Integrand size = 27, antiderivative size = 87 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {a^2 (A b-a B)}{5 b^4 (a+b x)^5}+\frac {a (2 A b-3 a B)}{4 b^4 (a+b x)^4}-\frac {A b-3 a B}{3 b^4 (a+b x)^3}-\frac {B}{2 b^4 (a+b x)^2} \] Output:
-1/5*a^2*(A*b-B*a)/b^4/(b*x+a)^5+1/4*a*(2*A*b-3*B*a)/b^4/(b*x+a)^4-1/3*(A* b-3*B*a)/b^4/(b*x+a)^3-1/2*B/b^4/(b*x+a)^2
Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.72 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {3 a^3 B+10 a b^2 x (A+3 B x)+10 b^3 x^2 (2 A+3 B x)+a^2 b (2 A+15 B x)}{60 b^4 (a+b x)^5} \] Input:
Integrate[(x^2*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^3,x]
Output:
-1/60*(3*a^3*B + 10*a*b^2*x*(A + 3*B*x) + 10*b^3*x^2*(2*A + 3*B*x) + a^2*b *(2*A + 15*B*x))/(b^4*(a + b*x)^5)
Time = 0.45 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1184, 27, 86, 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.
\(\displaystyle \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^6 \int \frac {x^2 (A+B x)}{b^6 (a+b x)^6}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^2 (A+B x)}{(a+b x)^6}dx\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \int \left (-\frac {a^2 (a B-A b)}{b^3 (a+b x)^6}+\frac {a (3 a B-2 A b)}{b^3 (a+b x)^5}+\frac {A b-3 a B}{b^3 (a+b x)^4}+\frac {B}{b^3 (a+b x)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^2 (A b-a B)}{5 b^4 (a+b x)^5}+\frac {a (2 A b-3 a B)}{4 b^4 (a+b x)^4}-\frac {A b-3 a B}{3 b^4 (a+b x)^3}-\frac {B}{2 b^4 (a+b x)^2}\) |
Input:
Int[(x^2*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^3,x]
Output:
-1/5*(a^2*(A*b - a*B))/(b^4*(a + b*x)^5) + (a*(2*A*b - 3*a*B))/(4*b^4*(a + b*x)^4) - (A*b - 3*a*B)/(3*b^4*(a + b*x)^3) - B/(2*b^4*(a + b*x)^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 1.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.90
method | result | size |
norman | \(\frac {-\frac {B \,x^{3}}{2 b}-\frac {\left (2 b^{2} A +3 a b B \right ) x^{2}}{6 b^{3}}-\frac {a \left (2 b^{2} A +3 a b B \right ) x}{12 b^{4}}-\frac {a^{2} \left (2 b^{2} A +3 a b B \right )}{60 b^{5}}}{\left (b x +a \right )^{5}}\) | \(78\) |
default | \(-\frac {a^{2} \left (A b -B a \right )}{5 b^{4} \left (b x +a \right )^{5}}+\frac {a \left (2 A b -3 B a \right )}{4 b^{4} \left (b x +a \right )^{4}}-\frac {A b -3 B a}{3 b^{4} \left (b x +a \right )^{3}}-\frac {B}{2 b^{4} \left (b x +a \right )^{2}}\) | \(80\) |
risch | \(\frac {-\frac {B \,x^{3}}{2 b}-\frac {\left (2 A b +3 B a \right ) x^{2}}{6 b^{2}}-\frac {a \left (2 A b +3 B a \right ) x}{12 b^{3}}-\frac {a^{2} \left (2 A b +3 B a \right )}{60 b^{4}}}{\left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}\) | \(87\) |
orering | \(-\frac {\left (30 x^{3} B \,b^{3}+20 A \,b^{3} x^{2}+30 B a \,b^{2} x^{2}+10 A a \,b^{2} x +15 B \,a^{2} b x +2 A \,a^{2} b +3 B \,a^{3}\right ) \left (b x +a \right )}{60 b^{4} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{3}}\) | \(87\) |
gosper | \(-\frac {30 x^{3} B \,b^{3}+20 A \,b^{3} x^{2}+30 B a \,b^{2} x^{2}+10 A a \,b^{2} x +15 B \,a^{2} b x +2 A \,a^{2} b +3 B \,a^{3}}{60 b^{4} \left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}\) | \(89\) |
parallelrisch | \(-\frac {30 b^{4} B \,x^{3}+20 A \,b^{4} x^{2}+30 B a \,b^{3} x^{2}+10 a A \,b^{3} x +15 B \,a^{2} b^{2} x +2 a^{2} A \,b^{2}+3 B \,a^{3} b}{60 b^{5} \left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2} \left (b x +a \right )}\) | \(94\) |
Input:
int(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x,method=_RETURNVERBOSE)
Output:
(-1/2*B*x^3/b-1/6*(2*A*b^2+3*B*a*b)/b^3*x^2-1/12*a*(2*A*b^2+3*B*a*b)/b^4*x -1/60*a^2*(2*A*b^2+3*B*a*b)/b^5)/(b*x+a)^5
Time = 0.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.37 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {30 \, B b^{3} x^{3} + 3 \, B a^{3} + 2 \, A a^{2} b + 10 \, {\left (3 \, B a b^{2} + 2 \, A b^{3}\right )} x^{2} + 5 \, {\left (3 \, B a^{2} b + 2 \, A a b^{2}\right )} x}{60 \, {\left (b^{9} x^{5} + 5 \, a b^{8} x^{4} + 10 \, a^{2} b^{7} x^{3} + 10 \, a^{3} b^{6} x^{2} + 5 \, a^{4} b^{5} x + a^{5} b^{4}\right )}} \] Input:
integrate(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")
Output:
-1/60*(30*B*b^3*x^3 + 3*B*a^3 + 2*A*a^2*b + 10*(3*B*a*b^2 + 2*A*b^3)*x^2 + 5*(3*B*a^2*b + 2*A*a*b^2)*x)/(b^9*x^5 + 5*a*b^8*x^4 + 10*a^2*b^7*x^3 + 10 *a^3*b^6*x^2 + 5*a^4*b^5*x + a^5*b^4)
Time = 0.53 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.45 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {- 2 A a^{2} b - 3 B a^{3} - 30 B b^{3} x^{3} + x^{2} \left (- 20 A b^{3} - 30 B a b^{2}\right ) + x \left (- 10 A a b^{2} - 15 B a^{2} b\right )}{60 a^{5} b^{4} + 300 a^{4} b^{5} x + 600 a^{3} b^{6} x^{2} + 600 a^{2} b^{7} x^{3} + 300 a b^{8} x^{4} + 60 b^{9} x^{5}} \] Input:
integrate(x**2*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**3,x)
Output:
(-2*A*a**2*b - 3*B*a**3 - 30*B*b**3*x**3 + x**2*(-20*A*b**3 - 30*B*a*b**2) + x*(-10*A*a*b**2 - 15*B*a**2*b))/(60*a**5*b**4 + 300*a**4*b**5*x + 600*a **3*b**6*x**2 + 600*a**2*b**7*x**3 + 300*a*b**8*x**4 + 60*b**9*x**5)
Time = 0.03 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.37 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {30 \, B b^{3} x^{3} + 3 \, B a^{3} + 2 \, A a^{2} b + 10 \, {\left (3 \, B a b^{2} + 2 \, A b^{3}\right )} x^{2} + 5 \, {\left (3 \, B a^{2} b + 2 \, A a b^{2}\right )} x}{60 \, {\left (b^{9} x^{5} + 5 \, a b^{8} x^{4} + 10 \, a^{2} b^{7} x^{3} + 10 \, a^{3} b^{6} x^{2} + 5 \, a^{4} b^{5} x + a^{5} b^{4}\right )}} \] Input:
integrate(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="maxima")
Output:
-1/60*(30*B*b^3*x^3 + 3*B*a^3 + 2*A*a^2*b + 10*(3*B*a*b^2 + 2*A*b^3)*x^2 + 5*(3*B*a^2*b + 2*A*a*b^2)*x)/(b^9*x^5 + 5*a*b^8*x^4 + 10*a^2*b^7*x^3 + 10 *a^3*b^6*x^2 + 5*a^4*b^5*x + a^5*b^4)
Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.80 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {30 \, B b^{3} x^{3} + 30 \, B a b^{2} x^{2} + 20 \, A b^{3} x^{2} + 15 \, B a^{2} b x + 10 \, A a b^{2} x + 3 \, B a^{3} + 2 \, A a^{2} b}{60 \, {\left (b x + a\right )}^{5} b^{4}} \] Input:
integrate(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")
Output:
-1/60*(30*B*b^3*x^3 + 30*B*a*b^2*x^2 + 20*A*b^3*x^2 + 15*B*a^2*b*x + 10*A* a*b^2*x + 3*B*a^3 + 2*A*a^2*b)/((b*x + a)^5*b^4)
Time = 0.05 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.30 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=-\frac {\frac {B\,x^3}{2\,b}+\frac {a^2\,\left (2\,A\,b+3\,B\,a\right )}{60\,b^4}+\frac {x^2\,\left (2\,A\,b+3\,B\,a\right )}{6\,b^2}+\frac {a\,x\,\left (2\,A\,b+3\,B\,a\right )}{12\,b^3}}{a^5+5\,a^4\,b\,x+10\,a^3\,b^2\,x^2+10\,a^2\,b^3\,x^3+5\,a\,b^4\,x^4+b^5\,x^5} \] Input:
int((x^2*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^3,x)
Output:
-((B*x^3)/(2*b) + (a^2*(2*A*b + 3*B*a))/(60*b^4) + (x^2*(2*A*b + 3*B*a))/( 6*b^2) + (a*x*(2*A*b + 3*B*a))/(12*b^3))/(a^5 + b^5*x^5 + 5*a*b^4*x^4 + 10 *a^3*b^2*x^2 + 10*a^2*b^3*x^3 + 5*a^4*b*x)
Time = 0.29 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.74 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx=\frac {-6 b^{2} x^{2}-4 a b x -a^{2}}{12 b^{3} \left (b^{4} x^{4}+4 a \,b^{3} x^{3}+6 a^{2} b^{2} x^{2}+4 a^{3} b x +a^{4}\right )} \] Input:
int(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x)
Output:
( - a**2 - 4*a*b*x - 6*b**2*x**2)/(12*b**3*(a**4 + 4*a**3*b*x + 6*a**2*b** 2*x**2 + 4*a*b**3*x**3 + b**4*x**4))