Integrand size = 24, antiderivative size = 71 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{13}} \, dx=-\frac {A b^3}{6 x^6}-\frac {b^2 (b B+3 A c)}{4 x^4}-\frac {3 b c (b B+A c)}{2 x^2}+\frac {1}{2} B c^3 x^2+c^2 (3 b B+A c) \log (x) \] Output:
-1/6*A*b^3/x^6-1/4*b^2*(3*A*c+B*b)/x^4-3/2*b*c*(A*c+B*b)/x^2+1/2*B*c^3*x^2 +c^2*(A*c+3*B*b)*ln(x)
Time = 0.03 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{13}} \, dx=-\frac {A b^3}{6 x^6}-\frac {b^2 (b B+3 A c)}{4 x^4}-\frac {3 b c (b B+A c)}{2 x^2}+\frac {1}{2} B c^3 x^2+c^2 (3 b B+A c) \log (x) \] Input:
Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^13,x]
Output:
-1/6*(A*b^3)/x^6 - (b^2*(b*B + 3*A*c))/(4*x^4) - (3*b*c*(b*B + A*c))/(2*x^ 2) + (B*c^3*x^2)/2 + c^2*(3*b*B + A*c)*Log[x]
Time = 0.36 (sec) , antiderivative size = 72, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{13}} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^7}dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\left (B x^2+A\right ) \left (c x^2+b\right )^3}{x^8}dx^2\) |
\(\Big \downarrow \) 85 |
\(\displaystyle \frac {1}{2} \int \left (\frac {A b^3}{x^8}+\frac {(b B+3 A c) b^2}{x^6}+\frac {3 c (b B+A c) b}{x^4}+B c^3+\frac {c^2 (3 b B+A c)}{x^2}\right )dx^2\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-\frac {A b^3}{3 x^6}-\frac {b^2 (3 A c+b B)}{2 x^4}+c^2 \log \left (x^2\right ) (A c+3 b B)-\frac {3 b c (A c+b B)}{x^2}+B c^3 x^2\right )\) |
Input:
Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^13,x]
Output:
(-1/3*(A*b^3)/x^6 - (b^2*(b*B + 3*A*c))/(2*x^4) - (3*b*c*(b*B + A*c))/x^2 + B*c^3*x^2 + c^2*(3*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 = 0.32 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {A \,b^{3}}{6 x^{6}}-\frac {b^{2} \left (3 A c +B b \right )}{4 x^{4}}-\frac {3 b c \left (A c +B b \right )}{2 x^{2}}+\frac {B \,c^{3} x^{2}}{2}+c^{2} \left (A c +3 B b \right ) \ln \left (x \right )\) | \(64\) |
risch | \(\frac {B \,c^{3} x^{2}}{2}+\frac {\left (-\frac {3}{2} A b \,c^{2}-\frac {3}{2} B \,b^{2} c \right ) x^{4}+\left (-\frac {3}{4} A \,b^{2} c -\frac {1}{4} B \,b^{3}\right ) x^{2}-\frac {A \,b^{3}}{6}}{x^{6}}+A \ln \left (x \right ) c^{3}+3 B \ln \left (x \right ) b \,c^{2}\) | \(75\) |
norman | \(\frac {\left (-\frac {3}{2} A b \,c^{2}-\frac {3}{2} B \,b^{2} c \right ) x^{10}+\left (-\frac {3}{4} A \,b^{2} c -\frac {1}{4} B \,b^{3}\right ) x^{8}-\frac {A \,b^{3} x^{6}}{6}+\frac {B \,c^{3} x^{14}}{2}}{x^{12}}+\left (A \,c^{3}+3 B b \,c^{2}\right ) \ln \left (x \right )\) | \(78\) |
parallelrisch | \(\frac {6 B \,c^{3} x^{8}+12 A \ln \left (x \right ) x^{6} c^{3}+36 B \ln \left (x \right ) x^{6} b \,c^{2}-18 A b \,c^{2} x^{4}-18 x^{4} B \,b^{2} c -9 A \,b^{2} c \,x^{2}-3 x^{2} B \,b^{3}-2 A \,b^{3}}{12 x^{6}}\) | \(84\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^13,x,method=_RETURNVERBOSE)
Output:
-1/6*A*b^3/x^6-1/4*b^2*(3*A*c+B*b)/x^4-3/2*b*c*(A*c+B*b)/x^2+1/2*B*c^3*x^2 +c^2*(A*c+3*B*b)*ln(x)
Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.08 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{13}} \, dx=\frac {6 \, B c^{3} x^{8} + 12 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} \log \left (x\right ) - 18 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} - 2 \, A b^{3} - 3 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{12 \, x^{6}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^13,x, algorithm="fricas")
Output:
1/12*(6*B*c^3*x^8 + 12*(3*B*b*c^2 + A*c^3)*x^6*log(x) - 18*(B*b^2*c + A*b* c^2)*x^4 - 2*A*b^3 - 3*(B*b^3 + 3*A*b^2*c)*x^2)/x^6
Time = 0.63 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.10 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{13}} \, dx=\frac {B c^{3} x^{2}}{2} + c^{2} \left (A c + 3 B b\right ) \log {\left (x \right )} + \frac {- 2 A b^{3} + x^{4} \left (- 18 A b c^{2} - 18 B b^{2} c\right ) + x^{2} \left (- 9 A b^{2} c - 3 B b^{3}\right )}{12 x^{6}} \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**13,x)
Output:
B*c**3*x**2/2 + c**2*(A*c + 3*B*b)*log(x) + (-2*A*b**3 + x**4*(-18*A*b*c** 2 - 18*B*b**2*c) + x**2*(-9*A*b**2*c - 3*B*b**3))/(12*x**6)
Time = 0.04 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.08 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{13}} \, dx=\frac {1}{2} \, B c^{3} x^{2} + \frac {1}{2} \, {\left (3 \, B b c^{2} + A c^{3}\right )} \log \left (x^{2}\right ) - \frac {18 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} + 2 \, A b^{3} + 3 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{12 \, x^{6}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^13,x, algorithm="maxima")
Output:
1/2*B*c^3*x^2 + 1/2*(3*B*b*c^2 + A*c^3)*log(x^2) - 1/12*(18*(B*b^2*c + A*b *c^2)*x^4 + 2*A*b^3 + 3*(B*b^3 + 3*A*b^2*c)*x^2)/x^6
Time = 0.19 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.39 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{13}} \, dx=\frac {1}{2} \, B c^{3} x^{2} + \frac {1}{2} \, {\left (3 \, B b c^{2} + A c^{3}\right )} \log \left (x^{2}\right ) - \frac {33 \, B b c^{2} x^{6} + 11 \, A c^{3} x^{6} + 18 \, B b^{2} c x^{4} + 18 \, A b c^{2} x^{4} + 3 \, B b^{3} x^{2} + 9 \, A b^{2} c x^{2} + 2 \, A b^{3}}{12 \, x^{6}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^13,x, algorithm="giac")
Output:
1/2*B*c^3*x^2 + 1/2*(3*B*b*c^2 + A*c^3)*log(x^2) - 1/12*(33*B*b*c^2*x^6 + 11*A*c^3*x^6 + 18*B*b^2*c*x^4 + 18*A*b*c^2*x^4 + 3*B*b^3*x^2 + 9*A*b^2*c*x ^2 + 2*A*b^3)/x^6
Time = 9.10 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.06 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{13}} \, dx=\ln \left (x\right )\,\left (A\,c^3+3\,B\,b\,c^2\right )-\frac {x^4\,\left (\frac {3\,B\,b^2\,c}{2}+\frac {3\,A\,b\,c^2}{2}\right )+\frac {A\,b^3}{6}+x^2\,\left (\frac {B\,b^3}{4}+\frac {3\,A\,c\,b^2}{4}\right )}{x^6}+\frac {B\,c^3\,x^2}{2} \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^13,x)
Output:
log(x)*(A*c^3 + 3*B*b*c^2) - (x^4*((3*A*b*c^2)/2 + (3*B*b^2*c)/2) + (A*b^3 )/6 + x^2*((B*b^3)/4 + (3*A*b^2*c)/4))/x^6 + (B*c^3*x^2)/2
Time = 0.22 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.15 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{13}} \, dx=\frac {12 \,\mathrm {log}\left (x \right ) a \,c^{3} x^{6}+36 \,\mathrm {log}\left (x \right ) b^{2} c^{2} x^{6}-2 a \,b^{3}-9 a \,b^{2} c \,x^{2}-18 a b \,c^{2} x^{4}-3 b^{4} x^{2}-18 b^{3} c \,x^{4}+6 b \,c^{3} x^{8}}{12 x^{6}} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^13,x)
Output:
(12*log(x)*a*c**3*x**6 + 36*log(x)*b**2*c**2*x**6 - 2*a*b**3 - 9*a*b**2*c* x**2 - 18*a*b*c**2*x**4 - 3*b**4*x**2 - 18*b**3*c*x**4 + 6*b*c**3*x**8)/(1 2*x**6)