3.494 \(\int \frac{x^9}{(a^2+2 a b x^2+b^2 x^4)^2} \, dx\)

Optimal. Leaf size=77 \[ -\frac{a^4}{6 b^5 \left (a+b x^2\right )^3}+\frac{a^3}{b^5 \left (a+b x^2\right )^2}-\frac{3 a^2}{b^5 \left (a+b x^2\right )}-\frac{2 a \log \left (a+b x^2\right )}{b^5}+\frac{x^2}{2 b^4} \]

[Out]

x^2/(2*b^4) - a^4/(6*b^5*(a + b*x^2)^3) + a^3/(b^5*(a + b*x^2)^2) - (3*a^2)/(b^5*(a + b*x^2)) - (2*a*Log[a + b
*x^2])/b^5

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Rubi [A]  time = 0.0733187, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {28, 266, 43} \[ -\frac{a^4}{6 b^5 \left (a+b x^2\right )^3}+\frac{a^3}{b^5 \left (a+b x^2\right )^2}-\frac{3 a^2}{b^5 \left (a+b x^2\right )}-\frac{2 a \log \left (a+b x^2\right )}{b^5}+\frac{x^2}{2 b^4} \]

Antiderivative was successfully verified.

[In]

Int[x^9/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

x^2/(2*b^4) - a^4/(6*b^5*(a + b*x^2)^3) + a^3/(b^5*(a + b*x^2)^2) - (3*a^2)/(b^5*(a + b*x^2)) - (2*a*Log[a + b
*x^2])/b^5

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx &=b^4 \int \frac{x^9}{\left (a b+b^2 x^2\right )^4} \, dx\\ &=\frac{1}{2} b^4 \operatorname{Subst}\left (\int \frac{x^4}{\left (a b+b^2 x\right )^4} \, dx,x,x^2\right )\\ &=\frac{1}{2} b^4 \operatorname{Subst}\left (\int \left (\frac{1}{b^8}+\frac{a^4}{b^8 (a+b x)^4}-\frac{4 a^3}{b^8 (a+b x)^3}+\frac{6 a^2}{b^8 (a+b x)^2}-\frac{4 a}{b^8 (a+b x)}\right ) \, dx,x,x^2\right )\\ &=\frac{x^2}{2 b^4}-\frac{a^4}{6 b^5 \left (a+b x^2\right )^3}+\frac{a^3}{b^5 \left (a+b x^2\right )^2}-\frac{3 a^2}{b^5 \left (a+b x^2\right )}-\frac{2 a \log \left (a+b x^2\right )}{b^5}\\ \end{align*}

Mathematica [A]  time = 0.0481209, size = 59, normalized size = 0.77 \[ -\frac{\frac{a^2 \left (13 a^2+30 a b x^2+18 b^2 x^4\right )}{\left (a+b x^2\right )^3}+12 a \log \left (a+b x^2\right )-3 b x^2}{6 b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^9/(a^2 + 2*a*b*x^2 + b^2*x^4)^2,x]

[Out]

-(-3*b*x^2 + (a^2*(13*a^2 + 30*a*b*x^2 + 18*b^2*x^4))/(a + b*x^2)^3 + 12*a*Log[a + b*x^2])/(6*b^5)

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Maple [A]  time = 0.051, size = 74, normalized size = 1. \begin{align*}{\frac{{x}^{2}}{2\,{b}^{4}}}-{\frac{{a}^{4}}{6\,{b}^{5} \left ( b{x}^{2}+a \right ) ^{3}}}+{\frac{{a}^{3}}{{b}^{5} \left ( b{x}^{2}+a \right ) ^{2}}}-3\,{\frac{{a}^{2}}{{b}^{5} \left ( b{x}^{2}+a \right ) }}-2\,{\frac{a\ln \left ( b{x}^{2}+a \right ) }{{b}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b^2*x^4+2*a*b*x^2+a^2)^2,x)

[Out]

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

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Maxima [A]  time = 0.997592, size = 119, normalized size = 1.55 \begin{align*} -\frac{18 \, a^{2} b^{2} x^{4} + 30 \, a^{3} b x^{2} + 13 \, a^{4}}{6 \,{\left (b^{8} x^{6} + 3 \, a b^{7} x^{4} + 3 \, a^{2} b^{6} x^{2} + a^{3} b^{5}\right )}} + \frac{x^{2}}{2 \, b^{4}} - \frac{2 \, a \log \left (b x^{2} + a\right )}{b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")

[Out]

-1/6*(18*a^2*b^2*x^4 + 30*a^3*b*x^2 + 13*a^4)/(b^8*x^6 + 3*a*b^7*x^4 + 3*a^2*b^6*x^2 + a^3*b^5) + 1/2*x^2/b^4
- 2*a*log(b*x^2 + a)/b^5

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Fricas [A]  time = 1.72482, size = 255, normalized size = 3.31 \begin{align*} \frac{3 \, b^{4} x^{8} + 9 \, a b^{3} x^{6} - 9 \, a^{2} b^{2} x^{4} - 27 \, a^{3} b x^{2} - 13 \, a^{4} - 12 \,{\left (a b^{3} x^{6} + 3 \, a^{2} b^{2} x^{4} + 3 \, a^{3} b x^{2} + a^{4}\right )} \log \left (b x^{2} + a\right )}{6 \,{\left (b^{8} x^{6} + 3 \, a b^{7} x^{4} + 3 \, a^{2} b^{6} x^{2} + a^{3} b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")

[Out]

1/6*(3*b^4*x^8 + 9*a*b^3*x^6 - 9*a^2*b^2*x^4 - 27*a^3*b*x^2 - 13*a^4 - 12*(a*b^3*x^6 + 3*a^2*b^2*x^4 + 3*a^3*b
*x^2 + a^4)*log(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)

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Sympy [A]  time = 0.741864, size = 88, normalized size = 1.14 \begin{align*} - \frac{2 a \log{\left (a + b x^{2} \right )}}{b^{5}} - \frac{13 a^{4} + 30 a^{3} b x^{2} + 18 a^{2} b^{2} x^{4}}{6 a^{3} b^{5} + 18 a^{2} b^{6} x^{2} + 18 a b^{7} x^{4} + 6 b^{8} x^{6}} + \frac{x^{2}}{2 b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(b**2*x**4+2*a*b*x**2+a**2)**2,x)

[Out]

-2*a*log(a + b*x**2)/b**5 - (13*a**4 + 30*a**3*b*x**2 + 18*a**2*b**2*x**4)/(6*a**3*b**5 + 18*a**2*b**6*x**2 +
18*a*b**7*x**4 + 6*b**8*x**6) + x**2/(2*b**4)

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Giac [A]  time = 1.11294, size = 99, normalized size = 1.29 \begin{align*} \frac{x^{2}}{2 \, b^{4}} - \frac{2 \, a \log \left ({\left | b x^{2} + a \right |}\right )}{b^{5}} + \frac{22 \, a b^{3} x^{6} + 48 \, a^{2} b^{2} x^{4} + 36 \, a^{3} b x^{2} + 9 \, a^{4}}{6 \,{\left (b x^{2} + a\right )}^{3} b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")

[Out]

1/2*x^2/b^4 - 2*a*log(abs(b*x^2 + a))/b^5 + 1/6*(22*a*b^3*x^6 + 48*a^2*b^2*x^4 + 36*a^3*b*x^2 + 9*a^4)/((b*x^2
+ a)^3*b^5)