Integrand size = 22, antiderivative size = 156 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^9} \, dx=\frac {(5 A b-8 a B) \sqrt {a+b x^2}}{48 a x^6}+\frac {b (5 A b-8 a B) \sqrt {a+b x^2}}{192 a^2 x^4}-\frac {b^2 (5 A b-8 a B) \sqrt {a+b x^2}}{128 a^3 x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{8 a x^8}+\frac {b^3 (5 A b-8 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{7/2}} \]
-1/8*A*(b*x^2+a)^(3/2)/a/x^8+1/128*b^3*(5*A*b-8*B*a)*arctanh((b*x^2+a)^(1/ 2)/a^(1/2))/a^(7/2)+1/48*(5*A*b-8*B*a)*(b*x^2+a)^(1/2)/a/x^6+1/192*b*(5*A* b-8*B*a)*(b*x^2+a)^(1/2)/a^2/x^4-1/128*b^2*(5*A*b-8*B*a)*(b*x^2+a)^(1/2)/a ^3/x^2
Time = 0.21 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^9} \, dx=\frac {\sqrt {a+b x^2} \left (-48 a^3 A-8 a^2 A b x^2-64 a^3 B x^2+10 a A b^2 x^4-16 a^2 b B x^4-15 A b^3 x^6+24 a b^2 B x^6\right )}{384 a^3 x^8}-\frac {b^3 (-5 A b+8 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{128 a^{7/2}} \]
(Sqrt[a + b*x^2]*(-48*a^3*A - 8*a^2*A*b*x^2 - 64*a^3*B*x^2 + 10*a*A*b^2*x^ 4 - 16*a^2*b*B*x^4 - 15*A*b^3*x^6 + 24*a*b^2*B*x^6))/(384*a^3*x^8) - (b^3* (-5*A*b + 8*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(128*a^(7/2))
Time = 0.23 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.91, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {354, 87, 51, 52, 52, 73, 221}
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.
| \(\displaystyle \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^9} \, dx\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {1}{2} \int \frac {\sqrt {b x^2+a} \left (B x^2+A\right )}{x^{10}}dx^2\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {1}{2} \left (-\frac {(5 A b-8 a B) \int \frac {\sqrt {b x^2+a}}{x^8}dx^2}{8 a}-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^8}\right )\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {1}{2} \left (-\frac {(5 A b-8 a B) \left (\frac {1}{6} b \int \frac {1}{x^6 \sqrt {b x^2+a}}dx^2-\frac {\sqrt {a+b x^2}}{3 x^6}\right )}{8 a}-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^8}\right )\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {1}{2} \left (-\frac {(5 A b-8 a B) \left (\frac {1}{6} b \left (-\frac {3 b \int \frac {1}{x^4 \sqrt {b x^2+a}}dx^2}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )-\frac {\sqrt {a+b x^2}}{3 x^6}\right )}{8 a}-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^8}\right )\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {1}{2} \left (-\frac {(5 A b-8 a B) \left (\frac {1}{6} b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2}{2 a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )-\frac {\sqrt {a+b x^2}}{3 x^6}\right )}{8 a}-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^8}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{2} \left (-\frac {(5 A b-8 a B) \left (\frac {1}{6} b \left (-\frac {3 b \left (-\frac {\int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )-\frac {\sqrt {a+b x^2}}{3 x^6}\right )}{8 a}-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^8}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} \left (-\frac {(5 A b-8 a B) \left (\frac {1}{6} b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x^2}}{a x^2}\right )}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )-\frac {\sqrt {a+b x^2}}{3 x^6}\right )}{8 a}-\frac {A \left (a+b x^2\right )^{3/2}}{4 a x^8}\right )\) |
(-1/4*(A*(a + b*x^2)^(3/2))/(a*x^8) - ((5*A*b - 8*a*B)*(-1/3*Sqrt[a + b*x^ 2]/x^6 + (b*(-1/2*Sqrt[a + b*x^2]/(a*x^4) - (3*b*(-(Sqrt[a + b*x^2]/(a*x^2 )) + (b*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/a^(3/2)))/(4*a)))/6))/(8*a))/2
3.6.18.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Time = 2.82 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.73
| method | result | size |
| pseudoelliptic | \(-\frac {\left (-\frac {15}{8} A \,b^{4}+3 B a \,b^{3}\right ) x^{8} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )+\sqrt {b \,x^{2}+a}\, \left (-\frac {5 x^{4} \left (\frac {12 x^{2} B}{5}+A \right ) b^{2} a^{\frac {3}{2}}}{4}+b \,x^{2} \left (2 x^{2} B +A \right ) a^{\frac {5}{2}}+\left (8 x^{2} B +6 A \right ) a^{\frac {7}{2}}+\frac {15 A \sqrt {a}\, b^{3} x^{6}}{8}\right )}{48 a^{\frac {7}{2}} x^{8}}\) | \(114\) |
| risch | \(-\frac {\sqrt {b \,x^{2}+a}\, \left (15 x^{6} b^{3} A -24 x^{6} a \,b^{2} B -10 A a \,b^{2} x^{4}+16 B \,a^{2} b \,x^{4}+8 A \,a^{2} b \,x^{2}+64 B \,a^{3} x^{2}+48 a^{3} A \right )}{384 x^{8} a^{3}}+\frac {\left (5 A b -8 B a \right ) b^{3} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{128 a^{\frac {7}{2}}}\) | \(124\) |
| default | \(B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 a \,x^{6}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{2 a}\right )}{4 a}\right )}{2 a}\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{8 a \,x^{8}}-\frac {5 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 a \,x^{6}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{2 a}\right )}{4 a}\right )}{2 a}\right )}{8 a}\right )\) | \(250\) |
-1/48*((-15/8*A*b^4+3*B*a*b^3)*x^8*arctanh((b*x^2+a)^(1/2)/a^(1/2))+(b*x^2 +a)^(1/2)*(-5/4*x^4*(12/5*x^2*B+A)*b^2*a^(3/2)+b*x^2*(2*B*x^2+A)*a^(5/2)+( 8*B*x^2+6*A)*a^(7/2)+15/8*A*a^(1/2)*b^3*x^6))/a^(7/2)/x^8
Time = 0.29 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.72 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^9} \, dx=\left [-\frac {3 \, {\left (8 \, B a b^{3} - 5 \, A b^{4}\right )} \sqrt {a} x^{8} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (3 \, {\left (8 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{6} - 48 \, A a^{4} - 2 \, {\left (8 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{4} - 8 \, {\left (8 \, B a^{4} + A a^{3} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{768 \, a^{4} x^{8}}, \frac {3 \, {\left (8 \, B a b^{3} - 5 \, A b^{4}\right )} \sqrt {-a} x^{8} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, {\left (8 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{6} - 48 \, A a^{4} - 2 \, {\left (8 \, B a^{3} b - 5 \, A a^{2} b^{2}\right )} x^{4} - 8 \, {\left (8 \, B a^{4} + A a^{3} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{384 \, a^{4} x^{8}}\right ] \]
[-1/768*(3*(8*B*a*b^3 - 5*A*b^4)*sqrt(a)*x^8*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(3*(8*B*a^2*b^2 - 5*A*a*b^3)*x^6 - 48*A*a^4 - 2 *(8*B*a^3*b - 5*A*a^2*b^2)*x^4 - 8*(8*B*a^4 + A*a^3*b)*x^2)*sqrt(b*x^2 + a ))/(a^4*x^8), 1/384*(3*(8*B*a*b^3 - 5*A*b^4)*sqrt(-a)*x^8*arctan(sqrt(-a)/ sqrt(b*x^2 + a)) + (3*(8*B*a^2*b^2 - 5*A*a*b^3)*x^6 - 48*A*a^4 - 2*(8*B*a^ 3*b - 5*A*a^2*b^2)*x^4 - 8*(8*B*a^4 + A*a^3*b)*x^2)*sqrt(b*x^2 + a))/(a^4* x^8)]
Time = 84.61 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.83 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^9} \, dx=- \frac {A a}{8 \sqrt {b} x^{9} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {7 A \sqrt {b}}{48 x^{7} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {A b^{\frac {3}{2}}}{192 a x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {5 A b^{\frac {5}{2}}}{384 a^{2} x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {5 A b^{\frac {7}{2}}}{128 a^{3} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {5 A b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{128 a^{\frac {7}{2}}} - \frac {B a}{6 \sqrt {b} x^{7} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {5 B \sqrt {b}}{24 x^{5} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B b^{\frac {3}{2}}}{48 a x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {B b^{\frac {5}{2}}}{16 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {B b^{3} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{16 a^{\frac {5}{2}}} \]
-A*a/(8*sqrt(b)*x**9*sqrt(a/(b*x**2) + 1)) - 7*A*sqrt(b)/(48*x**7*sqrt(a/( b*x**2) + 1)) + A*b**(3/2)/(192*a*x**5*sqrt(a/(b*x**2) + 1)) - 5*A*b**(5/2 )/(384*a**2*x**3*sqrt(a/(b*x**2) + 1)) - 5*A*b**(7/2)/(128*a**3*x*sqrt(a/( b*x**2) + 1)) + 5*A*b**4*asinh(sqrt(a)/(sqrt(b)*x))/(128*a**(7/2)) - B*a/( 6*sqrt(b)*x**7*sqrt(a/(b*x**2) + 1)) - 5*B*sqrt(b)/(24*x**5*sqrt(a/(b*x**2 ) + 1)) + B*b**(3/2)/(48*a*x**3*sqrt(a/(b*x**2) + 1)) + B*b**(5/2)/(16*a** 2*x*sqrt(a/(b*x**2) + 1)) - B*b**3*asinh(sqrt(a)/(sqrt(b)*x))/(16*a**(5/2) )
Time = 0.22 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^9} \, dx=-\frac {B b^{3} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{16 \, a^{\frac {5}{2}}} + \frac {5 \, A b^{4} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{128 \, a^{\frac {7}{2}}} + \frac {\sqrt {b x^{2} + a} B b^{3}}{16 \, a^{3}} - \frac {5 \, \sqrt {b x^{2} + a} A b^{4}}{128 \, a^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{2}}{16 \, a^{3} x^{2}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{3}}{128 \, a^{4} x^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B b}{8 \, a^{2} x^{4}} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{2}}{64 \, a^{3} x^{4}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{6 \, a x^{6}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b}{48 \, a^{2} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{8 \, a x^{8}} \]
-1/16*B*b^3*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) + 5/128*A*b^4*arcsinh(a/ (sqrt(a*b)*abs(x)))/a^(7/2) + 1/16*sqrt(b*x^2 + a)*B*b^3/a^3 - 5/128*sqrt( b*x^2 + a)*A*b^4/a^4 - 1/16*(b*x^2 + a)^(3/2)*B*b^2/(a^3*x^2) + 5/128*(b*x ^2 + a)^(3/2)*A*b^3/(a^4*x^2) + 1/8*(b*x^2 + a)^(3/2)*B*b/(a^2*x^4) - 5/64 *(b*x^2 + a)^(3/2)*A*b^2/(a^3*x^4) - 1/6*(b*x^2 + a)^(3/2)*B/(a*x^6) + 5/4 8*(b*x^2 + a)^(3/2)*A*b/(a^2*x^6) - 1/8*(b*x^2 + a)^(3/2)*A/(a*x^8)
Time = 0.29 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.24 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^9} \, dx=\frac {\frac {3 \, {\left (8 \, B a b^{4} - 5 \, A b^{5}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {24 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a b^{4} - 88 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{2} b^{4} + 40 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{3} b^{4} + 24 \, \sqrt {b x^{2} + a} B a^{4} b^{4} - 15 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A b^{5} + 55 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A a b^{5} - 73 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{2} b^{5} - 15 \, \sqrt {b x^{2} + a} A a^{3} b^{5}}{a^{3} b^{4} x^{8}}}{384 \, b} \]
1/384*(3*(8*B*a*b^4 - 5*A*b^5)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)* a^3) + (24*(b*x^2 + a)^(7/2)*B*a*b^4 - 88*(b*x^2 + a)^(5/2)*B*a^2*b^4 + 40 *(b*x^2 + a)^(3/2)*B*a^3*b^4 + 24*sqrt(b*x^2 + a)*B*a^4*b^4 - 15*(b*x^2 + a)^(7/2)*A*b^5 + 55*(b*x^2 + a)^(5/2)*A*a*b^5 - 73*(b*x^2 + a)^(3/2)*A*a^2 *b^5 - 15*sqrt(b*x^2 + a)*A*a^3*b^5)/(a^3*b^4*x^8))/b
Time = 6.90 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^9} \, dx=\frac {55\,A\,{\left (b\,x^2+a\right )}^{5/2}}{384\,a^2\,x^8}-\frac {B\,\sqrt {b\,x^2+a}}{16\,x^6}-\frac {73\,A\,{\left (b\,x^2+a\right )}^{3/2}}{384\,a\,x^8}-\frac {5\,A\,\sqrt {b\,x^2+a}}{128\,x^8}-\frac {5\,A\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a^3\,x^8}-\frac {B\,{\left (b\,x^2+a\right )}^{3/2}}{6\,a\,x^6}+\frac {B\,{\left (b\,x^2+a\right )}^{5/2}}{16\,a^2\,x^6}-\frac {A\,b^4\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{128\,a^{7/2}}+\frac {B\,b^3\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,1{}\mathrm {i}}{16\,a^{5/2}} \]
(B*b^3*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*1i)/(16*a^(5/2)) - (B*(a + b*x ^2)^(1/2))/(16*x^6) - (A*b^4*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*5i)/(128 *a^(7/2)) - (5*A*(a + b*x^2)^(1/2))/(128*x^8) - (73*A*(a + b*x^2)^(3/2))/( 384*a*x^8) + (55*A*(a + b*x^2)^(5/2))/(384*a^2*x^8) - (5*A*(a + b*x^2)^(7/ 2))/(128*a^3*x^8) - (B*(a + b*x^2)^(3/2))/(6*a*x^6) + (B*(a + b*x^2)^(5/2) )/(16*a^2*x^6)