Integrand size = 24, antiderivative size = 75 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^4} \, dx=\frac {1}{3} A b^3 x^3+\frac {1}{5} b^2 (b B+3 A c) x^5+\frac {3}{7} b c (b B+A c) x^7+\frac {1}{9} c^2 (3 b B+A c) x^9+\frac {1}{11} B c^3 x^{11} \] Output:
1/3*A*b^3*x^3+1/5*b^2*(3*A*c+B*b)*x^5+3/7*b*c*(A*c+B*b)*x^7+1/9*c^2*(A*c+3 *B*b)*x^9+1/11*B*c^3*x^11
Time = 0.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^4} \, dx=\frac {1}{3} A b^3 x^3+\frac {1}{5} b^2 (b B+3 A c) x^5+\frac {3}{7} b c (b B+A c) x^7+\frac {1}{9} c^2 (3 b B+A c) x^9+\frac {1}{11} B c^3 x^{11} \] Input:
Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^4,x]
Output:
(A*b^3*x^3)/3 + (b^2*(b*B + 3*A*c)*x^5)/5 + (3*b*c*(b*B + A*c)*x^7)/7 + (c ^2*(3*b*B + A*c)*x^9)/9 + (B*c^3*x^11)/11
Time = 0.35 (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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^4} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int x^2 \left (A+B x^2\right ) \left (b+c x^2\right )^3dx\) |
\(\Big \downarrow \) 355 |
\(\displaystyle \int \left (A b^3 x^2+b^2 x^4 (3 A c+b B)+c^2 x^8 (A c+3 b B)+3 b c x^6 (A c+b B)+B c^3 x^{10}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{3} A b^3 x^3+\frac {1}{5} b^2 x^5 (3 A c+b B)+\frac {1}{9} c^2 x^9 (A c+3 b B)+\frac {3}{7} b c x^7 (A c+b B)+\frac {1}{11} B c^3 x^{11}\) |
Input:
Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^4,x]
Output:
(A*b^3*x^3)/3 + (b^2*(b*B + 3*A*c)*x^5)/5 + (3*b*c*(b*B + A*c)*x^7)/7 + (c ^2*(3*b*B + A*c)*x^9)/9 + (B*c^3*x^11)/11
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 | \(\frac {B \,c^{3} x^{11}}{11}+\frac {\left (A \,c^{3}+3 B b \,c^{2}\right ) x^{9}}{9}+\frac {\left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{7}}{7}+\frac {\left (3 A \,b^{2} c +B \,b^{3}\right ) x^{5}}{5}+\frac {A \,b^{3} x^{3}}{3}\) | \(76\) |
risch | \(\frac {1}{11} B \,c^{3} x^{11}+\frac {1}{9} x^{9} A \,c^{3}+\frac {1}{3} x^{9} B b \,c^{2}+\frac {3}{7} x^{7} A b \,c^{2}+\frac {3}{7} x^{7} B \,b^{2} c +\frac {3}{5} x^{5} A \,b^{2} c +\frac {1}{5} B \,b^{3} x^{5}+\frac {1}{3} A \,b^{3} x^{3}\) | \(78\) |
parallelrisch | \(\frac {1}{11} B \,c^{3} x^{11}+\frac {1}{9} x^{9} A \,c^{3}+\frac {1}{3} x^{9} B b \,c^{2}+\frac {3}{7} x^{7} A b \,c^{2}+\frac {3}{7} x^{7} B \,b^{2} c +\frac {3}{5} x^{5} A \,b^{2} c +\frac {1}{5} B \,b^{3} x^{5}+\frac {1}{3} A \,b^{3} x^{3}\) | \(78\) |
norman | \(\frac {\left (\frac {1}{9} A \,c^{3}+\frac {1}{3} B b \,c^{2}\right ) x^{12}+\left (\frac {3}{7} A b \,c^{2}+\frac {3}{7} B \,b^{2} c \right ) x^{10}+\left (\frac {3}{5} A \,b^{2} c +\frac {1}{5} B \,b^{3}\right ) x^{8}+\frac {A \,b^{3} x^{6}}{3}+\frac {B \,c^{3} x^{14}}{11}}{x^{3}}\) | \(79\) |
gosper | \(\frac {x^{3} \left (315 B \,c^{3} x^{8}+385 A \,c^{3} x^{6}+1155 B b \,c^{2} x^{6}+1485 A b \,c^{2} x^{4}+1485 x^{4} B \,b^{2} c +2079 A \,b^{2} c \,x^{2}+693 x^{2} B \,b^{3}+1155 A \,b^{3}\right )}{3465}\) | \(80\) |
orering | \(\frac {\left (315 B \,c^{3} x^{8}+385 A \,c^{3} x^{6}+1155 B b \,c^{2} x^{6}+1485 A b \,c^{2} x^{4}+1485 x^{4} B \,b^{2} c +2079 A \,b^{2} c \,x^{2}+693 x^{2} B \,b^{3}+1155 A \,b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{3}}{3465 x^{3} \left (c \,x^{2}+b \right )^{3}}\) | \(102\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^4,x,method=_RETURNVERBOSE)
Output:
1/11*B*c^3*x^11+1/9*(A*c^3+3*B*b*c^2)*x^9+1/7*(3*A*b*c^2+3*B*b^2*c)*x^7+1/ 5*(3*A*b^2*c+B*b^3)*x^5+1/3*A*b^3*x^3
Time = 0.10 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^4} \, dx=\frac {1}{11} \, B c^{3} x^{11} + \frac {1}{9} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{9} + \frac {3}{7} \, {\left (B b^{2} c + A b c^{2}\right )} x^{7} + \frac {1}{3} \, A b^{3} x^{3} + \frac {1}{5} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{5} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^4,x, algorithm="fricas")
Output:
1/11*B*c^3*x^11 + 1/9*(3*B*b*c^2 + A*c^3)*x^9 + 3/7*(B*b^2*c + A*b*c^2)*x^ 7 + 1/3*A*b^3*x^3 + 1/5*(B*b^3 + 3*A*b^2*c)*x^5
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^4} \, dx=\frac {A b^{3} x^{3}}{3} + \frac {B c^{3} x^{11}}{11} + x^{9} \left (\frac {A c^{3}}{9} + \frac {B b c^{2}}{3}\right ) + x^{7} \cdot \left (\frac {3 A b c^{2}}{7} + \frac {3 B b^{2} c}{7}\right ) + x^{5} \cdot \left (\frac {3 A b^{2} c}{5} + \frac {B b^{3}}{5}\right ) \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**4,x)
Output:
A*b**3*x**3/3 + B*c**3*x**11/11 + x**9*(A*c**3/9 + B*b*c**2/3) + x**7*(3*A *b*c**2/7 + 3*B*b**2*c/7) + x**5*(3*A*b**2*c/5 + B*b**3/5)
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^4} \, dx=\frac {1}{11} \, B c^{3} x^{11} + \frac {1}{9} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{9} + \frac {3}{7} \, {\left (B b^{2} c + A b c^{2}\right )} x^{7} + \frac {1}{3} \, A b^{3} x^{3} + \frac {1}{5} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{5} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^4,x, algorithm="maxima")
Output:
1/11*B*c^3*x^11 + 1/9*(3*B*b*c^2 + A*c^3)*x^9 + 3/7*(B*b^2*c + A*b*c^2)*x^ 7 + 1/3*A*b^3*x^3 + 1/5*(B*b^3 + 3*A*b^2*c)*x^5
Time = 0.19 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.03 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^4} \, dx=\frac {1}{11} \, B c^{3} x^{11} + \frac {1}{3} \, B b c^{2} x^{9} + \frac {1}{9} \, A c^{3} x^{9} + \frac {3}{7} \, B b^{2} c x^{7} + \frac {3}{7} \, A b c^{2} x^{7} + \frac {1}{5} \, B b^{3} x^{5} + \frac {3}{5} \, A b^{2} c x^{5} + \frac {1}{3} \, A b^{3} x^{3} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^4,x, algorithm="giac")
Output:
1/11*B*c^3*x^11 + 1/3*B*b*c^2*x^9 + 1/9*A*c^3*x^9 + 3/7*B*b^2*c*x^7 + 3/7* A*b*c^2*x^7 + 1/5*B*b^3*x^5 + 3/5*A*b^2*c*x^5 + 1/3*A*b^3*x^3
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.92 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^4} \, dx=x^5\,\left (\frac {B\,b^3}{5}+\frac {3\,A\,c\,b^2}{5}\right )+x^9\,\left (\frac {A\,c^3}{9}+\frac {B\,b\,c^2}{3}\right )+\frac {A\,b^3\,x^3}{3}+\frac {B\,c^3\,x^{11}}{11}+\frac {3\,b\,c\,x^7\,\left (A\,c+B\,b\right )}{7} \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^4,x)
Output:
x^5*((B*b^3)/5 + (3*A*b^2*c)/5) + x^9*((A*c^3)/9 + (B*b*c^2)/3) + (A*b^3*x ^3)/3 + (B*c^3*x^11)/11 + (3*b*c*x^7*(A*c + B*b))/7
Time = 0.20 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.04 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^4} \, dx=\frac {x^{3} \left (315 b \,c^{3} x^{8}+385 a \,c^{3} x^{6}+1155 b^{2} c^{2} x^{6}+1485 a b \,c^{2} x^{4}+1485 b^{3} c \,x^{4}+2079 a \,b^{2} c \,x^{2}+693 b^{4} x^{2}+1155 a \,b^{3}\right )}{3465} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^4,x)
Output:
(x**3*(1155*a*b**3 + 2079*a*b**2*c*x**2 + 1485*a*b*c**2*x**4 + 385*a*c**3* x**6 + 693*b**4*x**2 + 1485*b**3*c*x**4 + 1155*b**2*c**2*x**6 + 315*b*c**3 *x**8))/3465