Integrand size = 24, antiderivative size = 72 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{11}} \, dx=-\frac {A b^3}{4 x^4}-\frac {b^2 (b B+3 A c)}{2 x^2}+\frac {1}{2} c^2 (3 b B+A c) x^2+\frac {1}{4} B c^3 x^4+3 b c (b B+A c) \log (x) \] Output:
-1/4*A*b^3/x^4-1/2*b^2*(3*A*c+B*b)/x^2+1/2*c^2*(A*c+3*B*b)*x^2+1/4*B*c^3*x ^4+3*b*c*(A*c+B*b)*ln(x)
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{11}} \, dx=\frac {-A \left (b^3+6 b^2 c x^2-2 c^3 x^6\right )+B x^2 \left (-2 b^3+6 b c^2 x^4+c^3 x^6\right )}{4 x^4}+3 b c (b B+A c) \log (x) \] Input:
Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^11,x]
Output:
(-(A*(b^3 + 6*b^2*c*x^2 - 2*c^3*x^6)) + B*x^2*(-2*b^3 + 6*b*c^2*x^4 + c^3* x^6))/(4*x^4) + 3*b*c*(b*B + A*c)*Log[x]
Time = 0.38 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{11}} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^5}dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\left (B x^2+A\right ) \left (c x^2+b\right )^3}{x^6}dx^2\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {1}{2} \int \left (\frac {A b^3}{x^6}+\frac {(b B+3 A c) b^2}{x^4}+\frac {3 c (b B+A c) b}{x^2}+B c^3 x^2+c^2 (3 b B+A c)\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {A b^3}{2 x^4}-\frac {b^2 (3 A c+b B)}{x^2}+c^2 x^2 (A c+3 b B)+3 b c \log \left (x^2\right ) (A c+b B)+\frac {1}{2} B c^3 x^4\right )\) |
Input:
Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^11,x]
Output:
(-1/2*(A*b^3)/x^4 - (b^2*(b*B + 3*A*c))/x^2 + c^2*(3*b*B + A*c)*x^2 + (B*c ^3*x^4)/2 + 3*b*c*(b*B + 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 = 0.41 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.94
method | result | size |
default | \(\frac {B \,c^{3} x^{4}}{4}+\frac {A \,c^{3} x^{2}}{2}+\frac {3 B b \,c^{2} x^{2}}{2}-\frac {b^{2} \left (3 A c +B b \right )}{2 x^{2}}-\frac {A \,b^{3}}{4 x^{4}}+3 b c \left (A c +B b \right ) \ln \left (x \right )\) | \(68\) |
norman | \(\frac {\left (\frac {1}{2} A \,c^{3}+\frac {3}{2} B b \,c^{2}\right ) x^{12}+\left (-\frac {3}{2} A \,b^{2} c -\frac {1}{2} B \,b^{3}\right ) x^{8}-\frac {A \,b^{3} x^{6}}{4}+\frac {B \,c^{3} x^{14}}{4}}{x^{10}}+\left (3 A b \,c^{2}+3 B \,b^{2} c \right ) \ln \left (x \right )\) | \(79\) |
parallelrisch | \(\frac {B \,c^{3} x^{8}+2 A \,c^{3} x^{6}+6 B b \,c^{2} x^{6}+12 A \ln \left (x \right ) x^{4} b \,c^{2}+12 B \ln \left (x \right ) x^{4} b^{2} c -6 A \,b^{2} c \,x^{2}-2 x^{2} B \,b^{3}-A \,b^{3}}{4 x^{4}}\) | \(83\) |
risch | \(\frac {B \,c^{3} x^{4}}{4}+\frac {A \,c^{3} x^{2}}{2}+\frac {3 B b \,c^{2} x^{2}}{2}+\frac {c^{3} A^{2}}{4 B}+\frac {3 A b \,c^{2}}{2}+\frac {9 B \,b^{2} c}{4}+\frac {\left (-\frac {3}{2} A \,b^{2} c -\frac {1}{2} B \,b^{3}\right ) x^{2}-\frac {A \,b^{3}}{4}}{x^{4}}+3 A \ln \left (x \right ) b \,c^{2}+3 B \ln \left (x \right ) b^{2} c\) | \(102\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^11,x,method=_RETURNVERBOSE)
Output:
1/4*B*c^3*x^4+1/2*A*c^3*x^2+3/2*B*b*c^2*x^2-1/2*b^2*(3*A*c+B*b)/x^2-1/4*A* b^3/x^4+3*b*c*(A*c+B*b)*ln(x)
Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{11}} \, dx=\frac {B c^{3} x^{8} + 2 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 12 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} \log \left (x\right ) - A b^{3} - 2 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{4 \, x^{4}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^11,x, algorithm="fricas")
Output:
1/4*(B*c^3*x^8 + 2*(3*B*b*c^2 + A*c^3)*x^6 + 12*(B*b^2*c + A*b*c^2)*x^4*lo g(x) - A*b^3 - 2*(B*b^3 + 3*A*b^2*c)*x^2)/x^4
Time = 0.31 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.04 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{11}} \, dx=\frac {B c^{3} x^{4}}{4} + 3 b c \left (A c + B b\right ) \log {\left (x \right )} + x^{2} \left (\frac {A c^{3}}{2} + \frac {3 B b c^{2}}{2}\right ) + \frac {- A b^{3} + x^{2} \left (- 6 A b^{2} c - 2 B b^{3}\right )}{4 x^{4}} \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**11,x)
Output:
B*c**3*x**4/4 + 3*b*c*(A*c + B*b)*log(x) + x**2*(A*c**3/2 + 3*B*b*c**2/2) + (-A*b**3 + x**2*(-6*A*b**2*c - 2*B*b**3))/(4*x**4)
Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{11}} \, dx=\frac {1}{4} \, B c^{3} x^{4} + \frac {1}{2} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{2} + \frac {3}{2} \, {\left (B b^{2} c + A b c^{2}\right )} \log \left (x^{2}\right ) - \frac {A b^{3} + 2 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{4 \, x^{4}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^11,x, algorithm="maxima")
Output:
1/4*B*c^3*x^4 + 1/2*(3*B*b*c^2 + A*c^3)*x^2 + 3/2*(B*b^2*c + A*b*c^2)*log( x^2) - 1/4*(A*b^3 + 2*(B*b^3 + 3*A*b^2*c)*x^2)/x^4
Time = 0.18 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.36 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{11}} \, dx=\frac {1}{4} \, B c^{3} x^{4} + \frac {3}{2} \, B b c^{2} x^{2} + \frac {1}{2} \, A c^{3} x^{2} + \frac {3}{2} \, {\left (B b^{2} c + A b c^{2}\right )} \log \left (x^{2}\right ) - \frac {9 \, B b^{2} c x^{4} + 9 \, A b c^{2} x^{4} + 2 \, B b^{3} x^{2} + 6 \, A b^{2} c x^{2} + A b^{3}}{4 \, x^{4}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^11,x, algorithm="giac")
Output:
1/4*B*c^3*x^4 + 3/2*B*b*c^2*x^2 + 1/2*A*c^3*x^2 + 3/2*(B*b^2*c + A*b*c^2)* log(x^2) - 1/4*(9*B*b^2*c*x^4 + 9*A*b*c^2*x^4 + 2*B*b^3*x^2 + 6*A*b^2*c*x^ 2 + A*b^3)/x^4
Time = 9.41 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{11}} \, dx=\ln \left (x\right )\,\left (3\,B\,b^2\,c+3\,A\,b\,c^2\right )-\frac {\frac {A\,b^3}{4}+x^2\,\left (\frac {B\,b^3}{2}+\frac {3\,A\,c\,b^2}{2}\right )}{x^4}+x^2\,\left (\frac {A\,c^3}{2}+\frac {3\,B\,b\,c^2}{2}\right )+\frac {B\,c^3\,x^4}{4} \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^11,x)
Output:
log(x)*(3*A*b*c^2 + 3*B*b^2*c) - ((A*b^3)/4 + x^2*((B*b^3)/2 + (3*A*b^2*c) /2))/x^4 + x^2*((A*c^3)/2 + (3*B*b*c^2)/2) + (B*c^3*x^4)/4
Time = 0.28 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.12 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{11}} \, dx=\frac {12 \,\mathrm {log}\left (x \right ) a b \,c^{2} x^{4}+12 \,\mathrm {log}\left (x \right ) b^{3} c \,x^{4}-a \,b^{3}-6 a \,b^{2} c \,x^{2}+2 a \,c^{3} x^{6}-2 b^{4} x^{2}+6 b^{2} c^{2} x^{6}+b \,c^{3} x^{8}}{4 x^{4}} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^11,x)
Output:
(12*log(x)*a*b*c**2*x**4 + 12*log(x)*b**3*c*x**4 - a*b**3 - 6*a*b**2*c*x** 2 + 2*a*c**3*x**6 - 2*b**4*x**2 + 6*b**2*c**2*x**6 + b*c**3*x**8)/(4*x**4)