3.1561 \(\int (a+\frac{b}{x})^2 x^3 \, dx\)

Optimal. Leaf size=30 \[ \frac{a^2 x^4}{4}+\frac{2}{3} a b x^3+\frac{b^2 x^2}{2} \]

[Out]

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

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Rubi [A]  time = 0.012121, antiderivative size = 30, 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, 43} \[ \frac{a^2 x^4}{4}+\frac{2}{3} a b x^3+\frac{b^2 x^2}{2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^2*x^3,x]

[Out]

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

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0012466, size = 30, normalized size = 1. \[ \frac{a^2 x^4}{4}+\frac{2}{3} a b x^3+\frac{b^2 x^2}{2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^2*x^3,x]

[Out]

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

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Maple [A]  time = 0.001, size = 25, normalized size = 0.8 \begin{align*}{\frac{{b}^{2}{x}^{2}}{2}}+{\frac{2\,{x}^{3}ab}{3}}+{\frac{{a}^{2}{x}^{4}}{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b^2*x^2+2/3*x^3*a*b+1/4*a^2*x^4

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Maxima [A]  time = 0.946022, size = 32, normalized size = 1.07 \begin{align*} \frac{1}{4} \, a^{2} x^{4} + \frac{2}{3} \, a b x^{3} + \frac{1}{2} \, b^{2} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*x^4 + 2/3*a*b*x^3 + 1/2*b^2*x^2

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Fricas [A]  time = 1.38786, size = 55, normalized size = 1.83 \begin{align*} \frac{1}{4} \, a^{2} x^{4} + \frac{2}{3} \, a b x^{3} + \frac{1}{2} \, b^{2} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*x^4 + 2/3*a*b*x^3 + 1/2*b^2*x^2

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Sympy [A]  time = 0.058638, size = 26, normalized size = 0.87 \begin{align*} \frac{a^{2} x^{4}}{4} + \frac{2 a b x^{3}}{3} + \frac{b^{2} x^{2}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*x**4/4 + 2*a*b*x**3/3 + b**2*x**2/2

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Giac [A]  time = 1.17525, size = 32, normalized size = 1.07 \begin{align*} \frac{1}{4} \, a^{2} x^{4} + \frac{2}{3} \, a b x^{3} + \frac{1}{2} \, b^{2} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a^2*x^4 + 2/3*a*b*x^3 + 1/2*b^2*x^2