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

Optimal. Leaf size=19 \[ \frac{x^4}{4 b \left (a x^2+b\right )^2} \]

[Out]

x^4/(4*b*(b + a*x^2)^2)

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Rubi [A]  time = 0.0054465, antiderivative size = 19, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {1593, 264} \[ \frac{x^4}{4 b \left (a x^2+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(b/x + a*x)^(-3),x]

[Out]

x^4/(4*b*(b + a*x^2)^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}+a x\right )^3} \, dx &=\int \frac{x^3}{\left (b+a x^2\right )^3} \, dx\\ &=\frac{x^4}{4 b \left (b+a x^2\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.0084775, size = 24, normalized size = 1.26 \[ -\frac{2 a x^2+b}{4 a^2 \left (a x^2+b\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(b/x + a*x)^(-3),x]

[Out]

-(b + 2*a*x^2)/(4*a^2*(b + a*x^2)^2)

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Maple [A]  time = 0.007, size = 31, normalized size = 1.6 \begin{align*}{\frac{b}{4\,{a}^{2} \left ( a{x}^{2}+b \right ) ^{2}}}-{\frac{1}{2\,{a}^{2} \left ( a{x}^{2}+b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.15915, size = 49, normalized size = 2.58 \begin{align*} -\frac{2 \, a x^{2} + b}{4 \,{\left (a^{4} x^{4} + 2 \, a^{3} b x^{2} + a^{2} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 0.784184, size = 73, normalized size = 3.84 \begin{align*} -\frac{2 \, a x^{2} + b}{4 \,{\left (a^{4} x^{4} + 2 \, a^{3} b x^{2} + a^{2} b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 0.482015, size = 36, normalized size = 1.89 \begin{align*} - \frac{2 a x^{2} + b}{4 a^{4} x^{4} + 8 a^{3} b x^{2} + 4 a^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15051, size = 30, normalized size = 1.58 \begin{align*} -\frac{2 \, a x^{2} + b}{4 \,{\left (a x^{2} + b\right )}^{2} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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