Integrand size = 22, antiderivative size = 126 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {3 x}{8 a^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x}{4 a \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {3 \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:
3/8*x/a^2/((b*x^2+a)^2)^(1/2)+1/4*x/a/(b*x^2+a)/((b*x^2+a)^2)^(1/2)+3/8*(b *x^2+a)*arctan(b^(1/2)*x/a^(1/2))/a^(5/2)/b^(1/2)/((b*x^2+a)^2)^(1/2)
Time = 1.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.66 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {\sqrt {a} \sqrt {b} x \left (5 a+3 b x^2\right )+3 \left (a+b x^2\right )^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{5/2} \sqrt {b} \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \] Input:
Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(-3/2),x]
Output:
(Sqrt[a]*Sqrt[b]*x*(5*a + 3*b*x^2) + 3*(a + b*x^2)^2*ArcTan[(Sqrt[b]*x)/Sq rt[a]])/(8*a^(5/2)*Sqrt[b]*(a + b*x^2)*Sqrt[(a + b*x^2)^2])
Time = 0.36 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {1384, 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.
\(\displaystyle \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^3 \left (a+b x^2\right ) \int \frac {1}{\left (b^2 x^2+a b\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {b^3 \left (a+b x^2\right ) \left (\frac {3 \int \frac {1}{\left (b^2 x^2+a b\right )^2}dx}{4 a b}+\frac {x}{4 a b^3 \left (a+b x^2\right )^2}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {b^3 \left (a+b x^2\right ) \left (\frac {3 \left (\frac {\int \frac {1}{b^2 x^2+a b}dx}{2 a b}+\frac {x}{2 a b^2 \left (a+b x^2\right )}\right )}{4 a b}+\frac {x}{4 a b^3 \left (a+b x^2\right )^2}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {b^3 \left (a+b x^2\right ) \left (\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} b^{5/2}}+\frac {x}{2 a b^2 \left (a+b x^2\right )}\right )}{4 a b}+\frac {x}{4 a b^3 \left (a+b x^2\right )^2}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
Input:
Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(-3/2),x]
Output:
(b^3*(a + b*x^2)*(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)))/Sqrt[a^2 + 2 *a*b*x^2 + b^2*x^4]
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[(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 = 0.67 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.77
method | result | size |
default | \(\frac {\left (3 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{2} x^{4}+3 \sqrt {a b}\, b \,x^{3}+6 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a b \,x^{2}+5 \sqrt {a b}\, a x +3 a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )\right ) \left (b \,x^{2}+a \right )}{8 \sqrt {a b}\, a^{2} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(97\) |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {3 b \,x^{3}}{8 a^{2}}+\frac {5 x}{8 a}\right )}{\left (b \,x^{2}+a \right )^{3}}-\frac {3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (b x +\sqrt {-a b}\right )}{16 \left (b \,x^{2}+a \right ) \sqrt {-a b}\, a^{2}}+\frac {3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (-b x +\sqrt {-a b}\right )}{16 \left (b \,x^{2}+a \right ) \sqrt {-a b}\, a^{2}}\) | \(124\) |
Input:
int(1/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
1/8*(3*arctan(b/(a*b)^(1/2)*x)*b^2*x^4+3*(a*b)^(1/2)*b*x^3+6*arctan(b/(a*b )^(1/2)*x)*a*b*x^2+5*(a*b)^(1/2)*a*x+3*a^2*arctan(b/(a*b)^(1/2)*x))*(b*x^2 +a)/(a*b)^(1/2)/a^2/((b*x^2+a)^2)^(3/2)
Time = 0.07 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.49 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\left [\frac {6 \, a b^{2} x^{3} + 10 \, a^{2} b x - 3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{16 \, {\left (a^{3} b^{3} x^{4} + 2 \, a^{4} b^{2} x^{2} + a^{5} b\right )}}, \frac {3 \, a b^{2} x^{3} + 5 \, a^{2} b x + 3 \, {\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{8 \, {\left (a^{3} b^{3} x^{4} + 2 \, a^{4} b^{2} x^{2} + a^{5} b\right )}}\right ] \] Input:
integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")
Output:
[1/16*(6*a*b^2*x^3 + 10*a^2*b*x - 3*(b^2*x^4 + 2*a*b*x^2 + a^2)*sqrt(-a*b) *log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a^3*b^3*x^4 + 2*a^4*b^2*x ^2 + a^5*b), 1/8*(3*a*b^2*x^3 + 5*a^2*b*x + 3*(b^2*x^4 + 2*a*b*x^2 + a^2)* sqrt(a*b)*arctan(sqrt(a*b)*x/a))/(a^3*b^3*x^4 + 2*a^4*b^2*x^2 + a^5*b)]
\[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {1}{\left (a^{2} + 2 a b x^{2} + b^{2} x^{4}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)
Output:
Integral((a**2 + 2*a*b*x**2 + b**2*x**4)**(-3/2), x)
Time = 0.11 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.46 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {3 \, b x^{3} + 5 \, a x}{8 \, {\left (a^{2} b^{2} x^{4} + 2 \, a^{3} b x^{2} + a^{4}\right )}} + \frac {3 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2}} \] Input:
integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")
Output:
1/8*(3*b*x^3 + 5*a*x)/(a^2*b^2*x^4 + 2*a^3*b*x^2 + a^4) + 3/8*arctan(b*x/s qrt(a*b))/(sqrt(a*b)*a^2)
Time = 0.13 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.52 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {3 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {3 \, b x^{3} + 5 \, a x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{2} \mathrm {sgn}\left (b x^{2} + a\right )} \] Input:
integrate(1/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")
Output:
3/8*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2*sgn(b*x^2 + a)) + 1/8*(3*b*x^3 + 5*a*x)/((b*x^2 + a)^2*a^2*sgn(b*x^2 + a))
Timed out. \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {1}{{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \] Input:
int(1/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2),x)
Output:
int(1/(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2), x)
Time = 0.17 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.90 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2}+6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,x^{2}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} x^{4}+5 a^{2} b x +3 a \,b^{2} x^{3}}{8 a^{3} b \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int(1/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x)
Output:
(3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2 + 6*sqrt(b)*sqrt(a)* atan((b*x)/(sqrt(b)*sqrt(a)))*a*b*x**2 + 3*sqrt(b)*sqrt(a)*atan((b*x)/(sqr t(b)*sqrt(a)))*b**2*x**4 + 5*a**2*b*x + 3*a*b**2*x**3)/(8*a**3*b*(a**2 + 2 *a*b*x**2 + b**2*x**4))