\(\int \frac {x^4}{(a^2+2 a b x^2+b^2 x^4)^{5/2}} \, dx\) [604]

Optimal result

Integrand size = 26, antiderivative size = 212 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {3 x}{128 a^2 b^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {x^3}{8 b \left (a+b x^2\right )^3 \sqrt {a^2+2 a b x^2+b^2 x^4}}-\frac {x}{16 b^2 \left (a+b x^2\right )^2 \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {x}{64 a b^2 \left (a+b x^2\right ) \sqrt {a^2+2 a b x^2+b^2 x^4}}+\frac {3 \left (a+b x^2\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 a^{5/2} b^{5/2} \sqrt {a^2+2 a b x^2+b^2 x^4}} \] Output:

3/128*x/a^2/b^2/((b*x^2+a)^2)^(1/2)-1/8*x^3/b/(b*x^2+a)^3/((b*x^2+a)^2)^(1 
/2)-1/16*x/b^2/(b*x^2+a)^2/((b*x^2+a)^2)^(1/2)+1/64*x/a/b^2/(b*x^2+a)/((b* 
x^2+a)^2)^(1/2)+3/128*(b*x^2+a)*arctan(b^(1/2)*x/a^(1/2))/a^(5/2)/b^(5/2)/ 
((b*x^2+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.50 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {\sqrt {a} \sqrt {b} x \left (-3 a^3-11 a^2 b x^2+11 a b^2 x^4+3 b^3 x^6\right )+3 \left (a+b x^2\right )^4 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{128 a^{5/2} b^{5/2} \left (a+b x^2\right )^3 \sqrt {\left (a+b x^2\right )^2}} \] Input:

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

Output:

(Sqrt[a]*Sqrt[b]*x*(-3*a^3 - 11*a^2*b*x^2 + 11*a*b^2*x^4 + 3*b^3*x^6) + 3* 
(a + b*x^2)^4*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(128*a^(5/2)*b^(5/2)*(a + b*x^2 
)^3*Sqrt[(a + b*x^2)^2])
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.72, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {1384, 27, 252, 252, 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.

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

\(\Big \downarrow \) 1384

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 215

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

\(\Big \downarrow \) 215

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

((a + b*x^2)*(-1/8*x^3/(b*(a + b*x^2)^4) + (3*(-1/6*x/(b*(a + b*x^2)^3) + 
(x/(4*a*(a + b*x^2)^2) + (3*(x/(2*a*(a + b*x^2)) + ArcTan[(Sqrt[b]*x)/Sqrt 
[a]]/(2*a^(3/2)*Sqrt[b])))/(4*a))/(6*b)))/(8*b)))/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 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 252

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]
 

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 = 1.50 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.69

method	result	size
risch	\(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (\frac {3 b \,x^{7}}{128 a^{2}}+\frac {11 x^{5}}{128 a}-\frac {11 x^{3}}{128 b}-\frac {3 a x}{128 b^{2}}\right )}{\left (b \,x^{2}+a \right )^{5}}-\frac {3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (b x +\sqrt {-a b}\right )}{256 \left (b \,x^{2}+a \right ) \sqrt {-a b}\, b^{2} a^{2}}+\frac {3 \sqrt {\left (b \,x^{2}+a \right )^{2}}\, \ln \left (-b x +\sqrt {-a b}\right )}{256 \left (b \,x^{2}+a \right ) \sqrt {-a b}\, b^{2} a^{2}}\)	\(147\)
default	\(-\frac {\left (-3 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) b^{4} x^{8}-3 \sqrt {a b}\, b^{3} x^{7}-12 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a \,b^{3} x^{6}-11 \sqrt {a b}\, a \,b^{2} x^{5}-18 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{2} b^{2} x^{4}+11 \sqrt {a b}\, a^{2} b \,x^{3}-12 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{3} b \,x^{2}+3 \sqrt {a b}\, a^{3} x -3 \arctan \left (\frac {b x}{\sqrt {a b}}\right ) a^{4}\right ) \left (b \,x^{2}+a \right )}{128 \sqrt {a b}\, b^{2} a^{2} {\left (\left (b \,x^{2}+a \right )^{2}\right )}^{\frac {5}{2}}}\)	\(172\)

Input:

int(x^4/(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*(3/128*b/a^2*x^7+11/128/a*x^5-11/128/b*x^3 
-3/128*a/b^2*x)-3/256*((b*x^2+a)^2)^(1/2)/(b*x^2+a)/(-a*b)^(1/2)/b^2/a^2*l 
n(b*x+(-a*b)^(1/2))+3/256*((b*x^2+a)^2)^(1/2)/(b*x^2+a)/(-a*b)^(1/2)/b^2/a 
^2*ln(-b*x+(-a*b)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.53 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\left [\frac {6 \, a b^{4} x^{7} + 22 \, a^{2} b^{3} x^{5} - 22 \, a^{3} b^{2} x^{3} - 6 \, a^{4} b x - 3 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{256 \, {\left (a^{3} b^{7} x^{8} + 4 \, a^{4} b^{6} x^{6} + 6 \, a^{5} b^{5} x^{4} + 4 \, a^{6} b^{4} x^{2} + a^{7} b^{3}\right )}}, \frac {3 \, a b^{4} x^{7} + 11 \, a^{2} b^{3} x^{5} - 11 \, a^{3} b^{2} x^{3} - 3 \, a^{4} b x + 3 \, {\left (b^{4} x^{8} + 4 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 4 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{128 \, {\left (a^{3} b^{7} x^{8} + 4 \, a^{4} b^{6} x^{6} + 6 \, a^{5} b^{5} x^{4} + 4 \, a^{6} b^{4} x^{2} + a^{7} b^{3}\right )}}\right ] \] Input:

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

Output:

[1/256*(6*a*b^4*x^7 + 22*a^2*b^3*x^5 - 22*a^3*b^2*x^3 - 6*a^4*b*x - 3*(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^3*b^7*x^8 + 4*a^4*b^6*x^6 + 6*a 
^5*b^5*x^4 + 4*a^6*b^4*x^2 + a^7*b^3), 1/128*(3*a*b^4*x^7 + 11*a^2*b^3*x^5 
 - 11*a^3*b^2*x^3 - 3*a^4*b*x + 3*(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^3*b^7*x^8 + 4*a^4* 
b^6*x^6 + 6*a^5*b^5*x^4 + 4*a^6*b^4*x^2 + a^7*b^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.52 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {3 \, b^{3} x^{7} + 11 \, a b^{2} x^{5} - 11 \, a^{2} b x^{3} - 3 \, a^{3} x}{128 \, {\left (a^{2} b^{6} x^{8} + 4 \, a^{3} b^{5} x^{6} + 6 \, a^{4} b^{4} x^{4} + 4 \, a^{5} b^{3} x^{2} + a^{6} b^{2}\right )}} + \frac {3 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} a^{2} b^{2}} \] Input:

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

Output:

1/128*(3*b^3*x^7 + 11*a*b^2*x^5 - 11*a^2*b*x^3 - 3*a^3*x)/(a^2*b^6*x^8 + 4 
*a^3*b^5*x^6 + 6*a^4*b^4*x^4 + 4*a^5*b^3*x^2 + a^6*b^2) + 3/128*arctan(b*x 
/sqrt(a*b))/(sqrt(a*b)*a^2*b^2)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.44 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {3 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{128 \, \sqrt {a b} a^{2} b^{2} \mathrm {sgn}\left (b x^{2} + a\right )} + \frac {3 \, b^{3} x^{7} + 11 \, a b^{2} x^{5} - 11 \, a^{2} b x^{3} - 3 \, a^{3} x}{128 \, {\left (b x^{2} + a\right )}^{4} a^{2} b^{2} \mathrm {sgn}\left (b x^{2} + a\right )} \] Input:

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

Output:

3/128*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^2*b^2*sgn(b*x^2 + a)) + 1/128*(3* 
b^3*x^7 + 11*a*b^2*x^5 - 11*a^2*b*x^3 - 3*a^3*x)/((b*x^2 + a)^4*a^2*b^2*sg 
n(b*x^2 + a))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.00 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^{5/2}} \, dx=\frac {3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{4}+12 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b \,x^{2}+18 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} x^{4}+12 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} x^{6}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{4} x^{8}-3 a^{4} b x -11 a^{3} b^{2} x^{3}+11 a^{2} b^{3} x^{5}+3 a \,b^{4} x^{7}}{128 a^{3} b^{3} \left (b^{4} x^{8}+4 a \,b^{3} x^{6}+6 a^{2} b^{2} x^{4}+4 a^{3} b \,x^{2}+a^{4}\right )} \] Input:

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

Output:

(3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4 + 12*sqrt(b)*sqrt(a) 
*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*x**2 + 18*sqrt(b)*sqrt(a)*atan((b*x) 
/(sqrt(b)*sqrt(a)))*a**2*b**2*x**4 + 12*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b 
)*sqrt(a)))*a*b**3*x**6 + 3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))* 
b**4*x**8 - 3*a**4*b*x - 11*a**3*b**2*x**3 + 11*a**2*b**3*x**5 + 3*a*b**4* 
x**7)/(128*a**3*b**3*(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))