Integrand size = 13, antiderivative size = 67 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^2 x^8} \, dx=-\frac {1}{3 b^2 x^3}+\frac {2 a}{b^3 x}+\frac {a^2 x}{2 b^3 \left (b+a x^2\right )}+\frac {5 a^{3/2} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{2 b^{7/2}} \] Output:
-1/3/b^2/x^3+2*a/b^3/x+1/2*a^2*x/b^3/(a*x^2+b)+5/2*a^(3/2)*arctan(a^(1/2)* x/b^(1/2))/b^(7/2)
Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^2 x^8} \, dx=-\frac {1}{3 b^2 x^3}+\frac {2 a}{b^3 x}+\frac {a^2 x}{2 b^3 \left (b+a x^2\right )}+\frac {5 a^{3/2} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{2 b^{7/2}} \] Input:
Integrate[1/((a + b/x^2)^2*x^8),x]
Output:
-1/3*1/(b^2*x^3) + (2*a)/(b^3*x) + (a^2*x)/(2*b^3*(b + a*x^2)) + (5*a^(3/2 )*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(2*b^(7/2))
Time = 0.30 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.16, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {795, 253, 264, 264, 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.
\(\displaystyle \int \frac {1}{x^8 \left (a+\frac {b}{x^2}\right )^2} \, dx\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \int \frac {1}{x^4 \left (a x^2+b\right )^2}dx\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {5 \int \frac {1}{x^4 \left (a x^2+b\right )}dx}{2 b}+\frac {1}{2 b x^3 \left (a x^2+b\right )}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {5 \left (-\frac {a \int \frac {1}{x^2 \left (a x^2+b\right )}dx}{b}-\frac {1}{3 b x^3}\right )}{2 b}+\frac {1}{2 b x^3 \left (a x^2+b\right )}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {5 \left (-\frac {a \left (-\frac {a \int \frac {1}{a x^2+b}dx}{b}-\frac {1}{b x}\right )}{b}-\frac {1}{3 b x^3}\right )}{2 b}+\frac {1}{2 b x^3 \left (a x^2+b\right )}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5 \left (-\frac {a \left (-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{b^{3/2}}-\frac {1}{b x}\right )}{b}-\frac {1}{3 b x^3}\right )}{2 b}+\frac {1}{2 b x^3 \left (a x^2+b\right )}\) |
Input:
Int[1/((a + b/x^2)^2*x^8),x]
Output:
1/(2*b*x^3*(b + a*x^2)) + (5*(-1/3*1/(b*x^3) - (a*(-(1/(b*x)) - (Sqrt[a]*A rcTan[(Sqrt[a]*x)/Sqrt[b]])/b^(3/2)))/b))/(2*b)
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*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)* (b + a/x^n)^p, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && NegQ[n]
Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.82
method | result | size |
default | \(\frac {a^{2} \left (\frac {x}{2 a \,x^{2}+2 b}+\frac {5 \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{3}}-\frac {1}{3 b^{2} x^{3}}+\frac {2 a}{b^{3} x}\) | \(55\) |
risch | \(\frac {\frac {5 a^{2} x^{4}}{2 b^{3}}+\frac {5 a \,x^{2}}{3 b^{2}}-\frac {1}{3 b}}{\left (a \,x^{2}+b \right ) x^{3}}+\frac {5 \sqrt {-a b}\, a \ln \left (-a x -\sqrt {-a b}\right )}{4 b^{4}}-\frac {5 \sqrt {-a b}\, a \ln \left (-a x +\sqrt {-a b}\right )}{4 b^{4}}\) | \(91\) |
Input:
int(1/(a+b/x^2)^2/x^8,x,method=_RETURNVERBOSE)
Output:
a^2/b^3*(1/2*x/(a*x^2+b)+5/2/(a*b)^(1/2)*arctan(a*x/(a*b)^(1/2)))-1/3/b^2/ x^3+2*a/b^3/x
Time = 0.08 (sec) , antiderivative size = 172, normalized size of antiderivative = 2.57 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^2 x^8} \, dx=\left [\frac {30 \, a^{2} x^{4} + 20 \, a b x^{2} + 15 \, {\left (a^{2} x^{5} + a b x^{3}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {a x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - b}{a x^{2} + b}\right ) - 4 \, b^{2}}{12 \, {\left (a b^{3} x^{5} + b^{4} x^{3}\right )}}, \frac {15 \, a^{2} x^{4} + 10 \, a b x^{2} + 15 \, {\left (a^{2} x^{5} + a b x^{3}\right )} \sqrt {\frac {a}{b}} \arctan \left (x \sqrt {\frac {a}{b}}\right ) - 2 \, b^{2}}{6 \, {\left (a b^{3} x^{5} + b^{4} x^{3}\right )}}\right ] \] Input:
integrate(1/(a+b/x^2)^2/x^8,x, algorithm="fricas")
Output:
[1/12*(30*a^2*x^4 + 20*a*b*x^2 + 15*(a^2*x^5 + a*b*x^3)*sqrt(-a/b)*log((a* x^2 + 2*b*x*sqrt(-a/b) - b)/(a*x^2 + b)) - 4*b^2)/(a*b^3*x^5 + b^4*x^3), 1 /6*(15*a^2*x^4 + 10*a*b*x^2 + 15*(a^2*x^5 + a*b*x^3)*sqrt(a/b)*arctan(x*sq rt(a/b)) - 2*b^2)/(a*b^3*x^5 + b^4*x^3)]
Time = 0.22 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.70 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^2 x^8} \, dx=- \frac {5 \sqrt {- \frac {a^{3}}{b^{7}}} \log {\left (x - \frac {b^{4} \sqrt {- \frac {a^{3}}{b^{7}}}}{a^{2}} \right )}}{4} + \frac {5 \sqrt {- \frac {a^{3}}{b^{7}}} \log {\left (x + \frac {b^{4} \sqrt {- \frac {a^{3}}{b^{7}}}}{a^{2}} \right )}}{4} + \frac {15 a^{2} x^{4} + 10 a b x^{2} - 2 b^{2}}{6 a b^{3} x^{5} + 6 b^{4} x^{3}} \] Input:
integrate(1/(a+b/x**2)**2/x**8,x)
Output:
-5*sqrt(-a**3/b**7)*log(x - b**4*sqrt(-a**3/b**7)/a**2)/4 + 5*sqrt(-a**3/b **7)*log(x + b**4*sqrt(-a**3/b**7)/a**2)/4 + (15*a**2*x**4 + 10*a*b*x**2 - 2*b**2)/(6*a*b**3*x**5 + 6*b**4*x**3)
Time = 0.11 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^2 x^8} \, dx=\frac {15 \, a^{2} x^{4} + 10 \, a b x^{2} - 2 \, b^{2}}{6 \, {\left (a b^{3} x^{5} + b^{4} x^{3}\right )}} + \frac {5 \, a^{2} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} \] Input:
integrate(1/(a+b/x^2)^2/x^8,x, algorithm="maxima")
Output:
1/6*(15*a^2*x^4 + 10*a*b*x^2 - 2*b^2)/(a*b^3*x^5 + b^4*x^3) + 5/2*a^2*arct an(a*x/sqrt(a*b))/(sqrt(a*b)*b^3)
Time = 0.12 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^2 x^8} \, dx=\frac {5 \, a^{2} \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} b^{3}} + \frac {a^{2} x}{2 \, {\left (a x^{2} + b\right )} b^{3}} + \frac {6 \, a x^{2} - b}{3 \, b^{3} x^{3}} \] Input:
integrate(1/(a+b/x^2)^2/x^8,x, algorithm="giac")
Output:
5/2*a^2*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*b^3) + 1/2*a^2*x/((a*x^2 + b)*b^3 ) + 1/3*(6*a*x^2 - b)/(b^3*x^3)
Time = 0.09 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.87 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^2 x^8} \, dx=\frac {\frac {5\,a\,x^2}{3\,b^2}-\frac {1}{3\,b}+\frac {5\,a^2\,x^4}{2\,b^3}}{a\,x^5+b\,x^3}+\frac {5\,a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{2\,b^{7/2}} \] Input:
int(1/(x^8*(a + b/x^2)^2),x)
Output:
((5*a*x^2)/(3*b^2) - 1/(3*b) + (5*a^2*x^4)/(2*b^3))/(a*x^5 + b*x^3) + (5*a ^(3/2)*atan((a^(1/2)*x)/b^(1/2)))/(2*b^(7/2))
Time = 0.22 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.31 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^2 x^8} \, dx=\frac {15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {a x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} x^{5}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {a x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,x^{3}+15 a^{2} b \,x^{4}+10 a \,b^{2} x^{2}-2 b^{3}}{6 b^{4} x^{3} \left (a \,x^{2}+b \right )} \] Input:
int(1/(a+b/x^2)^2/x^8,x)
Output:
(15*sqrt(b)*sqrt(a)*atan((a*x)/(sqrt(b)*sqrt(a)))*a**2*x**5 + 15*sqrt(b)*s qrt(a)*atan((a*x)/(sqrt(b)*sqrt(a)))*a*b*x**3 + 15*a**2*b*x**4 + 10*a*b**2 *x**2 - 2*b**3)/(6*b**4*x**3*(a*x**2 + b))