3.1832 \(\int (a+\frac{b}{x^2})^3 x^4 \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0145239, antiderivative size = 34, 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} \[ a^2 b x^3+\frac{a^3 x^5}{5}+3 a b^2 x-\frac{b^3}{x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^3*x^4,x]

[Out]

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

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

Mathematica [A]  time = 0.003811, size = 34, normalized size = 1. \[ a^2 b x^3+\frac{a^3 x^5}{5}+3 a b^2 x-\frac{b^3}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^3*x^4,x]

[Out]

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

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Maple [A]  time = 0.003, size = 33, normalized size = 1. \begin{align*} -{\frac{{b}^{3}}{x}}+3\,xa{b}^{2}+{a}^{2}b{x}^{3}+{\frac{{a}^{3}{x}^{5}}{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.957732, size = 43, normalized size = 1.26 \begin{align*} \frac{1}{5} \, a^{3} x^{5} + a^{2} b x^{3} + 3 \, a b^{2} x - \frac{b^{3}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)^3*x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.39149, size = 73, normalized size = 2.15 \begin{align*} \frac{a^{3} x^{6} + 5 \, a^{2} b x^{4} + 15 \, a b^{2} x^{2} - 5 \, b^{3}}{5 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)^3*x^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.25236, size = 29, normalized size = 0.85 \begin{align*} \frac{a^{3} x^{5}}{5} + a^{2} b x^{3} + 3 a b^{2} x - \frac{b^{3}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**2)**3*x**4,x)

[Out]

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

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Giac [A]  time = 1.21607, size = 43, normalized size = 1.26 \begin{align*} \frac{1}{5} \, a^{3} x^{5} + a^{2} b x^{3} + 3 \, a b^{2} x - \frac{b^{3}}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)^3*x^4,x, algorithm="giac")

[Out]

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