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]

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

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Rubi [A]  time = 0.01, antiderivative size = 34, normalized size of antiderivative = 1.00, 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 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])

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]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 1.01, size = 24, normalized size = 0.71 \[ \frac {2}{15} \, {\left (3 \, a^{2} x^{2} + 10 \, a b x + 15 \, b^{2}\right )} \sqrt {x} \]

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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giac [A]  time = 0.15, size = 24, normalized size = 0.71 \[ \frac {2}{5} \, a^{2} x^{\frac {5}{2}} + \frac {4}{3} \, a b x^{\frac {3}{2}} + 2 \, b^{2} \sqrt {x} \]

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)

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maple [A]  time = 0.00, size = 25, normalized size = 0.74 \[ \frac {2 \left (3 a^{2} x^{2}+10 a b x +15 b^{2}\right ) \sqrt {x}}{15} \]

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 = 1.02, size = 26, normalized size = 0.76 \[ \frac {2}{15} \, {\left (3 \, a^{2} + \frac {10 \, a b}{x} + \frac {15 \, b^{2}}{x^{2}}\right )} x^{\frac {5}{2}} \]

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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mupad [B]  time = 0.04, size = 24, normalized size = 0.71 \[ \frac {2\,\sqrt {x}\,\left (3\,a^2\,x^2+10\,a\,b\,x+15\,b^2\right )}{15} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(a + b/x)^2,x)

[Out]

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

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sympy [A]  time = 1.16, size = 32, normalized size = 0.94 \[ \frac {2 a^{2} x^{\frac {5}{2}}}{5} + \frac {4 a b x^{\frac {3}{2}}}{3} + 2 b^{2} \sqrt {x} \]

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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