Integrand size = 22, antiderivative size = 189 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^{11}} \, dx=\frac {(7 A b-10 a B) \sqrt {a+b x^2}}{80 a x^8}+\frac {b (7 A b-10 a B) \sqrt {a+b x^2}}{480 a^2 x^6}-\frac {b^2 (7 A b-10 a B) \sqrt {a+b x^2}}{384 a^3 x^4}+\frac {b^3 (7 A b-10 a B) \sqrt {a+b x^2}}{256 a^4 x^2}-\frac {A \left (a+b x^2\right )^{3/2}}{10 a x^{10}}-\frac {b^4 (7 A b-10 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{256 a^{9/2}} \]
-1/10*A*(b*x^2+a)^(3/2)/a/x^10-1/256*b^4*(7*A*b-10*B*a)*arctanh((b*x^2+a)^ (1/2)/a^(1/2))/a^(9/2)+1/80*(7*A*b-10*B*a)*(b*x^2+a)^(1/2)/a/x^8+1/480*b*( 7*A*b-10*B*a)*(b*x^2+a)^(1/2)/a^2/x^6-1/384*b^2*(7*A*b-10*B*a)*(b*x^2+a)^( 1/2)/a^3/x^4+1/256*b^3*(7*A*b-10*B*a)*(b*x^2+a)^(1/2)/a^4/x^2
Time = 0.25 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^{11}} \, dx=\frac {\frac {\sqrt {a} \sqrt {a+b x^2} \left (105 A b^4 x^8-16 a^3 b x^2 \left (3 A+5 B x^2\right )-96 a^4 \left (4 A+5 B x^2\right )-10 a b^3 x^6 \left (7 A+15 B x^2\right )+4 a^2 b^2 x^4 \left (14 A+25 B x^2\right )\right )}{x^{10}}-15 b^4 (7 A b-10 a B) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{3840 a^{9/2}} \]
((Sqrt[a]*Sqrt[a + b*x^2]*(105*A*b^4*x^8 - 16*a^3*b*x^2*(3*A + 5*B*x^2) - 96*a^4*(4*A + 5*B*x^2) - 10*a*b^3*x^6*(7*A + 15*B*x^2) + 4*a^2*b^2*x^4*(14 *A + 25*B*x^2)))/x^10 - 15*b^4*(7*A*b - 10*a*B)*ArcTanh[Sqrt[a + b*x^2]/Sq rt[a]])/(3840*a^(9/2))
Time = 0.25 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.91, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {354, 87, 51, 52, 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.
| \(\displaystyle \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^{11}} \, dx\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {1}{2} \int \frac {\sqrt {b x^2+a} \left (B x^2+A\right )}{x^{12}}dx^2\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {1}{2} \left (-\frac {(7 A b-10 a B) \int \frac {\sqrt {b x^2+a}}{x^{10}}dx^2}{10 a}-\frac {A \left (a+b x^2\right )^{3/2}}{5 a x^{10}}\right )\) |
| \(\Big \downarrow \) 51 |
| \(\displaystyle \frac {1}{2} \left (-\frac {(7 A b-10 a B) \left (\frac {1}{8} b \int \frac {1}{x^8 \sqrt {b x^2+a}}dx^2-\frac {\sqrt {a+b x^2}}{4 x^8}\right )}{10 a}-\frac {A \left (a+b x^2\right )^{3/2}}{5 a x^{10}}\right )\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {1}{2} \left (-\frac {(7 A b-10 a B) \left (\frac {1}{8} b \left (-\frac {5 b \int \frac {1}{x^6 \sqrt {b x^2+a}}dx^2}{6 a}-\frac {\sqrt {a+b x^2}}{3 a x^6}\right )-\frac {\sqrt {a+b x^2}}{4 x^8}\right )}{10 a}-\frac {A \left (a+b x^2\right )^{3/2}}{5 a x^{10}}\right )\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {1}{2} \left (-\frac {(7 A b-10 a B) \left (\frac {1}{8} b \left (-\frac {5 b \left (-\frac {3 b \int \frac {1}{x^4 \sqrt {b x^2+a}}dx^2}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )}{6 a}-\frac {\sqrt {a+b x^2}}{3 a x^6}\right )-\frac {\sqrt {a+b x^2}}{4 x^8}\right )}{10 a}-\frac {A \left (a+b x^2\right )^{3/2}}{5 a x^{10}}\right )\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {1}{2} \left (-\frac {(7 A b-10 a B) \left (\frac {1}{8} b \left (-\frac {5 b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{x^2 \sqrt {b x^2+a}}dx^2}{2 a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )}{6 a}-\frac {\sqrt {a+b x^2}}{3 a x^6}\right )-\frac {\sqrt {a+b x^2}}{4 x^8}\right )}{10 a}-\frac {A \left (a+b x^2\right )^{3/2}}{5 a x^{10}}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{2} \left (-\frac {(7 A b-10 a B) \left (\frac {1}{8} b \left (-\frac {5 b \left (-\frac {3 b \left (-\frac {\int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{a}-\frac {\sqrt {a+b x^2}}{a x^2}\right )}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )}{6 a}-\frac {\sqrt {a+b x^2}}{3 a x^6}\right )-\frac {\sqrt {a+b x^2}}{4 x^8}\right )}{10 a}-\frac {A \left (a+b x^2\right )^{3/2}}{5 a x^{10}}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{2} \left (-\frac {(7 A b-10 a B) \left (\frac {1}{8} b \left (-\frac {5 b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x^2}}{a x^2}\right )}{4 a}-\frac {\sqrt {a+b x^2}}{2 a x^4}\right )}{6 a}-\frac {\sqrt {a+b x^2}}{3 a x^6}\right )-\frac {\sqrt {a+b x^2}}{4 x^8}\right )}{10 a}-\frac {A \left (a+b x^2\right )^{3/2}}{5 a x^{10}}\right )\) |
(-1/5*(A*(a + b*x^2)^(3/2))/(a*x^10) - ((7*A*b - 10*a*B)*(-1/4*Sqrt[a + b* x^2]/x^8 + (b*(-1/3*Sqrt[a + b*x^2]/(a*x^6) - (5*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]/Sqr t[a]])/a^(3/2)))/(4*a)))/(6*a)))/8))/(10*a))/2
3.6.20.3.1 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 = 2.85 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.70
| method | result | size |
| pseudoelliptic | \(\frac {-\frac {7 x^{10} b^{4} \left (A b -\frac {10 B a}{7}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{\sqrt {a}}\right )}{256}+\frac {7 \sqrt {b \,x^{2}+a}\, \left (-\frac {5 x^{6} \left (\frac {15 x^{2} B}{7}+A \right ) b^{3} a^{\frac {3}{2}}}{4}+b^{2} x^{4} \left (\frac {25 x^{2} B}{14}+A \right ) a^{\frac {5}{2}}-\frac {6 x^{2} b \left (\frac {5 x^{2} B}{3}+A \right ) a^{\frac {7}{2}}}{7}+\frac {12 \left (-5 x^{2} B -4 A \right ) a^{\frac {9}{2}}}{7}+\frac {15 A \sqrt {a}\, b^{4} x^{8}}{8}\right )}{480}}{a^{\frac {9}{2}} x^{10}}\) | \(132\) |
| risch | \(-\frac {\sqrt {b \,x^{2}+a}\, \left (-105 A \,b^{4} x^{8}+150 B a \,b^{3} x^{8}+70 A a \,b^{3} x^{6}-100 B \,a^{2} b^{2} x^{6}-56 A \,a^{2} b^{2} x^{4}+80 B \,a^{3} b \,x^{4}+48 A \,a^{3} b \,x^{2}+480 B \,a^{4} x^{2}+384 A \,a^{4}\right )}{3840 x^{10} a^{4}}-\frac {\left (7 A b -10 B a \right ) b^{4} \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{256 a^{\frac {9}{2}}}\) | \(148\) |
| default | \(B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{8 a \,x^{8}}-\frac {5 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 a \,x^{6}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{2 a}\right )}{4 a}\right )}{2 a}\right )}{8 a}\right )+A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{10 a \,x^{10}}-\frac {7 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{8 a \,x^{8}}-\frac {5 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 a \,x^{6}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 a \,x^{4}}-\frac {b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{2 a \,x^{2}}+\frac {b \left (\sqrt {b \,x^{2}+a}-\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )\right )}{2 a}\right )}{4 a}\right )}{2 a}\right )}{8 a}\right )}{10 a}\right )\) | \(298\) |
7/480*(-15/8*x^10*b^4*(A*b-10/7*B*a)*arctanh((b*x^2+a)^(1/2)/a^(1/2))+(b*x ^2+a)^(1/2)*(-5/4*x^6*(15/7*x^2*B+A)*b^3*a^(3/2)+b^2*x^4*(25/14*x^2*B+A)*a ^(5/2)-6/7*x^2*b*(5/3*x^2*B+A)*a^(7/2)+12/7*(-5*B*x^2-4*A)*a^(9/2)+15/8*A* a^(1/2)*b^4*x^8))/a^(9/2)/x^10
Time = 0.31 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.68 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^{11}} \, dx=\left [-\frac {15 \, {\left (10 \, B a b^{4} - 7 \, A b^{5}\right )} \sqrt {a} x^{10} \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (15 \, {\left (10 \, B a^{2} b^{3} - 7 \, A a b^{4}\right )} x^{8} - 10 \, {\left (10 \, B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} x^{6} + 384 \, A a^{5} + 8 \, {\left (10 \, B a^{4} b - 7 \, A a^{3} b^{2}\right )} x^{4} + 48 \, {\left (10 \, B a^{5} + A a^{4} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{7680 \, a^{5} x^{10}}, -\frac {15 \, {\left (10 \, B a b^{4} - 7 \, A b^{5}\right )} \sqrt {-a} x^{10} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (15 \, {\left (10 \, B a^{2} b^{3} - 7 \, A a b^{4}\right )} x^{8} - 10 \, {\left (10 \, B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} x^{6} + 384 \, A a^{5} + 8 \, {\left (10 \, B a^{4} b - 7 \, A a^{3} b^{2}\right )} x^{4} + 48 \, {\left (10 \, B a^{5} + A a^{4} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{3840 \, a^{5} x^{10}}\right ] \]
[-1/7680*(15*(10*B*a*b^4 - 7*A*b^5)*sqrt(a)*x^10*log(-(b*x^2 - 2*sqrt(b*x^ 2 + a)*sqrt(a) + 2*a)/x^2) + 2*(15*(10*B*a^2*b^3 - 7*A*a*b^4)*x^8 - 10*(10 *B*a^3*b^2 - 7*A*a^2*b^3)*x^6 + 384*A*a^5 + 8*(10*B*a^4*b - 7*A*a^3*b^2)*x ^4 + 48*(10*B*a^5 + A*a^4*b)*x^2)*sqrt(b*x^2 + a))/(a^5*x^10), -1/3840*(15 *(10*B*a*b^4 - 7*A*b^5)*sqrt(-a)*x^10*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + ( 15*(10*B*a^2*b^3 - 7*A*a*b^4)*x^8 - 10*(10*B*a^3*b^2 - 7*A*a^2*b^3)*x^6 + 384*A*a^5 + 8*(10*B*a^4*b - 7*A*a^3*b^2)*x^4 + 48*(10*B*a^5 + A*a^4*b)*x^2 )*sqrt(b*x^2 + a))/(a^5*x^10)]
Timed out. \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^{11}} \, dx=\text {Timed out} \]
Time = 0.26 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.37 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^{11}} \, dx=\frac {5 \, B b^{4} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{128 \, a^{\frac {7}{2}}} - \frac {7 \, A b^{5} \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{256 \, a^{\frac {9}{2}}} - \frac {5 \, \sqrt {b x^{2} + a} B b^{4}}{128 \, a^{4}} + \frac {7 \, \sqrt {b x^{2} + a} A b^{5}}{256 \, a^{5}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{3}}{128 \, a^{4} x^{2}} - \frac {7 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{4}}{256 \, a^{5} x^{2}} - \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{2}}{64 \, a^{3} x^{4}} + \frac {7 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{3}}{128 \, a^{4} x^{4}} + \frac {5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b}{48 \, a^{2} x^{6}} - \frac {7 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b^{2}}{96 \, a^{3} x^{6}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B}{8 \, a x^{8}} + \frac {7 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A b}{80 \, a^{2} x^{8}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A}{10 \, a x^{10}} \]
5/128*B*b^4*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(7/2) - 7/256*A*b^5*arcsinh(a/ (sqrt(a*b)*abs(x)))/a^(9/2) - 5/128*sqrt(b*x^2 + a)*B*b^4/a^4 + 7/256*sqrt (b*x^2 + a)*A*b^5/a^5 + 5/128*(b*x^2 + a)^(3/2)*B*b^3/(a^4*x^2) - 7/256*(b *x^2 + a)^(3/2)*A*b^4/(a^5*x^2) - 5/64*(b*x^2 + a)^(3/2)*B*b^2/(a^3*x^4) + 7/128*(b*x^2 + a)^(3/2)*A*b^3/(a^4*x^4) + 5/48*(b*x^2 + a)^(3/2)*B*b/(a^2 *x^6) - 7/96*(b*x^2 + a)^(3/2)*A*b^2/(a^3*x^6) - 1/8*(b*x^2 + a)^(3/2)*B/( a*x^8) + 7/80*(b*x^2 + a)^(3/2)*A*b/(a^2*x^8) - 1/10*(b*x^2 + a)^(3/2)*A/( a*x^10)
Time = 0.30 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^{11}} \, dx=-\frac {\frac {15 \, {\left (10 \, B a b^{5} - 7 \, A b^{6}\right )} \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{4}} + \frac {150 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} B a b^{5} - 700 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a^{2} b^{5} + 1280 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{3} b^{5} - 580 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{4} b^{5} - 150 \, \sqrt {b x^{2} + a} B a^{5} b^{5} - 105 \, {\left (b x^{2} + a\right )}^{\frac {9}{2}} A b^{6} + 490 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A a b^{6} - 896 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A a^{2} b^{6} + 790 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a^{3} b^{6} + 105 \, \sqrt {b x^{2} + a} A a^{4} b^{6}}{a^{4} b^{5} x^{10}}}{3840 \, b} \]
-1/3840*(15*(10*B*a*b^5 - 7*A*b^6)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt( -a)*a^4) + (150*(b*x^2 + a)^(9/2)*B*a*b^5 - 700*(b*x^2 + a)^(7/2)*B*a^2*b^ 5 + 1280*(b*x^2 + a)^(5/2)*B*a^3*b^5 - 580*(b*x^2 + a)^(3/2)*B*a^4*b^5 - 1 50*sqrt(b*x^2 + a)*B*a^5*b^5 - 105*(b*x^2 + a)^(9/2)*A*b^6 + 490*(b*x^2 + a)^(7/2)*A*a*b^6 - 896*(b*x^2 + a)^(5/2)*A*a^2*b^6 + 790*(b*x^2 + a)^(3/2) *A*a^3*b^6 + 105*sqrt(b*x^2 + a)*A*a^4*b^6)/(a^4*b^5*x^10))/b
Time = 7.56 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.11 \[ \int \frac {\sqrt {a+b x^2} \left (A+B x^2\right )}{x^{11}} \, dx=\frac {7\,A\,{\left (b\,x^2+a\right )}^{5/2}}{30\,a^2\,x^{10}}-\frac {5\,B\,\sqrt {b\,x^2+a}}{128\,x^8}-\frac {79\,A\,{\left (b\,x^2+a\right )}^{3/2}}{384\,a\,x^{10}}-\frac {7\,A\,\sqrt {b\,x^2+a}}{256\,x^{10}}-\frac {49\,A\,{\left (b\,x^2+a\right )}^{7/2}}{384\,a^3\,x^{10}}+\frac {7\,A\,{\left (b\,x^2+a\right )}^{9/2}}{256\,a^4\,x^{10}}-\frac {73\,B\,{\left (b\,x^2+a\right )}^{3/2}}{384\,a\,x^8}+\frac {55\,B\,{\left (b\,x^2+a\right )}^{5/2}}{384\,a^2\,x^8}-\frac {5\,B\,{\left (b\,x^2+a\right )}^{7/2}}{128\,a^3\,x^8}+\frac {A\,b^5\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,7{}\mathrm {i}}{256\,a^{9/2}}-\frac {B\,b^4\,\mathrm {atan}\left (\frac {\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{128\,a^{7/2}} \]
(A*b^5*atan(((a + b*x^2)^(1/2)*1i)/a^(1/2))*7i)/(256*a^(9/2)) - (5*B*(a + b*x^2)^(1/2))/(128*x^8) - (7*A*(a + b*x^2)^(1/2))/(256*x^10) - (B*b^4*atan (((a + b*x^2)^(1/2)*1i)/a^(1/2))*5i)/(128*a^(7/2)) - (79*A*(a + b*x^2)^(3/ 2))/(384*a*x^10) + (7*A*(a + b*x^2)^(5/2))/(30*a^2*x^10) - (49*A*(a + b*x^ 2)^(7/2))/(384*a^3*x^10) + (7*A*(a + b*x^2)^(9/2))/(256*a^4*x^10) - (73*B* (a + b*x^2)^(3/2))/(384*a*x^8) + (55*B*(a + b*x^2)^(5/2))/(384*a^2*x^8) - (5*B*(a + b*x^2)^(7/2))/(128*a^3*x^8)