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

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

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.01, size = 34, normalized size = 1.00 \[ a^3 x-\frac {3 a^2 b}{x}-\frac {a b^2}{x^3}-\frac {b^3}{5 x^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^3,x]

[Out]

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

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fricas [A]  time = 1.18, size = 37, normalized size = 1.09 \[ \frac {5 \, a^{3} x^{6} - 15 \, a^{2} b x^{4} - 5 \, a b^{2} x^{2} - b^{3}}{5 \, x^{5}} \]

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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giac [A]  time = 0.16, size = 33, normalized size = 0.97 \[ a^{3} x - \frac {15 \, a^{2} b x^{4} + 5 \, a b^{2} x^{2} + b^{3}}{5 \, x^{5}} \]

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

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maple [A]  time = 0.01, size = 33, normalized size = 0.97 \[ a^{3} x -\frac {3 a^{2} b}{x}-\frac {a \,b^{2}}{x^{3}}-\frac {b^{3}}{5 x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^2)^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 = 0.88, size = 32, normalized size = 0.94 \[ a^{3} x - \frac {3 \, a^{2} b}{x} - \frac {a b^{2}}{x^{3}} - \frac {b^{3}}{5 \, x^{5}} \]

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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mupad [B]  time = 0.03, size = 34, normalized size = 1.00 \[ a^3\,x-\frac {3\,a^2\,b\,x^4+a\,b^2\,x^2+\frac {b^3}{5}}{x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^2)^3,x)

[Out]

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

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sympy [A]  time = 0.21, size = 34, normalized size = 1.00 \[ a^{3} x + \frac {- 15 a^{2} b x^{4} - 5 a b^{2} x^{2} - b^{3}}{5 x^{5}} \]

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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