3.1838 \(\int \frac{(a+\frac{b}{x^2})^3}{x^2} \, dx\)

Optimal. Leaf size=39 \[ -\frac{a^2 b}{x^3}-\frac{a^3}{x}-\frac{3 a b^2}{5 x^5}-\frac{b^3}{7 x^7} \]

[Out]

-b^3/(7*x^7) - (3*a*b^2)/(5*x^5) - (a^2*b)/x^3 - a^3/x

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Rubi [A]  time = 0.015652, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {263, 270} \[ -\frac{a^2 b}{x^3}-\frac{a^3}{x}-\frac{3 a b^2}{5 x^5}-\frac{b^3}{7 x^7} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^3/x^2,x]

[Out]

-b^3/(7*x^7) - (3*a*b^2)/(5*x^5) - (a^2*b)/x^3 - a^3/x

Rule 263

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

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 \frac{\left (a+\frac{b}{x^2}\right )^3}{x^2} \, dx &=\int \frac{\left (b+a x^2\right )^3}{x^8} \, dx\\ &=\int \left (\frac{b^3}{x^8}+\frac{3 a b^2}{x^6}+\frac{3 a^2 b}{x^4}+\frac{a^3}{x^2}\right ) \, dx\\ &=-\frac{b^3}{7 x^7}-\frac{3 a b^2}{5 x^5}-\frac{a^2 b}{x^3}-\frac{a^3}{x}\\ \end{align*}

Mathematica [A]  time = 0.0037484, size = 39, normalized size = 1. \[ -\frac{a^2 b}{x^3}-\frac{a^3}{x}-\frac{3 a b^2}{5 x^5}-\frac{b^3}{7 x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^3/x^2,x]

[Out]

-b^3/(7*x^7) - (3*a*b^2)/(5*x^5) - (a^2*b)/x^3 - a^3/x

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Maple [A]  time = 0.005, size = 36, normalized size = 0.9 \begin{align*} -{\frac{{b}^{3}}{7\,{x}^{7}}}-{\frac{3\,{b}^{2}a}{5\,{x}^{5}}}-{\frac{{a}^{2}b}{{x}^{3}}}-{\frac{{a}^{3}}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*b^3/x^7-3/5*a*b^2/x^5-a^2*b/x^3-a^3/x

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Maxima [A]  time = 1.01607, size = 50, normalized size = 1.28 \begin{align*} -\frac{35 \, a^{3} x^{6} + 35 \, a^{2} b x^{4} + 21 \, a b^{2} x^{2} + 5 \, b^{3}}{35 \, x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35*(35*a^3*x^6 + 35*a^2*b*x^4 + 21*a*b^2*x^2 + 5*b^3)/x^7

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Fricas [A]  time = 1.37189, size = 84, normalized size = 2.15 \begin{align*} -\frac{35 \, a^{3} x^{6} + 35 \, a^{2} b x^{4} + 21 \, a b^{2} x^{2} + 5 \, b^{3}}{35 \, x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35*(35*a^3*x^6 + 35*a^2*b*x^4 + 21*a*b^2*x^2 + 5*b^3)/x^7

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Sympy [A]  time = 0.383328, size = 39, normalized size = 1. \begin{align*} - \frac{35 a^{3} x^{6} + 35 a^{2} b x^{4} + 21 a b^{2} x^{2} + 5 b^{3}}{35 x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(35*a**3*x**6 + 35*a**2*b*x**4 + 21*a*b**2*x**2 + 5*b**3)/(35*x**7)

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Giac [A]  time = 1.16844, size = 50, normalized size = 1.28 \begin{align*} -\frac{35 \, a^{3} x^{6} + 35 \, a^{2} b x^{4} + 21 \, a b^{2} x^{2} + 5 \, b^{3}}{35 \, x^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35*(35*a^3*x^6 + 35*a^2*b*x^4 + 21*a*b^2*x^2 + 5*b^3)/x^7