3.526 \(\int \frac {x^{10}}{(a^2+2 a b x^2+b^2 x^4)^3} \, dx\)

Optimal. Leaf size=121 \[ \frac {63 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{256 \sqrt {a} b^{11/2}}-\frac {63 x}{256 b^5 \left (a+b x^2\right )}-\frac {21 x^3}{128 b^4 \left (a+b x^2\right )^2}-\frac {21 x^5}{160 b^3 \left (a+b x^2\right )^3}-\frac {9 x^7}{80 b^2 \left (a+b x^2\right )^4}-\frac {x^9}{10 b \left (a+b x^2\right )^5} \]

[Out]

-1/10*x^9/b/(b*x^2+a)^5-9/80*x^7/b^2/(b*x^2+a)^4-21/160*x^5/b^3/(b*x^2+a)^3-21/128*x^3/b^4/(b*x^2+a)^2-63/256*
x/b^5/(b*x^2+a)+63/256*arctan(x*b^(1/2)/a^(1/2))/b^(11/2)/a^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {28, 288, 205} \[ -\frac {9 x^7}{80 b^2 \left (a+b x^2\right )^4}-\frac {21 x^5}{160 b^3 \left (a+b x^2\right )^3}-\frac {21 x^3}{128 b^4 \left (a+b x^2\right )^2}-\frac {63 x}{256 b^5 \left (a+b x^2\right )}+\frac {63 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{256 \sqrt {a} b^{11/2}}-\frac {x^9}{10 b \left (a+b x^2\right )^5} \]

Antiderivative was successfully verified.

[In]

Int[x^10/(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]

[Out]

-x^9/(10*b*(a + b*x^2)^5) - (9*x^7)/(80*b^2*(a + b*x^2)^4) - (21*x^5)/(160*b^3*(a + b*x^2)^3) - (21*x^3)/(128*
b^4*(a + b*x^2)^2) - (63*x)/(256*b^5*(a + b*x^2)) + (63*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(256*Sqrt[a]*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]

Rubi steps

\begin {align*} \int \frac {x^{10}}{\left (a^2+2 a b x^2+b^2 x^4\right )^3} \, dx &=b^6 \int \frac {x^{10}}{\left (a b+b^2 x^2\right )^6} \, dx\\ &=-\frac {x^9}{10 b \left (a+b x^2\right )^5}+\frac {1}{10} \left (9 b^4\right ) \int \frac {x^8}{\left (a b+b^2 x^2\right )^5} \, dx\\ &=-\frac {x^9}{10 b \left (a+b x^2\right )^5}-\frac {9 x^7}{80 b^2 \left (a+b x^2\right )^4}+\frac {1}{80} \left (63 b^2\right ) \int \frac {x^6}{\left (a b+b^2 x^2\right )^4} \, dx\\ &=-\frac {x^9}{10 b \left (a+b x^2\right )^5}-\frac {9 x^7}{80 b^2 \left (a+b x^2\right )^4}-\frac {21 x^5}{160 b^3 \left (a+b x^2\right )^3}+\frac {21}{32} \int \frac {x^4}{\left (a b+b^2 x^2\right )^3} \, dx\\ &=-\frac {x^9}{10 b \left (a+b x^2\right )^5}-\frac {9 x^7}{80 b^2 \left (a+b x^2\right )^4}-\frac {21 x^5}{160 b^3 \left (a+b x^2\right )^3}-\frac {21 x^3}{128 b^4 \left (a+b x^2\right )^2}+\frac {63 \int \frac {x^2}{\left (a b+b^2 x^2\right )^2} \, dx}{128 b^2}\\ &=-\frac {x^9}{10 b \left (a+b x^2\right )^5}-\frac {9 x^7}{80 b^2 \left (a+b x^2\right )^4}-\frac {21 x^5}{160 b^3 \left (a+b x^2\right )^3}-\frac {21 x^3}{128 b^4 \left (a+b x^2\right )^2}-\frac {63 x}{256 b^5 \left (a+b x^2\right )}+\frac {63 \int \frac {1}{a b+b^2 x^2} \, dx}{256 b^4}\\ &=-\frac {x^9}{10 b \left (a+b x^2\right )^5}-\frac {9 x^7}{80 b^2 \left (a+b x^2\right )^4}-\frac {21 x^5}{160 b^3 \left (a+b x^2\right )^3}-\frac {21 x^3}{128 b^4 \left (a+b x^2\right )^2}-\frac {63 x}{256 b^5 \left (a+b x^2\right )}+\frac {63 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{256 \sqrt {a} b^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 88, normalized size = 0.73 \[ \frac {63 \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{256 \sqrt {a} b^{11/2}}-\frac {x \left (315 a^4+1470 a^3 b x^2+2688 a^2 b^2 x^4+2370 a b^3 x^6+965 b^4 x^8\right )}{1280 b^5 \left (a+b x^2\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^10/(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]

[Out]

-1/1280*(x*(315*a^4 + 1470*a^3*b*x^2 + 2688*a^2*b^2*x^4 + 2370*a*b^3*x^6 + 965*b^4*x^8))/(b^5*(a + b*x^2)^5) +
(63*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(256*Sqrt[a]*b^(11/2))

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fricas [A]  time = 0.96, size = 386, normalized size = 3.19 \[ \left [-\frac {1930 \, a b^{5} x^{9} + 4740 \, a^{2} b^{4} x^{7} + 5376 \, a^{3} b^{3} x^{5} + 2940 \, a^{4} b^{2} x^{3} + 630 \, a^{5} b x + 315 \, {\left (b^{5} x^{10} + 5 \, a b^{4} x^{8} + 10 \, a^{2} b^{3} x^{6} + 10 \, a^{3} b^{2} x^{4} + 5 \, a^{4} b x^{2} + a^{5}\right )} \sqrt {-a b} \log \left (\frac {b x^{2} - 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right )}{2560 \, {\left (a b^{11} x^{10} + 5 \, a^{2} b^{10} x^{8} + 10 \, a^{3} b^{9} x^{6} + 10 \, a^{4} b^{8} x^{4} + 5 \, a^{5} b^{7} x^{2} + a^{6} b^{6}\right )}}, -\frac {965 \, a b^{5} x^{9} + 2370 \, a^{2} b^{4} x^{7} + 2688 \, a^{3} b^{3} x^{5} + 1470 \, a^{4} b^{2} x^{3} + 315 \, a^{5} b x - 315 \, {\left (b^{5} x^{10} + 5 \, a b^{4} x^{8} + 10 \, a^{2} b^{3} x^{6} + 10 \, a^{3} b^{2} x^{4} + 5 \, a^{4} b x^{2} + a^{5}\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right )}{1280 \, {\left (a b^{11} x^{10} + 5 \, a^{2} b^{10} x^{8} + 10 \, a^{3} b^{9} x^{6} + 10 \, a^{4} b^{8} x^{4} + 5 \, a^{5} b^{7} x^{2} + a^{6} b^{6}\right )}}\right ] \]

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)^3,x, algorithm="fricas")

[Out]

[-1/2560*(1930*a*b^5*x^9 + 4740*a^2*b^4*x^7 + 5376*a^3*b^3*x^5 + 2940*a^4*b^2*x^3 + 630*a^5*b*x + 315*(b^5*x^1
0 + 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*b^11*x^10 + 5*a^2*b^10*x^8 + 10*a^3*b^9*x^6 + 10*a^4*b^8*x^4 + 5*a^5*b^7*x^2 + a^6*b^6),
-1/1280*(965*a*b^5*x^9 + 2370*a^2*b^4*x^7 + 2688*a^3*b^3*x^5 + 1470*a^4*b^2*x^3 + 315*a^5*b*x - 315*(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*b^11
*x^10 + 5*a^2*b^10*x^8 + 10*a^3*b^9*x^6 + 10*a^4*b^8*x^4 + 5*a^5*b^7*x^2 + a^6*b^6)]

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giac [A]  time = 0.19, size = 78, normalized size = 0.64 \[ \frac {63 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 \, \sqrt {a b} b^{5}} - \frac {965 \, b^{4} x^{9} + 2370 \, a b^{3} x^{7} + 2688 \, a^{2} b^{2} x^{5} + 1470 \, a^{3} b x^{3} + 315 \, a^{4} x}{1280 \, {\left (b x^{2} + a\right )}^{5} b^{5}} \]

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)^3,x, algorithm="giac")

[Out]

63/256*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^5) - 1/1280*(965*b^4*x^9 + 2370*a*b^3*x^7 + 2688*a^2*b^2*x^5 + 1470*
a^3*b*x^3 + 315*a^4*x)/((b*x^2 + a)^5*b^5)

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maple [A]  time = 0.01, size = 80, normalized size = 0.66 \[ \frac {63 \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 \sqrt {a b}\, b^{5}}+\frac {-\frac {193 x^{9}}{256 b}-\frac {237 a \,x^{7}}{128 b^{2}}-\frac {21 a^{2} x^{5}}{10 b^{3}}-\frac {147 a^{3} x^{3}}{128 b^{4}}-\frac {63 a^{4} x}{256 b^{5}}}{\left (b \,x^{2}+a \right )^{5}} \]

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)^3,x)

[Out]

(-193/256/b*x^9-237/128*a/b^2*x^7-21/10*a^2/b^3*x^5-147/128*a^3/b^4*x^3-63/256*a^4/b^5*x)/(b*x^2+a)^5+63/256/b
^5/(a*b)^(1/2)*arctan(1/(a*b)^(1/2)*b*x)

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maxima [A]  time = 3.07, size = 125, normalized size = 1.03 \[ -\frac {965 \, b^{4} x^{9} + 2370 \, a b^{3} x^{7} + 2688 \, a^{2} b^{2} x^{5} + 1470 \, a^{3} b x^{3} + 315 \, a^{4} x}{1280 \, {\left (b^{10} x^{10} + 5 \, a b^{9} x^{8} + 10 \, a^{2} b^{8} x^{6} + 10 \, a^{3} b^{7} x^{4} + 5 \, a^{4} b^{6} x^{2} + a^{5} b^{5}\right )}} + \frac {63 \, \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{256 \, \sqrt {a b} b^{5}} \]

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)^3,x, algorithm="maxima")

[Out]

-1/1280*(965*b^4*x^9 + 2370*a*b^3*x^7 + 2688*a^2*b^2*x^5 + 1470*a^3*b*x^3 + 315*a^4*x)/(b^10*x^10 + 5*a*b^9*x^
8 + 10*a^2*b^8*x^6 + 10*a^3*b^7*x^4 + 5*a^4*b^6*x^2 + a^5*b^5) + 63/256*arctan(b*x/sqrt(a*b))/(sqrt(a*b)*b^5)

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mupad [B]  time = 4.52, size = 122, normalized size = 1.01 \[ \frac {63\,\mathrm {atan}\left (\frac {\sqrt {b}\,x}{\sqrt {a}}\right )}{256\,\sqrt {a}\,b^{11/2}}-\frac {\frac {193\,x^9}{256\,b}+\frac {237\,a\,x^7}{128\,b^2}+\frac {63\,a^4\,x}{256\,b^5}+\frac {21\,a^2\,x^5}{10\,b^3}+\frac {147\,a^3\,x^3}{128\,b^4}}{a^5+5\,a^4\,b\,x^2+10\,a^3\,b^2\,x^4+10\,a^2\,b^3\,x^6+5\,a\,b^4\,x^8+b^5\,x^{10}} \]

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)^3,x)

[Out]

(63*atan((b^(1/2)*x)/a^(1/2)))/(256*a^(1/2)*b^(11/2)) - ((193*x^9)/(256*b) + (237*a*x^7)/(128*b^2) + (63*a^4*x
)/(256*b^5) + (21*a^2*x^5)/(10*b^3) + (147*a^3*x^3)/(128*b^4))/(a^5 + b^5*x^10 + 5*a^4*b*x^2 + 5*a*b^4*x^8 + 1
0*a^3*b^2*x^4 + 10*a^2*b^3*x^6)

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sympy [A]  time = 0.79, size = 182, normalized size = 1.50 \[ - \frac {63 \sqrt {- \frac {1}{a b^{11}}} \log {\left (- a b^{5} \sqrt {- \frac {1}{a b^{11}}} + x \right )}}{512} + \frac {63 \sqrt {- \frac {1}{a b^{11}}} \log {\left (a b^{5} \sqrt {- \frac {1}{a b^{11}}} + x \right )}}{512} + \frac {- 315 a^{4} x - 1470 a^{3} b x^{3} - 2688 a^{2} b^{2} x^{5} - 2370 a b^{3} x^{7} - 965 b^{4} x^{9}}{1280 a^{5} b^{5} + 6400 a^{4} b^{6} x^{2} + 12800 a^{3} b^{7} x^{4} + 12800 a^{2} b^{8} x^{6} + 6400 a b^{9} x^{8} + 1280 b^{10} x^{10}} \]

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)**3,x)

[Out]

-63*sqrt(-1/(a*b**11))*log(-a*b**5*sqrt(-1/(a*b**11)) + x)/512 + 63*sqrt(-1/(a*b**11))*log(a*b**5*sqrt(-1/(a*b
**11)) + x)/512 + (-315*a**4*x - 1470*a**3*b*x**3 - 2688*a**2*b**2*x**5 - 2370*a*b**3*x**7 - 965*b**4*x**9)/(1
280*a**5*b**5 + 6400*a**4*b**6*x**2 + 12800*a**3*b**7*x**4 + 12800*a**2*b**8*x**6 + 6400*a*b**9*x**8 + 1280*b*
*10*x**10)

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