Integrand size = 13, antiderivative size = 72 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^8} \, dx=-\frac {1}{b^3 x}-\frac {a x}{4 b^2 \left (b+a x^2\right )^2}-\frac {7 a x}{8 b^3 \left (b+a x^2\right )}-\frac {15 \sqrt {a} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 b^{7/2}} \] Output:
-1/b^3/x-1/4*a*x/b^2/(a*x^2+b)^2-7/8*a*x/b^3/(a*x^2+b)-15/8*a^(1/2)*arctan (a^(1/2)*x/b^(1/2))/b^(7/2)
Time = 0.03 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^8} \, dx=-\frac {8 b^2+25 a b x^2+15 a^2 x^4}{8 b^3 x \left (b+a x^2\right )^2}-\frac {15 \sqrt {a} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{8 b^{7/2}} \] Input:
Integrate[1/((a + b/x^2)^3*x^8),x]
Output:
-1/8*(8*b^2 + 25*a*b*x^2 + 15*a^2*x^4)/(b^3*x*(b + a*x^2)^2) - (15*Sqrt[a] *ArcTan[(Sqrt[a]*x)/Sqrt[b]])/(8*b^(7/2))
Time = 0.30 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.22, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {795, 253, 253, 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.
\(\displaystyle \int \frac {1}{x^8 \left (a+\frac {b}{x^2}\right )^3} \, dx\) |
\(\Big \downarrow \) 795 |
\(\displaystyle \int \frac {1}{x^2 \left (a x^2+b\right )^3}dx\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {5 \int \frac {1}{x^2 \left (a x^2+b\right )^2}dx}{4 b}+\frac {1}{4 b x \left (a x^2+b\right )^2}\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {5 \left (\frac {3 \int \frac {1}{x^2 \left (a x^2+b\right )}dx}{2 b}+\frac {1}{2 b x \left (a x^2+b\right )}\right )}{4 b}+\frac {1}{4 b x \left (a x^2+b\right )^2}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {a \int \frac {1}{a x^2+b}dx}{b}-\frac {1}{b x}\right )}{2 b}+\frac {1}{2 b x \left (a x^2+b\right )}\right )}{4 b}+\frac {1}{4 b x \left (a x^2+b\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5 \left (\frac {3 \left (-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right )}{b^{3/2}}-\frac {1}{b x}\right )}{2 b}+\frac {1}{2 b x \left (a x^2+b\right )}\right )}{4 b}+\frac {1}{4 b x \left (a x^2+b\right )^2}\) |
Input:
Int[1/((a + b/x^2)^3*x^8),x]
Output:
1/(4*b*x*(b + a*x^2)^2) + (5*(1/(2*b*x*(b + a*x^2)) + (3*(-(1/(b*x)) - (Sq rt[a]*ArcTan[(Sqrt[a]*x)/Sqrt[b]])/b^(3/2)))/(2*b)))/(4*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 = 0.10 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.75
method | result | size |
default | \(-\frac {a \left (\frac {\frac {7}{8} a \,x^{3}+\frac {9}{8} b x}{\left (a \,x^{2}+b \right )^{2}}+\frac {15 \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{b^{3}}-\frac {1}{b^{3} x}\) | \(54\) |
risch | \(\frac {-\frac {15 a^{2} x^{4}}{8 b^{3}}-\frac {25 a \,x^{2}}{8 b^{2}}-\frac {1}{b}}{x \left (a \,x^{2}+b \right )^{2}}+\frac {15 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b^{7} \textit {\_Z}^{2}+a \right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} b^{7}+2 a \right ) x +b^{4} \textit {\_R} \right )\right )}{16}\) | \(79\) |
Input:
int(1/(a+b/x^2)^3/x^8,x,method=_RETURNVERBOSE)
Output:
-1/b^3*a*((7/8*a*x^3+9/8*b*x)/(a*x^2+b)^2+15/8/(a*b)^(1/2)*arctan(a*x/(a*b )^(1/2)))-1/b^3/x
Time = 0.07 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.81 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^8} \, dx=\left [-\frac {30 \, a^{2} x^{4} + 50 \, a b x^{2} - 15 \, {\left (a^{2} x^{5} + 2 \, a b x^{3} + b^{2} x\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {a x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - b}{a x^{2} + b}\right ) + 16 \, b^{2}}{16 \, {\left (a^{2} b^{3} x^{5} + 2 \, a b^{4} x^{3} + b^{5} x\right )}}, -\frac {15 \, a^{2} x^{4} + 25 \, a b x^{2} + 15 \, {\left (a^{2} x^{5} + 2 \, a b x^{3} + b^{2} x\right )} \sqrt {\frac {a}{b}} \arctan \left (x \sqrt {\frac {a}{b}}\right ) + 8 \, b^{2}}{8 \, {\left (a^{2} b^{3} x^{5} + 2 \, a b^{4} x^{3} + b^{5} x\right )}}\right ] \] Input:
integrate(1/(a+b/x^2)^3/x^8,x, algorithm="fricas")
Output:
[-1/16*(30*a^2*x^4 + 50*a*b*x^2 - 15*(a^2*x^5 + 2*a*b*x^3 + b^2*x)*sqrt(-a /b)*log((a*x^2 - 2*b*x*sqrt(-a/b) - b)/(a*x^2 + b)) + 16*b^2)/(a^2*b^3*x^5 + 2*a*b^4*x^3 + b^5*x), -1/8*(15*a^2*x^4 + 25*a*b*x^2 + 15*(a^2*x^5 + 2*a *b*x^3 + b^2*x)*sqrt(a/b)*arctan(x*sqrt(a/b)) + 8*b^2)/(a^2*b^3*x^5 + 2*a* b^4*x^3 + b^5*x)]
Time = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.61 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^8} \, dx=\frac {15 \sqrt {- \frac {a}{b^{7}}} \log {\left (x - \frac {b^{4} \sqrt {- \frac {a}{b^{7}}}}{a} \right )}}{16} - \frac {15 \sqrt {- \frac {a}{b^{7}}} \log {\left (x + \frac {b^{4} \sqrt {- \frac {a}{b^{7}}}}{a} \right )}}{16} + \frac {- 15 a^{2} x^{4} - 25 a b x^{2} - 8 b^{2}}{8 a^{2} b^{3} x^{5} + 16 a b^{4} x^{3} + 8 b^{5} x} \] Input:
integrate(1/(a+b/x**2)**3/x**8,x)
Output:
15*sqrt(-a/b**7)*log(x - b**4*sqrt(-a/b**7)/a)/16 - 15*sqrt(-a/b**7)*log(x + b**4*sqrt(-a/b**7)/a)/16 + (-15*a**2*x**4 - 25*a*b*x**2 - 8*b**2)/(8*a* *2*b**3*x**5 + 16*a*b**4*x**3 + 8*b**5*x)
Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^8} \, dx=-\frac {15 \, a^{2} x^{4} + 25 \, a b x^{2} + 8 \, b^{2}}{8 \, {\left (a^{2} b^{3} x^{5} + 2 \, a b^{4} x^{3} + b^{5} x\right )}} - \frac {15 \, a \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} \] Input:
integrate(1/(a+b/x^2)^3/x^8,x, algorithm="maxima")
Output:
-1/8*(15*a^2*x^4 + 25*a*b*x^2 + 8*b^2)/(a^2*b^3*x^5 + 2*a*b^4*x^3 + b^5*x) - 15/8*a*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*b^3)
Time = 0.12 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.79 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^8} \, dx=-\frac {15 \, a \arctan \left (\frac {a x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} - \frac {7 \, a^{2} x^{3} + 9 \, a b x}{8 \, {\left (a x^{2} + b\right )}^{2} b^{3}} - \frac {1}{b^{3} x} \] Input:
integrate(1/(a+b/x^2)^3/x^8,x, algorithm="giac")
Output:
-15/8*a*arctan(a*x/sqrt(a*b))/(sqrt(a*b)*b^3) - 1/8*(7*a^2*x^3 + 9*a*b*x)/ ((a*x^2 + b)^2*b^3) - 1/(b^3*x)
Time = 0.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.92 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^8} \, dx=-\frac {\frac {1}{b}+\frac {25\,a\,x^2}{8\,b^2}+\frac {15\,a^2\,x^4}{8\,b^3}}{a^2\,x^5+2\,a\,b\,x^3+b^2\,x}-\frac {15\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {a}\,x}{\sqrt {b}}\right )}{8\,b^{7/2}} \] Input:
int(1/(x^8*(a + b/x^2)^3),x)
Output:
- (1/b + (25*a*x^2)/(8*b^2) + (15*a^2*x^4)/(8*b^3))/(b^2*x + a^2*x^5 + 2*a *b*x^3) - (15*a^(1/2)*atan((a^(1/2)*x)/b^(1/2)))/(8*b^(7/2))
Time = 0.21 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.68 \[ \int \frac {1}{\left (a+\frac {b}{x^2}\right )^3 x^8} \, dx=\frac {-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {a x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} x^{5}-30 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {a x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,x^{3}-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {a x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} x -15 a^{2} b \,x^{4}-25 a \,b^{2} x^{2}-8 b^{3}}{8 b^{4} x \left (a^{2} x^{4}+2 a b \,x^{2}+b^{2}\right )} \] Input:
int(1/(a+b/x^2)^3/x^8,x)
Output:
( - 15*sqrt(b)*sqrt(a)*atan((a*x)/(sqrt(b)*sqrt(a)))*a**2*x**5 - 30*sqrt(b )*sqrt(a)*atan((a*x)/(sqrt(b)*sqrt(a)))*a*b*x**3 - 15*sqrt(b)*sqrt(a)*atan ((a*x)/(sqrt(b)*sqrt(a)))*b**2*x - 15*a**2*b*x**4 - 25*a*b**2*x**2 - 8*b** 3)/(8*b**4*x*(a**2*x**4 + 2*a*b*x**2 + b**2))