\(\int \frac {x^4}{(a^2+2 a b x^2+b^2 x^4)^3} \, dx\) [413]

Optimal result

Integrand size = 24, antiderivative size = 124 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=-\frac {x^3}{10 b \left (a+b x^2\right )^5}-\frac {3 x}{80 b^2 \left (a+b x^2\right )^4}+\frac {x}{160 a b^2 \left (a+b x^2\right )^3}+\frac {x}{128 a^2 b^2 \left (a+b x^2\right )^2}+\frac {3 x}{256 a^3 b^2 \left (a+b x^2\right )}+\frac {3 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{256 a^{7/2} b^{5/2}} \] Output:

-1/10*x^3/b/(b*x^2+a)^5-3/80*x/b^2/(b*x^2+a)^4+1/160*x/a/b^2/(b*x^2+a)^3+1 
/128*x/a^2/b^2/(b*x^2+a)^2+3/256*x/a^3/b^2/(b*x^2+a)+3/256*arctan(b^(1/2)* 
x/a^(1/2))/a^(7/2)/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.73 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {-15 a^4 x-70 a^3 b x^3+128 a^2 b^2 x^5+70 a b^3 x^7+15 b^4 x^9}{1280 a^3 b^2 \left (a+b x^2\right )^5}+\frac {3 \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{256 a^{7/2} b^{5/2}} \] Input:

Integrate[x^4/(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]
 

Output:

(-15*a^4*x - 70*a^3*b*x^3 + 128*a^2*b^2*x^5 + 70*a*b^3*x^7 + 15*b^4*x^9)/( 
1280*a^3*b^2*(a + b*x^2)^5) + (3*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(256*a^(7/2) 
*b^(5/2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.19, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1380, 27, 252, 252, 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.

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

\(\Big \downarrow \) 1380

\(\displaystyle b^6 \int \frac {x^4}{b^6 \left (b x^2+a\right )^6}dx\)

\(\Big \downarrow \) 27

\(\displaystyle \int \frac {x^4}{\left (a+b x^2\right )^6}dx\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {3 \int \frac {x^2}{\left (b x^2+a\right )^5}dx}{10 b}-\frac {x^3}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 252

\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\left (b x^2+a\right )^4}dx}{8 b}-\frac {x}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^3}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {3 \left (\frac {\frac {5 \int \frac {1}{\left (b x^2+a\right )^3}dx}{6 a}+\frac {x}{6 a \left (a+b x^2\right )^3}}{8 b}-\frac {x}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^3}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {3 \left (\frac {\frac {5 \left (\frac {3 \int \frac {1}{\left (b x^2+a\right )^2}dx}{4 a}+\frac {x}{4 a \left (a+b x^2\right )^2}\right )}{6 a}+\frac {x}{6 a \left (a+b x^2\right )^3}}{8 b}-\frac {x}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^3}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {3 \left (\frac {\frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{b x^2+a}dx}{2 a}+\frac {x}{2 a \left (a+b x^2\right )}\right )}{4 a}+\frac {x}{4 a \left (a+b x^2\right )^2}\right )}{6 a}+\frac {x}{6 a \left (a+b x^2\right )^3}}{8 b}-\frac {x}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^3}{10 b \left (a+b x^2\right )^5}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {3 \left (\frac {\frac {5 \left (\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {b}}+\frac {x}{2 a \left (a+b x^2\right )}\right )}{4 a}+\frac {x}{4 a \left (a+b x^2\right )^2}\right )}{6 a}+\frac {x}{6 a \left (a+b x^2\right )^3}}{8 b}-\frac {x}{8 b \left (a+b x^2\right )^4}\right )}{10 b}-\frac {x^3}{10 b \left (a+b x^2\right )^5}\)

Input:

Int[x^4/(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]
 

Output:

-1/10*x^3/(b*(a + b*x^2)^5) + (3*(-1/8*x/(b*(a + b*x^2)^4) + (x/(6*a*(a + 
b*x^2)^3) + (5*(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*a))/(8*b)))/(10*b)
 

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 1380

Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S 
imp[1/c^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]
 

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.63

method	result	size
default	\(\frac {\frac {3 b^{2} x^{9}}{256 a^{3}}+\frac {7 b \,x^{7}}{128 a^{2}}+\frac {x^{5}}{10 a}-\frac {7 x^{3}}{128 b}-\frac {3 a x}{256 b^{2}}}{\left (b \,x^{2}+a \right )^{5}}+\frac {3 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 a^{3} b^{2} \sqrt {a b}}\)	\(78\)
risch	\(\frac {\frac {3 b^{2} x^{9}}{256 a^{3}}+\frac {7 b \,x^{7}}{128 a^{2}}+\frac {x^{5}}{10 a}-\frac {7 x^{3}}{128 b}-\frac {3 a x}{256 b^{2}}}{\left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )^{2}}-\frac {3 \ln \left (b x +\sqrt {-a b}\right )}{512 \sqrt {-a b}\, b^{2} a^{3}}+\frac {3 \ln \left (-b x +\sqrt {-a b}\right )}{512 \sqrt {-a b}\, b^{2} a^{3}}\)	\(127\)

Input:

int(x^4/(b^2*x^4+2*a*b*x^2+a^2)^3,x,method=_RETURNVERBOSE)
 

Output:

(3/256*b^2/a^3*x^9+7/128*b/a^2*x^7+1/10/a*x^5-7/128/b*x^3-3/256*a/b^2*x)/( 
b*x^2+a)^5+3/256/a^3/b^2/(a*b)^(1/2)*arctan(b/(a*b)^(1/2)*x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 390, normalized size of antiderivative = 3.15 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\left [\frac {30 \, a b^{5} x^{9} + 140 \, a^{2} b^{4} x^{7} + 256 \, a^{3} b^{3} x^{5} - 140 \, a^{4} b^{2} x^{3} - 30 \, a^{5} b x - 15 \, {\left (b^{5} x^{10} + 5 \, a b^{4} x^{8} + 10 \, a^{2} b^{3} x^{6} + 10 \, a^{3} b^{2} x^{4} + 5 \, a^{4} b x^{2} + a^{5}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{2560 \, {\left (a^{4} b^{8} x^{10} + 5 \, a^{5} b^{7} x^{8} + 10 \, a^{6} b^{6} x^{6} + 10 \, a^{7} b^{5} x^{4} + 5 \, a^{8} b^{4} x^{2} + a^{9} b^{3}\right )}}, \frac {15 \, a b^{5} x^{9} + 70 \, a^{2} b^{4} x^{7} + 128 \, a^{3} b^{3} x^{5} - 70 \, a^{4} b^{2} x^{3} - 15 \, a^{5} b x + 15 \, {\left (b^{5} x^{10} + 5 \, a b^{4} x^{8} + 10 \, a^{2} b^{3} x^{6} + 10 \, a^{3} b^{2} x^{4} + 5 \, a^{4} b x^{2} + a^{5}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{1280 \, {\left (a^{4} b^{8} x^{10} + 5 \, a^{5} b^{7} x^{8} + 10 \, a^{6} b^{6} x^{6} + 10 \, a^{7} b^{5} x^{4} + 5 \, a^{8} b^{4} x^{2} + a^{9} b^{3}\right )}}\right ] \] Input:

integrate(x^4/(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="fricas")
 

Output:

[1/2560*(30*a*b^5*x^9 + 140*a^2*b^4*x^7 + 256*a^3*b^3*x^5 - 140*a^4*b^2*x^ 
3 - 30*a^5*b*x - 15*(b^5*x^10 + 5*a*b^4*x^8 + 10*a^2*b^3*x^6 + 10*a^3*b^2* 
x^4 + 5*a^4*b*x^2 + a^5)*sqrt(-a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^ 
2 + a)))/(a^4*b^8*x^10 + 5*a^5*b^7*x^8 + 10*a^6*b^6*x^6 + 10*a^7*b^5*x^4 + 
 5*a^8*b^4*x^2 + a^9*b^3), 1/1280*(15*a*b^5*x^9 + 70*a^2*b^4*x^7 + 128*a^3 
*b^3*x^5 - 70*a^4*b^2*x^3 - 15*a^5*b*x + 15*(b^5*x^10 + 5*a*b^4*x^8 + 10*a 
^2*b^3*x^6 + 10*a^3*b^2*x^4 + 5*a^4*b*x^2 + a^5)*sqrt(a*b)*arctan(sqrt(a*b 
)*x/a))/(a^4*b^8*x^10 + 5*a^5*b^7*x^8 + 10*a^6*b^6*x^6 + 10*a^7*b^5*x^4 + 
5*a^8*b^4*x^2 + a^9*b^3)]
 

Sympy [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.58 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=- \frac {3 \sqrt {- \frac {1}{a^{7} b^{5}}} \log {\left (- a^{4} b^{2} \sqrt {- \frac {1}{a^{7} b^{5}}} + x \right )}}{512} + \frac {3 \sqrt {- \frac {1}{a^{7} b^{5}}} \log {\left (a^{4} b^{2} \sqrt {- \frac {1}{a^{7} b^{5}}} + x \right )}}{512} + \frac {- 15 a^{4} x - 70 a^{3} b x^{3} + 128 a^{2} b^{2} x^{5} + 70 a b^{3} x^{7} + 15 b^{4} x^{9}}{1280 a^{8} b^{2} + 6400 a^{7} b^{3} x^{2} + 12800 a^{6} b^{4} x^{4} + 12800 a^{5} b^{5} x^{6} + 6400 a^{4} b^{6} x^{8} + 1280 a^{3} b^{7} x^{10}} \] Input:

integrate(x**4/(b**2*x**4+2*a*b*x**2+a**2)**3,x)
 

Output:

-3*sqrt(-1/(a**7*b**5))*log(-a**4*b**2*sqrt(-1/(a**7*b**5)) + x)/512 + 3*s 
qrt(-1/(a**7*b**5))*log(a**4*b**2*sqrt(-1/(a**7*b**5)) + x)/512 + (-15*a** 
4*x - 70*a**3*b*x**3 + 128*a**2*b**2*x**5 + 70*a*b**3*x**7 + 15*b**4*x**9) 
/(1280*a**8*b**2 + 6400*a**7*b**3*x**2 + 12800*a**6*b**4*x**4 + 12800*a**5 
*b**5*x**6 + 6400*a**4*b**6*x**8 + 1280*a**3*b**7*x**10)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.07 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {15 \, b^{4} x^{9} + 70 \, a b^{3} x^{7} + 128 \, a^{2} b^{2} x^{5} - 70 \, a^{3} b x^{3} - 15 \, a^{4} x}{1280 \, {\left (a^{3} b^{7} x^{10} + 5 \, a^{4} b^{6} x^{8} + 10 \, a^{5} b^{5} x^{6} + 10 \, a^{6} b^{4} x^{4} + 5 \, a^{7} b^{3} x^{2} + a^{8} b^{2}\right )}} + \frac {3 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 \, \sqrt {a b} a^{3} b^{2}} \] Input:

integrate(x^4/(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="maxima")
 

Output:

1/1280*(15*b^4*x^9 + 70*a*b^3*x^7 + 128*a^2*b^2*x^5 - 70*a^3*b*x^3 - 15*a^ 
4*x)/(a^3*b^7*x^10 + 5*a^4*b^6*x^8 + 10*a^5*b^5*x^6 + 10*a^6*b^4*x^4 + 5*a 
^7*b^3*x^2 + a^8*b^2) + 3/256*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3*b^2)
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.68 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {3 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 \, \sqrt {a b} a^{3} b^{2}} + \frac {15 \, b^{4} x^{9} + 70 \, a b^{3} x^{7} + 128 \, a^{2} b^{2} x^{5} - 70 \, a^{3} b x^{3} - 15 \, a^{4} x}{1280 \, {\left (b x^{2} + a\right )}^{5} a^{3} b^{2}} \] Input:

integrate(x^4/(b^2*x^4+2*a*b*x^2+a^2)^3,x, algorithm="giac")
 

Output:

3/256*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*a^3*b^2) + 1/1280*(15*b^4*x^9 + 70* 
a*b^3*x^7 + 128*a^2*b^2*x^5 - 70*a^3*b*x^3 - 15*a^4*x)/((b*x^2 + a)^5*a^3* 
b^2)
 

Mupad [B] (verification not implemented)

Time = 18.43 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.94 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {\frac {x^5}{10\,a}-\frac {7\,x^3}{128\,b}+\frac {7\,b\,x^7}{128\,a^2}+\frac {3\,b^2\,x^9}{256\,a^3}-\frac {3\,a\,x}{256\,b^2}}{a^5+5\,a^4\,b\,x^2+10\,a^3\,b^2\,x^4+10\,a^2\,b^3\,x^6+5\,a\,b^4\,x^8+b^5\,x^{10}}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{256\,a^{7/2}\,b^{5/2}} \] Input:

int(x^4/(a^2 + b^2*x^4 + 2*a*b*x^2)^3,x)
 

Output:

(x^5/(10*a) - (7*x^3)/(128*b) + (7*b*x^7)/(128*a^2) + (3*b^2*x^9)/(256*a^3 
) - (3*a*x)/(256*b^2))/(a^5 + b^5*x^10 + 5*a^4*b*x^2 + 5*a*b^4*x^8 + 10*a^ 
3*b^2*x^4 + 10*a^2*b^3*x^6) + (3*atan((b^(1/2)*x)/a^(1/2)))/(256*a^(7/2)*b 
^(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 260, normalized size of antiderivative = 2.10 \[ \int \frac {x^4}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx=\frac {15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{5}+75 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{4} b \,x^{2}+150 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b^{2} x^{4}+150 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{3} x^{6}+75 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{4} x^{8}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{5} x^{10}-15 a^{5} b x -70 a^{4} b^{2} x^{3}+128 a^{3} b^{3} x^{5}+70 a^{2} b^{4} x^{7}+15 a \,b^{5} x^{9}}{1280 a^{4} b^{3} \left (b^{5} x^{10}+5 a \,b^{4} x^{8}+10 a^{2} b^{3} x^{6}+10 a^{3} b^{2} x^{4}+5 a^{4} b \,x^{2}+a^{5}\right )} \] Input:

int(x^4/(b^2*x^4+2*a*b*x^2+a^2)^3,x)
 

Output:

(15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**5 + 75*sqrt(b)*sqrt(a 
)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4*b*x**2 + 150*sqrt(b)*sqrt(a)*atan((b* 
x)/(sqrt(b)*sqrt(a)))*a**3*b**2*x**4 + 150*sqrt(b)*sqrt(a)*atan((b*x)/(sqr 
t(b)*sqrt(a)))*a**2*b**3*x**6 + 75*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqr 
t(a)))*a*b**4*x**8 + 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*b**5 
*x**10 - 15*a**5*b*x - 70*a**4*b**2*x**3 + 128*a**3*b**3*x**5 + 70*a**2*b* 
*4*x**7 + 15*a*b**5*x**9)/(1280*a**4*b**3*(a**5 + 5*a**4*b*x**2 + 10*a**3* 
b**2*x**4 + 10*a**2*b**3*x**6 + 5*a*b**4*x**8 + b**5*x**10))