Integrand size = 26, antiderivative size = 165 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {7 b x}{8 a^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b x}{4 a^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a+b x^2}{a^3 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {15 \sqrt {b} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:
-7/8*b*x/a^3/((b*x^2+a)^2)^(1/2)-1/4*b*x/a^2/(b*x^2+a)/((b*x^2+a)^2)^(1/2) -(b*x^2+a)/a^3/x/((b*x^2+a)^2)^(1/2)-15/8*b^(1/2)*(b*x^2+a)*arctan(b^(1/2) *x/a^(1/2))/a^(7/2)/((b*x^2+a)^2)^(1/2)
Time = 1.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.56 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {-\sqrt {a} \left (8 a^2+25 a b x^2+15 b^2 x^4\right )-15 \sqrt {b} x \left (a+b x^2\right )^2 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 a^{7/2} x \left (a+b x^2\right ) \sqrt {\left (a+b x^2\right )^2}} \] Input:
Integrate[1/(x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]
Output:
(-(Sqrt[a]*(8*a^2 + 25*a*b*x^2 + 15*b^2*x^4)) - 15*Sqrt[b]*x*(a + b*x^2)^2 *ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(8*a^(7/2)*x*(a + b*x^2)*Sqrt[(a + b*x^2)^2] )
Time = 0.38 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.72, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {1384, 27, 253, 253, 264, 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.
\(\displaystyle \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {b^3 \left (a+b x^2\right ) \int \frac {1}{b^3 x^2 \left (b x^2+a\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {1}{x^2 \left (b x^2+a\right )^3}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {5 \int \frac {1}{x^2 \left (b x^2+a\right )^2}dx}{4 a}+\frac {1}{4 a x \left (a+b x^2\right )^2}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 253 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {5 \left (\frac {3 \int \frac {1}{x^2 \left (b x^2+a\right )}dx}{2 a}+\frac {1}{2 a x \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x \left (a+b x^2\right )^2}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {5 \left (\frac {3 \left (-\frac {b \int \frac {1}{b x^2+a}dx}{a}-\frac {1}{a x}\right )}{2 a}+\frac {1}{2 a x \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x \left (a+b x^2\right )^2}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {5 \left (\frac {3 \left (-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}-\frac {1}{a x}\right )}{2 a}+\frac {1}{2 a x \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x \left (a+b x^2\right )^2}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\) |
Input:
Int[1/(x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)),x]
Output:
((a + b*x^2)*(1/(4*a*x*(a + b*x^2)^2) + (5*(1/(2*a*x*(a + b*x^2)) + (3*(-( 1/(a*x)) - (Sqrt[b]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/a^(3/2)))/(2*a)))/(4*a))) /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*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && 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)])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.01 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.67
method | result | size |
risch | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {15 b^{2} x^{4}}{8 a^{3}}-\frac {25 b \,x^{2}}{8 a^{2}}-\frac {1}{a}\right )}{\left (b \,x^{2}+a \right )^{3} x}+\frac {15 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a^{7} \textit {\_Z}^{2}+b \right )}{\sum }\textit {\_R} \ln \left (\left (3 \textit {\_R}^{2} a^{7}+2 b \right ) x +a^{4} \textit {\_R} \right )\right )}{16 \left (b \,x^{2}+a \right )}\) | \(110\) |
default | \(-\frac {\left (15 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{3} x^{5}+15 \sqrt {a b}\, b^{2} x^{4}+30 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a \,b^{2} x^{3}+25 \sqrt {a b}\, a b \,x^{2}+15 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2} b x +8 \sqrt {a b}\, a^{2}\right ) \left (b \,x^{2}+a \right )}{8 x \sqrt {a b}\, a^{3} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {3}{2}}}\) | \(119\) |
Input:
int(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
((b*x^2+a)^2)^(1/2)/(b*x^2+a)^3*(-15/8*b^2/a^3*x^4-25/8*b/a^2*x^2-1/a)/x+1 5/16*((b*x^2+a)^2)^(1/2)/(b*x^2+a)*sum(_R*ln((3*_R^2*a^7+2*b)*x+a^4*_R),_R =RootOf(_Z^2*a^7+b))
Time = 0.10 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\left [-\frac {30 \, b^{2} x^{4} + 50 \, a b x^{2} - 15 \, {\left (b^{2} x^{5} + 2 \, a b x^{3} + a^{2} x\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right ) + 16 \, a^{2}}{16 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}}, -\frac {15 \, b^{2} x^{4} + 25 \, a b x^{2} + 15 \, {\left (b^{2} x^{5} + 2 \, a b x^{3} + a^{2} x\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right ) + 8 \, a^{2}}{8 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}}\right ] \] Input:
integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="fricas")
Output:
[-1/16*(30*b^2*x^4 + 50*a*b*x^2 - 15*(b^2*x^5 + 2*a*b*x^3 + a^2*x)*sqrt(-b /a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)) + 16*a^2)/(a^3*b^2*x^5 + 2*a^4*b*x^3 + a^5*x), -1/8*(15*b^2*x^4 + 25*a*b*x^2 + 15*(b^2*x^5 + 2*a *b*x^3 + a^2*x)*sqrt(b/a)*arctan(x*sqrt(b/a)) + 8*a^2)/(a^3*b^2*x^5 + 2*a^ 4*b*x^3 + a^5*x)]
\[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {1}{x^{2} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/x**2/(b**2*x**4+2*a*b*x**2+a**2)**(3/2),x)
Output:
Integral(1/(x**2*((a + b*x**2)**2)**(3/2)), x)
Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.43 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {15 \, b^{2} x^{4} + 25 \, a b x^{2} + 8 \, a^{2}}{8 \, {\left (a^{3} b^{2} x^{5} + 2 \, a^{4} b x^{3} + a^{5} x\right )}} - \frac {15 \, b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3}} \] Input:
integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="maxima")
Output:
-1/8*(15*b^2*x^4 + 25*a*b*x^2 + 8*a^2)/(a^3*b^2*x^5 + 2*a^4*b*x^3 + a^5*x) - 15/8*b*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3)
Time = 0.13 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.53 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=-\frac {15 \, b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {7 \, b^{2} x^{3} + 9 \, a b x}{8 \, {\left (b x^{2} + a\right )}^{2} a^{3} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {1}{a^{3} x \mathrm {sgn}\left (b x^{2} + a\right )} \] Input:
integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(3/2),x, algorithm="giac")
Output:
-15/8*b*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3*sgn(b*x^2 + a)) - 1/8*(7*b^2* x^3 + 9*a*b*x)/((b*x^2 + a)^2*a^3*sgn(b*x^2 + a)) - 1/(a^3*x*sgn(b*x^2 + a ))
Timed out. \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\int \frac {1}{x^2\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{3/2}} \,d x \] Input:
int(1/(x^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)),x)
Output:
int(1/(x^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(3/2)), x)
Time = 0.17 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.73 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}} \, dx=\frac {-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} x -30 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a b \,x^{3}-15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{2} x^{5}-8 a^{3}-25 a^{2} b \,x^{2}-15 a \,b^{2} x^{4}}{8 a^{4} x \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:
int(1/x^2/(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*x - 30*sqrt(b)*s qrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a*b*x**3 - 15*sqrt(b)*sqrt(a)*atan((b *x)/(sqrt(b)*sqrt(a)))*b**2*x**5 - 8*a**3 - 25*a**2*b*x**2 - 15*a*b**2*x** 4)/(8*a**4*x*(a**2 + 2*a*b*x**2 + b**2*x**4))