3.519 \(\int \frac{x}{(a^2+2 a b x^2+b^2 x^4)^3} \, dx\)

Optimal. Leaf size=16 \[ -\frac{1}{10 b \left (a+b x^2\right )^5} \]

[Out]

-1/(10*b*(a + b*x^2)^5)

________________________________________________________________________________________

Rubi [A]  time = 0.0053067, 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} \[ -\frac{1}{10 b \left (a+b x^2\right )^5} \]

Antiderivative was successfully verified.

[In]

Int[x/(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]

[Out]

-1/(10*b*(a + b*x^2)^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 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 )^3} \, dx &=b^6 \int \frac{x}{\left (a b+b^2 x^2\right )^6} \, dx\\ &=-\frac{1}{10 b \left (a+b x^2\right )^5}\\ \end{align*}

Mathematica [A]  time = 0.0028531, size = 16, normalized size = 1. \[ -\frac{1}{10 b \left (a+b x^2\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a^2 + 2*a*b*x^2 + b^2*x^4)^3,x]

[Out]

-1/(10*b*(a + b*x^2)^5)

________________________________________________________________________________________

Maple [A]  time = 0.047, size = 15, normalized size = 0.9 \begin{align*} -{\frac{1}{10\,b \left ( b{x}^{2}+a \right ) ^{5}}} \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)^3,x)

[Out]

-1/10/b/(b*x^2+a)^5

________________________________________________________________________________________

Maxima [B]  time = 1.08085, size = 80, normalized size = 5. \begin{align*} -\frac{1}{10 \,{\left (b^{6} x^{10} + 5 \, a b^{5} x^{8} + 10 \, a^{2} b^{4} x^{6} + 10 \, a^{3} b^{3} x^{4} + 5 \, a^{4} b^{2} x^{2} + a^{5} 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)^3,x, algorithm="maxima")

[Out]

-1/10/(b^6*x^10 + 5*a*b^5*x^8 + 10*a^2*b^4*x^6 + 10*a^3*b^3*x^4 + 5*a^4*b^2*x^2 + a^5*b)

________________________________________________________________________________________

Fricas [B]  time = 1.68132, size = 122, normalized size = 7.62 \begin{align*} -\frac{1}{10 \,{\left (b^{6} x^{10} + 5 \, a b^{5} x^{8} + 10 \, a^{2} b^{4} x^{6} + 10 \, a^{3} b^{3} x^{4} + 5 \, a^{4} b^{2} x^{2} + a^{5} 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)^3,x, algorithm="fricas")

[Out]

-1/10/(b^6*x^10 + 5*a*b^5*x^8 + 10*a^2*b^4*x^6 + 10*a^3*b^3*x^4 + 5*a^4*b^2*x^2 + a^5*b)

________________________________________________________________________________________

Sympy [B]  time = 0.961724, size = 63, normalized size = 3.94 \begin{align*} - \frac{1}{10 a^{5} b + 50 a^{4} b^{2} x^{2} + 100 a^{3} b^{3} x^{4} + 100 a^{2} b^{4} x^{6} + 50 a b^{5} x^{8} + 10 b^{6} x^{10}} \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)**3,x)

[Out]

-1/(10*a**5*b + 50*a**4*b**2*x**2 + 100*a**3*b**3*x**4 + 100*a**2*b**4*x**6 + 50*a*b**5*x**8 + 10*b**6*x**10)

________________________________________________________________________________________

Giac [A]  time = 1.15108, size = 19, normalized size = 1.19 \begin{align*} -\frac{1}{10 \,{\left (b x^{2} + a\right )}^{5} 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)^3,x, algorithm="giac")

[Out]

-1/10/((b*x^2 + a)^5*b)