3.1664 \(\int \frac{(a+\frac{b}{x})^3}{x^{5/2}} \, dx\)

Optimal. Leaf size=51 \[ -\frac{6 a^2 b}{5 x^{5/2}}-\frac{2 a^3}{3 x^{3/2}}-\frac{6 a b^2}{7 x^{7/2}}-\frac{2 b^3}{9 x^{9/2}} \]

[Out]

(-2*b^3)/(9*x^(9/2)) - (6*a*b^2)/(7*x^(7/2)) - (6*a^2*b)/(5*x^(5/2)) - (2*a^3)/(3*x^(3/2))

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Rubi [A]  time = 0.0133382, antiderivative size = 51, 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} \[ -\frac{6 a^2 b}{5 x^{5/2}}-\frac{2 a^3}{3 x^{3/2}}-\frac{6 a b^2}{7 x^{7/2}}-\frac{2 b^3}{9 x^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^3/x^(5/2),x]

[Out]

(-2*b^3)/(9*x^(9/2)) - (6*a*b^2)/(7*x^(7/2)) - (6*a^2*b)/(5*x^(5/2)) - (2*a^3)/(3*x^(3/2))

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^{5/2}} \, dx &=\int \frac{(b+a x)^3}{x^{11/2}} \, dx\\ &=\int \left (\frac{b^3}{x^{11/2}}+\frac{3 a b^2}{x^{9/2}}+\frac{3 a^2 b}{x^{7/2}}+\frac{a^3}{x^{5/2}}\right ) \, dx\\ &=-\frac{2 b^3}{9 x^{9/2}}-\frac{6 a b^2}{7 x^{7/2}}-\frac{6 a^2 b}{5 x^{5/2}}-\frac{2 a^3}{3 x^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0112361, size = 39, normalized size = 0.76 \[ -\frac{2 \left (189 a^2 b x^2+105 a^3 x^3+135 a b^2 x+35 b^3\right )}{315 x^{9/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^3/x^(5/2),x]

[Out]

(-2*(35*b^3 + 135*a*b^2*x + 189*a^2*b*x^2 + 105*a^3*x^3))/(315*x^(9/2))

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Maple [A]  time = 0.004, size = 36, normalized size = 0.7 \begin{align*} -{\frac{210\,{a}^{3}{x}^{3}+378\,{a}^{2}b{x}^{2}+270\,xa{b}^{2}+70\,{b}^{3}}{315}{x}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^3/x^(5/2),x)

[Out]

-2/315*(105*a^3*x^3+189*a^2*b*x^2+135*a*b^2*x+35*b^3)/x^(9/2)

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Maxima [A]  time = 0.997705, size = 47, normalized size = 0.92 \begin{align*} -\frac{2 \, a^{3}}{3 \, x^{\frac{3}{2}}} - \frac{6 \, a^{2} b}{5 \, x^{\frac{5}{2}}} - \frac{6 \, a b^{2}}{7 \, x^{\frac{7}{2}}} - \frac{2 \, b^{3}}{9 \, x^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^3/x^(5/2),x, algorithm="maxima")

[Out]

-2/3*a^3/x^(3/2) - 6/5*a^2*b/x^(5/2) - 6/7*a*b^2/x^(7/2) - 2/9*b^3/x^(9/2)

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Fricas [A]  time = 1.79762, size = 93, normalized size = 1.82 \begin{align*} -\frac{2 \,{\left (105 \, a^{3} x^{3} + 189 \, a^{2} b x^{2} + 135 \, a b^{2} x + 35 \, b^{3}\right )}}{315 \, x^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^3/x^(5/2),x, algorithm="fricas")

[Out]

-2/315*(105*a^3*x^3 + 189*a^2*b*x^2 + 135*a*b^2*x + 35*b^3)/x^(9/2)

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Sympy [A]  time = 2.06731, size = 51, normalized size = 1. \begin{align*} - \frac{2 a^{3}}{3 x^{\frac{3}{2}}} - \frac{6 a^{2} b}{5 x^{\frac{5}{2}}} - \frac{6 a b^{2}}{7 x^{\frac{7}{2}}} - \frac{2 b^{3}}{9 x^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)**3/x**(5/2),x)

[Out]

-2*a**3/(3*x**(3/2)) - 6*a**2*b/(5*x**(5/2)) - 6*a*b**2/(7*x**(7/2)) - 2*b**3/(9*x**(9/2))

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Giac [A]  time = 1.1057, size = 47, normalized size = 0.92 \begin{align*} -\frac{2 \,{\left (105 \, a^{3} x^{3} + 189 \, a^{2} b x^{2} + 135 \, a b^{2} x + 35 \, b^{3}\right )}}{315 \, x^{\frac{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^3/x^(5/2),x, algorithm="giac")

[Out]

-2/315*(105*a^3*x^3 + 189*a^2*b*x^2 + 135*a*b^2*x + 35*b^3)/x^(9/2)