Integrand size = 24, antiderivative size = 63 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{15}} \, dx=-\frac {b^3 B}{6 x^6}-\frac {3 b^2 B c}{4 x^4}-\frac {3 b B c^2}{2 x^2}-\frac {A \left (b+c x^2\right )^4}{8 b x^8}+B c^3 \log (x) \] Output:
-1/6*b^3*B/x^6-3/4*b^2*B*c/x^4-3/2*b*B*c^2/x^2-1/8*A*(c*x^2+b)^4/b/x^8+B*c ^3*ln(x)
Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.22 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{15}} \, dx=-\frac {2 b B x^2 \left (2 b^2+9 b c x^2+18 c^2 x^4\right )+3 A \left (b^3+4 b^2 c x^2+6 b c^2 x^4+4 c^3 x^6\right )}{24 x^8}+B c^3 \log (x) \] Input:
Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^15,x]
Output:
-1/24*(2*b*B*x^2*(2*b^2 + 9*b*c*x^2 + 18*c^2*x^4) + 3*A*(b^3 + 4*b^2*c*x^2 + 6*b*c^2*x^4 + 4*c^3*x^6))/x^8 + B*c^3*Log[x]
Time = 0.33 (sec) , antiderivative size = 66, 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, 87, 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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{15}} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^9}dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\left (B x^2+A\right ) \left (c x^2+b\right )^3}{x^{10}}dx^2\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{2} \left (B \int \frac {\left (c x^2+b\right )^3}{x^8}dx^2-\frac {A \left (b+c x^2\right )^4}{4 b x^8}\right )\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{2} \left (B \int \left (\frac {b^3}{x^8}+\frac {3 c b^2}{x^6}+\frac {3 c^2 b}{x^4}+\frac {c^3}{x^2}\right )dx^2-\frac {A \left (b+c x^2\right )^4}{4 b x^8}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (B \left (-\frac {b^3}{3 x^6}-\frac {3 b^2 c}{2 x^4}-\frac {3 b c^2}{x^2}+c^3 \log \left (x^2\right )\right )-\frac {A \left (b+c x^2\right )^4}{4 b x^8}\right )\) |
Input:
Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^15,x]
Output:
(-1/4*(A*(b + c*x^2)^4)/(b*x^8) + B*(-1/3*b^3/x^6 - (3*b^2*c)/(2*x^4) - (3 *b*c^2)/x^2 + c^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*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02
method | result | size |
default | \(-\frac {b^{2} \left (3 A c +B b \right )}{6 x^{6}}-\frac {c^{2} \left (A c +3 B b \right )}{2 x^{2}}-\frac {A \,b^{3}}{8 x^{8}}-\frac {3 b c \left (A c +B b \right )}{4 x^{4}}+B \,c^{3} \ln \left (x \right )\) | \(64\) |
risch | \(\frac {\left (-\frac {1}{2} A \,c^{3}-\frac {3}{2} B b \,c^{2}\right ) x^{6}+\left (-\frac {3}{4} A b \,c^{2}-\frac {3}{4} B \,b^{2} c \right ) x^{4}+\left (-\frac {1}{2} A \,b^{2} c -\frac {1}{6} B \,b^{3}\right ) x^{2}-\frac {A \,b^{3}}{8}}{x^{8}}+B \,c^{3} \ln \left (x \right )\) | \(75\) |
norman | \(\frac {\left (-\frac {1}{2} A \,c^{3}-\frac {3}{2} B b \,c^{2}\right ) x^{12}+\left (-\frac {3}{4} A b \,c^{2}-\frac {3}{4} B \,b^{2} c \right ) x^{10}+\left (-\frac {1}{2} A \,b^{2} c -\frac {1}{6} B \,b^{3}\right ) x^{8}-\frac {A \,b^{3} x^{6}}{8}}{x^{14}}+B \,c^{3} \ln \left (x \right )\) | \(78\) |
parallelrisch | \(-\frac {-24 B \,c^{3} \ln \left (x \right ) x^{8}+12 A \,c^{3} x^{6}+36 B b \,c^{2} x^{6}+18 A b \,c^{2} x^{4}+18 x^{4} B \,b^{2} c +12 A \,b^{2} c \,x^{2}+4 x^{2} B \,b^{3}+3 A \,b^{3}}{24 x^{8}}\) | \(82\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^15,x,method=_RETURNVERBOSE)
Output:
-1/6*b^2*(3*A*c+B*b)/x^6-1/2*c^2*(A*c+3*B*b)/x^2-1/8*A*b^3/x^8-3/4*b*c*(A* c+B*b)/x^4+B*c^3*ln(x)
Time = 0.07 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.22 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{15}} \, dx=\frac {24 \, B c^{3} x^{8} \log \left (x\right ) - 12 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} - 18 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} - 3 \, A b^{3} - 4 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{24 \, x^{8}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^15,x, algorithm="fricas")
Output:
1/24*(24*B*c^3*x^8*log(x) - 12*(3*B*b*c^2 + A*c^3)*x^6 - 18*(B*b^2*c + A*b *c^2)*x^4 - 3*A*b^3 - 4*(B*b^3 + 3*A*b^2*c)*x^2)/x^8
Time = 1.10 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.30 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{15}} \, dx=B c^{3} \log {\left (x \right )} + \frac {- 3 A b^{3} + x^{6} \left (- 12 A c^{3} - 36 B b c^{2}\right ) + x^{4} \left (- 18 A b c^{2} - 18 B b^{2} c\right ) + x^{2} \left (- 12 A b^{2} c - 4 B b^{3}\right )}{24 x^{8}} \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**15,x)
Output:
B*c**3*log(x) + (-3*A*b**3 + x**6*(-12*A*c**3 - 36*B*b*c**2) + x**4*(-18*A *b*c**2 - 18*B*b**2*c) + x**2*(-12*A*b**2*c - 4*B*b**3))/(24*x**8)
Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.22 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{15}} \, dx=\frac {1}{2} \, B c^{3} \log \left (x^{2}\right ) - \frac {12 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 18 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} + 3 \, A b^{3} + 4 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{24 \, x^{8}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^15,x, algorithm="maxima")
Output:
1/2*B*c^3*log(x^2) - 1/24*(12*(3*B*b*c^2 + A*c^3)*x^6 + 18*(B*b^2*c + A*b* c^2)*x^4 + 3*A*b^3 + 4*(B*b^3 + 3*A*b^2*c)*x^2)/x^8
Time = 0.17 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.43 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{15}} \, dx=\frac {1}{2} \, B c^{3} \log \left (x^{2}\right ) - \frac {25 \, B c^{3} x^{8} + 36 \, B b c^{2} x^{6} + 12 \, A c^{3} x^{6} + 18 \, B b^{2} c x^{4} + 18 \, A b c^{2} x^{4} + 4 \, B b^{3} x^{2} + 12 \, A b^{2} c x^{2} + 3 \, A b^{3}}{24 \, x^{8}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^15,x, algorithm="giac")
Output:
1/2*B*c^3*log(x^2) - 1/24*(25*B*c^3*x^8 + 36*B*b*c^2*x^6 + 12*A*c^3*x^6 + 18*B*b^2*c*x^4 + 18*A*b*c^2*x^4 + 4*B*b^3*x^2 + 12*A*b^2*c*x^2 + 3*A*b^3)/ x^8
Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.19 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{15}} \, dx=B\,c^3\,\ln \left (x\right )-\frac {x^4\,\left (\frac {3\,B\,b^2\,c}{4}+\frac {3\,A\,b\,c^2}{4}\right )+\frac {A\,b^3}{8}+x^2\,\left (\frac {B\,b^3}{6}+\frac {A\,c\,b^2}{2}\right )+x^6\,\left (\frac {A\,c^3}{2}+\frac {3\,B\,b\,c^2}{2}\right )}{x^8} \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^15,x)
Output:
B*c^3*log(x) - (x^4*((3*A*b*c^2)/4 + (3*B*b^2*c)/4) + (A*b^3)/8 + x^2*((B* b^3)/6 + (A*b^2*c)/2) + x^6*((A*c^3)/2 + (3*B*b*c^2)/2))/x^8
Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.27 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{15}} \, dx=\frac {24 \,\mathrm {log}\left (x \right ) b \,c^{3} x^{8}-3 a \,b^{3}-12 a \,b^{2} c \,x^{2}-18 a b \,c^{2} x^{4}-12 a \,c^{3} x^{6}-4 b^{4} x^{2}-18 b^{3} c \,x^{4}-36 b^{2} c^{2} x^{6}}{24 x^{8}} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^15,x)
Output:
(24*log(x)*b*c**3*x**8 - 3*a*b**3 - 12*a*b**2*c*x**2 - 18*a*b*c**2*x**4 - 12*a*c**3*x**6 - 4*b**4*x**2 - 18*b**3*c*x**4 - 36*b**2*c**2*x**6)/(24*x** 8)