Integrand size = 24, antiderivative size = 73 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{16}} \, dx=-\frac {A b^3}{9 x^9}-\frac {b^2 (b B+3 A c)}{7 x^7}-\frac {3 b c (b B+A c)}{5 x^5}-\frac {c^2 (3 b B+A c)}{3 x^3}-\frac {B c^3}{x} \] Output:
-1/9*A*b^3/x^9-1/7*b^2*(3*A*c+B*b)/x^7-3/5*b*c*(A*c+B*b)/x^5-1/3*c^2*(A*c+ 3*B*b)/x^3-B*c^3/x
Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{16}} \, dx=-\frac {A b^3}{9 x^9}-\frac {b^2 (b B+3 A c)}{7 x^7}-\frac {3 b c (b B+A c)}{5 x^5}-\frac {c^2 (3 b B+A c)}{3 x^3}-\frac {B c^3}{x} \] Input:
Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^16,x]
Output:
-1/9*(A*b^3)/x^9 - (b^2*(b*B + 3*A*c))/(7*x^7) - (3*b*c*(b*B + A*c))/(5*x^ 5) - (c^2*(3*b*B + A*c))/(3*x^3) - (B*c^3)/x
Time = 0.37 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {9, 355, 2009}
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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{16}} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^{10}}dx\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \int \left (\frac {A b^3}{x^{10}}+\frac {b^2 (3 A c+b B)}{x^8}+\frac {c^2 (A c+3 b B)}{x^4}+\frac {3 b c (A c+b B)}{x^6}+\frac {B c^3}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {A b^3}{9 x^9}-\frac {b^2 (3 A c+b B)}{7 x^7}-\frac {c^2 (A c+3 b B)}{3 x^3}-\frac {3 b c (A c+b B)}{5 x^5}-\frac {B c^3}{x}\) |
Input:
Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^16,x]
Output:
-1/9*(A*b^3)/x^9 - (b^2*(b*B + 3*A*c))/(7*x^7) - (3*b*c*(b*B + A*c))/(5*x^ 5) - (c^2*(3*b*B + A*c))/(3*x^3) - (B*c^3)/x
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q _.), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(a + b*x^2)^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] & & IGtQ[q, 0]
Time = 0.31 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {A \,b^{3}}{9 x^{9}}-\frac {b^{2} \left (3 A c +B b \right )}{7 x^{7}}-\frac {3 b c \left (A c +B b \right )}{5 x^{5}}-\frac {c^{2} \left (A c +3 B b \right )}{3 x^{3}}-\frac {B \,c^{3}}{x}\) | \(66\) |
risch | \(\frac {-B \,c^{3} x^{8}+\left (-\frac {1}{3} A \,c^{3}-B b \,c^{2}\right ) x^{6}+\left (-\frac {3}{5} A b \,c^{2}-\frac {3}{5} B \,b^{2} c \right ) x^{4}+\left (-\frac {3}{7} A \,b^{2} c -\frac {1}{7} B \,b^{3}\right ) x^{2}-\frac {A \,b^{3}}{9}}{x^{9}}\) | \(76\) |
norman | \(\frac {\left (-\frac {1}{3} A \,c^{3}-B b \,c^{2}\right ) x^{12}+\left (-\frac {3}{5} A b \,c^{2}-\frac {3}{5} B \,b^{2} c \right ) x^{10}+\left (-\frac {3}{7} A \,b^{2} c -\frac {1}{7} B \,b^{3}\right ) x^{8}-\frac {A \,b^{3} x^{6}}{9}-B \,c^{3} x^{14}}{x^{15}}\) | \(79\) |
gosper | \(-\frac {315 B \,c^{3} x^{8}+105 A \,c^{3} x^{6}+315 B b \,c^{2} x^{6}+189 A b \,c^{2} x^{4}+189 x^{4} B \,b^{2} c +135 A \,b^{2} c \,x^{2}+45 x^{2} B \,b^{3}+35 A \,b^{3}}{315 x^{9}}\) | \(80\) |
parallelrisch | \(-\frac {315 B \,c^{3} x^{8}+105 A \,c^{3} x^{6}+315 B b \,c^{2} x^{6}+189 A b \,c^{2} x^{4}+189 x^{4} B \,b^{2} c +135 A \,b^{2} c \,x^{2}+45 x^{2} B \,b^{3}+35 A \,b^{3}}{315 x^{9}}\) | \(80\) |
orering | \(-\frac {\left (315 B \,c^{3} x^{8}+105 A \,c^{3} x^{6}+315 B b \,c^{2} x^{6}+189 A b \,c^{2} x^{4}+189 x^{4} B \,b^{2} c +135 A \,b^{2} c \,x^{2}+45 x^{2} B \,b^{3}+35 A \,b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{3}}{315 x^{15} \left (c \,x^{2}+b \right )^{3}}\) | \(102\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^16,x,method=_RETURNVERBOSE)
Output:
-1/9*A*b^3/x^9-1/7*b^2*(3*A*c+B*b)/x^7-3/5*b*c*(A*c+B*b)/x^5-1/3*c^2*(A*c+ 3*B*b)/x^3-B*c^3/x
Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{16}} \, dx=-\frac {315 \, B c^{3} x^{8} + 105 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 189 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} + 35 \, A b^{3} + 45 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{315 \, x^{9}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^16,x, algorithm="fricas")
Output:
-1/315*(315*B*c^3*x^8 + 105*(3*B*b*c^2 + A*c^3)*x^6 + 189*(B*b^2*c + A*b*c ^2)*x^4 + 35*A*b^3 + 45*(B*b^3 + 3*A*b^2*c)*x^2)/x^9
Time = 1.21 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.14 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{16}} \, dx=\frac {- 35 A b^{3} - 315 B c^{3} x^{8} + x^{6} \left (- 105 A c^{3} - 315 B b c^{2}\right ) + x^{4} \left (- 189 A b c^{2} - 189 B b^{2} c\right ) + x^{2} \left (- 135 A b^{2} c - 45 B b^{3}\right )}{315 x^{9}} \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**16,x)
Output:
(-35*A*b**3 - 315*B*c**3*x**8 + x**6*(-105*A*c**3 - 315*B*b*c**2) + x**4*( -189*A*b*c**2 - 189*B*b**2*c) + x**2*(-135*A*b**2*c - 45*B*b**3))/(315*x** 9)
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.03 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{16}} \, dx=-\frac {315 \, B c^{3} x^{8} + 105 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 189 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} + 35 \, A b^{3} + 45 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{315 \, x^{9}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^16,x, algorithm="maxima")
Output:
-1/315*(315*B*c^3*x^8 + 105*(3*B*b*c^2 + A*c^3)*x^6 + 189*(B*b^2*c + A*b*c ^2)*x^4 + 35*A*b^3 + 45*(B*b^3 + 3*A*b^2*c)*x^2)/x^9
Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.08 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{16}} \, dx=-\frac {315 \, B c^{3} x^{8} + 315 \, B b c^{2} x^{6} + 105 \, A c^{3} x^{6} + 189 \, B b^{2} c x^{4} + 189 \, A b c^{2} x^{4} + 45 \, B b^{3} x^{2} + 135 \, A b^{2} c x^{2} + 35 \, A b^{3}}{315 \, x^{9}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^16,x, algorithm="giac")
Output:
-1/315*(315*B*c^3*x^8 + 315*B*b*c^2*x^6 + 105*A*c^3*x^6 + 189*B*b^2*c*x^4 + 189*A*b*c^2*x^4 + 45*B*b^3*x^2 + 135*A*b^2*c*x^2 + 35*A*b^3)/x^9
Time = 0.05 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.01 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{16}} \, dx=-\frac {x^4\,\left (\frac {3\,B\,b^2\,c}{5}+\frac {3\,A\,b\,c^2}{5}\right )+\frac {A\,b^3}{9}+x^2\,\left (\frac {B\,b^3}{7}+\frac {3\,A\,c\,b^2}{7}\right )+x^6\,\left (\frac {A\,c^3}{3}+B\,b\,c^2\right )+B\,c^3\,x^8}{x^9} \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^16,x)
Output:
-(x^4*((3*A*b*c^2)/5 + (3*B*b^2*c)/5) + (A*b^3)/9 + x^2*((B*b^3)/7 + (3*A* b^2*c)/7) + x^6*((A*c^3)/3 + B*b*c^2) + B*c^3*x^8)/x^9
Time = 0.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.07 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{16}} \, dx=\frac {-315 b \,c^{3} x^{8}-105 a \,c^{3} x^{6}-315 b^{2} c^{2} x^{6}-189 a b \,c^{2} x^{4}-189 b^{3} c \,x^{4}-135 a \,b^{2} c \,x^{2}-45 b^{4} x^{2}-35 a \,b^{3}}{315 x^{9}} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^16,x)
Output:
( - 35*a*b**3 - 135*a*b**2*c*x**2 - 189*a*b*c**2*x**4 - 105*a*c**3*x**6 - 45*b**4*x**2 - 189*b**3*c*x**4 - 315*b**2*c**2*x**6 - 315*b*c**3*x**8)/(31 5*x**9)