Integrand size = 24, antiderivative size = 49 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{17}} \, dx=-\frac {A \left (b+c x^2\right )^4}{10 b x^{10}}-\frac {(5 b B-A c) \left (b+c x^2\right )^4}{40 b^2 x^8} \] Output:
-1/10*A*(c*x^2+b)^4/b/x^10-1/40*(-A*c+5*B*b)*(c*x^2+b)^4/b^2/x^8
Time = 0.02 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.59 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{17}} \, dx=-\frac {5 B x^2 \left (b^3+4 b^2 c x^2+6 b c^2 x^4+4 c^3 x^6\right )+A \left (4 b^3+15 b^2 c x^2+20 b c^2 x^4+10 c^3 x^6\right )}{40 x^{10}} \] Input:
Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^17,x]
Output:
-1/40*(5*B*x^2*(b^3 + 4*b^2*c*x^2 + 6*b*c^2*x^4 + 4*c^3*x^6) + A*(4*b^3 + 15*b^2*c*x^2 + 20*b*c^2*x^4 + 10*c^3*x^6))/x^10
Time = 0.29 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {9, 354, 87, 48}
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.
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{17}} \, dx\) |
\(\Big \downarrow \) 9 |
\(\displaystyle \int \frac {\left (A+B x^2\right ) \left (b+c x^2\right )^3}{x^{11}}dx\) |
\(\Big \downarrow \) 354 |
\(\displaystyle \frac {1}{2} \int \frac {\left (B x^2+A\right ) \left (c x^2+b\right )^3}{x^{12}}dx^2\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {1}{2} \left (\frac {(5 b B-A c) \int \frac {\left (c x^2+b\right )^3}{x^{10}}dx^2}{5 b}-\frac {A \left (b+c x^2\right )^4}{5 b x^{10}}\right )\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {1}{2} \left (-\frac {\left (b+c x^2\right )^4 (5 b B-A c)}{20 b^2 x^8}-\frac {A \left (b+c x^2\right )^4}{5 b x^{10}}\right )\) |
Input:
Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^17,x]
Output:
(-1/5*(A*(b + c*x^2)^4)/(b*x^10) - ((5*b*B - A*c)*(b + c*x^2)^4)/(20*b^2*x ^8))/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] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
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 = 66, normalized size of antiderivative = 1.35
method | result | size |
default | \(-\frac {b c \left (A c +B b \right )}{2 x^{6}}-\frac {B \,c^{3}}{2 x^{2}}-\frac {b^{2} \left (3 A c +B b \right )}{8 x^{8}}-\frac {c^{2} \left (A c +3 B b \right )}{4 x^{4}}-\frac {A \,b^{3}}{10 x^{10}}\) | \(66\) |
risch | \(\frac {-\frac {B \,c^{3} x^{8}}{2}+\left (-\frac {1}{4} A \,c^{3}-\frac {3}{4} B b \,c^{2}\right ) x^{6}+\left (-\frac {1}{2} A b \,c^{2}-\frac {1}{2} B \,b^{2} c \right ) x^{4}+\left (-\frac {3}{8} A \,b^{2} c -\frac {1}{8} B \,b^{3}\right ) x^{2}-\frac {A \,b^{3}}{10}}{x^{10}}\) | \(76\) |
norman | \(\frac {\left (-\frac {1}{4} A \,c^{3}-\frac {3}{4} B b \,c^{2}\right ) x^{12}+\left (-\frac {1}{2} A b \,c^{2}-\frac {1}{2} B \,b^{2} c \right ) x^{10}+\left (-\frac {3}{8} A \,b^{2} c -\frac {1}{8} B \,b^{3}\right ) x^{8}-\frac {A \,b^{3} x^{6}}{10}-\frac {B \,c^{3} x^{14}}{2}}{x^{16}}\) | \(79\) |
gosper | \(-\frac {20 B \,c^{3} x^{8}+10 A \,c^{3} x^{6}+30 B b \,c^{2} x^{6}+20 A b \,c^{2} x^{4}+20 x^{4} B \,b^{2} c +15 A \,b^{2} c \,x^{2}+5 x^{2} B \,b^{3}+4 A \,b^{3}}{40 x^{10}}\) | \(80\) |
parallelrisch | \(-\frac {20 B \,c^{3} x^{8}+10 A \,c^{3} x^{6}+30 B b \,c^{2} x^{6}+20 A b \,c^{2} x^{4}+20 x^{4} B \,b^{2} c +15 A \,b^{2} c \,x^{2}+5 x^{2} B \,b^{3}+4 A \,b^{3}}{40 x^{10}}\) | \(80\) |
orering | \(-\frac {\left (20 B \,c^{3} x^{8}+10 A \,c^{3} x^{6}+30 B b \,c^{2} x^{6}+20 A b \,c^{2} x^{4}+20 x^{4} B \,b^{2} c +15 A \,b^{2} c \,x^{2}+5 x^{2} B \,b^{3}+4 A \,b^{3}\right ) \left (c \,x^{4}+b \,x^{2}\right )^{3}}{40 x^{16} \left (c \,x^{2}+b \right )^{3}}\) | \(102\) |
Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^17,x,method=_RETURNVERBOSE)
Output:
-1/2*b*c*(A*c+B*b)/x^6-1/2*B*c^3/x^2-1/8*b^2*(3*A*c+B*b)/x^8-1/4*c^2*(A*c+ 3*B*b)/x^4-1/10*A*b^3/x^10
Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.53 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{17}} \, dx=-\frac {20 \, B c^{3} x^{8} + 10 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 20 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} + 4 \, A b^{3} + 5 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{40 \, x^{10}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^17,x, algorithm="fricas")
Output:
-1/40*(20*B*c^3*x^8 + 10*(3*B*b*c^2 + A*c^3)*x^6 + 20*(B*b^2*c + A*b*c^2)* x^4 + 4*A*b^3 + 5*(B*b^3 + 3*A*b^2*c)*x^2)/x^10
Time = 1.71 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.69 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{17}} \, dx=\frac {- 4 A b^{3} - 20 B c^{3} x^{8} + x^{6} \left (- 10 A c^{3} - 30 B b c^{2}\right ) + x^{4} \left (- 20 A b c^{2} - 20 B b^{2} c\right ) + x^{2} \left (- 15 A b^{2} c - 5 B b^{3}\right )}{40 x^{10}} \] Input:
integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**17,x)
Output:
(-4*A*b**3 - 20*B*c**3*x**8 + x**6*(-10*A*c**3 - 30*B*b*c**2) + x**4*(-20* A*b*c**2 - 20*B*b**2*c) + x**2*(-15*A*b**2*c - 5*B*b**3))/(40*x**10)
Time = 0.03 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.53 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{17}} \, dx=-\frac {20 \, B c^{3} x^{8} + 10 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{6} + 20 \, {\left (B b^{2} c + A b c^{2}\right )} x^{4} + 4 \, A b^{3} + 5 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x^{2}}{40 \, x^{10}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^17,x, algorithm="maxima")
Output:
-1/40*(20*B*c^3*x^8 + 10*(3*B*b*c^2 + A*c^3)*x^6 + 20*(B*b^2*c + A*b*c^2)* x^4 + 4*A*b^3 + 5*(B*b^3 + 3*A*b^2*c)*x^2)/x^10
Time = 0.18 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.61 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{17}} \, dx=-\frac {20 \, B c^{3} x^{8} + 30 \, B b c^{2} x^{6} + 10 \, A c^{3} x^{6} + 20 \, B b^{2} c x^{4} + 20 \, A b c^{2} x^{4} + 5 \, B b^{3} x^{2} + 15 \, A b^{2} c x^{2} + 4 \, A b^{3}}{40 \, x^{10}} \] Input:
integrate((B*x^2+A)*(c*x^4+b*x^2)^3/x^17,x, algorithm="giac")
Output:
-1/40*(20*B*c^3*x^8 + 30*B*b*c^2*x^6 + 10*A*c^3*x^6 + 20*B*b^2*c*x^4 + 20* A*b*c^2*x^4 + 5*B*b^3*x^2 + 15*A*b^2*c*x^2 + 4*A*b^3)/x^10
Time = 8.68 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.55 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{17}} \, dx=-\frac {x^4\,\left (\frac {B\,b^2\,c}{2}+\frac {A\,b\,c^2}{2}\right )+\frac {A\,b^3}{10}+x^2\,\left (\frac {B\,b^3}{8}+\frac {3\,A\,c\,b^2}{8}\right )+x^6\,\left (\frac {A\,c^3}{4}+\frac {3\,B\,b\,c^2}{4}\right )+\frac {B\,c^3\,x^8}{2}}{x^{10}} \] Input:
int(((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^17,x)
Output:
-(x^4*((A*b*c^2)/2 + (B*b^2*c)/2) + (A*b^3)/10 + x^2*((B*b^3)/8 + (3*A*b^2 *c)/8) + x^6*((A*c^3)/4 + (3*B*b*c^2)/4) + (B*c^3*x^8)/2)/x^10
Time = 0.21 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.59 \[ \int \frac {\left (A+B x^2\right ) \left (b x^2+c x^4\right )^3}{x^{17}} \, dx=\frac {-20 b \,c^{3} x^{8}-10 a \,c^{3} x^{6}-30 b^{2} c^{2} x^{6}-20 a b \,c^{2} x^{4}-20 b^{3} c \,x^{4}-15 a \,b^{2} c \,x^{2}-5 b^{4} x^{2}-4 a \,b^{3}}{40 x^{10}} \] Input:
int((B*x^2+A)*(c*x^4+b*x^2)^3/x^17,x)
Output:
( - 4*a*b**3 - 15*a*b**2*c*x**2 - 20*a*b*c**2*x**4 - 10*a*c**3*x**6 - 5*b* *4*x**2 - 20*b**3*c*x**4 - 30*b**2*c**2*x**6 - 20*b*c**3*x**8)/(40*x**10)