Integrand size = 13, antiderivative size = 91 \[ \int \frac {1}{x^8 \left (a+b x^2\right )^2} \, dx=-\frac {1}{7 a^2 x^7}+\frac {2 b}{5 a^3 x^5}-\frac {b^2}{a^4 x^3}+\frac {4 b^3}{a^5 x}+\frac {b^4 x}{2 a^5 \left (a+b x^2\right )}+\frac {9 b^{7/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{11/2}} \] Output:
-1/7/a^2/x^7+2/5*b/a^3/x^5-b^2/a^4/x^3+4*b^3/a^5/x+1/2*b^4*x/a^5/(b*x^2+a) +9/2*b^(7/2)*arctan(b^(1/2)*x/a^(1/2))/a^(11/2)
Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^8 \left (a+b x^2\right )^2} \, dx=-\frac {1}{7 a^2 x^7}+\frac {2 b}{5 a^3 x^5}-\frac {b^2}{a^4 x^3}+\frac {4 b^3}{a^5 x}+\frac {b^4 x}{2 a^5 \left (a+b x^2\right )}+\frac {9 b^{7/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{11/2}} \] Input:
Integrate[1/(x^8*(a + b*x^2)^2),x]
Output:
-1/7*1/(a^2*x^7) + (2*b)/(5*a^3*x^5) - b^2/(a^4*x^3) + (4*b^3)/(a^5*x) + ( b^4*x)/(2*a^5*(a + b*x^2)) + (9*b^(7/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(2*a^ (11/2))
Time = 0.19 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.23, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {253, 264, 264, 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.
\(\displaystyle \int \frac {1}{x^8 \left (a+b x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {9 \int \frac {1}{x^8 \left (b x^2+a\right )}dx}{2 a}+\frac {1}{2 a x^7 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {9 \left (-\frac {b \int \frac {1}{x^6 \left (b x^2+a\right )}dx}{a}-\frac {1}{7 a x^7}\right )}{2 a}+\frac {1}{2 a x^7 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {9 \left (-\frac {b \left (-\frac {b \int \frac {1}{x^4 \left (b x^2+a\right )}dx}{a}-\frac {1}{5 a x^5}\right )}{a}-\frac {1}{7 a x^7}\right )}{2 a}+\frac {1}{2 a x^7 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {9 \left (-\frac {b \left (-\frac {b \left (-\frac {b \int \frac {1}{x^2 \left (b x^2+a\right )}dx}{a}-\frac {1}{3 a x^3}\right )}{a}-\frac {1}{5 a x^5}\right )}{a}-\frac {1}{7 a x^7}\right )}{2 a}+\frac {1}{2 a x^7 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {9 \left (-\frac {b \left (-\frac {b \left (-\frac {b \left (-\frac {b \int \frac {1}{b x^2+a}dx}{a}-\frac {1}{a x}\right )}{a}-\frac {1}{3 a x^3}\right )}{a}-\frac {1}{5 a x^5}\right )}{a}-\frac {1}{7 a x^7}\right )}{2 a}+\frac {1}{2 a x^7 \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {9 \left (-\frac {b \left (-\frac {b \left (-\frac {b \left (-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}-\frac {1}{a x}\right )}{a}-\frac {1}{3 a x^3}\right )}{a}-\frac {1}{5 a x^5}\right )}{a}-\frac {1}{7 a x^7}\right )}{2 a}+\frac {1}{2 a x^7 \left (a+b x^2\right )}\) |
Input:
Int[1/(x^8*(a + b*x^2)^2),x]
Output:
1/(2*a*x^7*(a + b*x^2)) + (9*(-1/7*1/(a*x^7) - (b*(-1/5*1/(a*x^5) - (b*(-1 /3*1/(a*x^3) - (b*(-(1/(a*x)) - (Sqrt[b]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/a^(3 /2)))/a))/a))/a))/(2*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*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]
Time = 0.31 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.85
method | result | size |
default | \(\frac {b^{4} \left (\frac {x}{2 b \,x^{2}+2 a}+\frac {9 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{5}}-\frac {1}{7 a^{2} x^{7}}+\frac {2 b}{5 a^{3} x^{5}}-\frac {b^{2}}{a^{4} x^{3}}+\frac {4 b^{3}}{a^{5} x}\) | \(77\) |
risch | \(\frac {\frac {9 b^{4} x^{8}}{2 a^{5}}+\frac {3 b^{3} x^{6}}{a^{4}}-\frac {3 b^{2} x^{4}}{5 a^{3}}+\frac {9 b \,x^{2}}{35 a^{2}}-\frac {1}{7 a}}{x^{7} \left (b \,x^{2}+a \right )}+\frac {9 \sqrt {-a b}\, b^{3} \ln \left (-b x -\sqrt {-a b}\right )}{4 a^{6}}-\frac {9 \sqrt {-a b}\, b^{3} \ln \left (-b x +\sqrt {-a b}\right )}{4 a^{6}}\) | \(117\) |
Input:
int(1/x^8/(b*x^2+a)^2,x,method=_RETURNVERBOSE)
Output:
b^4/a^5*(1/2*x/(b*x^2+a)+9/2/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))-1/7/a^2/ x^7+2/5*b/a^3/x^5-b^2/a^4/x^3+4*b^3/a^5/x
Time = 0.07 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.42 \[ \int \frac {1}{x^8 \left (a+b x^2\right )^2} \, dx=\left [\frac {630 \, b^{4} x^{8} + 420 \, a b^{3} x^{6} - 84 \, a^{2} b^{2} x^{4} + 36 \, a^{3} b x^{2} - 20 \, a^{4} + 315 \, {\left (b^{4} x^{9} + a b^{3} x^{7}\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} + 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{140 \, {\left (a^{5} b x^{9} + a^{6} x^{7}\right )}}, \frac {315 \, b^{4} x^{8} + 210 \, a b^{3} x^{6} - 42 \, a^{2} b^{2} x^{4} + 18 \, a^{3} b x^{2} - 10 \, a^{4} + 315 \, {\left (b^{4} x^{9} + a b^{3} x^{7}\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right )}{70 \, {\left (a^{5} b x^{9} + a^{6} x^{7}\right )}}\right ] \] Input:
integrate(1/x^8/(b*x^2+a)^2,x, algorithm="fricas")
Output:
[1/140*(630*b^4*x^8 + 420*a*b^3*x^6 - 84*a^2*b^2*x^4 + 36*a^3*b*x^2 - 20*a ^4 + 315*(b^4*x^9 + a*b^3*x^7)*sqrt(-b/a)*log((b*x^2 + 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/(a^5*b*x^9 + a^6*x^7), 1/70*(315*b^4*x^8 + 210*a*b^3*x^6 - 42*a^2*b^2*x^4 + 18*a^3*b*x^2 - 10*a^4 + 315*(b^4*x^9 + a*b^3*x^7)*sqrt( b/a)*arctan(x*sqrt(b/a)))/(a^5*b*x^9 + a^6*x^7)]
Time = 0.24 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.52 \[ \int \frac {1}{x^8 \left (a+b x^2\right )^2} \, dx=- \frac {9 \sqrt {- \frac {b^{7}}{a^{11}}} \log {\left (- \frac {a^{6} \sqrt {- \frac {b^{7}}{a^{11}}}}{b^{4}} + x \right )}}{4} + \frac {9 \sqrt {- \frac {b^{7}}{a^{11}}} \log {\left (\frac {a^{6} \sqrt {- \frac {b^{7}}{a^{11}}}}{b^{4}} + x \right )}}{4} + \frac {- 10 a^{4} + 18 a^{3} b x^{2} - 42 a^{2} b^{2} x^{4} + 210 a b^{3} x^{6} + 315 b^{4} x^{8}}{70 a^{6} x^{7} + 70 a^{5} b x^{9}} \] Input:
integrate(1/x**8/(b*x**2+a)**2,x)
Output:
-9*sqrt(-b**7/a**11)*log(-a**6*sqrt(-b**7/a**11)/b**4 + x)/4 + 9*sqrt(-b** 7/a**11)*log(a**6*sqrt(-b**7/a**11)/b**4 + x)/4 + (-10*a**4 + 18*a**3*b*x* *2 - 42*a**2*b**2*x**4 + 210*a*b**3*x**6 + 315*b**4*x**8)/(70*a**6*x**7 + 70*a**5*b*x**9)
Time = 0.11 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^8 \left (a+b x^2\right )^2} \, dx=\frac {315 \, b^{4} x^{8} + 210 \, a b^{3} x^{6} - 42 \, a^{2} b^{2} x^{4} + 18 \, a^{3} b x^{2} - 10 \, a^{4}}{70 \, {\left (a^{5} b x^{9} + a^{6} x^{7}\right )}} + \frac {9 \, b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{5}} \] Input:
integrate(1/x^8/(b*x^2+a)^2,x, algorithm="maxima")
Output:
1/70*(315*b^4*x^8 + 210*a*b^3*x^6 - 42*a^2*b^2*x^4 + 18*a^3*b*x^2 - 10*a^4 )/(a^5*b*x^9 + a^6*x^7) + 9/2*b^4*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^5)
Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^8 \left (a+b x^2\right )^2} \, dx=\frac {9 \, b^{4} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {a b} a^{5}} + \frac {b^{4} x}{2 \, {\left (b x^{2} + a\right )} a^{5}} + \frac {140 \, b^{3} x^{6} - 35 \, a b^{2} x^{4} + 14 \, a^{2} b x^{2} - 5 \, a^{3}}{35 \, a^{5} x^{7}} \] Input:
integrate(1/x^8/(b*x^2+a)^2,x, algorithm="giac")
Output:
9/2*b^4*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^5) + 1/2*b^4*x/((b*x^2 + a)*a^5 ) + 1/35*(140*b^3*x^6 - 35*a*b^2*x^4 + 14*a^2*b*x^2 - 5*a^3)/(a^5*x^7)
Time = 0.23 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^8 \left (a+b x^2\right )^2} \, dx=\frac {\frac {9\,b\,x^2}{35\,a^2}-\frac {1}{7\,a}-\frac {3\,b^2\,x^4}{5\,a^3}+\frac {3\,b^3\,x^6}{a^4}+\frac {9\,b^4\,x^8}{2\,a^5}}{b\,x^9+a\,x^7}+\frac {9\,b^{7/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{2\,a^{11/2}} \] Input:
int(1/(x^8*(a + b*x^2)^2),x)
Output:
((9*b*x^2)/(35*a^2) - 1/(7*a) - (3*b^2*x^4)/(5*a^3) + (3*b^3*x^6)/a^4 + (9 *b^4*x^8)/(2*a^5))/(a*x^7 + b*x^9) + (9*b^(7/2)*atan((b^(1/2)*x)/a^(1/2))) /(2*a^(11/2))
Time = 0.25 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.23 \[ \int \frac {1}{x^8 \left (a+b x^2\right )^2} \, dx=\frac {315 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} x^{7}+315 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{4} x^{9}-10 a^{5}+18 a^{4} b \,x^{2}-42 a^{3} b^{2} x^{4}+210 a^{2} b^{3} x^{6}+315 a \,b^{4} x^{8}}{70 a^{6} x^{7} \left (b \,x^{2}+a \right )} \] Input:
int(1/x^8/(b*x^2+a)^2,x)
Output:
(315*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b**3*x**7 + 315*sqrt( b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**4*x**9 - 10*a**5 + 18*a**4*b*x **2 - 42*a**3*b**2*x**4 + 210*a**2*b**3*x**6 + 315*a*b**4*x**8)/(70*a**6*x **7*(a + b*x**2))