Integrand size = 24, antiderivative size = 85 \[ \int \frac {x^2}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {x}{6 b \left (a+b x^2\right )^3}+\frac {x}{24 a b \left (a+b x^2\right )^2}+\frac {x}{16 a^2 b \left (a+b x^2\right )}+\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{5/2} b^{3/2}} \] Output:
-1/6*x/b/(b*x^2+a)^3+1/24*x/a/b/(b*x^2+a)^2+1/16*x/a^2/b/(b*x^2+a)+1/16*ar ctan(b^(1/2)*x/a^(1/2))/a^(5/2)/b^(3/2)
Time = 0.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.81 \[ \int \frac {x^2}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {-3 a^2 x+8 a b x^3+3 b^2 x^5}{48 a^2 b \left (a+b x^2\right )^3}+\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{5/2} b^{3/2}} \] Input:
Integrate[x^2/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]
Output:
(-3*a^2*x + 8*a*b*x^3 + 3*b^2*x^5)/(48*a^2*b*(a + b*x^2)^3) + ArcTan[(Sqrt [b]*x)/Sqrt[a]]/(16*a^(5/2)*b^(3/2))
Time = 0.33 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1380, 27, 252, 215, 215, 218}
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.
\(\displaystyle \int \frac {x^2}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle b^4 \int \frac {x^2}{b^4 \left (b x^2+a\right )^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^2}{\left (a+b x^2\right )^4}dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\int \frac {1}{\left (b x^2+a\right )^3}dx}{6 b}-\frac {x}{6 b \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {3 \int \frac {1}{\left (b x^2+a\right )^2}dx}{4 a}+\frac {x}{4 a \left (a+b x^2\right )^2}}{6 b}-\frac {x}{6 b \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {1}{b x^2+a}dx}{2 a}+\frac {x}{2 a \left (a+b x^2\right )}\right )}{4 a}+\frac {x}{4 a \left (a+b x^2\right )^2}}{6 b}-\frac {x}{6 b \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {x}{2 a \left (a+b x^2\right )}\right )}{4 a}+\frac {x}{4 a \left (a+b x^2\right )^2}}{6 b}-\frac {x}{6 b \left (a+b x^2\right )^3}\) |
Input:
Int[x^2/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]
Output:
-1/6*x/(b*(a + b*x^2)^3) + (x/(4*a*(a + b*x^2)^2) + (3*(x/(2*a*(a + b*x^2) ) + ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(2*a^(3/2)*Sqrt[b])))/(4*a))/(6*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_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.68
method | result | size |
default | \(\frac {\frac {b \,x^{5}}{16 a^{2}}+\frac {x^{3}}{6 a}-\frac {x}{16 b}}{\left (b \,x^{2}+a \right )^{3}}+\frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 a^{2} b \sqrt {a b}}\) | \(58\) |
risch | \(\frac {\frac {b \,x^{5}}{16 a^{2}}+\frac {x^{3}}{6 a}-\frac {x}{16 b}}{\left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}-\frac {\ln \left (b x +\sqrt {-a b}\right )}{32 \sqrt {-a b}\, b \,a^{2}}+\frac {\ln \left (-b x +\sqrt {-a b}\right )}{32 \sqrt {-a b}\, b \,a^{2}}\) | \(107\) |
Input:
int(x^2/(b^2*x^4+2*a*b*x^2+a^2)^2,x,method=_RETURNVERBOSE)
Output:
(1/16*b/a^2*x^5+1/6/a*x^3-1/16/b*x)/(b*x^2+a)^3+1/16/a^2/b/(a*b)^(1/2)*arc tan(b/(a*b)^(1/2)*x)
Time = 0.08 (sec) , antiderivative size = 258, normalized size of antiderivative = 3.04 \[ \int \frac {x^2}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\left [\frac {6 \, a b^{3} x^{5} + 16 \, a^{2} b^{2} x^{3} - 6 \, a^{3} b x - 3 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{96 \, {\left (a^{3} b^{5} x^{6} + 3 \, a^{4} b^{4} x^{4} + 3 \, a^{5} b^{3} x^{2} + a^{6} b^{2}\right )}}, \frac {3 \, a b^{3} x^{5} + 8 \, a^{2} b^{2} x^{3} - 3 \, a^{3} b x + 3 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{48 \, {\left (a^{3} b^{5} x^{6} + 3 \, a^{4} b^{4} x^{4} + 3 \, a^{5} b^{3} x^{2} + a^{6} b^{2}\right )}}\right ] \] Input:
integrate(x^2/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")
Output:
[1/96*(6*a*b^3*x^5 + 16*a^2*b^2*x^3 - 6*a^3*b*x - 3*(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^2 + a^3)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a^3*b^5*x^6 + 3*a^4*b^4*x^4 + 3*a^5*b^3*x^2 + a^6*b^2), 1/48*(3*a*b ^3*x^5 + 8*a^2*b^2*x^3 - 3*a^3*b*x + 3*(b^3*x^6 + 3*a*b^2*x^4 + 3*a^2*b*x^ 2 + a^3)*sqrt(a*b)*arctan(sqrt(a*b)*x/a))/(a^3*b^5*x^6 + 3*a^4*b^4*x^4 + 3 *a^5*b^3*x^2 + a^6*b^2)]
Leaf count of result is larger than twice the leaf count of optimal. 139 vs. \(2 (68) = 136\).
Time = 0.18 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.64 \[ \int \frac {x^2}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=- \frac {\sqrt {- \frac {1}{a^{5} b^{3}}} \log {\left (- a^{3} b \sqrt {- \frac {1}{a^{5} b^{3}}} + x \right )}}{32} + \frac {\sqrt {- \frac {1}{a^{5} b^{3}}} \log {\left (a^{3} b \sqrt {- \frac {1}{a^{5} b^{3}}} + x \right )}}{32} + \frac {- 3 a^{2} x + 8 a b x^{3} + 3 b^{2} x^{5}}{48 a^{5} b + 144 a^{4} b^{2} x^{2} + 144 a^{3} b^{3} x^{4} + 48 a^{2} b^{4} x^{6}} \] Input:
integrate(x**2/(b**2*x**4+2*a*b*x**2+a**2)**2,x)
Output:
-sqrt(-1/(a**5*b**3))*log(-a**3*b*sqrt(-1/(a**5*b**3)) + x)/32 + sqrt(-1/( a**5*b**3))*log(a**3*b*sqrt(-1/(a**5*b**3)) + x)/32 + (-3*a**2*x + 8*a*b*x **3 + 3*b**2*x**5)/(48*a**5*b + 144*a**4*b**2*x**2 + 144*a**3*b**3*x**4 + 48*a**2*b**4*x**6)
Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.02 \[ \int \frac {x^2}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {3 \, b^{2} x^{5} + 8 \, a b x^{3} - 3 \, a^{2} x}{48 \, {\left (a^{2} b^{4} x^{6} + 3 \, a^{3} b^{3} x^{4} + 3 \, a^{4} b^{2} x^{2} + a^{5} b\right )}} + \frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{2} b} \] Input:
integrate(x^2/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")
Output:
1/48*(3*b^2*x^5 + 8*a*b*x^3 - 3*a^2*x)/(a^2*b^4*x^6 + 3*a^3*b^3*x^4 + 3*a^ 4*b^2*x^2 + a^5*b) + 1/16*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2*b)
Time = 0.17 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.73 \[ \int \frac {x^2}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{2} b} + \frac {3 \, b^{2} x^{5} + 8 \, a b x^{3} - 3 \, a^{2} x}{48 \, {\left (b x^{2} + a\right )}^{3} a^{2} b} \] Input:
integrate(x^2/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")
Output:
1/16*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2*b) + 1/48*(3*b^2*x^5 + 8*a*b*x^3 - 3*a^2*x)/((b*x^2 + a)^3*a^2*b)
Time = 17.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.87 \[ \int \frac {x^2}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\frac {x^3}{6\,a}-\frac {x}{16\,b}+\frac {b\,x^5}{16\,a^2}}{a^3+3\,a^2\,b\,x^2+3\,a\,b^2\,x^4+b^3\,x^6}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{16\,a^{5/2}\,b^{3/2}} \] Input:
int(x^2/(a^2 + b^2*x^4 + 2*a*b*x^2)^2,x)
Output:
(x^3/(6*a) - x/(16*b) + (b*x^5)/(16*a^2))/(a^3 + b^3*x^6 + 3*a^2*b*x^2 + 3 *a*b^2*x^4) + atan((b^(1/2)*x)/a^(1/2))/(16*a^(5/2)*b^(3/2))
Time = 0.16 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.91 \[ \int \frac {x^2}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3}+9 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b \,x^{2}+9 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} x^{4}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} x^{6}-3 a^{3} b x +8 a^{2} b^{2} x^{3}+3 a \,b^{3} x^{5}}{48 a^{3} b^{2} \left (b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}\right )} \] Input:
int(x^2/(b^2*x^4+2*a*b*x^2+a^2)^2,x)
Output:
(3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3 + 9*sqrt(b)*sqrt(a)* atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b*x**2 + 9*sqrt(b)*sqrt(a)*atan((b*x)/( sqrt(b)*sqrt(a)))*a*b**2*x**4 + 3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt (a)))*b**3*x**6 - 3*a**3*b*x + 8*a**2*b**2*x**3 + 3*a*b**3*x**5)/(48*a**3* b**2*(a**3 + 3*a**2*b*x**2 + 3*a*b**2*x**4 + b**3*x**6))