\(\int \frac {x^{12}}{(a^2+2 a b x^2+b^2 x^4)^2} \, dx\) [502]

Optimal result

Integrand size = 24, antiderivative size = 117 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {231 a^2 x}{16 b^6}-\frac {77 a x^3}{16 b^5}+\frac {231 x^5}{80 b^4}-\frac {x^{11}}{6 b \left (a+b x^2\right )^3}-\frac {11 x^9}{24 b^2 \left (a+b x^2\right )^2}-\frac {33 x^7}{16 b^3 \left (a+b x^2\right )}-\frac {231 a^{5/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 b^{13/2}} \]

[Out]

231/16*a^2*x/b^6-77/16*a*x^3/b^5+231/80*x^5/b^4-1/6*x^11/b/(b*x^2+a)^3-11/24*x^9/b^2/(b*x^2+a)^2-33/16*x^7/b^3
/(b*x^2+a)-231/16*a^(5/2)*arctan(x*b^(1/2)/a^(1/2))/b^(13/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {28, 294, 308, 211} \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {231 a^{5/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 b^{13/2}}+\frac {231 a^2 x}{16 b^6}-\frac {77 a x^3}{16 b^5}-\frac {33 x^7}{16 b^3 \left (a+b x^2\right )}-\frac {11 x^9}{24 b^2 \left (a+b x^2\right )^2}-\frac {x^{11}}{6 b \left (a+b x^2\right )^3}+\frac {231 x^5}{80 b^4} \]

[In]

Int[x^12/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

(231*a^2*x)/(16*b^6) - (77*a*x^3)/(16*b^5) + (231*x^5)/(80*b^4) - x^11/(6*b*(a + b*x^2)^3) - (11*x^9)/(24*b^2*
(a + b*x^2)^2) - (33*x^7)/(16*b^3*(a + b*x^2)) - (231*a^(5/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(16*b^(13/2))

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rubi steps \begin{align*} \text {integral}& = b^4 \int \frac {x^{12}}{\left (a b+b^2 x^2\right )^4} \, dx \\ & = -\frac {x^{11}}{6 b \left (a+b x^2\right )^3}+\frac {1}{6} \left (11 b^2\right ) \int \frac {x^{10}}{\left (a b+b^2 x^2\right )^3} \, dx \\ & = -\frac {x^{11}}{6 b \left (a+b x^2\right )^3}-\frac {11 x^9}{24 b^2 \left (a+b x^2\right )^2}+\frac {33}{8} \int \frac {x^8}{\left (a b+b^2 x^2\right )^2} \, dx \\ & = -\frac {x^{11}}{6 b \left (a+b x^2\right )^3}-\frac {11 x^9}{24 b^2 \left (a+b x^2\right )^2}-\frac {33 x^7}{16 b^3 \left (a+b x^2\right )}+\frac {231 \int \frac {x^6}{a b+b^2 x^2} \, dx}{16 b^2} \\ & = -\frac {x^{11}}{6 b \left (a+b x^2\right )^3}-\frac {11 x^9}{24 b^2 \left (a+b x^2\right )^2}-\frac {33 x^7}{16 b^3 \left (a+b x^2\right )}+\frac {231 \int \left (\frac {a^2}{b^4}-\frac {a x^2}{b^3}+\frac {x^4}{b^2}-\frac {a^3}{b^3 \left (a b+b^2 x^2\right )}\right ) \, dx}{16 b^2} \\ & = \frac {231 a^2 x}{16 b^6}-\frac {77 a x^3}{16 b^5}+\frac {231 x^5}{80 b^4}-\frac {x^{11}}{6 b \left (a+b x^2\right )^3}-\frac {11 x^9}{24 b^2 \left (a+b x^2\right )^2}-\frac {33 x^7}{16 b^3 \left (a+b x^2\right )}-\frac {\left (231 a^3\right ) \int \frac {1}{a b+b^2 x^2} \, dx}{16 b^5} \\ & = \frac {231 a^2 x}{16 b^6}-\frac {77 a x^3}{16 b^5}+\frac {231 x^5}{80 b^4}-\frac {x^{11}}{6 b \left (a+b x^2\right )^3}-\frac {11 x^9}{24 b^2 \left (a+b x^2\right )^2}-\frac {33 x^7}{16 b^3 \left (a+b x^2\right )}-\frac {231 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 b^{13/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.85 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {3465 a^5 x+9240 a^4 b x^3+7623 a^3 b^2 x^5+1584 a^2 b^3 x^7-176 a b^4 x^9+48 b^5 x^{11}}{240 b^6 \left (a+b x^2\right )^3}-\frac {231 a^{5/2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 b^{13/2}} \]

[In]

Integrate[x^12/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

(3465*a^5*x + 9240*a^4*b*x^3 + 7623*a^3*b^2*x^5 + 1584*a^2*b^3*x^7 - 176*a*b^4*x^9 + 48*b^5*x^11)/(240*b^6*(a
+ b*x^2)^3) - (231*a^(5/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(16*b^(13/2))

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.73

method	result	size
default	\(\frac {\frac {1}{5} b^{2} x^{5}-\frac {4}{3} a b \,x^{3}+10 a^{2} x}{b^{6}}-\frac {a^{3} \left (\frac {-\frac {89}{16} b^{2} x^{5}-\frac {59}{6} a b \,x^{3}-\frac {71}{16} a^{2} x}{\left (b \,x^{2}+a \right )^{3}}+\frac {231 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \sqrt {a b}}\right )}{b^{6}}\)	\(85\)
risch	\(\frac {x^{5}}{5 b^{4}}-\frac {4 a \,x^{3}}{3 b^{5}}+\frac {10 a^{2} x}{b^{6}}+\frac {\frac {89}{16} a^{3} b^{2} x^{5}+\frac {59}{6} a^{4} b \,x^{3}+\frac {71}{16} a^{5} x}{b^{6} \left (b \,x^{2}+a \right ) \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}+\frac {231 \sqrt {-a b}\, a^{2} \ln \left (-\sqrt {-a b}\, x -a \right )}{32 b^{7}}-\frac {231 \sqrt {-a b}\, a^{2} \ln \left (\sqrt {-a b}\, x -a \right )}{32 b^{7}}\)	\(143\)

[In]

int(x^12/(b^2*x^4+2*a*b*x^2+a^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/b^6*(1/5*b^2*x^5-4/3*a*b*x^3+10*a^2*x)-1/b^6*a^3*((-89/16*b^2*x^5-59/6*a*b*x^3-71/16*a^2*x)/(b*x^2+a)^3+231/
16/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 322, normalized size of antiderivative = 2.75 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\left [\frac {96 \, b^{5} x^{11} - 352 \, a b^{4} x^{9} + 3168 \, a^{2} b^{3} x^{7} + 15246 \, a^{3} b^{2} x^{5} + 18480 \, a^{4} b x^{3} + 6930 \, a^{5} x + 3465 \, {\left (a^{2} b^{3} x^{6} + 3 \, a^{3} b^{2} x^{4} + 3 \, a^{4} b x^{2} + a^{5}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} - 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right )}{480 \, {\left (b^{9} x^{6} + 3 \, a b^{8} x^{4} + 3 \, a^{2} b^{7} x^{2} + a^{3} b^{6}\right )}}, \frac {48 \, b^{5} x^{11} - 176 \, a b^{4} x^{9} + 1584 \, a^{2} b^{3} x^{7} + 7623 \, a^{3} b^{2} x^{5} + 9240 \, a^{4} b x^{3} + 3465 \, a^{5} x - 3465 \, {\left (a^{2} b^{3} x^{6} + 3 \, a^{3} b^{2} x^{4} + 3 \, a^{4} b x^{2} + a^{5}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right )}{240 \, {\left (b^{9} x^{6} + 3 \, a b^{8} x^{4} + 3 \, a^{2} b^{7} x^{2} + a^{3} b^{6}\right )}}\right ] \]

[In]

integrate(x^12/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")

[Out]

[1/480*(96*b^5*x^11 - 352*a*b^4*x^9 + 3168*a^2*b^3*x^7 + 15246*a^3*b^2*x^5 + 18480*a^4*b*x^3 + 6930*a^5*x + 34
65*(a^2*b^3*x^6 + 3*a^3*b^2*x^4 + 3*a^4*b*x^2 + a^5)*sqrt(-a/b)*log((b*x^2 - 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)
))/(b^9*x^6 + 3*a*b^8*x^4 + 3*a^2*b^7*x^2 + a^3*b^6), 1/240*(48*b^5*x^11 - 176*a*b^4*x^9 + 1584*a^2*b^3*x^7 +
7623*a^3*b^2*x^5 + 9240*a^4*b*x^3 + 3465*a^5*x - 3465*(a^2*b^3*x^6 + 3*a^3*b^2*x^4 + 3*a^4*b*x^2 + a^5)*sqrt(a
/b)*arctan(b*x*sqrt(a/b)/a))/(b^9*x^6 + 3*a*b^8*x^4 + 3*a^2*b^7*x^2 + a^3*b^6)]

Sympy [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.47 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {10 a^{2} x}{b^{6}} - \frac {4 a x^{3}}{3 b^{5}} + \frac {231 \sqrt {- \frac {a^{5}}{b^{13}}} \log {\left (x - \frac {b^{6} \sqrt {- \frac {a^{5}}{b^{13}}}}{a^{2}} \right )}}{32} - \frac {231 \sqrt {- \frac {a^{5}}{b^{13}}} \log {\left (x + \frac {b^{6} \sqrt {- \frac {a^{5}}{b^{13}}}}{a^{2}} \right )}}{32} + \frac {213 a^{5} x + 472 a^{4} b x^{3} + 267 a^{3} b^{2} x^{5}}{48 a^{3} b^{6} + 144 a^{2} b^{7} x^{2} + 144 a b^{8} x^{4} + 48 b^{9} x^{6}} + \frac {x^{5}}{5 b^{4}} \]

[In]

integrate(x**12/(b**2*x**4+2*a*b*x**2+a**2)**2,x)

[Out]

10*a**2*x/b**6 - 4*a*x**3/(3*b**5) + 231*sqrt(-a**5/b**13)*log(x - b**6*sqrt(-a**5/b**13)/a**2)/32 - 231*sqrt(
-a**5/b**13)*log(x + b**6*sqrt(-a**5/b**13)/a**2)/32 + (213*a**5*x + 472*a**4*b*x**3 + 267*a**3*b**2*x**5)/(48
*a**3*b**6 + 144*a**2*b**7*x**2 + 144*a*b**8*x**4 + 48*b**9*x**6) + x**5/(5*b**4)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.99 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {267 \, a^{3} b^{2} x^{5} + 472 \, a^{4} b x^{3} + 213 \, a^{5} x}{48 \, {\left (b^{9} x^{6} + 3 \, a b^{8} x^{4} + 3 \, a^{2} b^{7} x^{2} + a^{3} b^{6}\right )}} - \frac {231 \, a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} b^{6}} + \frac {3 \, b^{2} x^{5} - 20 \, a b x^{3} + 150 \, a^{2} x}{15 \, b^{6}} \]

[In]

integrate(x^12/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")

[Out]

1/48*(267*a^3*b^2*x^5 + 472*a^4*b*x^3 + 213*a^5*x)/(b^9*x^6 + 3*a*b^8*x^4 + 3*a^2*b^7*x^2 + a^3*b^6) - 231/16*
a^3*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^6) + 1/15*(3*b^2*x^5 - 20*a*b*x^3 + 150*a^2*x)/b^6

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.82 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=-\frac {231 \, a^{3} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} b^{6}} + \frac {267 \, a^{3} b^{2} x^{5} + 472 \, a^{4} b x^{3} + 213 \, a^{5} x}{48 \, {\left (b x^{2} + a\right )}^{3} b^{6}} + \frac {3 \, b^{16} x^{5} - 20 \, a b^{15} x^{3} + 150 \, a^{2} b^{14} x}{15 \, b^{20}} \]

[In]

integrate(x^12/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")

[Out]

-231/16*a^3*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^6) + 1/48*(267*a^3*b^2*x^5 + 472*a^4*b*x^3 + 213*a^5*x)/((b*x^2
+ a)^3*b^6) + 1/15*(3*b^16*x^5 - 20*a*b^15*x^3 + 150*a^2*b^14*x)/b^20

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.93 \[ \int \frac {x^{12}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx=\frac {\frac {71\,a^5\,x}{16}+\frac {59\,a^4\,b\,x^3}{6}+\frac {89\,a^3\,b^2\,x^5}{16}}{a^3\,b^6+3\,a^2\,b^7\,x^2+3\,a\,b^8\,x^4+b^9\,x^6}+\frac {x^5}{5\,b^4}-\frac {4\,a\,x^3}{3\,b^5}+\frac {10\,a^2\,x}{b^6}-\frac {231\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{16\,b^{13/2}} \]

[In]

int(x^12/(a^2 + b^2*x^4 + 2*a*b*x^2)^2,x)

[Out]

((71*a^5*x)/16 + (59*a^4*b*x^3)/6 + (89*a^3*b^2*x^5)/16)/(a^3*b^6 + b^9*x^6 + 3*a*b^8*x^4 + 3*a^2*b^7*x^2) + x
^5/(5*b^4) - (4*a*x^3)/(3*b^5) + (10*a^2*x)/b^6 - (231*a^(5/2)*atan((b^(1/2)*x)/a^(1/2)))/(16*b^(13/2))