Integrand size = 24, antiderivative size = 71 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^9} \, dx=-\frac {A b^3}{2 x^2}+\frac {3}{2} b c (b B+A c) x^2+\frac {1}{4} c^2 (3 b B+A c) x^4+\frac {1}{6} B c^3 x^6+b^2 (b B+3 A c) \log (x) \] Output:
-1/2*A*b^3/x^2+3/2*b*c*(A*c+B*b)*x^2+1/4*c^2*(A*c+3*B*b)*x^4+1/6*B*c^3*x^6 +b^2*(3*A*c+B*b)*ln(x)
Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.03 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^9} \, dx=-\frac {A b^3}{2 x^2}+\frac {3}{2} b c (b B+A c) x^2+\frac {1}{4} c^2 (3 b B+A c) x^4+\frac {1}{6} B c^3 x^6+\left (b^3 B+3 A b^2 c\right ) \log (x) \] Input:
Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^9,x]
Output:
-1/2*(A*b^3)/x^2 + (3*b*c*(b*B + A*c)*x^2)/2 + (c^2*(3*b*B + A*c)*x^4)/4 + (B*c^3*x^6)/6 + (b^3*B + 3*A*b^2*c)*Log[x]
Time = 0.37 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.03, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^9} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^3}dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\left (B x^2+A\right ) \left (c x^2+b\right )^3}{x^4}dx^2\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {1}{2} \int \left (B c^3 x^4+c^2 (3 b B+A c) x^2+3 b c (b B+A c)+\frac {b^2 (b B+3 A c)}{x^2}+\frac {A b^3}{x^4}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {A b^3}{x^2}+b^2 \log \left (x^2\right ) (3 A c+b B)+\frac {1}{2} c^2 x^4 (A c+3 b B)+3 b c x^2 (A c+b B)+\frac {1}{3} B c^3 x^6\right )\) |
Input:
Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^9,x]
Output:
(-((A*b^3)/x^2) + 3*b*c*(b*B + A*c)*x^2 + (c^2*(3*b*B + A*c)*x^4)/2 + (B*c ^3*x^6)/3 + b^2*(b*B + 3*A*c)*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[((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 = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.03
method | result | size |
default | \(\frac {B \,c^{3} x^{6}}{6}+\frac {A \,c^{3} x^{4}}{4}+\frac {3 B b \,c^{2} x^{4}}{4}+\frac {3 A b \,c^{2} x^{2}}{2}+\frac {3 x^{2} B \,b^{2} c}{2}-\frac {A \,b^{3}}{2 x^{2}}+b^{2} \left (3 A c +B b \right ) \ln \left (x \right )\) | \(73\) |
risch | \(\frac {B \,c^{3} x^{6}}{6}+\frac {A \,c^{3} x^{4}}{4}+\frac {3 B b \,c^{2} x^{4}}{4}+\frac {3 A b \,c^{2} x^{2}}{2}+\frac {3 x^{2} B \,b^{2} c}{2}-\frac {A \,b^{3}}{2 x^{2}}+3 A \ln \left (x \right ) b^{2} c +B \ln \left (x \right ) b^{3}\) | \(75\) |
norman | \(\frac {\left (\frac {1}{4} A \,c^{3}+\frac {3}{4} B b \,c^{2}\right ) x^{12}+\left (\frac {3}{2} A b \,c^{2}+\frac {3}{2} B \,b^{2} c \right ) x^{10}-\frac {A \,b^{3} x^{6}}{2}+\frac {B \,c^{3} x^{14}}{6}}{x^{8}}+\left (3 A \,b^{2} c +B \,b^{3}\right ) \ln \left (x \right )\) | \(78\) |
parallelrisch | \(\frac {2 B \,c^{3} x^{8}+3 A \,c^{3} x^{6}+9 B b \,c^{2} x^{6}+18 A b \,c^{2} x^{4}+18 x^{4} B \,b^{2} c +36 A \ln \left (x \right ) x^{2} b^{2} c +12 B \ln \left (x \right ) x^{2} b^{3}-6 A \,b^{3}}{12 x^{2}}\) | \(84\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^9,x,method=_RETURNVERBOSE)
Output:
1/6*B*c^3*x^6+1/4*A*c^3*x^4+3/4*B*b*c^2*x^4+3/2*A*b*c^2*x^2+3/2*x^2*B*b^2* c-1/2*A*b^3/x^2+b^2*(3*A*c+B*b)*ln(x)
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.08 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^9} \, dx=\frac {2 \, B c^{3} x^{8} + 3 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 18 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} - 6 \, A b^{3} + 12 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2} \log \left (x\right )}{12 \, x^{2}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^9,x, algorithm="fricas")
Output:
1/12*(2*B*c^3*x^8 + 3*(3*B*b*c^2 + A*c^3)*x^6 + 18*(B*b^2*c + A*b*c^2)*x^4 - 6*A*b^3 + 12*(B*b^3 + 3*A*b^2*c)*x^2*log(x))/x^2
Time = 0.13 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.10 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^9} \, dx=- \frac {A b^{3}}{2 x^{2}} + \frac {B c^{3} x^{6}}{6} + b^{2} \cdot \left (3 A c + B b\right ) \log {\left (x \right )} + x^{4} \left (\frac {A c^{3}}{4} + \frac {3 B b c^{2}}{4}\right ) + x^{2} \cdot \left (\frac {3 A b c^{2}}{2} + \frac {3 B b^{2} c}{2}\right ) \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**9,x)
Output:
-A*b**3/(2*x**2) + B*c**3*x**6/6 + b**2*(3*A*c + B*b)*log(x) + x**4*(A*c** 3/4 + 3*B*b*c**2/4) + x**2*(3*A*b*c**2/2 + 3*B*b**2*c/2)
Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.04 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^9} \, dx=\frac {1}{6} \, B c^{3} x^{6} + \frac {1}{4} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{4} + \frac {3}{2} \, {\left (B b^{2} c + A b c^{2}\right )} x^{2} - \frac {A b^{3}}{2 \, x^{2}} + \frac {1}{2} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} \log \left (x^{2}\right ) \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^9,x, algorithm="maxima")
Output:
1/6*B*c^3*x^6 + 1/4*(3*B*b*c^2 + A*c^3)*x^4 + 3/2*(B*b^2*c + A*b*c^2)*x^2 - 1/2*A*b^3/x^2 + 1/2*(B*b^3 + 3*A*b^2*c)*log(x^2)
Time = 0.16 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.37 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^9} \, dx=\frac {1}{6} \, B c^{3} x^{6} + \frac {3}{4} \, B b c^{2} x^{4} + \frac {1}{4} \, A c^{3} x^{4} + \frac {3}{2} \, B b^{2} c x^{2} + \frac {3}{2} \, A b c^{2} x^{2} + \frac {1}{2} \, {\left (B b^{3} + 3 \, A b^{2} c\right )} \log \left (x^{2}\right ) - \frac {B b^{3} x^{2} + 3 \, A b^{2} c x^{2} + A b^{3}}{2 \, x^{2}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^9,x, algorithm="giac")
Output:
1/6*B*c^3*x^6 + 3/4*B*b*c^2*x^4 + 1/4*A*c^3*x^4 + 3/2*B*b^2*c*x^2 + 3/2*A* b*c^2*x^2 + 1/2*(B*b^3 + 3*A*b^2*c)*log(x^2) - 1/2*(B*b^3*x^2 + 3*A*b^2*c* x^2 + A*b^3)/x^2
Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.94 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^9} \, dx=x^4\,\left (\frac {A\,c^3}{4}+\frac {3\,B\,b\,c^2}{4}\right )+\ln \left (x\right )\,\left (B\,b^3+3\,A\,c\,b^2\right )-\frac {A\,b^3}{2\,x^2}+\frac {B\,c^3\,x^6}{6}+\frac {3\,b\,c\,x^2\,\left (A\,c+B\,b\right )}{2} \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^9,x)
Output:
x^4*((A*c^3)/4 + (3*B*b*c^2)/4) + log(x)*(B*b^3 + 3*A*b^2*c) - (A*b^3)/(2* x^2) + (B*c^3*x^6)/6 + (3*b*c*x^2*(A*c + B*b))/2
Time = 0.26 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.15 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^9} \, dx=\frac {36 \,\mathrm {log}\left (x \right ) a \,b^{2} c \,x^{2}+12 \,\mathrm {log}\left (x \right ) b^{4} x^{2}-6 a \,b^{3}+18 a b \,c^{2} x^{4}+3 a \,c^{3} x^{6}+18 b^{3} c \,x^{4}+9 b^{2} c^{2} x^{6}+2 b \,c^{3} x^{8}}{12 x^{2}} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^9,x)
Output:
(36*log(x)*a*b**2*c*x**2 + 12*log(x)*b**4*x**2 - 6*a*b**3 + 18*a*b*c**2*x* *4 + 3*a*c**3*x**6 + 18*b**3*c*x**4 + 9*b**2*c**2*x**6 + 2*b*c**3*x**8)/(1 2*x**2)