Integrand size = 20, antiderivative size = 79 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {x}{6 a \left (a+b x^2\right )^3}+\frac {5 x}{24 a^2 \left (a+b x^2\right )^2}+\frac {5 x}{16 a^3 \left (a+b x^2\right )}+\frac {5 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{7/2} \sqrt {b}} \] Output:
1/6*x/a/(b*x^2+a)^3+5/24*x/a^2/(b*x^2+a)^2+5/16*x/a^3/(b*x^2+a)+5/16*arcta n(b^(1/2)*x/a^(1/2))/a^(7/2)/b^(1/2)
Time = 0.01 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {33 a^2 x+40 a b x^3+15 b^2 x^5}{48 a^3 \left (a+b x^2\right )^3}+\frac {5 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 a^{7/2} \sqrt {b}} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(-2),x]
Output:
(33*a^2*x + 40*a*b*x^3 + 15*b^2*x^5)/(48*a^3*(a + b*x^2)^3) + (5*ArcTan[(S qrt[b]*x)/Sqrt[a]])/(16*a^(7/2)*Sqrt[b])
Time = 0.34 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.44, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {1379, 215, 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.
\(\displaystyle \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 1379 |
\(\displaystyle b^4 \int \frac {1}{\left (b^2 x^2+a b\right )^4}dx\) |
\(\Big \downarrow \) 215 |
\(\displaystyle b^4 \left (\frac {5 \int \frac {1}{\left (b^2 x^2+a b\right )^3}dx}{6 a b}+\frac {x}{6 a b^4 \left (a+b x^2\right )^3}\right )\) |
\(\Big \downarrow \) 215 |
\(\displaystyle b^4 \left (\frac {5 \left (\frac {3 \int \frac {1}{\left (b^2 x^2+a b\right )^2}dx}{4 a b}+\frac {x}{4 a b^3 \left (a+b x^2\right )^2}\right )}{6 a b}+\frac {x}{6 a b^4 \left (a+b x^2\right )^3}\right )\) |
\(\Big \downarrow \) 215 |
\(\displaystyle b^4 \left (\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{b^2 x^2+a b}dx}{2 a b}+\frac {x}{2 a b^2 \left (a+b x^2\right )}\right )}{4 a b}+\frac {x}{4 a b^3 \left (a+b x^2\right )^2}\right )}{6 a b}+\frac {x}{6 a b^4 \left (a+b x^2\right )^3}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle b^4 \left (\frac {5 \left (\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} b^{5/2}}+\frac {x}{2 a b^2 \left (a+b x^2\right )}\right )}{4 a b}+\frac {x}{4 a b^3 \left (a+b x^2\right )^2}\right )}{6 a b}+\frac {x}{6 a b^4 \left (a+b x^2\right )^3}\right )\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(-2),x]
Output:
b^4*(x/(6*a*b^4*(a + b*x^2)^3) + (5*(x/(4*a*b^3*(a + b*x^2)^2) + (3*(x/(2* a*b^2*(a + b*x^2)) + ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(2*a^(3/2)*b^(5/2))))/(4* a*b)))/(6*a*b))
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[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/ c^p Int[(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n 2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && NeQ[p, 1]
Time = 0.05 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.99
method | result | size |
default | \(\frac {x}{6 a \left (b \,x^{2}+a \right )^{3}}+\frac {\frac {5 x}{24 a \left (b \,x^{2}+a \right )^{2}}+\frac {5 \left (\frac {3 x}{8 a \left (b \,x^{2}+a \right )}+\frac {3 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 a \sqrt {a b}}\right )}{6 a}}{a}\) | \(78\) |
risch | \(\frac {\frac {5 b^{2} x^{5}}{16 a^{3}}+\frac {5 b \,x^{3}}{6 a^{2}}+\frac {11 x}{16 a}}{\left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right ) \left (b \,x^{2}+a \right )}-\frac {5 \ln \left (b x +\sqrt {-a b}\right )}{32 \sqrt {-a b}\, a^{3}}+\frac {5 \ln \left (-b x +\sqrt {-a b}\right )}{32 \sqrt {-a b}\, a^{3}}\) | \(104\) |
Input:
int(1/(b^2*x^4+2*a*b*x^2+a^2)^2,x,method=_RETURNVERBOSE)
Output:
1/6*x/a/(b*x^2+a)^3+5/6/a*(1/4*x/a/(b*x^2+a)^2+3/4/a*(1/2*x/a/(b*x^2+a)+1/ 2/a/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))))
Time = 0.11 (sec) , antiderivative size = 254, normalized size of antiderivative = 3.22 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\left [\frac {30 \, a b^{3} x^{5} + 80 \, a^{2} b^{2} x^{3} + 66 \, a^{3} b x - 15 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{96 \, {\left (a^{4} b^{4} x^{6} + 3 \, a^{5} b^{3} x^{4} + 3 \, a^{6} b^{2} x^{2} + a^{7} b\right )}}, \frac {15 \, a b^{3} x^{5} + 40 \, a^{2} b^{2} x^{3} + 33 \, a^{3} b x + 15 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{48 \, {\left (a^{4} b^{4} x^{6} + 3 \, a^{5} b^{3} x^{4} + 3 \, a^{6} b^{2} x^{2} + a^{7} b\right )}}\right ] \] Input:
integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")
Output:
[1/96*(30*a*b^3*x^5 + 80*a^2*b^2*x^3 + 66*a^3*b*x - 15*(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^4*b^4*x^6 + 3*a^5*b^3*x^4 + 3*a^6*b^2*x^2 + a^7*b), 1/48*(15*a *b^3*x^5 + 40*a^2*b^2*x^3 + 33*a^3*b*x + 15*(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^4*b^4*x^6 + 3*a^5*b^3*x^ 4 + 3*a^6*b^2*x^2 + a^7*b)]
Time = 0.22 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.63 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=- \frac {5 \sqrt {- \frac {1}{a^{7} b}} \log {\left (- a^{4} \sqrt {- \frac {1}{a^{7} b}} + x \right )}}{32} + \frac {5 \sqrt {- \frac {1}{a^{7} b}} \log {\left (a^{4} \sqrt {- \frac {1}{a^{7} b}} + x \right )}}{32} + \frac {33 a^{2} x + 40 a b x^{3} + 15 b^{2} x^{5}}{48 a^{6} + 144 a^{5} b x^{2} + 144 a^{4} b^{2} x^{4} + 48 a^{3} b^{3} x^{6}} \] Input:
integrate(1/(b**2*x**4+2*a*b*x**2+a**2)**2,x)
Output:
-5*sqrt(-1/(a**7*b))*log(-a**4*sqrt(-1/(a**7*b)) + x)/32 + 5*sqrt(-1/(a**7 *b))*log(a**4*sqrt(-1/(a**7*b)) + x)/32 + (33*a**2*x + 40*a*b*x**3 + 15*b* *2*x**5)/(48*a**6 + 144*a**5*b*x**2 + 144*a**4*b**2*x**4 + 48*a**3*b**3*x* *6)
Time = 0.11 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {15 \, b^{2} x^{5} + 40 \, a b x^{3} + 33 \, a^{2} x}{48 \, {\left (a^{3} b^{3} x^{6} + 3 \, a^{4} b^{2} x^{4} + 3 \, a^{5} b x^{2} + a^{6}\right )}} + \frac {5 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{3}} \] Input:
integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")
Output:
1/48*(15*b^2*x^5 + 40*a*b*x^3 + 33*a^2*x)/(a^3*b^3*x^6 + 3*a^4*b^2*x^4 + 3 *a^5*b*x^2 + a^6) + 5/16*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3)
Time = 0.11 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.71 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {5 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} a^{3}} + \frac {15 \, b^{2} x^{5} + 40 \, a b x^{3} + 33 \, a^{2} x}{48 \, {\left (b x^{2} + a\right )}^{3} a^{3}} \] Input:
integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")
Output:
5/16*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3) + 1/48*(15*b^2*x^5 + 40*a*b*x^3 + 33*a^2*x)/((b*x^2 + a)^3*a^3)
Time = 21.88 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\frac {11\,x}{16\,a}+\frac {5\,b\,x^3}{6\,a^2}+\frac {5\,b^2\,x^5}{16\,a^3}}{a^3+3\,a^2\,b\,x^2+3\,a\,b^2\,x^4+b^3\,x^6}+\frac {5\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{16\,a^{7/2}\,\sqrt {b}} \] Input:
int(1/(a^2 + b^2*x^4 + 2*a*b*x^2)^2,x)
Output:
((11*x)/(16*a) + (5*b*x^3)/(6*a^2) + (5*b^2*x^5)/(16*a^3))/(a^3 + b^3*x^6 + 3*a^2*b*x^2 + 3*a*b^2*x^4) + (5*atan((b^(1/2)*x)/a^(1/2)))/(16*a^(7/2)*b ^(1/2))
Time = 0.16 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.05 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3}+45 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b \,x^{2}+45 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} x^{4}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} x^{6}+33 a^{3} b x +40 a^{2} b^{2} x^{3}+15 a \,b^{3} x^{5}}{48 a^{4} b \left (b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}\right )} \] Input:
int(1/(b^2*x^4+2*a*b*x^2+a^2)^2,x)
Output:
(15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3 + 45*sqrt(b)*sqrt(a )*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b*x**2 + 45*sqrt(b)*sqrt(a)*atan((b*x )/(sqrt(b)*sqrt(a)))*a*b**2*x**4 + 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)* sqrt(a)))*b**3*x**6 + 33*a**3*b*x + 40*a**2*b**2*x**3 + 15*a*b**3*x**5)/(4 8*a**4*b*(a**3 + 3*a**2*b*x**2 + 3*a*b**2*x**4 + b**3*x**6))