3.2138 \(\int \frac{(a+b \sqrt{x})^3}{x^4} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0201946, 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 = {266, 43} \[ -\frac{6 a^2 b}{5 x^{5/2}}-\frac{a^3}{3 x^3}-\frac{3 a b^2}{2 x^2}-\frac{2 b^3}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[x])^3/x^4,x]

[Out]

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/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+b \sqrt{x}\right )^3}{x^4} \, dx &=2 \operatorname{Subst}\left (\int \frac{(a+b x)^3}{x^7} \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (\frac{a^3}{x^7}+\frac{3 a^2 b}{x^6}+\frac{3 a b^2}{x^5}+\frac{b^3}{x^4}\right ) \, dx,x,\sqrt{x}\right )\\ &=-\frac{a^3}{3 x^3}-\frac{6 a^2 b}{5 x^{5/2}}-\frac{3 a b^2}{2 x^2}-\frac{2 b^3}{3 x^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0159845, size = 41, normalized size = 0.87 \[ -\frac{36 a^2 b \sqrt{x}+10 a^3+45 a b^2 x+20 b^3 x^{3/2}}{30 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[x])^3/x^4,x]

[Out]

-(10*a^3 + 36*a^2*b*Sqrt[x] + 45*a*b^2*x + 20*b^3*x^(3/2))/(30*x^3)

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Maple [A]  time = 0., size = 36, normalized size = 0.8 \begin{align*} -{\frac{{a}^{3}}{3\,{x}^{3}}}-{\frac{6\,b{a}^{2}}{5}{x}^{-{\frac{5}{2}}}}-{\frac{3\,{b}^{2}a}{2\,{x}^{2}}}-{\frac{2\,{b}^{3}}{3}{x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*a^3/x^3-6/5*a^2*b/x^(5/2)-3/2*a*b^2/x^2-2/3*b^3/x^(3/2)

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Maxima [A]  time = 0.977509, size = 47, normalized size = 1. \begin{align*} -\frac{20 \, b^{3} x^{\frac{3}{2}} + 45 \, a b^{2} x + 36 \, a^{2} b \sqrt{x} + 10 \, a^{3}}{30 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/2))^3/x^4,x, algorithm="maxima")

[Out]

-1/30*(20*b^3*x^(3/2) + 45*a*b^2*x + 36*a^2*b*sqrt(x) + 10*a^3)/x^3

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Fricas [A]  time = 1.4627, size = 88, normalized size = 1.87 \begin{align*} -\frac{45 \, a b^{2} x + 10 \, a^{3} + 4 \,{\left (5 \, b^{3} x + 9 \, a^{2} b\right )} \sqrt{x}}{30 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/2))^3/x^4,x, algorithm="fricas")

[Out]

-1/30*(45*a*b^2*x + 10*a^3 + 4*(5*b^3*x + 9*a^2*b)*sqrt(x))/x^3

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Sympy [A]  time = 1.35406, size = 46, normalized size = 0.98 \begin{align*} - \frac{a^{3}}{3 x^{3}} - \frac{6 a^{2} b}{5 x^{\frac{5}{2}}} - \frac{3 a b^{2}}{2 x^{2}} - \frac{2 b^{3}}{3 x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x**(1/2))**3/x**4,x)

[Out]

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

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Giac [A]  time = 1.08464, size = 47, normalized size = 1. \begin{align*} -\frac{20 \, b^{3} x^{\frac{3}{2}} + 45 \, a b^{2} x + 36 \, a^{2} b \sqrt{x} + 10 \, a^{3}}{30 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*x^(1/2))^3/x^4,x, algorithm="giac")

[Out]

-1/30*(20*b^3*x^(3/2) + 45*a*b^2*x + 36*a^2*b*sqrt(x) + 10*a^3)/x^3