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

Optimal. Leaf size=34 \[ \frac{a}{10 b^2 \left (a+b x^2\right )^5}-\frac{1}{8 b^2 \left (a+b x^2\right )^4} \]

[Out]

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

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Rubi [A]  time = 0.0307099, 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} \[ \frac{a}{10 b^2 \left (a+b x^2\right )^5}-\frac{1}{8 b^2 \left (a+b x^2\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 )^3} \, dx &=b^6 \int \frac{x^3}{\left (a b+b^2 x^2\right )^6} \, dx\\ &=\frac{1}{2} b^6 \operatorname{Subst}\left (\int \frac{x}{\left (a b+b^2 x\right )^6} \, dx,x,x^2\right )\\ &=\frac{1}{2} b^6 \operatorname{Subst}\left (\int \left (-\frac{a}{b^7 (a+b x)^6}+\frac{1}{b^7 (a+b x)^5}\right ) \, dx,x,x^2\right )\\ &=\frac{a}{10 b^2 \left (a+b x^2\right )^5}-\frac{1}{8 b^2 \left (a+b x^2\right )^4}\\ \end{align*}

Mathematica [A]  time = 0.0077173, size = 24, normalized size = 0.71 \[ -\frac{a+5 b x^2}{40 b^2 \left (a+b x^2\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a + 5*b*x^2)/(40*b^2*(a + b*x^2)^5)

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Maple [A]  time = 0.049, size = 31, normalized size = 0.9 \begin{align*}{\frac{a}{10\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{5}}}-{\frac{1}{8\,{b}^{2} \left ( b{x}^{2}+a \right ) ^{4}}} \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)^3,x)

[Out]

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

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Maxima [B]  time = 1.42217, size = 93, normalized size = 2.74 \begin{align*} -\frac{5 \, b x^{2} + a}{40 \,{\left (b^{7} x^{10} + 5 \, a b^{6} x^{8} + 10 \, a^{2} b^{5} x^{6} + 10 \, a^{3} b^{4} x^{4} + 5 \, a^{4} b^{3} x^{2} + a^{5} 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)^3,x, algorithm="maxima")

[Out]

-1/40*(5*b*x^2 + a)/(b^7*x^10 + 5*a*b^6*x^8 + 10*a^2*b^5*x^6 + 10*a^3*b^4*x^4 + 5*a^4*b^3*x^2 + a^5*b^2)

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Fricas [B]  time = 1.92577, size = 143, normalized size = 4.21 \begin{align*} -\frac{5 \, b x^{2} + a}{40 \,{\left (b^{7} x^{10} + 5 \, a b^{6} x^{8} + 10 \, a^{2} b^{5} x^{6} + 10 \, a^{3} b^{4} x^{4} + 5 \, a^{4} b^{3} x^{2} + a^{5} 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)^3,x, algorithm="fricas")

[Out]

-1/40*(5*b*x^2 + a)/(b^7*x^10 + 5*a*b^6*x^8 + 10*a^2*b^5*x^6 + 10*a^3*b^4*x^4 + 5*a^4*b^3*x^2 + a^5*b^2)

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Sympy [B]  time = 1.01158, size = 71, normalized size = 2.09 \begin{align*} - \frac{a + 5 b x^{2}}{40 a^{5} b^{2} + 200 a^{4} b^{3} x^{2} + 400 a^{3} b^{4} x^{4} + 400 a^{2} b^{5} x^{6} + 200 a b^{6} x^{8} + 40 b^{7} x^{10}} \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)**3,x)

[Out]

-(a + 5*b*x**2)/(40*a**5*b**2 + 200*a**4*b**3*x**2 + 400*a**3*b**4*x**4 + 400*a**2*b**5*x**6 + 200*a*b**6*x**8
+ 40*b**7*x**10)

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Giac [A]  time = 1.14657, size = 30, normalized size = 0.88 \begin{align*} -\frac{5 \, b x^{2} + a}{40 \,{\left (b x^{2} + a\right )}^{5} 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)^3,x, algorithm="giac")

[Out]

-1/40*(5*b*x^2 + a)/((b*x^2 + a)^5*b^2)