3.1836 \(\int (a+\frac{b}{x^2})^3 \, dx\)

Optimal. Leaf size=34 \[ -\frac{3 a^2 b}{x}+a^3 x-\frac{a b^2}{x^3}-\frac{b^3}{5 x^5} \]

[Out]

-b^3/(5*x^5) - (a*b^2)/x^3 - (3*a^2*b)/x + a^3*x

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Rubi [A]  time = 0.0153526, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 9, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {193, 270} \[ -\frac{3 a^2 b}{x}+a^3 x-\frac{a b^2}{x^3}-\frac{b^3}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^3,x]

[Out]

-b^3/(5*x^5) - (a*b^2)/x^3 - (3*a^2*b)/x + a^3*x

Rule 193

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && LtQ[n, 0]
&& IntegerQ[p]

Rule 270

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

Rubi steps

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

Mathematica [A]  time = 0.0046173, size = 34, normalized size = 1. \[ -\frac{3 a^2 b}{x}+a^3 x-\frac{a b^2}{x^3}-\frac{b^3}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^3,x]

[Out]

-b^3/(5*x^5) - (a*b^2)/x^3 - (3*a^2*b)/x + a^3*x

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Maple [A]  time = 0.005, size = 33, normalized size = 1. \begin{align*} -{\frac{{b}^{3}}{5\,{x}^{5}}}-{\frac{{b}^{2}a}{{x}^{3}}}-3\,{\frac{{a}^{2}b}{x}}+{a}^{3}x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+1/x^2*b)^3,x)

[Out]

-1/5*b^3/x^5-a*b^2/x^3-3*a^2*b/x+a^3*x

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Maxima [A]  time = 0.968912, size = 43, normalized size = 1.26 \begin{align*} a^{3} x - \frac{3 \, a^{2} b}{x} - \frac{a b^{2}}{x^{3}} - \frac{b^{3}}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x - 3*a^2*b/x - a*b^2/x^3 - 1/5*b^3/x^5

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Fricas [A]  time = 1.40245, size = 76, normalized size = 2.24 \begin{align*} \frac{5 \, a^{3} x^{6} - 15 \, a^{2} b x^{4} - 5 \, a b^{2} x^{2} - b^{3}}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(5*a^3*x^6 - 15*a^2*b*x^4 - 5*a*b^2*x^2 - b^3)/x^5

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Sympy [A]  time = 0.336675, size = 32, normalized size = 0.94 \begin{align*} a^{3} x - \frac{15 a^{2} b x^{4} + 5 a b^{2} x^{2} + b^{3}}{5 x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**3*x - (15*a**2*b*x**4 + 5*a*b**2*x**2 + b**3)/(5*x**5)

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Giac [A]  time = 1.16378, size = 45, normalized size = 1.32 \begin{align*} a^{3} x - \frac{15 \, a^{2} b x^{4} + 5 \, a b^{2} x^{2} + b^{3}}{5 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x - 1/5*(15*a^2*b*x^4 + 5*a*b^2*x^2 + b^3)/x^5