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

Optimal. Leaf size=34 \[ \frac{a}{6 b^2 \left (a+b x^2\right )^3}-\frac{1}{4 b^2 \left (a+b x^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.0292993, antiderivative size = 34, normalized size of antiderivative = 1., 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} \[ \frac{a}{6 b^2 \left (a+b x^2\right )^3}-\frac{1}{4 b^2 \left (a+b x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rubi steps

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

Mathematica [A]  time = 0.0074374, size = 24, normalized size = 0.71 \[ -\frac{a+3 b x^2}{12 b^2 \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a + 3*b*x^2)/(12*b^2*(a + b*x^2)^3)

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Maple [A]  time = 0.047, size = 31, normalized size = 0.9 \begin{align*}{\frac{a}{6\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{3}}}-{\frac{1}{4\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(b^2*x^4+2*a*b*x^2+a^2)^2,x)

[Out]

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

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Maxima [A]  time = 0.993504, size = 63, normalized size = 1.85 \begin{align*} -\frac{3 \, b x^{2} + a}{12 \,{\left (b^{5} x^{6} + 3 \, a b^{4} x^{4} + 3 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.64251, size = 96, normalized size = 2.82 \begin{align*} -\frac{3 \, b x^{2} + a}{12 \,{\left (b^{5} x^{6} + 3 \, a b^{4} x^{4} + 3 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.556278, size = 48, normalized size = 1.41 \begin{align*} - \frac{a + 3 b x^{2}}{12 a^{3} b^{2} + 36 a^{2} b^{3} x^{2} + 36 a b^{4} x^{4} + 12 b^{5} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3/(b**2*x**4+2*a*b*x**2+a**2)**2,x)

[Out]

-(a + 3*b*x**2)/(12*a**3*b**2 + 36*a**2*b**3*x**2 + 36*a*b**4*x**4 + 12*b**5*x**6)

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Giac [A]  time = 1.21101, size = 30, normalized size = 0.88 \begin{align*} -\frac{3 \, b x^{2} + a}{12 \,{\left (b x^{2} + a\right )}^{3} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(b^2*x^4+2*a*b*x^2+a^2)^2,x, algorithm="giac")

[Out]

-1/12*(3*b*x^2 + a)/((b*x^2 + a)^3*b^2)