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

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

[Out]

-1/(2*b*(a + b*x^2))

________________________________________________________________________________________

Rubi [A]  time = 0.0053895, antiderivative size = 16, normalized size of antiderivative = 1., 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} \[ -\frac{1}{2 b \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[x/(a^2 + 2*a*b*x^2 + b^2*x^4),x]

[Out]

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

Mathematica [A]  time = 0.0028788, size = 16, normalized size = 1. \[ -\frac{1}{2 b \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(2*b*(a + b*x^2))

________________________________________________________________________________________

Maple [A]  time = 0.046, size = 15, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,b \left ( b{x}^{2}+a \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.04055, size = 20, normalized size = 1.25 \begin{align*} -\frac{1}{2 \,{\left (b^{2} x^{2} + a b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(b^2*x^2 + a*b)

________________________________________________________________________________________

Fricas [A]  time = 1.57589, size = 30, normalized size = 1.88 \begin{align*} -\frac{1}{2 \,{\left (b^{2} x^{2} + a b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(b^2*x^2 + a*b)

________________________________________________________________________________________

Sympy [A]  time = 0.293952, size = 15, normalized size = 0.94 \begin{align*} - \frac{1}{2 a b + 2 b^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(2*a*b + 2*b**2*x**2)

________________________________________________________________________________________

Giac [A]  time = 1.15217, size = 19, normalized size = 1.19 \begin{align*} -\frac{1}{2 \,{\left (b x^{2} + a\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/((b*x^2 + a)*b)