a^3-1/4*b^2*x/a^3/(a*x^2+b)^2+9/8*b*x/a^3/(a*x^2+b)-15/8*b^(1/2)*arctan( a^(1/2)*x/b^(1/2))/a^(7/2)
Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3} \, dx=\frac {15 b^2 x+25 a b x^3+8 a^2 x^5}{8 a^3 \left (b+a x^2\right )^2}-\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 a^{7/2}} \] Input:
Integrate[(a + b/x^2)^(-3),x]
Output:
(15*b^2*x + 25*a*b*x^3 + 8*a^2*x^5)/(8*a^3*(b + a*x^2)^2) - (15*Sqrt[b]*Ar cTan[(Sqrt[a]*x)/Sqrt[b]])/(8*a^(7/2))
Time = 0.31 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.20, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {772, 252, 252, 262, 218}
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.
\(\displaystyle \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3} \, dx\) |
\(\Big \downarrow \) 772 |
\(\displaystyle \int \frac {x^6}{\left (a x^2+b\right )^3}dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {5 \int \frac {x^4}{\left (a x^2+b\right )^2}dx}{4 a}-\frac {x^5}{4 a \left (a x^2+b\right )^2}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {5 \left (\frac {3 \int \frac {x^2}{a x^2+b}dx}{2 a}-\frac {x^3}{2 a \left (a x^2+b\right )}\right )}{4 a}-\frac {x^5}{4 a \left (a x^2+b\right )^2}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {x}{a}-\frac {b \int \frac {1}{a x^2+b}dx}{a}\right )}{2 a}-\frac {x^3}{2 a \left (a x^2+b\right )}\right )}{4 a}-\frac {x^5}{4 a \left (a x^2+b\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {x}{a}-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{a^{3/2}}\right )}{2 a}-\frac {x^3}{2 a \left (a x^2+b\right )}\right )}{4 a}-\frac {x^5}{4 a \left (a x^2+b\right )^2}\) |
Input:
Int[(a + b/x^2)^(-3),x]
Output:
-1/4*x^5/(a*(b + a*x^2)^2) + (5*(-1/2*x^3/(a*(b + a*x^2)) + (3*(x/a - (Sqr t[b]*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/a^(3/2)))/(2*a)))/(4*a)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && ILtQ[n, 0] && IntegerQ[p]
Time = 0.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.72
method | result | size |
default | \(\frac {x}{a^{3}}-\frac {b \left (\frac {-\frac {9}{8} a \,x^{3}-\frac {7}{8} b x}{\left (a \,x^{2}+b \right )^{2}}+\frac {15 \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}\) | \(51\) |
risch | \(\frac {x}{a^{3}}+\frac {\frac {9}{8} a b \,x^{3}+\frac {7}{8} b^{2} x}{a^{3} \left (a \,x^{2}+b \right )^{2}}+\frac {15 \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x -b \right )}{16 a^{4}}-\frac {15 \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x -b \right )}{16 a^{4}}\) | \(83\) |
Input:
int(1/(a+b/x^2)^3,x,method=_RETURNVERBOSE)
Output:
x/a^3-1/a^3*b*((-9/8*a*x^3-7/8*b*x)/(a*x^2+b)^2+15/8/(a*b)^(1/2)*arctan(a* x/(a*b)^(1/2)))
Time = 0.07 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.85 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3} \, dx=\left [\frac {16 \, a^{2} x^{5} + 50 \, a b x^{3} + 30 \, b^{2} x + 15 \, {\left (a^{2} x^{4} + 2 \, a b x^{2} + b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {a x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - b}{a x^{2} + b}\right )}{16 \, {\left (a^{5} x^{4} + 2 \, a^{4} b x^{2} + a^{3} b^{2}\right )}}, \frac {8 \, a^{2} x^{5} + 25 \, a b x^{3} + 15 \, b^{2} x - 15 \, {\left (a^{2} x^{4} + 2 \, a b x^{2} + b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a x \sqrt {\frac {b}{a}}}{b}\right )}{8 \, {\left (a^{5} x^{4} + 2 \, a^{4} b x^{2} + a^{3} b^{2}\right )}}\right ] \] Input:
integrate(1/(a+b/x^2)^3,x, algorithm="fricas")
Output:
[1/16*(16*a^2*x^5 + 50*a*b*x^3 + 30*b^2*x + 15*(a^2*x^4 + 2*a*b*x^2 + b^2) *sqrt(-b/a)*log((a*x^2 - 2*a*x*sqrt(-b/a) - b)/(a*x^2 + b)))/(a^5*x^4 + 2* a^4*b*x^2 + a^3*b^2), 1/8*(8*a^2*x^5 + 25*a*b*x^3 + 15*b^2*x - 15*(a^2*x^4 + 2*a*b*x^2 + b^2)*sqrt(b/a)*arctan(a*x*sqrt(b/a)/b))/(a^5*x^4 + 2*a^4*b* x^2 + a^3*b^2)]
Time = 0.26 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.51 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3} \, dx=\frac {15 \sqrt {- \frac {b}{a^{7}}} \log {\left (- a^{3} \sqrt {- \frac {b}{a^{7}}} + x \right )}}{16} - \frac {15 \sqrt {- \frac {b}{a^{7}}} \log {\left (a^{3} \sqrt {- \frac {b}{a^{7}}} + x \right )}}{16} + \frac {9 a b x^{3} + 7 b^{2} x}{8 a^{5} x^{4} + 16 a^{4} b x^{2} + 8 a^{3} b^{2}} + \frac {x}{a^{3}} \] Input:
integrate(1/(a+b/x**2)**3,x)
Output:
15*sqrt(-b/a**7)*log(-a**3*sqrt(-b/a**7) + x)/16 - 15*sqrt(-b/a**7)*log(a* *3*sqrt(-b/a**7) + x)/16 + (9*a*b*x**3 + 7*b**2*x)/(8*a**5*x**4 + 16*a**4* b*x**2 + 8*a**3*b**2) + x/a**3
Time = 0.11 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3} \, dx=\frac {9 \, a b x^{3} + 7 \, b^{2} x}{8 \, {\left (a^{5} x^{4} + 2 \, a^{4} b x^{2} + a^{3} b^{2}\right )}} - \frac {15 \, b \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} + \frac {x}{a^{3}} \] Input:
integrate(1/(a+b/x^2)^3,x, algorithm="maxima")
Output:
1/8*(9*a*b*x^3 + 7*b^2*x)/(a^5*x^4 + 2*a^4*b*x^2 + a^3*b^2) - 15/8*b*arcta n(a*x/sqrt(a*b))/(sqrt(a*b)*a^3) + x/a^3
Time = 0.12 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3} \, dx=-\frac {15 \, b \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} + \frac {x}{a^{3}} + \frac {9 \, a b x^{3} + 7 \, b^{2} x}{8 \, {\left (a x^{2} + b\right )}^{2} a^{3}} \] Input:
integrate(1/(a+b/x^2)^3,x, algorithm="giac")
Output:
-15/8*b*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*a^3) + x/a^3 + 1/8*(9*a*b*x^3 + 7 *b^2*x)/((a*x^2 + b)^2*a^3)
Time = 0.34 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3} \, dx=\frac {\frac {7\,b^2\,x}{8}+\frac {9\,a\,b\,x^3}{8}}{a^5\,x^4+2\,a^4\,b\,x^2+a^3\,b^2}+\frac {x}{a^3}-\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{8\,a^{7/2}} \] Input:
int(1/(a + b/x^2)^3,x)
Output:
((7*b^2*x)/8 + (9*a*b*x^3)/8)/(a^3*b^2 + a^5*x^4 + 2*a^4*b*x^2) + x/a^3 - (15*b^(1/2)*atan((a^(1/2)*x)/b^(1/2)))/(8*a^(7/2))
Time = 0.26 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.66 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3} \, dx=\frac {-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {a x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} x^{4}-30 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {a x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,x^{2}-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {a x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2}+8 a^{3} x^{5}+25 a^{2} b \,x^{3}+15 a \,b^{2} x}{8 a^{4} \left (a^{2} x^{4}+2 a b \,x^{2}+b^{2}\right )} \] Input:
int(1/(a+b/x^2)^3,x)
Output:
( - 15*sqrt(b)*sqrt(a)*atan((a*x)/(sqrt(b)*sqrt(a)))*a**2*x**4 - 30*sqrt(b )*sqrt(a)*atan((a*x)/(sqrt(b)*sqrt(a)))*a*b*x**2 - 15*sqrt(b)*sqrt(a)*atan ((a*x)/(sqrt(b)*sqrt(a)))*b**2 + 8*a**3*x**5 + 25*a**2*b*x**3 + 15*a*b**2* x)/(8*a**4*(a**2*x**4 + 2*a*b*x**2 + b**2))