Integrand size = 24, antiderivative size = 60 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^7} \, dx=\frac {3}{2} A b^2 c x^2+\frac {3}{4} A b c^2 x^4+\frac {1}{6} A c^3 x^6+\frac {B \left (b+c x^2\right )^4}{8 c}+A b^3 \log (x) \] Output:
3/2*A*b^2*c*x^2+3/4*A*b*c^2*x^4+1/6*A*c^3*x^6+1/8*B*(c*x^2+b)^4/c+A*b^3*ln (x)
Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.18 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^7} \, dx=\frac {1}{2} b^2 (b B+3 A c) x^2+\frac {3}{4} b c (b B+A c) x^4+\frac {1}{6} c^2 (3 b B+A c) x^6+\frac {1}{8} B c^3 x^8+A b^3 \log (x) \] Input:
Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^7,x]
Output:
(b^2*(b*B + 3*A*c)*x^2)/2 + (3*b*c*(b*B + A*c)*x^4)/4 + (c^2*(3*b*B + A*c) *x^6)/6 + (B*c^3*x^8)/8 + A*b^3*Log[x]
Time = 0.32 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {9, 354, 90, 49, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^7} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x}dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\left (B x^2+A\right ) \left (c x^2+b\right )^3}{x^2}dx^2\) |
\(\Big \downarrow \) 90 |
\(\displaystyle \frac {1}{2} \left (A \int \frac {\left (c x^2+b\right )^3}{x^2}dx^2+\frac {B \left (b+c x^2\right )^4}{4 c}\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{2} \left (A \int \left (c^3 x^4+3 b c^2 x^2+3 b^2 c+\frac {b^3}{x^2}\right )dx^2+\frac {B \left (b+c x^2\right )^4}{4 c}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (A \left (b^3 \log \left (x^2\right )+3 b^2 c x^2+\frac {3}{2} b c^2 x^4+\frac {c^3 x^6}{3}\right )+\frac {B \left (b+c x^2\right )^4}{4 c}\right )\) |
Input:
Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^7,x]
Output:
((B*(b + c*x^2)^4)/(4*c) + A*(3*b^2*c*x^2 + (3*b*c^2*x^4)/2 + (c^3*x^6)/3 + b^3*Log[x^2]))/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[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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 = 0.32 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.27
method | result | size |
default | \(\frac {B \,c^{3} x^{8}}{8}+\frac {A \,c^{3} x^{6}}{6}+\frac {B b \,c^{2} x^{6}}{2}+\frac {3 A b \,c^{2} x^{4}}{4}+\frac {3 x^{4} B \,b^{2} c}{4}+\frac {3 A \,b^{2} c \,x^{2}}{2}+\frac {x^{2} B \,b^{3}}{2}+A \,b^{3} \ln \left (x \right )\) | \(76\) |
risch | \(\frac {B \,c^{3} x^{8}}{8}+\frac {A \,c^{3} x^{6}}{6}+\frac {B b \,c^{2} x^{6}}{2}+\frac {3 A b \,c^{2} x^{4}}{4}+\frac {3 x^{4} B \,b^{2} c}{4}+\frac {3 A \,b^{2} c \,x^{2}}{2}+\frac {x^{2} B \,b^{3}}{2}+A \,b^{3} \ln \left (x \right )\) | \(76\) |
parallelrisch | \(\frac {B \,c^{3} x^{8}}{8}+\frac {A \,c^{3} x^{6}}{6}+\frac {B b \,c^{2} x^{6}}{2}+\frac {3 A b \,c^{2} x^{4}}{4}+\frac {3 x^{4} B \,b^{2} c}{4}+\frac {3 A \,b^{2} c \,x^{2}}{2}+\frac {x^{2} B \,b^{3}}{2}+A \,b^{3} \ln \left (x \right )\) | \(76\) |
norman | \(\frac {\left (\frac {1}{6} A \,c^{3}+\frac {1}{2} B b \,c^{2}\right ) x^{12}+\left (\frac {3}{4} A b \,c^{2}+\frac {3}{4} B \,b^{2} c \right ) x^{10}+\left (\frac {3}{2} A \,b^{2} c +\frac {1}{2} B \,b^{3}\right ) x^{8}+\frac {B \,c^{3} x^{14}}{8}}{x^{6}}+A \,b^{3} \ln \left (x \right )\) | \(78\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^7,x,method=_RETURNVERBOSE)
Output:
1/8*B*c^3*x^8+1/6*A*c^3*x^6+1/2*B*b*c^2*x^6+3/4*A*b*c^2*x^4+3/4*x^4*B*b^2* c+3/2*A*b^2*c*x^2+1/2*x^2*B*b^3+A*b^3*ln(x)
Time = 0.07 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.18 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^7} \, dx=\frac {1}{8} \, B c^{3} x^{8} + \frac {1}{6} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + \frac {3}{4} \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} + A b^{3} \log \left (x\right ) + \frac {1}{2} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^7,x, algorithm="fricas")
Output:
1/8*B*c^3*x^8 + 1/6*(3*B*b*c^2 + A*c^3)*x^6 + 3/4*(B*b^2*c + A*b*c^2)*x^4 + A*b^3*log(x) + 1/2*(B*b^3 + 3*A*b^2*c)*x^2
Time = 0.08 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.33 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^7} \, dx=A b^{3} \log {\left (x \right )} + \frac {B c^{3} x^{8}}{8} + x^{6} \left (\frac {A c^{3}}{6} + \frac {B b c^{2}}{2}\right ) + x^{4} \cdot \left (\frac {3 A b c^{2}}{4} + \frac {3 B b^{2} c}{4}\right ) + x^{2} \cdot \left (\frac {3 A b^{2} c}{2} + \frac {B b^{3}}{2}\right ) \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**7,x)
Output:
A*b**3*log(x) + B*c**3*x**8/8 + x**6*(A*c**3/6 + B*b*c**2/2) + x**4*(3*A*b *c**2/4 + 3*B*b**2*c/4) + x**2*(3*A*b**2*c/2 + B*b**3/2)
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.23 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^7} \, dx=\frac {1}{8} \, B c^{3} x^{8} + \frac {1}{6} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + \frac {3}{4} \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} + \frac {1}{2} \, A b^{3} \log \left (x^{2}\right ) + \frac {1}{2} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^7,x, algorithm="maxima")
Output:
1/8*B*c^3*x^8 + 1/6*(3*B*b*c^2 + A*c^3)*x^6 + 3/4*(B*b^2*c + A*b*c^2)*x^4 + 1/2*A*b^3*log(x^2) + 1/2*(B*b^3 + 3*A*b^2*c)*x^2
Time = 0.18 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.30 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^7} \, dx=\frac {1}{8} \, B c^{3} x^{8} + \frac {1}{2} \, B b c^{2} x^{6} + \frac {1}{6} \, A c^{3} x^{6} + \frac {3}{4} \, B b^{2} c x^{4} + \frac {3}{4} \, A b c^{2} x^{4} + \frac {1}{2} \, B b^{3} x^{2} + \frac {3}{2} \, A b^{2} c x^{2} + \frac {1}{2} \, A b^{3} \log \left (x^{2}\right ) \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^7,x, algorithm="giac")
Output:
1/8*B*c^3*x^8 + 1/2*B*b*c^2*x^6 + 1/6*A*c^3*x^6 + 3/4*B*b^2*c*x^4 + 3/4*A* b*c^2*x^4 + 1/2*B*b^3*x^2 + 3/2*A*b^2*c*x^2 + 1/2*A*b^3*log(x^2)
Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.12 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^7} \, dx=x^2\,\left (\frac {B\,b^3}{2}+\frac {3\,A\,c\,b^2}{2}\right )+x^6\,\left (\frac {A\,c^3}{6}+\frac {B\,b\,c^2}{2}\right )+\frac {B\,c^3\,x^8}{8}+A\,b^3\,\ln \left (x\right )+\frac {3\,b\,c\,x^4\,\left (A\,c+B\,b\right )}{4} \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^7,x)
Output:
x^2*((B*b^3)/2 + (3*A*b^2*c)/2) + x^6*((A*c^3)/6 + (B*b*c^2)/2) + (B*c^3*x ^8)/8 + A*b^3*log(x) + (3*b*c*x^4*(A*c + B*b))/4
Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.23 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^7} \, dx=\mathrm {log}\left (x \right ) a \,b^{3}+\frac {3 a \,b^{2} c \,x^{2}}{2}+\frac {3 a b \,c^{2} x^{4}}{4}+\frac {a \,c^{3} x^{6}}{6}+\frac {b^{4} x^{2}}{2}+\frac {3 b^{3} c \,x^{4}}{4}+\frac {b^{2} c^{2} x^{6}}{2}+\frac {b \,c^{3} x^{8}}{8} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^7,x)
Output:
(24*log(x)*a*b**3 + 36*a*b**2*c*x**2 + 18*a*b*c**2*x**4 + 4*a*c**3*x**6 + 12*b**4*x**2 + 18*b**3*c*x**4 + 12*b**2*c**2*x**6 + 3*b*c**3*x**8)/24