\(\int \frac {1}{x^2 (a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\) [607]

Optimal result

Integrand size = 26, antiderivative size = 245 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {187 b x}{128 a^5 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {b x}{8 a^2 \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {5 b x}{16 a^3 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {41 b x}{64 a^4 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {a+b x^2}{a^5 x \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {315 \sqrt {b} \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 a^{11/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:

-187/128*b*x/a^5/((b*x^2+a)^2)^(1/2)-1/8*b*x/a^2/(b*x^2+a)^3/((b*x^2+a)^2) 
^(1/2)-5/16*b*x/a^3/(b*x^2+a)^2/((b*x^2+a)^2)^(1/2)-41/64*b*x/a^4/(b*x^2+a 
)/((b*x^2+a)^2)^(1/2)-(b*x^2+a)/a^5/x/((b*x^2+a)^2)^(1/2)-315/128*b^(1/2)* 
(b*x^2+a)*arctan(b^(1/2)*x/a^(1/2))/a^(11/2)/((b*x^2+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {-\sqrt {a} \left (128 a^4+837 a^3 b x^2+1533 a^2 b^2 x^4+1155 a b^3 x^6+315 b^4 x^8\right )-315 \sqrt {b} x \left (a+b x^2\right )^4 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 a^{11/2} x \left (a+b x^2\right )^3 \sqrt {\left (a+b x^2\right )^2}} \] Input:

Integrate[1/(x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)),x]
 

Output:

(-(Sqrt[a]*(128*a^4 + 837*a^3*b*x^2 + 1533*a^2*b^2*x^4 + 1155*a*b^3*x^6 + 
315*b^4*x^8)) - 315*Sqrt[b]*x*(a + b*x^2)^4*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/( 
128*a^(11/2)*x*(a + b*x^2)^3*Sqrt[(a + b*x^2)^2])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.70, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {1384, 27, 253, 253, 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.

\(\displaystyle \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 1384

\(\displaystyle \frac {b^5 \left (a+b x^2\right ) \int \frac {1}{b^5 x^2 \left (b x^2+a\right )^5}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (a+b x^2\right ) \int \frac {1}{x^2 \left (b x^2+a\right )^5}dx}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 253

\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {9 \int \frac {1}{x^2 \left (b x^2+a\right )^4}dx}{8 a}+\frac {1}{8 a x \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 253

\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {9 \left (\frac {7 \int \frac {1}{x^2 \left (b x^2+a\right )^3}dx}{6 a}+\frac {1}{6 a x \left (a+b x^2\right )^3}\right )}{8 a}+\frac {1}{8 a x \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 253

\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {9 \left (\frac {7 \left (\frac {5 \int \frac {1}{x^2 \left (b x^2+a\right )^2}dx}{4 a}+\frac {1}{4 a x \left (a+b x^2\right )^2}\right )}{6 a}+\frac {1}{6 a x \left (a+b x^2\right )^3}\right )}{8 a}+\frac {1}{8 a x \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 253

\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \int \frac {1}{x^2 \left (b x^2+a\right )}dx}{2 a}+\frac {1}{2 a x \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x \left (a+b x^2\right )^2}\right )}{6 a}+\frac {1}{6 a x \left (a+b x^2\right )^3}\right )}{8 a}+\frac {1}{8 a x \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 264

\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (-\frac {b \int \frac {1}{b x^2+a}dx}{a}-\frac {1}{a x}\right )}{2 a}+\frac {1}{2 a x \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x \left (a+b x^2\right )^2}\right )}{6 a}+\frac {1}{6 a x \left (a+b x^2\right )^3}\right )}{8 a}+\frac {1}{8 a x \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\left (a+b x^2\right ) \left (\frac {9 \left (\frac {7 \left (\frac {5 \left (\frac {3 \left (-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a^{3/2}}-\frac {1}{a x}\right )}{2 a}+\frac {1}{2 a x \left (a+b x^2\right )}\right )}{4 a}+\frac {1}{4 a x \left (a+b x^2\right )^2}\right )}{6 a}+\frac {1}{6 a x \left (a+b x^2\right )^3}\right )}{8 a}+\frac {1}{8 a x \left (a+b x^2\right )^4}\right )}{\sqrt {a^2+2 a b x^2+b^2 x^4}}\)

Input:

Int[1/(x^2*(a^2 + 2*a*b*x^2 + b^2*x^4)^(5/2)),x]
 

Output:

((a + b*x^2)*(1/(8*a*x*(a + b*x^2)^4) + (9*(1/(6*a*x*(a + b*x^2)^3) + (7*( 
1/(4*a*x*(a + b*x^2)^2) + (5*(1/(2*a*x*(a + b*x^2)) + (3*(-(1/(a*x)) - (Sq 
rt[b]*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/a^(3/2)))/(2*a)))/(4*a)))/(6*a)))/(8*a) 
))/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

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]
 

rule 253

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]
 

rule 264

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]
 

rule 1384

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)])
 

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.66

method	result	size
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (-\frac {315 b^{4} x^{8}}{128 a^{5}}-\frac {1155 b^{3} x^{6}}{128 a^{4}}-\frac {1533 b^{2} x^{4}}{128 a^{3}}-\frac {837 b \,x^{2}}{128 a^{2}}-\frac {1}{a}\right )}{\left (b \,x^{2}+a \right )^{5} x}+\frac {315 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, \ln \left (-b x +\sqrt {-a b}\right )}{256 \left (b \,x^{2}+a \right ) a^{6}}-\frac {315 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \sqrt {-a b}\, \ln \left (-b x -\sqrt {-a b}\right )}{256 \left (b \,x^{2}+a \right ) a^{6}}\)	\(162\)
default	\(-\frac {\left (315 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{5} x^{9}+315 \sqrt {a b}\, b^{4} x^{8}+1260 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a \,b^{4} x^{7}+1155 \sqrt {a b}\, a \,b^{3} x^{6}+1890 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2} b^{3} x^{5}+1533 \sqrt {a b}\, a^{2} b^{2} x^{4}+1260 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{3} b^{2} x^{3}+837 \sqrt {a b}\, a^{3} b \,x^{2}+315 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{4} b x +128 \sqrt {a b}\, a^{4}\right ) \left (b \,x^{2}+a \right )}{128 x \sqrt {a b}\, a^{5} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(191\)

Input:

int(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

((b*x^2+a)^2)^(1/2)/(b*x^2+a)^5*(-315/128*b^4/a^5*x^8-1155/128*b^3/a^4*x^6 
-1533/128*b^2/a^3*x^4-837/128*b/a^2*x^2-1/a)/x+315/256*((b*x^2+a)^2)^(1/2) 
/(b*x^2+a)/a^6*(-a*b)^(1/2)*ln(-b*x+(-a*b)^(1/2))-315/256*((b*x^2+a)^2)^(1 
/2)/(b*x^2+a)/a^6*(-a*b)^(1/2)*ln(-b*x-(-a*b)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.36 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\left [-\frac {630 \, b^{4} x^{8} + 2310 \, a b^{3} x^{6} + 3066 \, a^{2} b^{2} x^{4} + 1674 \, a^{3} b x^{2} + 256 \, a^{4} - 315 \, {\left (b^{4} x^{9} + 4 \, a b^{3} x^{7} + 6 \, a^{2} b^{2} x^{5} + 4 \, a^{3} b x^{3} + a^{4} x\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b x^{2} - 2 \, a x \sqrt {-\frac {b}{a}} - a}{b x^{2} + a}\right )}{256 \, {\left (a^{5} b^{4} x^{9} + 4 \, a^{6} b^{3} x^{7} + 6 \, a^{7} b^{2} x^{5} + 4 \, a^{8} b x^{3} + a^{9} x\right )}}, -\frac {315 \, b^{4} x^{8} + 1155 \, a b^{3} x^{6} + 1533 \, a^{2} b^{2} x^{4} + 837 \, a^{3} b x^{2} + 128 \, a^{4} + 315 \, {\left (b^{4} x^{9} + 4 \, a b^{3} x^{7} + 6 \, a^{2} b^{2} x^{5} + 4 \, a^{3} b x^{3} + a^{4} x\right )} \sqrt {\frac {b}{a}} \arctan \left (x \sqrt {\frac {b}{a}}\right )}{128 \, {\left (a^{5} b^{4} x^{9} + 4 \, a^{6} b^{3} x^{7} + 6 \, a^{7} b^{2} x^{5} + 4 \, a^{8} b x^{3} + a^{9} x\right )}}\right ] \] Input:

integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="fricas")
 

Output:

[-1/256*(630*b^4*x^8 + 2310*a*b^3*x^6 + 3066*a^2*b^2*x^4 + 1674*a^3*b*x^2 
+ 256*a^4 - 315*(b^4*x^9 + 4*a*b^3*x^7 + 6*a^2*b^2*x^5 + 4*a^3*b*x^3 + a^4 
*x)*sqrt(-b/a)*log((b*x^2 - 2*a*x*sqrt(-b/a) - a)/(b*x^2 + a)))/(a^5*b^4*x 
^9 + 4*a^6*b^3*x^7 + 6*a^7*b^2*x^5 + 4*a^8*b*x^3 + a^9*x), -1/128*(315*b^4 
*x^8 + 1155*a*b^3*x^6 + 1533*a^2*b^2*x^4 + 837*a^3*b*x^2 + 128*a^4 + 315*( 
b^4*x^9 + 4*a*b^3*x^7 + 6*a^2*b^2*x^5 + 4*a^3*b*x^3 + a^4*x)*sqrt(b/a)*arc 
tan(x*sqrt(b/a)))/(a^5*b^4*x^9 + 4*a^6*b^3*x^7 + 6*a^7*b^2*x^5 + 4*a^8*b*x 
^3 + a^9*x)]
 

Sympy [F]

\[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {1}{x^{2} \left (\left (a + b x^{2}\right )^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(1/x**2/(b**2*x**4+2*a*b*x**2+a**2)**(5/2),x)
 

Output:

Integral(1/(x**2*((a + b*x**2)**2)**(5/2)), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {315 \, b^{4} x^{8} + 1155 \, a b^{3} x^{6} + 1533 \, a^{2} b^{2} x^{4} + 837 \, a^{3} b x^{2} + 128 \, a^{4}}{128 \, {\left (a^{5} b^{4} x^{9} + 4 \, a^{6} b^{3} x^{7} + 6 \, a^{7} b^{2} x^{5} + 4 \, a^{8} b x^{3} + a^{9} x\right )}} - \frac {315 \, b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} a^{5}} \] Input:

integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="maxima")
 

Output:

-1/128*(315*b^4*x^8 + 1155*a*b^3*x^6 + 1533*a^2*b^2*x^4 + 837*a^3*b*x^2 + 
128*a^4)/(a^5*b^4*x^9 + 4*a^6*b^3*x^7 + 6*a^7*b^2*x^5 + 4*a^8*b*x^3 + a^9* 
x) - 315/128*b*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^5)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=-\frac {315 \, b \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} a^{5} \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {1}{a^{5} x \mathrm {sgn}\left (b x^{2} + a\right )} - \frac {187 \, b^{4} x^{7} + 643 \, a b^{3} x^{5} + 765 \, a^{2} b^{2} x^{3} + 325 \, a^{3} b x}{128 \, {\left (b x^{2} + a\right )}^{4} a^{5} \mathrm {sgn}\left (b x^{2} + a\right )} \] Input:

integrate(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x, algorithm="giac")
 

Output:

-315/128*b*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^5*sgn(b*x^2 + a)) - 1/(a^5*x 
*sgn(b*x^2 + a)) - 1/128*(187*b^4*x^7 + 643*a*b^3*x^5 + 765*a^2*b^2*x^3 + 
325*a^3*b*x)/((b*x^2 + a)^4*a^5*sgn(b*x^2 + a))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\int \frac {1}{x^2\,{\left (a^2+2\,a\,b\,x^2+b^2\,x^4\right )}^{5/2}} \,d x \] Input:

int(1/(x^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)),x)
 

Output:

int(1/(x^2*(a^2 + b^2*x^4 + 2*a*b*x^2)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^2 \left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {-315 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{4} x -1260 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b \,x^{3}-1890 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} x^{5}-1260 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} x^{7}-315 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{4} x^{9}-128 a^{5}-837 a^{4} b \,x^{2}-1533 a^{3} b^{2} x^{4}-1155 a^{2} b^{3} x^{6}-315 a \,b^{4} x^{8}}{128 a^{6} x \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:

int(1/x^2/(b^2*x^4+2*a*b*x^2+a^2)^(5/2),x)
 

Output:

( - 315*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4*x - 1260*sqrt(b 
)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*x**3 - 1890*sqrt(b)*sqrt(a) 
*atan((b*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*x**5 - 1260*sqrt(b)*sqrt(a)*atan( 
(b*x)/(sqrt(b)*sqrt(a)))*a*b**3*x**7 - 315*sqrt(b)*sqrt(a)*atan((b*x)/(sqr 
t(b)*sqrt(a)))*b**4*x**9 - 128*a**5 - 837*a**4*b*x**2 - 1533*a**3*b**2*x** 
4 - 1155*a**2*b**3*x**6 - 315*a*b**4*x**8)/(128*a**6*x*(a**4 + 4*a**3*b*x* 
*2 + 6*a**2*b**2*x**4 + 4*a*b**3*x**6 + b**4*x**8))