Integrand size = 26, antiderivative size = 211 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {5 x}{128 a b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {x^5}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 x^3}{48 b^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 x}{64 b^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {5 \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 a^{3/2} b^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \]
5/128*x/a/b^3/((b*x^2+a)^2)^(1/2)-1/8*x^5/b/(b*x^2+a)^3/((b*x^2+a)^2)^(1/2 )-5/48*x^3/b^2/(b*x^2+a)^2/((b*x^2+a)^2)^(1/2)-5/64*x/b^3/(b*x^2+a)/((b*x^ 2+a)^2)^(1/2)+5/128*(b*x^2+a)*arctan(x*b^(1/2)/a^(1/2))/a^(3/2)/b^(7/2)/(( b*x^2+a)^2)^(1/2)
Time = 1.05 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.50 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {\sqrt {a} \sqrt {b} x \left (-15 a^3-55 a^2 b x^2-73 a b^2 x^4+15 b^3 x^6\right )+15 \left (a+b x^2\right )^4 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{384 a^{3/2} b^{7/2} \left (a+b x^2\right )^3 \sqrt {\left (a+b x^2\right )^2}} \]
(Sqrt[a]*Sqrt[b]*x*(-15*a^3 - 55*a^2*b*x^2 - 73*a*b^2*x^4 + 15*b^3*x^6) + 15*(a + b*x^2)^4*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(384*a^(3/2)*b^(7/2)*(a + b* x^2)^3*Sqrt[(a + b*x^2)^2])
Time = 0.25 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.73, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {1384, 27, 252, 252, 252, 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.
| \(\displaystyle \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1384 |
| \(\displaystyle \frac {b^5 \left (a+b x^2\right ) \int \frac {x^6}{b^5 \left (b x^2+a\right )^5}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {x^6}{\left (b x^2+a\right )^5}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {5 \int \frac {x^4}{\left (b x^2+a\right )^4}dx}{8 b}-\frac {x^5}{8 b \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {5 \left (\frac {\int \frac {x^2}{\left (b x^2+a\right )^3}dx}{2 b}-\frac {x^3}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^5}{8 b \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {5 \left (\frac {\frac {\int \frac {1}{\left (b x^2+a\right )^2}dx}{4 b}-\frac {x}{4 b \left (a+b x^2\right )^2}}{2 b}-\frac {x^3}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^5}{8 b \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {5 \left (\frac {\frac {\frac {\int \frac {1}{b x^2+a}dx}{2 a}+\frac {x}{2 a \left (a+b x^2\right )}}{4 b}-\frac {x}{4 b \left (a+b x^2\right )^2}}{2 b}-\frac {x^3}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^5}{8 b \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {5 \left (\frac {\frac {\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {x}{2 a \left (a+b x^2\right )}}{4 b}-\frac {x}{4 b \left (a+b x^2\right )^2}}{2 b}-\frac {x^3}{6 b \left (a+b x^2\right )^3}\right )}{8 b}-\frac {x^5}{8 b \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
((a + b*x^2)*(-1/8*x^5/(b*(a + b*x^2)^4) + (5*(-1/6*x^3/(b*(a + b*x^2)^3) + (-1/4*x/(b*(a + b*x^2)^2) + (x/(2*a*(a + b*x^2)) + ArcTan[(Sqrt[b]*x)/Sq rt[a]]/(2*a^(3/2)*Sqrt[b]))/(4*b))/(2*b)))/(8*b)))/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]
3.7.53.3.1 Defintions of rubi rules used
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[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac Part[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 - 1/2] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)] && !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
Time = 1.91 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.71
| method | result | size |
| risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {5 x^{7}}{128 a}-\frac {73 x^{5}}{384 b}-\frac {55 a \,x^{3}}{384 b^{2}}-\frac {5 a^{2} x}{128 b^{3}}\right )}{\left (b \,x^{2}+a \right )^{5}}-\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (b x +\sqrt {-a b}\right )}{256 \left (b \,x^{2}+a \right ) \sqrt {-a b}\, b^{3} a}+\frac {5 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (-b x +\sqrt {-a b}\right )}{256 \left (b \,x^{2}+a \right ) \sqrt {-a b}\, b^{3} a}\) | \(149\) |
| default | \(-\frac {\left (-15 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{4} x^{8}-15 \sqrt {a b}\, b^{3} x^{7}-60 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a \,b^{3} x^{6}+73 \sqrt {a b}\, a \,b^{2} x^{5}-90 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2} b^{2} x^{4}+55 \sqrt {a b}\, a^{2} b \,x^{3}-60 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{3} b \,x^{2}+15 \sqrt {a b}\, a^{3} x -15 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{4}\right ) \left (b \,x^{2}+a \right )}{384 \sqrt {a b}\, b^{3} a {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\) | \(172\) |
((b*x^2+a)^2)^(1/2)/(b*x^2+a)^5*(5/128/a*x^7-73/384/b*x^5-55/384*a/b^2*x^3 -5/128*a^2/b^3*x)-5/256*((b*x^2+a)^2)^(1/2)/(b*x^2+a)/(-a*b)^(1/2)/b^3/a*l n(b*x+(-a*b)^(1/2))+5/256*((b*x^2+a)^2)^(1/2)/(b*x^2+a)/(-a*b)^(1/2)/b^3/a *ln(-b*x+(-a*b)^(1/2))
Time = 0.26 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.54 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\left [\frac {30 \, a b^{4} x^{7} - 146 \, a^{2} b^{3} x^{5} - 110 \, a^{3} b^{2} x^{3} - 30 \, a^{4} b x - 15 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{768 \, {\left (a^{2} b^{8} x^{8} + 4 \, a^{3} b^{7} x^{6} + 6 \, a^{4} b^{6} x^{4} + 4 \, a^{5} b^{5} x^{2} + a^{6} b^{4}\right )}}, \frac {15 \, a b^{4} x^{7} - 73 \, a^{2} b^{3} x^{5} - 55 \, a^{3} b^{2} x^{3} - 15 \, a^{4} b x + 15 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{384 \, {\left (a^{2} b^{8} x^{8} + 4 \, a^{3} b^{7} x^{6} + 6 \, a^{4} b^{6} x^{4} + 4 \, a^{5} b^{5} x^{2} + a^{6} b^{4}\right )}}\right ] \]
[1/768*(30*a*b^4*x^7 - 146*a^2*b^3*x^5 - 110*a^3*b^2*x^3 - 30*a^4*b*x - 15 *(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b^2*x^4 + 4*a^3*b*x^2 + a^4)*sqrt(-a*b)*lo g((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a^2*b^8*x^8 + 4*a^3*b^7*x^6 + 6*a^4*b^6*x^4 + 4*a^5*b^5*x^2 + a^6*b^4), 1/384*(15*a*b^4*x^7 - 73*a^2*b ^3*x^5 - 55*a^3*b^2*x^3 - 15*a^4*b*x + 15*(b^4*x^8 + 4*a*b^3*x^6 + 6*a^2*b ^2*x^4 + 4*a^3*b*x^2 + a^4)*sqrt(a*b)*arctan(sqrt(a*b)*x/a))/(a^2*b^8*x^8 + 4*a^3*b^7*x^6 + 6*a^4*b^6*x^4 + 4*a^5*b^5*x^2 + a^6*b^4)]
\[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {x^{6}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.52 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {15 \, b^{3} x^{7} - 73 \, a b^{2} x^{5} - 55 \, a^{2} b x^{3} - 15 \, a^{3} x}{384 \, {\left (a b^{7} x^{8} + 4 \, a^{2} b^{6} x^{6} + 6 \, a^{3} b^{5} x^{4} + 4 \, a^{4} b^{4} x^{2} + a^{5} b^{3}\right )}} + \frac {5 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} a b^{3}} \]
1/384*(15*b^3*x^7 - 73*a*b^2*x^5 - 55*a^2*b*x^3 - 15*a^3*x)/(a*b^7*x^8 + 4 *a^2*b^6*x^6 + 6*a^3*b^5*x^4 + 4*a^4*b^4*x^2 + a^5*b^3) + 5/128*arctan(b*x /sqrt(a*b))/(sqrt(a*b)*a*b^3)
Time = 0.28 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.44 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {5 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} a b^{3} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {15 \, b^{3} x^{7} - 73 \, a b^{2} x^{5} - 55 \, a^{2} b x^{3} - 15 \, a^{3} x}{384 \, {\left (b x^{2} + a\right )}^{4} a b^{3} \mathrm {sgn}\left (b x^{2} + a\right )} \]
5/128*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a*b^3*sgn(b*x^2 + a)) + 1/384*(15*b ^3*x^7 - 73*a*b^2*x^5 - 55*a^2*b*x^3 - 15*a^3*x)/((b*x^2 + a)^4*a*b^3*sgn( b*x^2 + a))
Timed out. \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {x^6}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}} \,d x \]