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

Optimal. Leaf size=70 \[ -\frac {a^4}{2 b^5 \left (a+b x^2\right )}-\frac {2 a^3 \log \left (a+b x^2\right )}{b^5}+\frac {3 a^2 x^2}{2 b^4}-\frac {a x^4}{2 b^3}+\frac {x^6}{6 b^2} \]

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Rubi [A]  time = 0.06, antiderivative size = 70, 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 {3 a^2 x^2}{2 b^4}-\frac {a^4}{2 b^5 \left (a+b x^2\right )}-\frac {2 a^3 \log \left (a+b x^2\right )}{b^5}-\frac {a x^4}{2 b^3}+\frac {x^6}{6 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.02, size = 60, normalized size = 0.86 \begin {gather*} \frac {-\frac {3 a^4}{a+b x^2}-12 a^3 \log \left (a+b x^2\right )+9 a^2 b x^2-3 a b^2 x^4+b^3 x^6}{6 b^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

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

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fricas [A]  time = 0.59, size = 81, normalized size = 1.16 \begin {gather*} \frac {b^{4} x^{8} - 2 \, a b^{3} x^{6} + 6 \, a^{2} b^{2} x^{4} + 9 \, a^{3} b x^{2} - 3 \, a^{4} - 12 \, {\left (a^{3} b x^{2} + a^{4}\right )} \log \left (b x^{2} + a\right )}{6 \, {\left (b^{6} x^{2} + a 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),x, algorithm="fricas")

[Out]

1/6*(b^4*x^8 - 2*a*b^3*x^6 + 6*a^2*b^2*x^4 + 9*a^3*b*x^2 - 3*a^4 - 12*(a^3*b*x^2 + a^4)*log(b*x^2 + a))/(b^6*x
^2 + a*b^5)

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giac [A]  time = 0.17, size = 80, normalized size = 1.14 \begin {gather*} -\frac {2 \, a^{3} \log \left ({\left | b x^{2} + a \right |}\right )}{b^{5}} + \frac {b^{4} x^{6} - 3 \, a b^{3} x^{4} + 9 \, a^{2} b^{2} x^{2}}{6 \, b^{6}} + \frac {4 \, a^{3} b x^{2} + 3 \, a^{4}}{2 \, {\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),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 63, normalized size = 0.90 \begin {gather*} \frac {x^{6}}{6 b^{2}}-\frac {a \,x^{4}}{2 b^{3}}+\frac {3 a^{2} x^{2}}{2 b^{4}}-\frac {a^{4}}{2 \left (b \,x^{2}+a \right ) b^{5}}-\frac {2 a^{3} \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),x)

[Out]

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

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maxima [A]  time = 1.35, size = 65, normalized size = 0.93 \begin {gather*} -\frac {a^{4}}{2 \, {\left (b^{6} x^{2} + a b^{5}\right )}} - \frac {2 \, a^{3} \log \left (b x^{2} + a\right )}{b^{5}} + \frac {b^{2} x^{6} - 3 \, a b x^{4} + 9 \, a^{2} x^{2}}{6 \, 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),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.04, size = 68, normalized size = 0.97 \begin {gather*} \frac {x^6}{6\,b^2}-\frac {a^4}{2\,b\,\left (b^5\,x^2+a\,b^4\right )}-\frac {a\,x^4}{2\,b^3}-\frac {2\,a^3\,\ln \left (b\,x^2+a\right )}{b^5}+\frac {3\,a^2\,x^2}{2\,b^4} \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),x)

[Out]

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

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sympy [A]  time = 0.29, size = 66, normalized size = 0.94 \begin {gather*} - \frac {a^{4}}{2 a b^{5} + 2 b^{6} x^{2}} - \frac {2 a^{3} \log {\left (a + b x^{2} \right )}}{b^{5}} + \frac {3 a^{2} x^{2}}{2 b^{4}} - \frac {a x^{4}}{2 b^{3}} + \frac {x^{6}}{6 b^{2}} \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),x)

[Out]

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

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