3.503 \(\int \frac {x^{10}}{(a^2+2 a b x^2+b^2 x^4)^2} \, dx\)

Optimal. Leaf size=104 \[ \frac {105 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 b^{11/2}}-\frac {105 a x}{16 b^5}-\frac {21 x^5}{16 b^3 \left (a+b x^2\right )}-\frac {3 x^7}{8 b^2 \left (a+b x^2\right )^2}-\frac {x^9}{6 b \left (a+b x^2\right )^3}+\frac {35 x^3}{16 b^4} \]

[Out]

-105/16*a*x/b^5+35/16*x^3/b^4-1/6*x^9/b/(b*x^2+a)^3-3/8*x^7/b^2/(b*x^2+a)^2-21/16*x^5/b^3/(b*x^2+a)+105/16*a^(
3/2)*arctan(x*b^(1/2)/a^(1/2))/b^(11/2)

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Rubi [A]  time = 0.06, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {28, 288, 302, 205} \[ \frac {105 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 b^{11/2}}-\frac {3 x^7}{8 b^2 \left (a+b x^2\right )^2}-\frac {21 x^5}{16 b^3 \left (a+b x^2\right )}-\frac {105 a x}{16 b^5}-\frac {x^9}{6 b \left (a+b x^2\right )^3}+\frac {35 x^3}{16 b^4} \]

Antiderivative was successfully verified.

[In]

Int[x^10/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

(-105*a*x)/(16*b^5) + (35*x^3)/(16*b^4) - x^9/(6*b*(a + b*x^2)^3) - (3*x^7)/(8*b^2*(a + b*x^2)^2) - (21*x^5)/(
16*b^3*(a + b*x^2)) + (105*a^(3/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(16*b^(11/2))

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 205

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

Rule 288

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 302

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*} \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx &=b^4 \int \frac {x^{10}}{\left (a b+b^2 x^2\right )^4} \, dx\\ &=-\frac {x^9}{6 b \left (a+b x^2\right )^3}+\frac {1}{2} \left (3 b^2\right ) \int \frac {x^8}{\left (a b+b^2 x^2\right )^3} \, dx\\ &=-\frac {x^9}{6 b \left (a+b x^2\right )^3}-\frac {3 x^7}{8 b^2 \left (a+b x^2\right )^2}+\frac {21}{8} \int \frac {x^6}{\left (a b+b^2 x^2\right )^2} \, dx\\ &=-\frac {x^9}{6 b \left (a+b x^2\right )^3}-\frac {3 x^7}{8 b^2 \left (a+b x^2\right )^2}-\frac {21 x^5}{16 b^3 \left (a+b x^2\right )}+\frac {105 \int \frac {x^4}{a b+b^2 x^2} \, dx}{16 b^2}\\ &=-\frac {x^9}{6 b \left (a+b x^2\right )^3}-\frac {3 x^7}{8 b^2 \left (a+b x^2\right )^2}-\frac {21 x^5}{16 b^3 \left (a+b x^2\right )}+\frac {105 \int \left (-\frac {a}{b^3}+\frac {x^2}{b^2}+\frac {a^2}{b^2 \left (a b+b^2 x^2\right )}\right ) \, dx}{16 b^2}\\ &=-\frac {105 a x}{16 b^5}+\frac {35 x^3}{16 b^4}-\frac {x^9}{6 b \left (a+b x^2\right )^3}-\frac {3 x^7}{8 b^2 \left (a+b x^2\right )^2}-\frac {21 x^5}{16 b^3 \left (a+b x^2\right )}+\frac {\left (105 a^2\right ) \int \frac {1}{a b+b^2 x^2} \, dx}{16 b^4}\\ &=-\frac {105 a x}{16 b^5}+\frac {35 x^3}{16 b^4}-\frac {x^9}{6 b \left (a+b x^2\right )^3}-\frac {3 x^7}{8 b^2 \left (a+b x^2\right )^2}-\frac {21 x^5}{16 b^3 \left (a+b x^2\right )}+\frac {105 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{16 b^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 89, normalized size = 0.86 \[ \frac {315 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )+\frac {\sqrt {b} x \left (-315 a^4-840 a^3 b x^2-693 a^2 b^2 x^4-144 a b^3 x^6+16 b^4 x^8\right )}{\left (a+b x^2\right )^3}}{48 b^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^10/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

((Sqrt[b]*x*(-315*a^4 - 840*a^3*b*x^2 - 693*a^2*b^2*x^4 - 144*a*b^3*x^6 + 16*b^4*x^8))/(a + b*x^2)^3 + 315*a^(
3/2)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(48*b^(11/2))

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fricas [A]  time = 1.00, size = 296, normalized size = 2.85 \[ \left [\frac {32 \, b^{4} x^{9} - 288 \, a b^{3} x^{7} - 1386 \, a^{2} b^{2} x^{5} - 1680 \, a^{3} b x^{3} - 630 \, a^{4} x + 315 \, {\left (a b^{3} x^{6} + 3 \, a^{2} b^{2} x^{4} + 3 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {-\frac {a}{b}} \log \left (\frac {b x^{2} + 2 \, b x \sqrt {-\frac {a}{b}} - a}{b x^{2} + a}\right )}{96 \, {\left (b^{8} x^{6} + 3 \, a b^{7} x^{4} + 3 \, a^{2} b^{6} x^{2} + a^{3} b^{5}\right )}}, \frac {16 \, b^{4} x^{9} - 144 \, a b^{3} x^{7} - 693 \, a^{2} b^{2} x^{5} - 840 \, a^{3} b x^{3} - 315 \, a^{4} x + 315 \, {\left (a b^{3} x^{6} + 3 \, a^{2} b^{2} x^{4} + 3 \, a^{3} b x^{2} + a^{4}\right )} \sqrt {\frac {a}{b}} \arctan \left (\frac {b x \sqrt {\frac {a}{b}}}{a}\right )}{48 \, {\left (b^{8} x^{6} + 3 \, a b^{7} x^{4} + 3 \, a^{2} b^{6} x^{2} + a^{3} b^{5}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")

[Out]

[1/96*(32*b^4*x^9 - 288*a*b^3*x^7 - 1386*a^2*b^2*x^5 - 1680*a^3*b*x^3 - 630*a^4*x + 315*(a*b^3*x^6 + 3*a^2*b^2
*x^4 + 3*a^3*b*x^2 + a^4)*sqrt(-a/b)*log((b*x^2 + 2*b*x*sqrt(-a/b) - a)/(b*x^2 + a)))/(b^8*x^6 + 3*a*b^7*x^4 +
3*a^2*b^6*x^2 + a^3*b^5), 1/48*(16*b^4*x^9 - 144*a*b^3*x^7 - 693*a^2*b^2*x^5 - 840*a^3*b*x^3 - 315*a^4*x + 31
5*(a*b^3*x^6 + 3*a^2*b^2*x^4 + 3*a^3*b*x^2 + a^4)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a))/(b^8*x^6 + 3*a*b^7*x^4 +
3*a^2*b^6*x^2 + a^3*b^5)]

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giac [A]  time = 0.16, size = 84, normalized size = 0.81 \[ \frac {105 \, a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} b^{5}} - \frac {165 \, a^{2} b^{2} x^{5} + 280 \, a^{3} b x^{3} + 123 \, a^{4} x}{48 \, {\left (b x^{2} + a\right )}^{3} b^{5}} + \frac {b^{8} x^{3} - 12 \, a b^{7} x}{3 \, b^{12}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")

[Out]

105/16*a^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^5) - 1/48*(165*a^2*b^2*x^5 + 280*a^3*b*x^3 + 123*a^4*x)/((b*x^2
+ a)^3*b^5) + 1/3*(b^8*x^3 - 12*a*b^7*x)/b^12

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maple [A]  time = 0.01, size = 97, normalized size = 0.93 \[ -\frac {55 a^{2} x^{5}}{16 \left (b \,x^{2}+a \right )^{3} b^{3}}-\frac {35 a^{3} x^{3}}{6 \left (b \,x^{2}+a \right )^{3} b^{4}}-\frac {41 a^{4} x}{16 \left (b \,x^{2}+a \right )^{3} b^{5}}+\frac {x^{3}}{3 b^{4}}+\frac {105 a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \sqrt {a b}\, b^{5}}-\frac {4 a x}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10/(b^2*x^4+2*a*b*x^2+a^2)^2,x)

[Out]

1/3*x^3/b^4-4*a*x/b^5-55/16/b^3*a^2/(b*x^2+a)^3*x^5-35/6/b^4*a^3/(b*x^2+a)^3*x^3-41/16/b^5*a^4/(b*x^2+a)^3*x+1
05/16/b^5*a^2/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)

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maxima [A]  time = 3.01, size = 104, normalized size = 1.00 \[ -\frac {165 \, a^{2} b^{2} x^{5} + 280 \, a^{3} b x^{3} + 123 \, a^{4} x}{48 \, {\left (b^{8} x^{6} + 3 \, a b^{7} x^{4} + 3 \, a^{2} b^{6} x^{2} + a^{3} b^{5}\right )}} + \frac {105 \, a^{2} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {a b} b^{5}} + \frac {b x^{3} - 12 \, a x}{3 \, b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")

[Out]

-1/48*(165*a^2*b^2*x^5 + 280*a^3*b*x^3 + 123*a^4*x)/(b^8*x^6 + 3*a*b^7*x^4 + 3*a^2*b^6*x^2 + a^3*b^5) + 105/16
*a^2*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^5) + 1/3*(b*x^3 - 12*a*x)/b^5

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mupad [B]  time = 4.36, size = 99, normalized size = 0.95 \[ \frac {x^3}{3\,b^4}-\frac {\frac {41\,a^4\,x}{16}+\frac {35\,a^3\,b\,x^3}{6}+\frac {55\,a^2\,b^2\,x^5}{16}}{a^3\,b^5+3\,a^2\,b^6\,x^2+3\,a\,b^7\,x^4+b^8\,x^6}+\frac {105\,a^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{16\,b^{11/2}}-\frac {4\,a\,x}{b^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10/(a^2 + b^2*x^4 + 2*a*b*x^2)^2,x)

[Out]

x^3/(3*b^4) - ((41*a^4*x)/16 + (35*a^3*b*x^3)/6 + (55*a^2*b^2*x^5)/16)/(a^3*b^5 + b^8*x^6 + 3*a*b^7*x^4 + 3*a^
2*b^6*x^2) + (105*a^(3/2)*atan((b^(1/2)*x)/a^(1/2)))/(16*b^(11/2)) - (4*a*x)/b^5

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sympy [A]  time = 0.63, size = 156, normalized size = 1.50 \[ - \frac {4 a x}{b^{5}} - \frac {105 \sqrt {- \frac {a^{3}}{b^{11}}} \log {\left (x - \frac {b^{5} \sqrt {- \frac {a^{3}}{b^{11}}}}{a} \right )}}{32} + \frac {105 \sqrt {- \frac {a^{3}}{b^{11}}} \log {\left (x + \frac {b^{5} \sqrt {- \frac {a^{3}}{b^{11}}}}{a} \right )}}{32} + \frac {- 123 a^{4} x - 280 a^{3} b x^{3} - 165 a^{2} b^{2} x^{5}}{48 a^{3} b^{5} + 144 a^{2} b^{6} x^{2} + 144 a b^{7} x^{4} + 48 b^{8} x^{6}} + \frac {x^{3}}{3 b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10/(b**2*x**4+2*a*b*x**2+a**2)**2,x)

[Out]

-4*a*x/b**5 - 105*sqrt(-a**3/b**11)*log(x - b**5*sqrt(-a**3/b**11)/a)/32 + 105*sqrt(-a**3/b**11)*log(x + b**5*
sqrt(-a**3/b**11)/a)/32 + (-123*a**4*x - 280*a**3*b*x**3 - 165*a**2*b**2*x**5)/(48*a**3*b**5 + 144*a**2*b**6*x
**2 + 144*a*b**7*x**4 + 48*b**8*x**6) + x**3/(3*b**4)

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