Integrand size = 26, antiderivative size = 251 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{16}} \, dx=-\frac {c (12 b B-7 A c) \sqrt {b x^2+c x^4}}{320 b x^9}-\frac {c^2 (12 b B-7 A c) \sqrt {b x^2+c x^4}}{1920 b^2 x^7}+\frac {c^3 (12 b B-7 A c) \sqrt {b x^2+c x^4}}{1536 b^3 x^5}-\frac {c^4 (12 b B-7 A c) \sqrt {b x^2+c x^4}}{1024 b^4 x^3}-\frac {(12 b B-7 A c) \left (b x^2+c x^4\right )^{3/2}}{120 b x^{13}}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{12 b x^{17}}+\frac {c^5 (12 b B-7 A c) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{1024 b^{9/2}} \]
-1/120*(-7*A*c+12*B*b)*(c*x^4+b*x^2)^(3/2)/b/x^13-1/12*A*(c*x^4+b*x^2)^(5/ 2)/b/x^17+1/1024*c^5*(-7*A*c+12*B*b)*arctanh(x*b^(1/2)/(c*x^4+b*x^2)^(1/2) )/b^(9/2)-1/320*c*(-7*A*c+12*B*b)*(c*x^4+b*x^2)^(1/2)/b/x^9-1/1920*c^2*(-7 *A*c+12*B*b)*(c*x^4+b*x^2)^(1/2)/b^2/x^7+1/1536*c^3*(-7*A*c+12*B*b)*(c*x^4 +b*x^2)^(1/2)/b^3/x^5-1/1024*c^4*(-7*A*c+12*B*b)*(c*x^4+b*x^2)^(1/2)/b^4/x ^3
Time = 0.54 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.79 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{16}} \, dx=-\frac {\sqrt {x^2 \left (b+c x^2\right )} \left (\sqrt {b} \sqrt {b+c x^2} \left (12 b B x^2 \left (128 b^4+176 b^3 c x^2+8 b^2 c^2 x^4-10 b c^3 x^6+15 c^4 x^8\right )+A \left (1280 b^5+1664 b^4 c x^2+48 b^3 c^2 x^4-56 b^2 c^3 x^6+70 b c^4 x^8-105 c^5 x^{10}\right )\right )+15 c^5 (-12 b B+7 A c) x^{12} \text {arctanh}\left (\frac {\sqrt {b+c x^2}}{\sqrt {b}}\right )\right )}{15360 b^{9/2} x^{13} \sqrt {b+c x^2}} \]
-1/15360*(Sqrt[x^2*(b + c*x^2)]*(Sqrt[b]*Sqrt[b + c*x^2]*(12*b*B*x^2*(128* b^4 + 176*b^3*c*x^2 + 8*b^2*c^2*x^4 - 10*b*c^3*x^6 + 15*c^4*x^8) + A*(1280 *b^5 + 1664*b^4*c*x^2 + 48*b^3*c^2*x^4 - 56*b^2*c^3*x^6 + 70*b*c^4*x^8 - 1 05*c^5*x^10)) + 15*c^5*(-12*b*B + 7*A*c)*x^12*ArcTanh[Sqrt[b + c*x^2]/Sqrt [b]]))/(b^(9/2)*x^13*Sqrt[b + c*x^2])
Time = 0.44 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.90, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {1944, 1425, 1425, 1430, 1430, 1430, 1400, 219}
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.
| \(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{16}} \, dx\) |
| \(\Big \downarrow \) 1944 |
| \(\displaystyle \frac {(12 b B-7 A c) \int \frac {\left (c x^4+b x^2\right )^{3/2}}{x^{14}}dx}{12 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{12 b x^{17}}\) |
| \(\Big \downarrow \) 1425 |
| \(\displaystyle \frac {(12 b B-7 A c) \left (\frac {3}{10} c \int \frac {\sqrt {c x^4+b x^2}}{x^{10}}dx-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}\right )}{12 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{12 b x^{17}}\) |
| \(\Big \downarrow \) 1425 |
| \(\displaystyle \frac {(12 b B-7 A c) \left (\frac {3}{10} c \left (\frac {1}{8} c \int \frac {1}{x^6 \sqrt {c x^4+b x^2}}dx-\frac {\sqrt {b x^2+c x^4}}{8 x^9}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}\right )}{12 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{12 b x^{17}}\) |
| \(\Big \downarrow \) 1430 |
| \(\displaystyle \frac {(12 b B-7 A c) \left (\frac {3}{10} c \left (\frac {1}{8} c \left (-\frac {5 c \int \frac {1}{x^4 \sqrt {c x^4+b x^2}}dx}{6 b}-\frac {\sqrt {b x^2+c x^4}}{6 b x^7}\right )-\frac {\sqrt {b x^2+c x^4}}{8 x^9}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}\right )}{12 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{12 b x^{17}}\) |
| \(\Big \downarrow \) 1430 |
| \(\displaystyle \frac {(12 b B-7 A c) \left (\frac {3}{10} c \left (\frac {1}{8} c \left (-\frac {5 c \left (-\frac {3 c \int \frac {1}{x^2 \sqrt {c x^4+b x^2}}dx}{4 b}-\frac {\sqrt {b x^2+c x^4}}{4 b x^5}\right )}{6 b}-\frac {\sqrt {b x^2+c x^4}}{6 b x^7}\right )-\frac {\sqrt {b x^2+c x^4}}{8 x^9}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}\right )}{12 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{12 b x^{17}}\) |
| \(\Big \downarrow \) 1430 |
| \(\displaystyle \frac {(12 b B-7 A c) \left (\frac {3}{10} c \left (\frac {1}{8} c \left (-\frac {5 c \left (-\frac {3 c \left (-\frac {c \int \frac {1}{\sqrt {c x^4+b x^2}}dx}{2 b}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )}{4 b}-\frac {\sqrt {b x^2+c x^4}}{4 b x^5}\right )}{6 b}-\frac {\sqrt {b x^2+c x^4}}{6 b x^7}\right )-\frac {\sqrt {b x^2+c x^4}}{8 x^9}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}\right )}{12 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{12 b x^{17}}\) |
| \(\Big \downarrow \) 1400 |
| \(\displaystyle \frac {(12 b B-7 A c) \left (\frac {3}{10} c \left (\frac {1}{8} c \left (-\frac {5 c \left (-\frac {3 c \left (\frac {c \int \frac {1}{1-\frac {b x^2}{c x^4+b x^2}}d\frac {x}{\sqrt {c x^4+b x^2}}}{2 b}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )}{4 b}-\frac {\sqrt {b x^2+c x^4}}{4 b x^5}\right )}{6 b}-\frac {\sqrt {b x^2+c x^4}}{6 b x^7}\right )-\frac {\sqrt {b x^2+c x^4}}{8 x^9}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}\right )}{12 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{12 b x^{17}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {(12 b B-7 A c) \left (\frac {3}{10} c \left (\frac {1}{8} c \left (-\frac {5 c \left (-\frac {3 c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {b x^2+c x^4}}\right )}{2 b^{3/2}}-\frac {\sqrt {b x^2+c x^4}}{2 b x^3}\right )}{4 b}-\frac {\sqrt {b x^2+c x^4}}{4 b x^5}\right )}{6 b}-\frac {\sqrt {b x^2+c x^4}}{6 b x^7}\right )-\frac {\sqrt {b x^2+c x^4}}{8 x^9}\right )-\frac {\left (b x^2+c x^4\right )^{3/2}}{10 x^{13}}\right )}{12 b}-\frac {A \left (b x^2+c x^4\right )^{5/2}}{12 b x^{17}}\) |
-1/12*(A*(b*x^2 + c*x^4)^(5/2))/(b*x^17) + ((12*b*B - 7*A*c)*(-1/10*(b*x^2 + c*x^4)^(3/2)/x^13 + (3*c*(-1/8*Sqrt[b*x^2 + c*x^4]/x^9 + (c*(-1/6*Sqrt[ b*x^2 + c*x^4]/(b*x^7) - (5*c*(-1/4*Sqrt[b*x^2 + c*x^4]/(b*x^5) - (3*c*(-1 /2*Sqrt[b*x^2 + c*x^4]/(b*x^3) + (c*ArcTanh[(Sqrt[b]*x)/Sqrt[b*x^2 + c*x^4 ]])/(2*b^(3/2))))/(4*b)))/(6*b)))/8))/10))/(12*b)
3.2.29.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> -Subst[Int[1/(1 - b*x ^2), x], x, x/Sqrt[b*x^2 + c*x^4]] /; FreeQ[{b, c}, x]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [(d*x)^(m + 1)*((b*x^2 + c*x^4)^p/(d*(m + 2*p + 1))), x] - Simp[2*c*(p/(d^4 *(m + 2*p + 1))) Int[(d*x)^(m + 4)*(b*x^2 + c*x^4)^(p - 1), x], x] /; Fre eQ[{b, c, d, m, p}, x] && !IntegerQ[p] && GtQ[p, 0] && LtQ[m + 2*p + 1, 0]
Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp [d*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[c*( (m + 4*p + 3)/(b*d^2*(m + 2*p + 1))) Int[(d*x)^(m + 2)*(b*x^2 + c*x^4)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] && !IntegerQ[p] && LtQ[m + 2*p + 1, 0 ]
Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + (d_.)*(x_)^(n_.)), x_Symbol] :> Simp[c*e^(j - 1)*(e*x)^(m - j + 1)*((a*x^j + b*x^(j + n))^(p + 1)/(a*(m + j*p + 1))), x] + Simp[(a*d*(m + j*p + 1) - b *c*(m + n + p*(j + n) + 1))/(a*e^n*(m + j*p + 1)) Int[(e*x)^(m + n)*(a*x^ j + b*x^(j + n))^p, x], x] /; FreeQ[{a, b, c, d, e, j, p}, x] && EqQ[jn, j + n] && !IntegerQ[p] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && (LtQ[m + j*p, -1 ] || (IntegersQ[m - 1/2, p - 1/2] && LtQ[p, 0] && LtQ[m, (-n)*p - 1])) && ( GtQ[e, 0] || IntegersQ[j, n]) && NeQ[m + j*p + 1, 0] && NeQ[m - n + j*p + 1 , 0]
Time = 3.64 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.80
| method | result | size |
| risch | \(-\frac {\left (-105 A \,c^{5} x^{10}+180 B b \,c^{4} x^{10}+70 A \,x^{8} b \,c^{4}-120 B \,b^{2} c^{3} x^{8}-56 A \,b^{2} c^{3} x^{6}+96 B \,b^{3} c^{2} x^{6}+48 A \,b^{3} c^{2} x^{4}+2112 B \,b^{4} c \,x^{4}+1664 A \,b^{4} c \,x^{2}+1536 b^{5} B \,x^{2}+1280 b^{5} A \right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{15360 x^{13} b^{4}}-\frac {\left (7 A c -12 B b \right ) c^{5} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) \sqrt {x^{2} \left (c \,x^{2}+b \right )}}{1024 b^{\frac {9}{2}} x \sqrt {c \,x^{2}+b}}\) | \(201\) |
| default | \(-\frac {\left (x^{4} c +b \,x^{2}\right )^{\frac {3}{2}} \left (-35 A \left (c \,x^{2}+b \right )^{\frac {3}{2}} c^{6} x^{12}+105 A \,b^{\frac {3}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c^{6} x^{12}+60 B \left (c \,x^{2}+b \right )^{\frac {3}{2}} b \,c^{5} x^{12}-180 B \,b^{\frac {5}{2}} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {c \,x^{2}+b}}{x}\right ) c^{5} x^{12}+35 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} c^{5} x^{10}-105 A \sqrt {c \,x^{2}+b}\, b \,c^{6} x^{12}-60 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \,c^{4} x^{10}+180 B \sqrt {c \,x^{2}+b}\, b^{2} c^{5} x^{12}+70 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} b \,c^{4} x^{8}-120 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{2} c^{3} x^{8}-280 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{2} c^{3} x^{6}+480 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{3} c^{2} x^{6}+560 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{3} c^{2} x^{4}-960 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{4} c \,x^{4}-896 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{4} c \,x^{2}+1536 B \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{5} x^{2}+1280 A \left (c \,x^{2}+b \right )^{\frac {5}{2}} b^{5}\right )}{15360 x^{15} \left (c \,x^{2}+b \right )^{\frac {3}{2}} b^{6}}\) | \(386\) |
-1/15360*(-105*A*c^5*x^10+180*B*b*c^4*x^10+70*A*b*c^4*x^8-120*B*b^2*c^3*x^ 8-56*A*b^2*c^3*x^6+96*B*b^3*c^2*x^6+48*A*b^3*c^2*x^4+2112*B*b^4*c*x^4+1664 *A*b^4*c*x^2+1536*B*b^5*x^2+1280*A*b^5)/x^13/b^4*(x^2*(c*x^2+b))^(1/2)-1/1 024*(7*A*c-12*B*b)*c^5/b^(9/2)*ln((2*b+2*b^(1/2)*(c*x^2+b)^(1/2))/x)*(x^2* (c*x^2+b))^(1/2)/x/(c*x^2+b)^(1/2)
Time = 0.36 (sec) , antiderivative size = 393, normalized size of antiderivative = 1.57 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{16}} \, dx=\left [-\frac {15 \, {\left (12 \, B b c^{5} - 7 \, A c^{6}\right )} \sqrt {b} x^{13} \log \left (-\frac {c x^{3} + 2 \, b x - 2 \, \sqrt {c x^{4} + b x^{2}} \sqrt {b}}{x^{3}}\right ) + 2 \, {\left (15 \, {\left (12 \, B b^{2} c^{4} - 7 \, A b c^{5}\right )} x^{10} - 10 \, {\left (12 \, B b^{3} c^{3} - 7 \, A b^{2} c^{4}\right )} x^{8} + 1280 \, A b^{6} + 8 \, {\left (12 \, B b^{4} c^{2} - 7 \, A b^{3} c^{3}\right )} x^{6} + 48 \, {\left (44 \, B b^{5} c + A b^{4} c^{2}\right )} x^{4} + 128 \, {\left (12 \, B b^{6} + 13 \, A b^{5} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{30720 \, b^{5} x^{13}}, -\frac {15 \, {\left (12 \, B b c^{5} - 7 \, A c^{6}\right )} \sqrt {-b} x^{13} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2}} \sqrt {-b}}{c x^{3} + b x}\right ) + {\left (15 \, {\left (12 \, B b^{2} c^{4} - 7 \, A b c^{5}\right )} x^{10} - 10 \, {\left (12 \, B b^{3} c^{3} - 7 \, A b^{2} c^{4}\right )} x^{8} + 1280 \, A b^{6} + 8 \, {\left (12 \, B b^{4} c^{2} - 7 \, A b^{3} c^{3}\right )} x^{6} + 48 \, {\left (44 \, B b^{5} c + A b^{4} c^{2}\right )} x^{4} + 128 \, {\left (12 \, B b^{6} + 13 \, A b^{5} c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{15360 \, b^{5} x^{13}}\right ] \]
[-1/30720*(15*(12*B*b*c^5 - 7*A*c^6)*sqrt(b)*x^13*log(-(c*x^3 + 2*b*x - 2* sqrt(c*x^4 + b*x^2)*sqrt(b))/x^3) + 2*(15*(12*B*b^2*c^4 - 7*A*b*c^5)*x^10 - 10*(12*B*b^3*c^3 - 7*A*b^2*c^4)*x^8 + 1280*A*b^6 + 8*(12*B*b^4*c^2 - 7*A *b^3*c^3)*x^6 + 48*(44*B*b^5*c + A*b^4*c^2)*x^4 + 128*(12*B*b^6 + 13*A*b^5 *c)*x^2)*sqrt(c*x^4 + b*x^2))/(b^5*x^13), -1/15360*(15*(12*B*b*c^5 - 7*A*c ^6)*sqrt(-b)*x^13*arctan(sqrt(c*x^4 + b*x^2)*sqrt(-b)/(c*x^3 + b*x)) + (15 *(12*B*b^2*c^4 - 7*A*b*c^5)*x^10 - 10*(12*B*b^3*c^3 - 7*A*b^2*c^4)*x^8 + 1 280*A*b^6 + 8*(12*B*b^4*c^2 - 7*A*b^3*c^3)*x^6 + 48*(44*B*b^5*c + A*b^4*c^ 2)*x^4 + 128*(12*B*b^6 + 13*A*b^5*c)*x^2)*sqrt(c*x^4 + b*x^2))/(b^5*x^13)]
\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{16}} \, dx=\int \frac {\left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}} \left (A + B x^{2}\right )}{x^{16}}\, dx \]
\[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{16}} \, dx=\int { \frac {{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} {\left (B x^{2} + A\right )}}{x^{16}} \,d x } \]
Time = 0.33 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.17 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{16}} \, dx=-\frac {\frac {15 \, {\left (12 \, B b c^{6} \mathrm {sgn}\left (x\right ) - 7 \, A c^{7} \mathrm {sgn}\left (x\right )\right )} \arctan \left (\frac {\sqrt {c x^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{4}} + \frac {180 \, {\left (c x^{2} + b\right )}^{\frac {11}{2}} B b c^{6} \mathrm {sgn}\left (x\right ) - 1020 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} B b^{2} c^{6} \mathrm {sgn}\left (x\right ) + 2376 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} B b^{3} c^{6} \mathrm {sgn}\left (x\right ) - 696 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} B b^{4} c^{6} \mathrm {sgn}\left (x\right ) - 1020 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} B b^{5} c^{6} \mathrm {sgn}\left (x\right ) + 180 \, \sqrt {c x^{2} + b} B b^{6} c^{6} \mathrm {sgn}\left (x\right ) - 105 \, {\left (c x^{2} + b\right )}^{\frac {11}{2}} A c^{7} \mathrm {sgn}\left (x\right ) + 595 \, {\left (c x^{2} + b\right )}^{\frac {9}{2}} A b c^{7} \mathrm {sgn}\left (x\right ) - 1386 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} A b^{2} c^{7} \mathrm {sgn}\left (x\right ) + 1686 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} A b^{3} c^{7} \mathrm {sgn}\left (x\right ) + 595 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} A b^{4} c^{7} \mathrm {sgn}\left (x\right ) - 105 \, \sqrt {c x^{2} + b} A b^{5} c^{7} \mathrm {sgn}\left (x\right )}{b^{4} c^{6} x^{12}}}{15360 \, c} \]
-1/15360*(15*(12*B*b*c^6*sgn(x) - 7*A*c^7*sgn(x))*arctan(sqrt(c*x^2 + b)/s qrt(-b))/(sqrt(-b)*b^4) + (180*(c*x^2 + b)^(11/2)*B*b*c^6*sgn(x) - 1020*(c *x^2 + b)^(9/2)*B*b^2*c^6*sgn(x) + 2376*(c*x^2 + b)^(7/2)*B*b^3*c^6*sgn(x) - 696*(c*x^2 + b)^(5/2)*B*b^4*c^6*sgn(x) - 1020*(c*x^2 + b)^(3/2)*B*b^5*c ^6*sgn(x) + 180*sqrt(c*x^2 + b)*B*b^6*c^6*sgn(x) - 105*(c*x^2 + b)^(11/2)* A*c^7*sgn(x) + 595*(c*x^2 + b)^(9/2)*A*b*c^7*sgn(x) - 1386*(c*x^2 + b)^(7/ 2)*A*b^2*c^7*sgn(x) + 1686*(c*x^2 + b)^(5/2)*A*b^3*c^7*sgn(x) + 595*(c*x^2 + b)^(3/2)*A*b^4*c^7*sgn(x) - 105*sqrt(c*x^2 + b)*A*b^5*c^7*sgn(x))/(b^4* c^6*x^12))/c
Timed out. \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^{3/2}}{x^{16}} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (c\,x^4+b\,x^2\right )}^{3/2}}{x^{16}} \,d x \]