Integrand size = 24, antiderivative size = 75 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^2} \, dx=\frac {1}{5} A b^3 x^5+\frac {1}{7} b^2 (b B+3 A c) x^7+\frac {1}{3} b c (b B+A c) x^9+\frac {1}{11} c^2 (3 b B+A c) x^{11}+\frac {1}{13} B c^3 x^{13} \] Output:
1/5*A*b^3*x^5+1/7*b^2*(3*A*c+B*b)*x^7+1/3*b*c*(A*c+B*b)*x^9+1/11*c^2*(A*c+ 3*B*b)*x^11+1/13*B*c^3*x^13
Time = 0.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^2} \, dx=\frac {1}{5} A b^3 x^5+\frac {1}{7} b^2 (b B+3 A c) x^7+\frac {1}{3} b c (b B+A c) x^9+\frac {1}{11} c^2 (3 b B+A c) x^{11}+\frac {1}{13} B c^3 x^{13} \] Input:
Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^2,x]
Output:
(A*b^3*x^5)/5 + (b^2*(b*B + 3*A*c)*x^7)/7 + (b*c*(b*B + A*c)*x^9)/3 + (c^2 *(3*b*B + A*c)*x^11)/11 + (B*c^3*x^13)/13
Time = 0.37 (sec) , antiderivative size = 75, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^2} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int x^4 \left (A+B x^2\right ) \left (b+c x^2\right )^3dx\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \int \left (A b^3 x^4+b^2 x^6 (3 A c+b B)+c^2 x^{10} (A c+3 b B)+3 b c x^8 (A c+b B)+B c^3 x^{12}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{5} A b^3 x^5+\frac {1}{7} b^2 x^7 (3 A c+b B)+\frac {1}{11} c^2 x^{11} (A c+3 b B)+\frac {1}{3} b c x^9 (A c+b B)+\frac {1}{13} B c^3 x^{13}\) |
Input:
Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^2,x]
Output:
(A*b^3*x^5)/5 + (b^2*(b*B + 3*A*c)*x^7)/7 + (b*c*(b*B + A*c)*x^9)/3 + (c^2 *(3*b*B + A*c)*x^11)/11 + (B*c^3*x^13)/13
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 = 0.37 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.01
method | result | size |
default | \(\frac {B \,c^{3} x^{13}}{13}+\frac {\left (A \,c^{3}+3 B b \,c^{2}\right ) x^{11}}{11}+\frac {\left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{9}}{9}+\frac {\left (3 A \,b^{2} c +B \,b^{3}\right ) x^{7}}{7}+\frac {A \,b^{3} x^{5}}{5}\) | \(76\) |
risch | \(\frac {1}{13} B \,c^{3} x^{13}+\frac {1}{11} x^{11} A \,c^{3}+\frac {3}{11} x^{11} B b \,c^{2}+\frac {1}{3} x^{9} A b \,c^{2}+\frac {1}{3} x^{9} B \,b^{2} c +\frac {3}{7} x^{7} A \,b^{2} c +\frac {1}{7} B \,b^{3} x^{7}+\frac {1}{5} A \,b^{3} x^{5}\) | \(78\) |
parallelrisch | \(\frac {1}{13} B \,c^{3} x^{13}+\frac {1}{11} x^{11} A \,c^{3}+\frac {3}{11} x^{11} B b \,c^{2}+\frac {1}{3} x^{9} A b \,c^{2}+\frac {1}{3} x^{9} B \,b^{2} c +\frac {3}{7} x^{7} A \,b^{2} c +\frac {1}{7} B \,b^{3} x^{7}+\frac {1}{5} A \,b^{3} x^{5}\) | \(78\) |
norman | \(\frac {\left (\frac {1}{11} A \,c^{3}+\frac {3}{11} B b \,c^{2}\right ) x^{12}+\left (\frac {1}{3} A b \,c^{2}+\frac {1}{3} B \,b^{2} c \right ) x^{10}+\left (\frac {3}{7} A \,b^{2} c +\frac {1}{7} B \,b^{3}\right ) x^{8}+\frac {A \,b^{3} x^{6}}{5}+\frac {B \,c^{3} x^{14}}{13}}{x}\) | \(79\) |
gosper | \(\frac {x^{5} \left (1155 B \,c^{3} x^{8}+1365 A \,c^{3} x^{6}+4095 B b \,c^{2} x^{6}+5005 A b \,c^{2} x^{4}+5005 x^{4} B \,b^{2} c +6435 A \,b^{2} c \,x^{2}+2145 x^{2} B \,b^{3}+3003 A \,b^{3}\right )}{15015}\) | \(80\) |
orering | \(\frac {\left (1155 B \,c^{3} x^{8}+1365 A \,c^{3} x^{6}+4095 B b \,c^{2} x^{6}+5005 A b \,c^{2} x^{4}+5005 x^{4} B \,b^{2} c +6435 A \,b^{2} c \,x^{2}+2145 x^{2} B \,b^{3}+3003 A \,b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{3}}{15015 x \left (c \,x^{2}+b \right )^{3}}\) | \(102\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^2,x,method=_RETURNVERBOSE)
Output:
1/13*B*c^3*x^13+1/11*(A*c^3+3*B*b*c^2)*x^11+1/9*(3*A*b*c^2+3*B*b^2*c)*x^9+ 1/7*(3*A*b^2*c+B*b^3)*x^7+1/5*A*b^3*x^5
Time = 0.06 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^2} \, dx=\frac {1}{13} \, B c^{3} x^{13} + \frac {1}{11} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{11} + \frac {1}{3} \, {\left (B b^{2} c + A b c^{2}\right )} x^{9} + \frac {1}{5} \, A b^{3} x^{5} + \frac {1}{7} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{7} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^2,x, algorithm="fricas")
Output:
1/13*B*c^3*x^13 + 1/11*(3*B*b*c^2 + A*c^3)*x^11 + 1/3*(B*b^2*c + A*b*c^2)* x^9 + 1/5*A*b^3*x^5 + 1/7*(B*b^3 + 3*A*b^2*c)*x^7
Time = 0.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.07 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^2} \, dx=\frac {A b^{3} x^{5}}{5} + \frac {B c^{3} x^{13}}{13} + x^{11} \left (\frac {A c^{3}}{11} + \frac {3 B b c^{2}}{11}\right ) + x^{9} \left (\frac {A b c^{2}}{3} + \frac {B b^{2} c}{3}\right ) + x^{7} \cdot \left (\frac {3 A b^{2} c}{7} + \frac {B b^{3}}{7}\right ) \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**2,x)
Output:
A*b**3*x**5/5 + B*c**3*x**13/13 + x**11*(A*c**3/11 + 3*B*b*c**2/11) + x**9 *(A*b*c**2/3 + B*b**2*c/3) + x**7*(3*A*b**2*c/7 + B*b**3/7)
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^2} \, dx=\frac {1}{13} \, B c^{3} x^{13} + \frac {1}{11} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{11} + \frac {1}{3} \, {\left (B b^{2} c + A b c^{2}\right )} x^{9} + \frac {1}{5} \, A b^{3} x^{5} + \frac {1}{7} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{7} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^2,x, algorithm="maxima")
Output:
1/13*B*c^3*x^13 + 1/11*(3*B*b*c^2 + A*c^3)*x^11 + 1/3*(B*b^2*c + A*b*c^2)* x^9 + 1/5*A*b^3*x^5 + 1/7*(B*b^3 + 3*A*b^2*c)*x^7
Time = 0.21 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^2} \, dx=\frac {1}{13} \, B c^{3} x^{13} + \frac {3}{11} \, B b c^{2} x^{11} + \frac {1}{11} \, A c^{3} x^{11} + \frac {1}{3} \, B b^{2} c x^{9} + \frac {1}{3} \, A b c^{2} x^{9} + \frac {1}{7} \, B b^{3} x^{7} + \frac {3}{7} \, A b^{2} c x^{7} + \frac {1}{5} \, A b^{3} x^{5} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^2,x, algorithm="giac")
Output:
1/13*B*c^3*x^13 + 3/11*B*b*c^2*x^11 + 1/11*A*c^3*x^11 + 1/3*B*b^2*c*x^9 + 1/3*A*b*c^2*x^9 + 1/7*B*b^3*x^7 + 3/7*A*b^2*c*x^7 + 1/5*A*b^3*x^5
Time = 9.49 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.92 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^2} \, dx=x^7\,\left (\frac {B\,b^3}{7}+\frac {3\,A\,c\,b^2}{7}\right )+x^{11}\,\left (\frac {A\,c^3}{11}+\frac {3\,B\,b\,c^2}{11}\right )+\frac {A\,b^3\,x^5}{5}+\frac {B\,c^3\,x^{13}}{13}+\frac {b\,c\,x^9\,\left (A\,c+B\,b\right )}{3} \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^2,x)
Output:
x^7*((B*b^3)/7 + (3*A*b^2*c)/7) + x^11*((A*c^3)/11 + (3*B*b*c^2)/11) + (A* b^3*x^5)/5 + (B*c^3*x^13)/13 + (b*c*x^9*(A*c + B*b))/3
Time = 0.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.04 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^2} \, dx=\frac {x^{5} \left (1155 b \,c^{3} x^{8}+1365 a \,c^{3} x^{6}+4095 b^{2} c^{2} x^{6}+5005 a b \,c^{2} x^{4}+5005 b^{3} c \,x^{4}+6435 a \,b^{2} c \,x^{2}+2145 b^{4} x^{2}+3003 a \,b^{3}\right )}{15015} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^2,x)
Output:
(x**5*(3003*a*b**3 + 6435*a*b**2*c*x**2 + 5005*a*b*c**2*x**4 + 1365*a*c**3 *x**6 + 2145*b**4*x**2 + 5005*b**3*c*x**4 + 4095*b**2*c**2*x**6 + 1155*b*c **3*x**8))/15015