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

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 77, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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])

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]]

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 = 0.05, size = 59, normalized size = 0.77 \begin {gather*} -\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*(-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])/b^5

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^9}{\left (a^2+2 a b x^2+b^2 x^4\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.78, size = 124, normalized size = 1.61 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

giac [A]  time = 0.16, size = 73, normalized size = 0.95 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 74, normalized size = 0.96 \begin {gather*} -\frac {a^{4}}{6 \left (b \,x^{2}+a \right )^{3} b^{5}}+\frac {a^{3}}{\left (b \,x^{2}+a \right )^{2} b^{5}}+\frac {x^{2}}{2 b^{4}}-\frac {3 a^{2}}{\left (b \,x^{2}+a \right ) b^{5}}-\frac {2 a \ln \left (b \,x^{2}+a \right )}{b^{5}} \end {gather*}

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

________________________________________________________________________________________

maxima [A]  time = 1.40, size = 88, normalized size = 1.14 \begin {gather*} -\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 {gather*}

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

________________________________________________________________________________________

mupad [B]  time = 4.51, size = 88, normalized size = 1.14 \begin {gather*} \frac {x^2}{2\,b^4}-\frac {\frac {13\,a^4}{6\,b}+5\,a^3\,x^2+3\,a^2\,b\,x^4}{a^3\,b^4+3\,a^2\,b^5\,x^2+3\,a\,b^6\,x^4+b^7\,x^6}-\frac {2\,a\,\ln \left (b\,x^2+a\right )}{b^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.59, size = 90, normalized size = 1.17 \begin {gather*} - \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 {gather*}

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)

________________________________________________________________________________________