Integrand size = 24, antiderivative size = 83 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {x^5}{6 b \left (a+b x^2\right )^3}-\frac {5 x^3}{24 b^2 \left (a+b x^2\right )^2}-\frac {5 x}{16 b^3 \left (a+b x^2\right )}+\frac {5 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 \sqrt {a} b^{7/2}} \] Output:
-1/6*x^5/b/(b*x^2+a)^3-5/24*x^3/b^2/(b*x^2+a)^2-5/16*x/b^3/(b*x^2+a)+5/16* arctan(b^(1/2)*x/a^(1/2))/a^(1/2)/b^(7/2)
Time = 0.03 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.80 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {x \left (15 a^2+40 a b x^2+33 b^2 x^4\right )}{48 b^3 \left (a+b x^2\right )^3}+\frac {5 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 \sqrt {a} b^{7/2}} \] Input:
Integrate[x^6/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]
Output:
-1/48*(x*(15*a^2 + 40*a*b*x^2 + 33*b^2*x^4))/(b^3*(a + b*x^2)^3) + (5*ArcT an[(Sqrt[b]*x)/Sqrt[a]])/(16*Sqrt[a]*b^(7/2))
Time = 0.36 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.19, 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, 252, 252, 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.
\(\displaystyle \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle b^4 \int \frac {x^6}{b^4 \left (b x^2+a\right )^4}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {x^6}{\left (a+b x^2\right )^4}dx\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {5 \int \frac {x^4}{\left (b x^2+a\right )^3}dx}{6 b}-\frac {x^5}{6 b \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {5 \left (\frac {3 \int \frac {x^2}{\left (b x^2+a\right )^2}dx}{4 b}-\frac {x^3}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^5}{6 b \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{b x^2+a}dx}{2 b}-\frac {x}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^3}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^5}{6 b \left (a+b x^2\right )^3}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 \sqrt {a} b^{3/2}}-\frac {x}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^3}{4 b \left (a+b x^2\right )^2}\right )}{6 b}-\frac {x^5}{6 b \left (a+b x^2\right )^3}\) |
Input:
Int[x^6/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]
Output:
-1/6*x^5/(b*(a + b*x^2)^3) + (5*(-1/4*x^3/(b*(a + b*x^2)^2) + (3*(-1/2*x/( b*(a + b*x^2)) + ArcTan[(Sqrt[b]*x)/Sqrt[a]]/(2*Sqrt[a]*b^(3/2))))/(4*b))) /(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)^(-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 = 0.70
method | result | size |
default | \(\frac {-\frac {11 x^{5}}{16 b}-\frac {5 a \,x^{3}}{6 b^{2}}-\frac {5 a^{2} x}{16 b^{3}}}{\left (b \,x^{2}+a \right )^{3}}+\frac {5 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 b^{3} \sqrt {a b}}\) | \(58\) |
risch | \(\frac {-\frac {11 x^{5}}{16 b}-\frac {5 a \,x^{3}}{6 b^{2}}-\frac {5 a^{2} x}{16 b^{3}}}{\left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}-\frac {5 \ln \left (b x +\sqrt {-a b}\right )}{32 \sqrt {-a b}\, b^{3}}+\frac {5 \ln \left (-b x +\sqrt {-a b}\right )}{32 \sqrt {-a b}\, b^{3}}\) | \(104\) |
Input:
int(x^6/(b^2*x^4+2*a*b*x^2+a^2)^2,x,method=_RETURNVERBOSE)
Output:
(-11/16/b*x^5-5/6*a/b^2*x^3-5/16*a^2/b^3*x)/(b*x^2+a)^3+5/16/b^3/(a*b)^(1/ 2)*arctan(b/(a*b)^(1/2)*x)
Time = 0.07 (sec) , antiderivative size = 254, normalized size of antiderivative = 3.06 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\left [-\frac {66 \, a b^{3} x^{5} + 80 \, a^{2} b^{2} x^{3} + 30 \, a^{3} b x + 15 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{96 \, {\left (a b^{7} x^{6} + 3 \, a^{2} b^{6} x^{4} + 3 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )}}, -\frac {33 \, a b^{3} x^{5} + 40 \, a^{2} b^{2} x^{3} + 15 \, a^{3} b x - 15 \, {\left (b^{3} x^{6} + 3 \, a b^{2} x^{4} + 3 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{48 \, {\left (a b^{7} x^{6} + 3 \, a^{2} b^{6} x^{4} + 3 \, a^{3} b^{5} x^{2} + a^{4} b^{4}\right )}}\right ] \] Input:
integrate(x^6/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")
Output:
[-1/96*(66*a*b^3*x^5 + 80*a^2*b^2*x^3 + 30*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*b^7*x^6 + 3*a^2*b^6*x^4 + 3*a^3*b^5*x^2 + a^4*b^4), -1/48*(33 *a*b^3*x^5 + 40*a^2*b^2*x^3 + 15*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*b^7*x^6 + 3*a^2*b^6*x^ 4 + 3*a^3*b^5*x^2 + a^4*b^4)]
Time = 0.21 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.61 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=- \frac {5 \sqrt {- \frac {1}{a b^{7}}} \log {\left (- a b^{3} \sqrt {- \frac {1}{a b^{7}}} + x \right )}}{32} + \frac {5 \sqrt {- \frac {1}{a b^{7}}} \log {\left (a b^{3} \sqrt {- \frac {1}{a b^{7}}} + x \right )}}{32} + \frac {- 15 a^{2} x - 40 a b x^{3} - 33 b^{2} x^{5}}{48 a^{3} b^{3} + 144 a^{2} b^{4} x^{2} + 144 a b^{5} x^{4} + 48 b^{6} x^{6}} \] Input:
integrate(x**6/(b**2*x**4+2*a*b*x**2+a**2)**2,x)
Output:
-5*sqrt(-1/(a*b**7))*log(-a*b**3*sqrt(-1/(a*b**7)) + x)/32 + 5*sqrt(-1/(a* b**7))*log(a*b**3*sqrt(-1/(a*b**7)) + x)/32 + (-15*a**2*x - 40*a*b*x**3 - 33*b**2*x**5)/(48*a**3*b**3 + 144*a**2*b**4*x**2 + 144*a*b**5*x**4 + 48*b* *6*x**6)
Time = 0.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.98 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {33 \, b^{2} x^{5} + 40 \, a b x^{3} + 15 \, a^{2} x}{48 \, {\left (b^{6} x^{6} + 3 \, a b^{5} x^{4} + 3 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )}} + \frac {5 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} b^{3}} \] Input:
integrate(x^6/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")
Output:
-1/48*(33*b^2*x^5 + 40*a*b*x^3 + 15*a^2*x)/(b^6*x^6 + 3*a*b^5*x^4 + 3*a^2* b^4*x^2 + a^3*b^3) + 5/16*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3)
Time = 0.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.67 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {5 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} b^{3}} - \frac {33 \, b^{2} x^{5} + 40 \, a b x^{3} + 15 \, a^{2} x}{48 \, {\left (b x^{2} + a\right )}^{3} b^{3}} \] Input:
integrate(x^6/(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)*b^3) - 1/48*(33*b^2*x^5 + 40*a*b*x^3 + 15*a^2*x)/((b*x^2 + a)^3*b^3)
Time = 17.20 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.94 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {5\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{16\,\sqrt {a}\,b^{7/2}}-\frac {\frac {11\,x^5}{16\,b}+\frac {5\,a\,x^3}{6\,b^2}+\frac {5\,a^2\,x}{16\,b^3}}{a^3+3\,a^2\,b\,x^2+3\,a\,b^2\,x^4+b^3\,x^6} \] Input:
int(x^6/(a^2 + b^2*x^4 + 2*a*b*x^2)^2,x)
Output:
(5*atan((b^(1/2)*x)/a^(1/2)))/(16*a^(1/2)*b^(7/2)) - ((11*x^5)/(16*b) + (5 *a*x^3)/(6*b^2) + (5*a^2*x)/(16*b^3))/(a^3 + b^3*x^6 + 3*a^2*b*x^2 + 3*a*b ^2*x^4)
Time = 0.16 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.95 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3}+45 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b \,x^{2}+45 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{2} x^{4}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{3} x^{6}-15 a^{3} b x -40 a^{2} b^{2} x^{3}-33 a \,b^{3} x^{5}}{48 a \,b^{4} \left (b^{3} x^{6}+3 a \,b^{2} x^{4}+3 a^{2} b \,x^{2}+a^{3}\right )} \] Input:
int(x^6/(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 - 15*a**3*b*x - 40*a**2*b**2*x**3 - 33*a*b**3*x**5)/(4 8*a*b**4*(a**3 + 3*a**2*b*x**2 + 3*a*b**2*x**4 + b**3*x**6))