3.1654 \(\int (a+\frac{b}{x})^2 x^{3/2} \, dx\)

Optimal. Leaf size=34 \[ \frac{2}{5} a^2 x^{5/2}+\frac{4}{3} a b x^{3/2}+2 b^2 \sqrt{x} \]

[Out]

2*b^2*Sqrt[x] + (4*a*b*x^(3/2))/3 + (2*a^2*x^(5/2))/5

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Rubi [A]  time = 0.0092085, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {263, 43} \[ \frac{2}{5} a^2 x^{5/2}+\frac{4}{3} a b x^{3/2}+2 b^2 \sqrt{x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^2*x^(3/2),x]

[Out]

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

Mathematica [A]  time = 0.0069384, size = 28, normalized size = 0.82 \[ \frac{2}{15} \sqrt{x} \left (3 a^2 x^2+10 a b x+15 b^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^2*x^(3/2),x]

[Out]

(2*Sqrt[x]*(15*b^2 + 10*a*b*x + 3*a^2*x^2))/15

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Maple [A]  time = 0.003, size = 25, normalized size = 0.7 \begin{align*}{\frac{6\,{a}^{2}{x}^{2}+20\,xab+30\,{b}^{2}}{15}\sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*a^2*x^2+10*a*b*x+15*b^2)*x^(1/2)

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Maxima [A]  time = 0.983316, size = 35, normalized size = 1.03 \begin{align*} \frac{2}{15} \,{\left (3 \, a^{2} + \frac{10 \, a b}{x} + \frac{15 \, b^{2}}{x^{2}}\right )} x^{\frac{5}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*a^2 + 10*a*b/x + 15*b^2/x^2)*x^(5/2)

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Fricas [A]  time = 1.68799, size = 62, normalized size = 1.82 \begin{align*} \frac{2}{15} \,{\left (3 \, a^{2} x^{2} + 10 \, a b x + 15 \, b^{2}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(3*a^2*x^2 + 10*a*b*x + 15*b^2)*sqrt(x)

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Sympy [A]  time = 1.00118, size = 32, normalized size = 0.94 \begin{align*} \frac{2 a^{2} x^{\frac{5}{2}}}{5} + \frac{4 a b x^{\frac{3}{2}}}{3} + 2 b^{2} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16179, size = 32, normalized size = 0.94 \begin{align*} \frac{2}{5} \, a^{2} x^{\frac{5}{2}} + \frac{4}{3} \, a b x^{\frac{3}{2}} + 2 \, b^{2} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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