3.1663 \(\int \frac{(a+\frac{b}{x})^3}{x^{3/2}} \, dx\)

Optimal. Leaf size=47 \[ -\frac{2 a^2 b}{x^{3/2}}-\frac{2 a^3}{\sqrt{x}}-\frac{6 a b^2}{5 x^{5/2}}-\frac{2 b^3}{7 x^{7/2}} \]

[Out]

(-2*b^3)/(7*x^(7/2)) - (6*a*b^2)/(5*x^(5/2)) - (2*a^2*b)/x^(3/2) - (2*a^3)/Sqrt[x]

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Rubi [A]  time = 0.0126803, antiderivative size = 47, 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} \[ -\frac{2 a^2 b}{x^{3/2}}-\frac{2 a^3}{\sqrt{x}}-\frac{6 a b^2}{5 x^{5/2}}-\frac{2 b^3}{7 x^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^3/x^(3/2),x]

[Out]

(-2*b^3)/(7*x^(7/2)) - (6*a*b^2)/(5*x^(5/2)) - (2*a^2*b)/x^(3/2) - (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}{x^{3/2}} \, dx &=\int \frac{(b+a x)^3}{x^{9/2}} \, dx\\ &=\int \left (\frac{b^3}{x^{9/2}}+\frac{3 a b^2}{x^{7/2}}+\frac{3 a^2 b}{x^{5/2}}+\frac{a^3}{x^{3/2}}\right ) \, dx\\ &=-\frac{2 b^3}{7 x^{7/2}}-\frac{6 a b^2}{5 x^{5/2}}-\frac{2 a^2 b}{x^{3/2}}-\frac{2 a^3}{\sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.011938, size = 39, normalized size = 0.83 \[ -\frac{2 \left (35 a^2 b x^2+35 a^3 x^3+21 a b^2 x+5 b^3\right )}{35 x^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^3/x^(3/2),x]

[Out]

(-2*(5*b^3 + 21*a*b^2*x + 35*a^2*b*x^2 + 35*a^3*x^3))/(35*x^(7/2))

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Maple [A]  time = 0.005, size = 36, normalized size = 0.8 \begin{align*} -{\frac{70\,{a}^{3}{x}^{3}+70\,{a}^{2}b{x}^{2}+42\,xa{b}^{2}+10\,{b}^{3}}{35}{x}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(35*a^3*x^3+35*a^2*b*x^2+21*a*b^2*x+5*b^3)/x^(7/2)

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Maxima [A]  time = 0.976587, size = 47, normalized size = 1. \begin{align*} -\frac{2 \, a^{3}}{\sqrt{x}} - \frac{2 \, a^{2} b}{x^{\frac{3}{2}}} - \frac{6 \, a b^{2}}{5 \, x^{\frac{5}{2}}} - \frac{2 \, b^{3}}{7 \, x^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a^3/sqrt(x) - 2*a^2*b/x^(3/2) - 6/5*a*b^2/x^(5/2) - 2/7*b^3/x^(7/2)

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Fricas [A]  time = 1.75888, size = 86, normalized size = 1.83 \begin{align*} -\frac{2 \,{\left (35 \, a^{3} x^{3} + 35 \, a^{2} b x^{2} + 21 \, a b^{2} x + 5 \, b^{3}\right )}}{35 \, x^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(35*a^3*x^3 + 35*a^2*b*x^2 + 21*a*b^2*x + 5*b^3)/x^(7/2)

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Sympy [A]  time = 1.1246, size = 48, normalized size = 1.02 \begin{align*} - \frac{2 a^{3}}{\sqrt{x}} - \frac{2 a^{2} b}{x^{\frac{3}{2}}} - \frac{6 a b^{2}}{5 x^{\frac{5}{2}}} - \frac{2 b^{3}}{7 x^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a**3/sqrt(x) - 2*a**2*b/x**(3/2) - 6*a*b**2/(5*x**(5/2)) - 2*b**3/(7*x**(7/2))

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Giac [A]  time = 1.11551, size = 47, normalized size = 1. \begin{align*} -\frac{2 \,{\left (35 \, a^{3} x^{3} + 35 \, a^{2} b x^{2} + 21 \, a b^{2} x + 5 \, b^{3}\right )}}{35 \, x^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35*(35*a^3*x^3 + 35*a^2*b*x^2 + 21*a*b^2*x + 5*b^3)/x^(7/2)