3.343 \(\int \frac{1}{(\frac{b}{x^5}+a x^3)^3} \, dx\)

Optimal. Leaf size=19 \[ \frac{x^{16}}{16 b \left (a x^8+b\right )^2} \]

[Out]

x^16/(16*b*(b + a*x^8)^2)

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Rubi [A]  time = 0.0057949, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1593, 264} \[ \frac{x^{16}}{16 b \left (a x^8+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(b/x^5 + a*x^3)^(-3),x]

[Out]

x^16/(16*b*(b + a*x^8)^2)

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{1}{\left (\frac{b}{x^5}+a x^3\right )^3} \, dx &=\int \frac{x^{15}}{\left (b+a x^8\right )^3} \, dx\\ &=\frac{x^{16}}{16 b \left (b+a x^8\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.0111089, size = 24, normalized size = 1.26 \[ -\frac{2 a x^8+b}{16 a^2 \left (a x^8+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(b/x^5 + a*x^3)^(-3),x]

[Out]

-(b + 2*a*x^8)/(16*a^2*(b + a*x^8)^2)

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Maple [A]  time = 0.008, size = 31, normalized size = 1.6 \begin{align*}{\frac{b}{16\,{a}^{2} \left ( a{x}^{8}+b \right ) ^{2}}}-{\frac{1}{8\,{a}^{2} \left ( a{x}^{8}+b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b/x^5+a*x^3)^3,x)

[Out]

1/16*b/a^2/(a*x^8+b)^2-1/8/a^2/(a*x^8+b)

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Maxima [B]  time = 1.10908, size = 49, normalized size = 2.58 \begin{align*} -\frac{2 \, a x^{8} + b}{16 \,{\left (a^{4} x^{16} + 2 \, a^{3} b x^{8} + a^{2} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b/x^5+a*x^3)^3,x, algorithm="maxima")

[Out]

-1/16*(2*a*x^8 + b)/(a^4*x^16 + 2*a^3*b*x^8 + a^2*b^2)

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Fricas [B]  time = 0.820367, size = 76, normalized size = 4. \begin{align*} -\frac{2 \, a x^{8} + b}{16 \,{\left (a^{4} x^{16} + 2 \, a^{3} b x^{8} + a^{2} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b/x^5+a*x^3)^3,x, algorithm="fricas")

[Out]

-1/16*(2*a*x^8 + b)/(a^4*x^16 + 2*a^3*b*x^8 + a^2*b^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b/x**5+a*x**3)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.17288, size = 30, normalized size = 1.58 \begin{align*} -\frac{2 \, a x^{8} + b}{16 \,{\left (a x^{8} + b\right )}^{2} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b/x^5+a*x^3)^3,x, algorithm="giac")

[Out]

-1/16*(2*a*x^8 + b)/((a*x^8 + b)^2*a^2)