Integrand size = 27, antiderivative size = 72 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {(A b-a B) x^3}{3 a b (a+b x)^3}-\frac {a^2 B}{2 b^4 (a+b x)^2}+\frac {2 a B}{b^4 (a+b x)}+\frac {B \log (a+b x)}{b^4} \] Output:
1/3*(A*b-B*a)*x^3/a/b/(b*x+a)^3-1/2*a^2*B/b^4/(b*x+a)^2+2*a*B/b^4/(b*x+a)+ B*ln(b*x+a)/b^4
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.01 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {11 a^3 B-6 A b^3 x^2-6 a b^2 x (A-3 B x)+a^2 (-2 A b+27 b B x)+6 B (a+b x)^3 \log (a+b x)}{6 b^4 (a+b x)^3} \] Input:
Integrate[(x^2*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
(11*a^3*B - 6*A*b^3*x^2 - 6*a*b^2*x*(A - 3*B*x) + a^2*(-2*A*b + 27*b*B*x) + 6*B*(a + b*x)^3*Log[a + b*x])/(6*b^4*(a + b*x)^3)
Time = 0.40 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1184, 27, 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.
\(\displaystyle \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 1184 |
\(\displaystyle b^4 \int \frac {x^2 (A+B x)}{b^4 (a+b x)^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^2 (A+B x)}{(a+b x)^4}dx\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {B \int \frac {x^2}{(a+b x)^3}dx}{b}+\frac {x^3 (A b-a B)}{3 a b (a+b x)^3}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {B \int \left (\frac {a^2}{b^2 (a+b x)^3}-\frac {2 a}{b^2 (a+b x)^2}+\frac {1}{b^2 (a+b x)}\right )dx}{b}+\frac {x^3 (A b-a B)}{3 a b (a+b x)^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {B \left (-\frac {a^2}{2 b^3 (a+b x)^2}+\frac {2 a}{b^3 (a+b x)}+\frac {\log (a+b x)}{b^3}\right )}{b}+\frac {x^3 (A b-a B)}{3 a b (a+b x)^3}\) |
Input:
Int[(x^2*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^2,x]
Output:
((A*b - a*B)*x^3)/(3*a*b*(a + b*x)^3) + (B*(-1/2*a^2/(b^3*(a + b*x)^2) + ( 2*a)/(b^3*(a + b*x)) + Log[a + b*x]/b^3))/b
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_))^(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[((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 = 71, normalized size of antiderivative = 0.99
method | result | size |
norman | \(\frac {-\frac {a^{2} \left (2 A b -11 B a \right )}{6 b^{4}}-\frac {\left (A b -3 B a \right ) x^{2}}{b^{2}}-\frac {a \left (2 A b -9 B a \right ) x}{2 b^{3}}}{\left (b x +a \right )^{3}}+\frac {B \ln \left (b x +a \right )}{b^{4}}\) | \(71\) |
default | \(-\frac {a^{2} \left (A b -B a \right )}{3 b^{4} \left (b x +a \right )^{3}}-\frac {A b -3 B a}{b^{4} \left (b x +a \right )}+\frac {B \ln \left (b x +a \right )}{b^{4}}+\frac {a \left (2 A b -3 B a \right )}{2 b^{4} \left (b x +a \right )^{2}}\) | \(78\) |
risch | \(\frac {-\frac {a^{2} \left (2 A b -11 B a \right )}{6 b^{4}}-\frac {\left (A b -3 B a \right ) x^{2}}{b^{2}}-\frac {a \left (2 A b -9 B a \right ) x}{2 b^{3}}}{\left (b x +a \right ) \left (b^{2} x^{2}+2 a b x +a^{2}\right )}+\frac {B \ln \left (b x +a \right )}{b^{4}}\) | \(89\) |
parallelrisch | \(-\frac {-6 B \ln \left (b x +a \right ) x^{3} b^{3}-18 B \ln \left (b x +a \right ) x^{2} a \,b^{2}+6 A \,b^{3} x^{2}-18 B \ln \left (b x +a \right ) x \,a^{2} b -18 B a \,b^{2} x^{2}+6 A a \,b^{2} x -6 B \ln \left (b x +a \right ) a^{3}-27 B \,a^{2} b x +2 A \,a^{2} b -11 B \,a^{3}}{6 b^{4} \left (b^{2} x^{2}+2 a b x +a^{2}\right ) \left (b x +a \right )}\) | \(137\) |
Input:
int(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x,method=_RETURNVERBOSE)
Output:
(-1/6*a^2*(2*A*b-11*B*a)/b^4-(A*b-3*B*a)/b^2*x^2-1/2*a*(2*A*b-9*B*a)/b^3*x )/(b*x+a)^3+B*ln(b*x+a)/b^4
Time = 0.07 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.78 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {11 \, B a^{3} - 2 \, A a^{2} b + 6 \, {\left (3 \, B a b^{2} - A b^{3}\right )} x^{2} + 3 \, {\left (9 \, B a^{2} b - 2 \, A a b^{2}\right )} x + 6 \, {\left (B b^{3} x^{3} + 3 \, B a b^{2} x^{2} + 3 \, B a^{2} b x + B a^{3}\right )} \log \left (b x + a\right )}{6 \, {\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}} \] Input:
integrate(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")
Output:
1/6*(11*B*a^3 - 2*A*a^2*b + 6*(3*B*a*b^2 - A*b^3)*x^2 + 3*(9*B*a^2*b - 2*A *a*b^2)*x + 6*(B*b^3*x^3 + 3*B*a*b^2*x^2 + 3*B*a^2*b*x + B*a^3)*log(b*x + a))/(b^7*x^3 + 3*a*b^6*x^2 + 3*a^2*b^5*x + a^3*b^4)
Time = 0.34 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.39 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {B \log {\left (a + b x \right )}}{b^{4}} + \frac {- 2 A a^{2} b + 11 B a^{3} + x^{2} \left (- 6 A b^{3} + 18 B a b^{2}\right ) + x \left (- 6 A a b^{2} + 27 B a^{2} b\right )}{6 a^{3} b^{4} + 18 a^{2} b^{5} x + 18 a b^{6} x^{2} + 6 b^{7} x^{3}} \] Input:
integrate(x**2*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**2,x)
Output:
B*log(a + b*x)/b**4 + (-2*A*a**2*b + 11*B*a**3 + x**2*(-6*A*b**3 + 18*B*a* b**2) + x*(-6*A*a*b**2 + 27*B*a**2*b))/(6*a**3*b**4 + 18*a**2*b**5*x + 18* a*b**6*x**2 + 6*b**7*x**3)
Time = 0.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.39 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {11 \, B a^{3} - 2 \, A a^{2} b + 6 \, {\left (3 \, B a b^{2} - A b^{3}\right )} x^{2} + 3 \, {\left (9 \, B a^{2} b - 2 \, A a b^{2}\right )} x}{6 \, {\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\right )}} + \frac {B \log \left (b x + a\right )}{b^{4}} \] Input:
integrate(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")
Output:
1/6*(11*B*a^3 - 2*A*a^2*b + 6*(3*B*a*b^2 - A*b^3)*x^2 + 3*(9*B*a^2*b - 2*A *a*b^2)*x)/(b^7*x^3 + 3*a*b^6*x^2 + 3*a^2*b^5*x + a^3*b^4) + B*log(b*x + a )/b^4
Time = 0.20 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.06 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {B \log \left ({\left | b x + a \right |}\right )}{b^{4}} + \frac {6 \, {\left (3 \, B a b - A b^{2}\right )} x^{2} + 3 \, {\left (9 \, B a^{2} - 2 \, A a b\right )} x + \frac {11 \, B a^{3} - 2 \, A a^{2} b}{b}}{6 \, {\left (b x + a\right )}^{3} b^{3}} \] Input:
integrate(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")
Output:
B*log(abs(b*x + a))/b^4 + 1/6*(6*(3*B*a*b - A*b^2)*x^2 + 3*(9*B*a^2 - 2*A* a*b)*x + (11*B*a^3 - 2*A*a^2*b)/b)/((b*x + a)^3*b^3)
Time = 0.08 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.33 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {\frac {11\,B\,a^3-2\,A\,a^2\,b}{6\,b^4}-\frac {x^2\,\left (A\,b-3\,B\,a\right )}{b^2}+\frac {x\,\left (9\,B\,a^2-2\,A\,a\,b\right )}{2\,b^3}}{a^3+3\,a^2\,b\,x+3\,a\,b^2\,x^2+b^3\,x^3}+\frac {B\,\ln \left (a+b\,x\right )}{b^4} \] Input:
int((x^2*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^2,x)
Output:
((11*B*a^3 - 2*A*a^2*b)/(6*b^4) - (x^2*(A*b - 3*B*a))/b^2 + (x*(9*B*a^2 - 2*A*a*b))/(2*b^3))/(a^3 + b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x) + (B*log(a + b*x))/b^4
Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.99 \[ \int \frac {x^2 (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {2 \,\mathrm {log}\left (b x +a \right ) a^{2}+4 \,\mathrm {log}\left (b x +a \right ) a b x +2 \,\mathrm {log}\left (b x +a \right ) b^{2} x^{2}+a^{2}-2 b^{2} x^{2}}{2 b^{3} \left (b^{2} x^{2}+2 a b x +a^{2}\right )} \] Input:
int(x^2*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x)
Output:
(2*log(a + b*x)*a**2 + 4*log(a + b*x)*a*b*x + 2*log(a + b*x)*b**2*x**2 + a **2 - 2*b**2*x**2)/(2*b**3*(a**2 + 2*a*b*x + b**2*x**2))