Integrand size = 24, antiderivative size = 68 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^3} \, dx=\frac {b (b B-A c) \left (b+c x^2\right )^4}{8 c^3}-\frac {(2 b B-A c) \left (b+c x^2\right )^5}{10 c^3}+\frac {B \left (b+c x^2\right )^6}{12 c^3} \] Output:
1/8*b*(-A*c+B*b)*(c*x^2+b)^4/c^3-1/10*(-A*c+2*B*b)*(c*x^2+b)^5/c^3+1/12*B* (c*x^2+b)^6/c^3
Time = 0.02 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^3} \, dx=\frac {1}{120} x^4 \left (30 A b^3+20 b^2 (b B+3 A c) x^2+45 b c (b B+A c) x^4+12 c^2 (3 b B+A c) x^6+10 B c^3 x^8\right ) \] Input:
Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^3,x]
Output:
(x^4*(30*A*b^3 + 20*b^2*(b*B + 3*A*c)*x^2 + 45*b*c*(b*B + A*c)*x^4 + 12*c^ 2*(3*b*B + A*c)*x^6 + 10*B*c^3*x^8))/120
Time = 0.38 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {9, 354, 85, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^3} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int x^3 \left (A+B x^2\right ) \left (b+c x^2\right )^3dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int x^2 \left (B x^2+A\right ) \left (c x^2+b\right )^3dx^2\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {1}{2} \int \left (\frac {B \left (c x^2+b\right )^5}{c^2}+\frac {(A c-2 b B) \left (c x^2+b\right )^4}{c^2}+\frac {b (b B-A c) \left (c x^2+b\right )^3}{c^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {\left (b+c x^2\right )^5 (2 b B-A c)}{5 c^3}+\frac {b \left (b+c x^2\right )^4 (b B-A c)}{4 c^3}+\frac {B \left (b+c x^2\right )^6}{6 c^3}\right )\) |
Input:
Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^3,x]
Output:
((b*(b*B - A*c)*(b + c*x^2)^4)/(4*c^3) - ((2*b*B - A*c)*(b + c*x^2)^5)/(5* c^3) + (B*(b + c*x^2)^6)/(6*c^3))/2
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[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_] : > Int[ExpandIntegrand[(a + b*x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && NeQ[b*e + a* f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])
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 = 0.36 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.12
method | result | size |
default | \(\frac {B \,c^{3} x^{12}}{12}+\frac {\left (A \,c^{3}+3 B b \,c^{2}\right ) x^{10}}{10}+\frac {\left (3 A b \,c^{2}+3 B \,b^{2} c \right ) x^{8}}{8}+\frac {\left (3 A \,b^{2} c +B \,b^{3}\right ) x^{6}}{6}+\frac {A \,b^{3} x^{4}}{4}\) | \(76\) |
risch | \(\frac {1}{12} B \,c^{3} x^{12}+\frac {1}{10} x^{10} A \,c^{3}+\frac {3}{10} x^{10} B b \,c^{2}+\frac {3}{8} x^{8} A b \,c^{2}+\frac {3}{8} x^{8} B \,b^{2} c +\frac {1}{2} x^{6} A \,b^{2} c +\frac {1}{6} B \,b^{3} x^{6}+\frac {1}{4} A \,b^{3} x^{4}\) | \(78\) |
parallelrisch | \(\frac {1}{12} B \,c^{3} x^{12}+\frac {1}{10} x^{10} A \,c^{3}+\frac {3}{10} x^{10} B b \,c^{2}+\frac {3}{8} x^{8} A b \,c^{2}+\frac {3}{8} x^{8} B \,b^{2} c +\frac {1}{2} x^{6} A \,b^{2} c +\frac {1}{6} B \,b^{3} x^{6}+\frac {1}{4} A \,b^{3} x^{4}\) | \(78\) |
norman | \(\frac {\left (\frac {1}{10} A \,c^{3}+\frac {3}{10} B b \,c^{2}\right ) x^{12}+\left (\frac {3}{8} A b \,c^{2}+\frac {3}{8} B \,b^{2} c \right ) x^{10}+\left (\frac {1}{2} A \,b^{2} c +\frac {1}{6} B \,b^{3}\right ) x^{8}+\frac {A \,b^{3} x^{6}}{4}+\frac {B \,c^{3} x^{14}}{12}}{x^{2}}\) | \(79\) |
gosper | \(\frac {x^{4} \left (10 B \,c^{3} x^{8}+12 A \,c^{3} x^{6}+36 B b \,c^{2} x^{6}+45 A b \,c^{2} x^{4}+45 x^{4} B \,b^{2} c +60 A \,b^{2} c \,x^{2}+20 x^{2} B \,b^{3}+30 A \,b^{3}\right )}{120}\) | \(80\) |
orering | \(\frac {\left (10 B \,c^{3} x^{8}+12 A \,c^{3} x^{6}+36 B b \,c^{2} x^{6}+45 A b \,c^{2} x^{4}+45 x^{4} B \,b^{2} c +60 A \,b^{2} c \,x^{2}+20 x^{2} B \,b^{3}+30 A \,b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{3}}{120 x^{2} \left (c \,x^{2}+b \right )^{3}}\) | \(102\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^3,x,method=_RETURNVERBOSE)
Output:
1/12*B*c^3*x^12+1/10*(A*c^3+3*B*b*c^2)*x^10+1/8*(3*A*b*c^2+3*B*b^2*c)*x^8+ 1/6*(3*A*b^2*c+B*b^3)*x^6+1/4*A*b^3*x^4
Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.07 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^3} \, dx=\frac {1}{12} \, B c^{3} x^{12} + \frac {1}{10} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{10} + \frac {3}{8} \, {\left (B b^{2} c + A b c^{2}\right )} x^{8} + \frac {1}{4} \, A b^{3} x^{4} + \frac {1}{6} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{6} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^3,x, algorithm="fricas")
Output:
1/12*B*c^3*x^12 + 1/10*(3*B*b*c^2 + A*c^3)*x^10 + 3/8*(B*b^2*c + A*b*c^2)* x^8 + 1/4*A*b^3*x^4 + 1/6*(B*b^3 + 3*A*b^2*c)*x^6
Time = 0.03 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.21 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^3} \, dx=\frac {A b^{3} x^{4}}{4} + \frac {B c^{3} x^{12}}{12} + x^{10} \left (\frac {A c^{3}}{10} + \frac {3 B b c^{2}}{10}\right ) + x^{8} \cdot \left (\frac {3 A b c^{2}}{8} + \frac {3 B b^{2} c}{8}\right ) + x^{6} \left (\frac {A b^{2} c}{2} + \frac {B b^{3}}{6}\right ) \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**3,x)
Output:
A*b**3*x**4/4 + B*c**3*x**12/12 + x**10*(A*c**3/10 + 3*B*b*c**2/10) + x**8 *(3*A*b*c**2/8 + 3*B*b**2*c/8) + x**6*(A*b**2*c/2 + B*b**3/6)
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.07 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^3} \, dx=\frac {1}{12} \, B c^{3} x^{12} + \frac {1}{10} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{10} + \frac {3}{8} \, {\left (B b^{2} c + A b c^{2}\right )} x^{8} + \frac {1}{4} \, A b^{3} x^{4} + \frac {1}{6} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{6} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^3,x, algorithm="maxima")
Output:
1/12*B*c^3*x^12 + 1/10*(3*B*b*c^2 + A*c^3)*x^10 + 3/8*(B*b^2*c + A*b*c^2)* x^8 + 1/4*A*b^3*x^4 + 1/6*(B*b^3 + 3*A*b^2*c)*x^6
Time = 0.16 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.13 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^3} \, dx=\frac {1}{12} \, B c^{3} x^{12} + \frac {3}{10} \, B b c^{2} x^{10} + \frac {1}{10} \, A c^{3} x^{10} + \frac {3}{8} \, B b^{2} c x^{8} + \frac {3}{8} \, A b c^{2} x^{8} + \frac {1}{6} \, B b^{3} x^{6} + \frac {1}{2} \, A b^{2} c x^{6} + \frac {1}{4} \, A b^{3} x^{4} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^3,x, algorithm="giac")
Output:
1/12*B*c^3*x^12 + 3/10*B*b*c^2*x^10 + 1/10*A*c^3*x^10 + 3/8*B*b^2*c*x^8 + 3/8*A*b*c^2*x^8 + 1/6*B*b^3*x^6 + 1/2*A*b^2*c*x^6 + 1/4*A*b^3*x^4
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.01 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^3} \, dx=x^6\,\left (\frac {B\,b^3}{6}+\frac {A\,c\,b^2}{2}\right )+x^{10}\,\left (\frac {A\,c^3}{10}+\frac {3\,B\,b\,c^2}{10}\right )+\frac {A\,b^3\,x^4}{4}+\frac {B\,c^3\,x^{12}}{12}+\frac {3\,b\,c\,x^8\,\left (A\,c+B\,b\right )}{8} \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^3,x)
Output:
x^6*((B*b^3)/6 + (A*b^2*c)/2) + x^10*((A*c^3)/10 + (3*B*b*c^2)/10) + (A*b^ 3*x^4)/4 + (B*c^3*x^12)/12 + (3*b*c*x^8*(A*c + B*b))/8
Time = 0.24 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.15 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^3} \, dx=\frac {x^{4} \left (10 b \,c^{3} x^{8}+12 a \,c^{3} x^{6}+36 b^{2} c^{2} x^{6}+45 a b \,c^{2} x^{4}+45 b^{3} c \,x^{4}+60 a \,b^{2} c \,x^{2}+20 b^{4} x^{2}+30 a \,b^{3}\right )}{120} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^3,x)
Output:
(x**4*(30*a*b**3 + 60*a*b**2*c*x**2 + 45*a*b*c**2*x**4 + 12*a*c**3*x**6 + 20*b**4*x**2 + 45*b**3*c*x**4 + 36*b**2*c**2*x**6 + 10*b*c**3*x**8))/120