3.1659 \(\int (a+\frac{b}{x})^3 x^{5/2} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0127937, 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} \[ \frac{6}{5} a^2 b x^{5/2}+\frac{2}{7} a^3 x^{7/2}+2 a b^2 x^{3/2}+2 b^3 \sqrt{x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^3*x^(5/2),x]

[Out]

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

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

Mathematica [A]  time = 0.0108839, size = 39, normalized size = 0.83 \[ \frac{2}{35} \sqrt{x} \left (21 a^2 b x^2+5 a^3 x^3+35 a b^2 x+35 b^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^3*x^(5/2),x]

[Out]

(2*Sqrt[x]*(35*b^3 + 35*a*b^2*x + 21*a^2*b*x^2 + 5*a^3*x^3))/35

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Maple [A]  time = 0.003, size = 36, normalized size = 0.8 \begin{align*}{\frac{10\,{a}^{3}{x}^{3}+42\,{a}^{2}b{x}^{2}+70\,xa{b}^{2}+70\,{b}^{3}}{35}\sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(5*a^3*x^3+21*a^2*b*x^2+35*a*b^2*x+35*b^3)*x^(1/2)

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Maxima [A]  time = 0.975012, size = 50, normalized size = 1.06 \begin{align*} \frac{2}{35} \,{\left (5 \, a^{3} + \frac{21 \, a^{2} b}{x} + \frac{35 \, a b^{2}}{x^{2}} + \frac{35 \, b^{3}}{x^{3}}\right )} x^{\frac{7}{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]

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

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Fricas [A]  time = 1.7565, size = 85, normalized size = 1.81 \begin{align*} \frac{2}{35} \,{\left (5 \, a^{3} x^{3} + 21 \, a^{2} b x^{2} + 35 \, a b^{2} x + 35 \, b^{3}\right )} \sqrt{x} \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]

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

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Sympy [A]  time = 3.68115, size = 46, normalized size = 0.98 \begin{align*} \frac{2 a^{3} x^{\frac{7}{2}}}{7} + \frac{6 a^{2} b x^{\frac{5}{2}}}{5} + 2 a b^{2} x^{\frac{3}{2}} + 2 b^{3} \sqrt{x} \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*x**(7/2)/7 + 6*a**2*b*x**(5/2)/5 + 2*a*b**2*x**(3/2) + 2*b**3*sqrt(x)

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Giac [A]  time = 1.13408, size = 47, normalized size = 1. \begin{align*} \frac{2}{7} \, a^{3} x^{\frac{7}{2}} + \frac{6}{5} \, a^{2} b x^{\frac{5}{2}} + 2 \, a b^{2} x^{\frac{3}{2}} + 2 \, b^{3} \sqrt{x} \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]

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