3.1662 \(\int \frac{(a+\frac{b}{x})^3}{\sqrt{x}} \, dx\)

Optimal. Leaf size=45 \[ -\frac{6 a^2 b}{\sqrt{x}}+2 a^3 \sqrt{x}-\frac{2 a b^2}{x^{3/2}}-\frac{2 b^3}{5 x^{5/2}} \]

[Out]

(-2*b^3)/(5*x^(5/2)) - (2*a*b^2)/x^(3/2) - (6*a^2*b)/Sqrt[x] + 2*a^3*Sqrt[x]

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Rubi [A]  time = 0.0131865, antiderivative size = 45, 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} \[ -\frac{6 a^2 b}{\sqrt{x}}+2 a^3 \sqrt{x}-\frac{2 a b^2}{x^{3/2}}-\frac{2 b^3}{5 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^3/Sqrt[x],x]

[Out]

(-2*b^3)/(5*x^(5/2)) - (2*a*b^2)/x^(3/2) - (6*a^2*b)/Sqrt[x] + 2*a^3*Sqrt[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 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 \frac{\left (a+\frac{b}{x}\right )^3}{\sqrt{x}} \, dx &=\int \frac{(b+a x)^3}{x^{7/2}} \, dx\\ &=\int \left (\frac{b^3}{x^{7/2}}+\frac{3 a b^2}{x^{5/2}}+\frac{3 a^2 b}{x^{3/2}}+\frac{a^3}{\sqrt{x}}\right ) \, dx\\ &=-\frac{2 b^3}{5 x^{5/2}}-\frac{2 a b^2}{x^{3/2}}-\frac{6 a^2 b}{\sqrt{x}}+2 a^3 \sqrt{x}\\ \end{align*}

Mathematica [A]  time = 0.0102485, size = 39, normalized size = 0.87 \[ \frac{2 \left (-15 a^2 b x^2+5 a^3 x^3-5 a b^2 x-b^3\right )}{5 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^3/Sqrt[x],x]

[Out]

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

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Maple [A]  time = 0.004, size = 36, normalized size = 0.8 \begin{align*}{\frac{10\,{a}^{3}{x}^{3}-30\,{a}^{2}b{x}^{2}-10\,xa{b}^{2}-2\,{b}^{3}}{5}{x}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.981312, size = 47, normalized size = 1.04 \begin{align*} 2 \, a^{3} \sqrt{x} - \frac{6 \, a^{2} b}{\sqrt{x}} - \frac{2 \, a b^{2}}{x^{\frac{3}{2}}} - \frac{2 \, b^{3}}{5 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^3/x^(1/2),x, algorithm="maxima")

[Out]

2*a^3*sqrt(x) - 6*a^2*b/sqrt(x) - 2*a*b^2/x^(3/2) - 2/5*b^3/x^(5/2)

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Fricas [A]  time = 1.73576, size = 78, normalized size = 1.73 \begin{align*} \frac{2 \,{\left (5 \, a^{3} x^{3} - 15 \, a^{2} b x^{2} - 5 \, a b^{2} x - b^{3}\right )}}{5 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^3/x^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.773901, size = 44, normalized size = 0.98 \begin{align*} 2 a^{3} \sqrt{x} - \frac{6 a^{2} b}{\sqrt{x}} - \frac{2 a b^{2}}{x^{\frac{3}{2}}} - \frac{2 b^{3}}{5 x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a**3*sqrt(x) - 6*a**2*b/sqrt(x) - 2*a*b**2/x**(3/2) - 2*b**3/(5*x**(5/2))

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Giac [A]  time = 1.09222, size = 46, normalized size = 1.02 \begin{align*} 2 \, a^{3} \sqrt{x} - \frac{2 \,{\left (15 \, a^{2} b x^{2} + 5 \, a b^{2} x + b^{3}\right )}}{5 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^3/x^(1/2),x, algorithm="giac")

[Out]

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