\(\int \frac {1}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\) [606]

Optimal result

Integrand size = 22, antiderivative size = 204 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {35 x}{128 a^4 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x}{8 a \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {7 x}{48 a^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 x}{192 a^3 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {35 \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 a^{9/2} \sqrt {b} \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:

35/128*x/a^4/((b*x^2+a)^2)^(1/2)+1/8*x/a/(b*x^2+a)^3/((b*x^2+a)^2)^(1/2)+7 
/48*x/a^2/(b*x^2+a)^2/((b*x^2+a)^2)^(1/2)+35/192*x/a^3/(b*x^2+a)/((b*x^2+a 
)^2)^(1/2)+35/128*(b*x^2+a)*arctan(b^(1/2)*x/a^(1/2))/a^(9/2)/b^(1/2)/((b* 
x^2+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.51 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {\sqrt {a} \sqrt {b} x \left (279 a^3+511 a^2 b x^2+385 a b^2 x^4+105 b^3 x^6\right )+105 \left (a+b x^2\right )^4 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{384 a^{9/2} \sqrt {b} \left (a+b x^2\right )^3 \sqrt {\left (a+b x^2\right )^2}} \] Input:

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

Output:

(Sqrt[a]*Sqrt[b]*x*(279*a^3 + 511*a^2*b*x^2 + 385*a*b^2*x^4 + 105*b^3*x^6) 
 + 105*(a + b*x^2)^4*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(384*a^(9/2)*Sqrt[b]*(a 
+ b*x^2)^3*Sqrt[(a + b*x^2)^2])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.85, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1384, 215, 215, 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.

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

\(\Big \downarrow \) 1384

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

\(\Big \downarrow \) 215

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

\(\Big \downarrow \) 215

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

\(\Big \downarrow \) 215

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

\(\Big \downarrow \) 215

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 215

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])
 

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 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 = 0.68 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.72

method	result	size
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {35 b^{3} x^{7}}{128 a^{4}}+\frac {385 b^{2} x^{5}}{384 a^{3}}+\frac {511 b \,x^{3}}{384 a^{2}}+\frac {93 x}{128 a}\right )}{\left (b \,x^{2}+a \right )^{5}}-\frac {35 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (b x +\sqrt {-a b}\right )}{256 \left (b \,x^{2}+a \right ) \sqrt {-a b}\, a^{4}}+\frac {35 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (-b x +\sqrt {-a b}\right )}{256 \left (b \,x^{2}+a \right ) \sqrt {-a b}\, a^{4}}\)	\(146\)
default	\(\frac {\left (105 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{4} x^{8}+105 \sqrt {a b}\, b^{3} x^{7}+420 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a \,b^{3} x^{6}+385 \sqrt {a b}\, a \,b^{2} x^{5}+630 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2} b^{2} x^{4}+511 \sqrt {a b}\, a^{2} b \,x^{3}+420 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{3} b \,x^{2}+279 \sqrt {a b}\, a^{3} x +105 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{4}\right ) \left (b \,x^{2}+a \right )}{384 \sqrt {a b}\, a^{4} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(169\)

Input:

int(1/(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*(35/128*b^3/a^4*x^7+385/384*b^2/a^3*x^5+51 
1/384*b/a^2*x^3+93/128*x/a)-35/256*((b*x^2+a)^2)^(1/2)/(b*x^2+a)/(-a*b)^(1 
/2)/a^4*ln(b*x+(-a*b)^(1/2))+35/256*((b*x^2+a)^2)^(1/2)/(b*x^2+a)/(-a*b)^( 
1/2)/a^4*ln(-b*x+(-a*b)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.57 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\left [\frac {210 \, a b^{4} x^{7} + 770 \, a^{2} b^{3} x^{5} + 1022 \, a^{3} b^{2} x^{3} + 558 \, a^{4} b x - 105 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{768 \, {\left (a^{5} b^{5} x^{8} + 4 \, a^{6} b^{4} x^{6} + 6 \, a^{7} b^{3} x^{4} + 4 \, a^{8} b^{2} x^{2} + a^{9} b\right )}}, \frac {105 \, a b^{4} x^{7} + 385 \, a^{2} b^{3} x^{5} + 511 \, a^{3} b^{2} x^{3} + 279 \, a^{4} b x + 105 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{384 \, {\left (a^{5} b^{5} x^{8} + 4 \, a^{6} b^{4} x^{6} + 6 \, a^{7} b^{3} x^{4} + 4 \, a^{8} b^{2} x^{2} + a^{9} b\right )}}\right ] \] Input:

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

Output:

[1/768*(210*a*b^4*x^7 + 770*a^2*b^3*x^5 + 1022*a^3*b^2*x^3 + 558*a^4*b*x - 
 105*(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 
)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)))/(a^5*b^5*x^8 + 4*a^6*b^4* 
x^6 + 6*a^7*b^3*x^4 + 4*a^8*b^2*x^2 + a^9*b), 1/384*(105*a*b^4*x^7 + 385*a 
^2*b^3*x^5 + 511*a^3*b^2*x^3 + 279*a^4*b*x + 105*(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^5*b 
^5*x^8 + 4*a^6*b^4*x^6 + 6*a^7*b^3*x^4 + 4*a^8*b^2*x^2 + a^9*b)]
 

Sympy [F]

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

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

Output:

Integral((a**2 + 2*a*b*x**2 + b**2*x**4)**(-5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.50 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {105 \, b^{3} x^{7} + 385 \, a b^{2} x^{5} + 511 \, a^{2} b x^{3} + 279 \, a^{3} x}{384 \, {\left (a^{4} b^{4} x^{8} + 4 \, a^{5} b^{3} x^{6} + 6 \, a^{6} b^{2} x^{4} + 4 \, a^{7} b x^{2} + a^{8}\right )}} + \frac {35 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} a^{4}} \] Input:

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

Output:

1/384*(105*b^3*x^7 + 385*a*b^2*x^5 + 511*a^2*b*x^3 + 279*a^3*x)/(a^4*b^4*x 
^8 + 4*a^5*b^3*x^6 + 6*a^6*b^2*x^4 + 4*a^7*b*x^2 + a^8) + 35/128*arctan(b* 
x/sqrt(a*b))/(sqrt(a*b)*a^4)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.43 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {35 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} a^{4} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {105 \, b^{3} x^{7} + 385 \, a b^{2} x^{5} + 511 \, a^{2} b x^{3} + 279 \, a^{3} x}{384 \, {\left (b x^{2} + a\right )}^{4} a^{4} \mathrm {sgn}\left (b x^{2} + a\right )} \] Input:

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

Output:

35/128*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^4*sgn(b*x^2 + a)) + 1/384*(105*b 
^3*x^7 + 385*a*b^2*x^5 + 511*a^2*b*x^3 + 279*a^3*x)/((b*x^2 + a)^4*a^4*sgn 
(b*x^2 + a))
 

Mupad [F(-1)]

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

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

Output:

int(1/(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 = 211, normalized size of antiderivative = 1.03 \[ \int \frac {1}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{4}+420 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b \,x^{2}+630 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} x^{4}+420 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} x^{6}+105 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{4} x^{8}+279 a^{4} b x +511 a^{3} b^{2} x^{3}+385 a^{2} b^{3} x^{5}+105 a \,b^{4} x^{7}}{384 a^{5} b \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:

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

Output:

(105*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4 + 420*sqrt(b)*sqrt 
(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*x**2 + 630*sqrt(b)*sqrt(a)*atan(( 
b*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*x**4 + 420*sqrt(b)*sqrt(a)*atan((b*x)/(s 
qrt(b)*sqrt(a)))*a*b**3*x**6 + 105*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqr 
t(a)))*b**4*x**8 + 279*a**4*b*x + 511*a**3*b**2*x**3 + 385*a**2*b**3*x**5 
+ 105*a*b**4*x**7)/(384*a**5*b*(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))