Integrand size = 26, antiderivative size = 164 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {9 a x}{8 b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a^2 x}{4 b^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x \left (a+b x^2\right )}{b^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {15 \sqrt {a} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:
9/8*a*x/b^3/((b*x^2+a)^2)^(1/2)-1/4*a^2*x/b^3/(b*x^2+a)/((b*x^2+a)^2)^(1/2 )+x*(b*x^2+a)/b^3/((b*x^2+a)^2)^(1/2)-15/8*a^(1/2)*(b*x^2+a)*arctan(b^(1/2 )*x/a^(1/2))/b^(7/2)/((b*x^2+a)^2)^(1/2)
Time = 1.04 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.54 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {\sqrt {b} x \left (15 a^2+25 a b x^2+8 b^2 x^4\right )-15 \sqrt {a} \left (a+b x^2\right )^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{7/2} \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \] Input:
Integrate[x^6/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]
Output:
(Sqrt[b]*x*(15*a^2 + 25*a*b*x^2 + 8*b^2*x^4) - 15*Sqrt[a]*(a + b*x^2)^2*Ar cTan[(Sqrt[b]*x)/Sqrt[a]])/(8*b^(7/2)*(a + b*x^2)*Sqrt[(a + b*x^2)^2])
Time = 0.39 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.70, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1384, 27, 252, 252, 262, 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.
\(\displaystyle \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^3 \left (a+b x^2\right ) \int \frac {x^6}{b^3 \left (b x^2+a\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {x^6}{\left (b x^2+a\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {5 \int \frac {x^4}{\left (b x^2+a\right )^2}dx}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {5 \left (\frac {3 \int \frac {x^2}{b x^2+a}dx}{2 b}-\frac {x^3}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {5 \left (\frac {3 \left (\frac {x}{b}-\frac {a \int \frac {1}{b x^2+a}dx}{b}\right )}{2 b}-\frac {x^3}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {5 \left (\frac {3 \left (\frac {x}{b}-\frac {\sqrt {a} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{3/2}}\right )}{2 b}-\frac {x^3}{2 b \left (a+b x^2\right )}\right )}{4 b}-\frac {x^5}{4 b \left (a+b x^2\right )^2}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
Input:
Int[x^6/(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2),x]
Output:
((a + b*x^2)*(-1/4*x^5/(b*(a + b*x^2)^2) + (5*(-1/2*x^3/(b*(a + b*x^2)) + (3*(x/b - (Sqrt[a]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/b^(3/2)))/(2*b)))/(4*b)))/ Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[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 = 2.10 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.71
method | result | size |
default | \(\frac {\left (8 \sqrt {a b}\, b^{2} x^{5}-15 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a \,b^{2} x^{4}+25 \sqrt {a b}\, a b \,x^{3}-30 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2} b \,x^{2}+15 \sqrt {a b}\, a^{2} x -15 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{3}\right ) \left (b \,x^{2}+a \right )}{8 \sqrt {a b}\, b^{3} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(116\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, x}{\left (b \,x^{2}+a \right ) b^{3}}+\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {9}{8} a b \,x^{3}+\frac {7}{8} a^{2} x \right )}{\left (b \,x^{2}+a \right )^{3} b^{3}}+\frac {15 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, \ln \left (-\sqrt {-a b}\, x -a \right )}{16 \left (b \,x^{2}+a \right ) b^{4}}-\frac {15 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, \ln \left (\sqrt {-a b}\, x -a \right )}{16 \left (b \,x^{2}+a \right ) b^{4}}\) | \(154\) |
Input:
int(x^6/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/8*(8*(a*b)^(1/2)*b^2*x^5-15*arctan(b/(a*b)^(1/2)*x)*a*b^2*x^4+25*(a*b)^( 1/2)*a*b*x^3-30*arctan(b/(a*b)^(1/2)*x)*a^2*b*x^2+15*(a*b)^(1/2)*a^2*x-15* arctan(b/(a*b)^(1/2)*x)*a^3)*(b*x^2+a)/(a*b)^(1/2)/b^3/((b*x^2+a)^2)^(3/2)
Time = 0.08 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.23 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\left [\frac {16 \, b^{2} x^{5} + 50 \, a b x^{3} + 30 \, a^{2} x + 15 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right )}{16 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}}, \frac {8 \, b^{2} x^{5} + 25 \, a b x^{3} + 15 \, a^{2} x - 15 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right )}{8 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}}\right ] \] Input:
integrate(x^6/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")
Output:
[1/16*(16*b^2*x^5 + 50*a*b*x^3 + 30*a^2*x + 15*(b^2*x^4 + 2*a*b*x^2 + a^2) *sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)))/(b^5*x^4 + 2* a*b^4*x^2 + a^2*b^3), 1/8*(8*b^2*x^5 + 25*a*b*x^3 + 15*a^2*x - 15*(b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a))/(b^5*x^4 + 2*a*b^4* x^2 + a^2*b^3)]
\[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {x^{6}}{\left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**6/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)
Output:
Integral(x**6/((a + b*x**2)**2)**(3/2), x)
Time = 0.12 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.41 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {9 \, a b x^{3} + 7 \, a^{2} x}{8 \, {\left (b^{5} x^{4} + 2 \, a b^{4} x^{2} + a^{2} b^{3}\right )}} - \frac {15 \, a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3}} + \frac {x}{b^{3}} \] Input:
integrate(x^6/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")
Output:
1/8*(9*a*b*x^3 + 7*a^2*x)/(b^5*x^4 + 2*a*b^4*x^2 + a^2*b^3) - 15/8*a*arcta n(b*x/sqrt(a*b))/(sqrt(a*b)*b^3) + x/b^3
Time = 0.14 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.51 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {15 \, a \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} b^{3} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {x}{b^{3} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {9 \, a b x^{3} + 7 \, a^{2} x}{8 \, {\left (b x^{2} + a\right )}^{2} b^{3} \mathrm {sgn}\left (b x^{2} + a\right )} \] Input:
integrate(x^6/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")
Output:
-15/8*a*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^3*sgn(b*x^2 + a)) + x/(b^3*sgn( b*x^2 + a)) + 1/8*(9*a*b*x^3 + 7*a^2*x)/((b*x^2 + a)^2*b^3*sgn(b*x^2 + a))
Timed out. \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {x^6}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \] Input:
int(x^6/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)
Output:
int(x^6/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)
Time = 0.16 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.72 \[ \int \frac {x^6}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2}-30 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,x^{2}-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} x^{4}+15 a^{2} b x +25 a \,b^{2} x^{3}+8 b^{3} x^{5}}{8 b^{4} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int(x^6/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)
Output:
( - 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2 - 30*sqrt(b)*sqr t(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b*x**2 - 15*sqrt(b)*sqrt(a)*atan((b*x )/(sqrt(b)*sqrt(a)))*b**2*x**4 + 15*a**2*b*x + 25*a*b**2*x**3 + 8*b**3*x** 5)/(8*b**4*(a**2 + 2*a*b*x**2 + b**2*x**4))