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

Optimal. Leaf size=16 \[ -\frac {1}{6 b \left (a+b x^2\right )^3} \]

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Rubi [A]  time = 0.00, antiderivative size = 16, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {28, 261} \begin {gather*} -\frac {1}{6 b \left (a+b x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(6*b*(a + b*x^2)^3)

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 261

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

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 16, normalized size = 1.00 \begin {gather*} -\frac {1}{6 b \left (a+b x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*1/(b*(a + b*x^2)^3)

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

[Out]

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

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fricas [B]  time = 0.80, size = 37, normalized size = 2.31 \begin {gather*} -\frac {1}{6 \, {\left (b^{4} x^{6} + 3 \, a b^{3} x^{4} + 3 \, a^{2} b^{2} x^{2} + a^{3} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 14, normalized size = 0.88 \begin {gather*} -\frac {1}{6 \, {\left (b x^{2} + a\right )}^{3} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/((b*x^2 + a)^3*b)

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maple [A]  time = 0.00, size = 15, normalized size = 0.94 \begin {gather*} -\frac {1}{6 \left (b \,x^{2}+a \right )^{3} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.34, size = 37, normalized size = 2.31 \begin {gather*} -\frac {1}{6 \, {\left (b^{4} x^{6} + 3 \, a b^{3} x^{4} + 3 \, a^{2} b^{2} x^{2} + a^{3} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.28, size = 39, normalized size = 2.44 \begin {gather*} -\frac {1}{6\,a^3\,b+18\,a^2\,b^2\,x^2+18\,a\,b^3\,x^4+6\,b^4\,x^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(6*a^3*b + 6*b^4*x^6 + 18*a*b^3*x^4 + 18*a^2*b^2*x^2)

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sympy [B]  time = 0.33, size = 39, normalized size = 2.44 \begin {gather*} - \frac {1}{6 a^{3} b + 18 a^{2} b^{2} x^{2} + 18 a b^{3} x^{4} + 6 b^{4} x^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(6*a**3*b + 18*a**2*b**2*x**2 + 18*a*b**3*x**4 + 6*b**4*x**6)

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