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

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

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Rubi [A]  time = 0.01, antiderivative size = 19, normalized size of antiderivative = 1.00, 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} \begin {gather*} \frac {x^4}{4 b \left (a x^2+b\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b/x + a*x)^(-3),x]

[Out]

x^4/(4*b*(b + a*x^2)^2)

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]

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]

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*}

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Mathematica [A]  time = 0.01, size = 24, normalized size = 1.26 \begin {gather*} -\frac {2 a x^2+b}{4 a^2 \left (a x^2+b\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b/x + a*x)^(-3),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (\frac {b}{x}+a x\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(b/x + a*x)^(-3),x]

[Out]

IntegrateAlgebraic[(b/x + a*x)^(-3), x]

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fricas [B]  time = 0.37, size = 36, normalized size = 1.89 \begin {gather*} -\frac {2 \, a x^{2} + b}{4 \, {\left (a^{4} x^{4} + 2 \, a^{3} b x^{2} + a^{2} b^{2}\right )}} \end {gather*}

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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giac [A]  time = 0.19, size = 22, normalized size = 1.16 \begin {gather*} -\frac {2 \, a x^{2} + b}{4 \, {\left (a x^{2} + b\right )}^{2} a^{2}} \end {gather*}

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)

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maple [A]  time = 0.05, size = 31, normalized size = 1.63 \begin {gather*} \frac {b}{4 \left (a \,x^{2}+b \right )^{2} a^{2}}-\frac {1}{2 \left (a \,x^{2}+b \right ) a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.33, size = 36, normalized size = 1.89 \begin {gather*} -\frac {2 \, a x^{2} + b}{4 \, {\left (a^{4} x^{4} + 2 \, a^{3} b x^{2} + a^{2} b^{2}\right )}} \end {gather*}

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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mupad [B]  time = 0.04, size = 37, normalized size = 1.95 \begin {gather*} -\frac {\frac {b}{4\,a^2}+\frac {x^2}{2\,a}}{a^2\,x^4+2\,a\,b\,x^2+b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x + b/x)^3,x)

[Out]

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

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sympy [B]  time = 0.28, size = 36, normalized size = 1.89 \begin {gather*} \frac {- 2 a x^{2} - b}{4 a^{4} x^{4} + 8 a^{3} b x^{2} + 4 a^{2} b^{2}} \end {gather*}

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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