3.2210 \(\int \frac{1}{(a+b \sqrt{x})^3} \, dx\)

Optimal. Leaf size=16 \[ \frac{x}{a \left (a+b \sqrt{x}\right )^2} \]

[Out]

x/(a*(a + b*Sqrt[x])^2)

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Rubi [A]  time = 0.0050682, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {190, 37} \[ \frac{x}{a \left (a+b \sqrt{x}\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sqrt[x])^(-3),x]

[Out]

x/(a*(a + b*Sqrt[x])^2)

Rule 190

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(1/n - 1)*(a + b*x)^p, x], x, x^n], x] /
; FreeQ[{a, b, p}, x] && FractionQ[n] && IntegerQ[1/n]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b \sqrt{x}\right )^3} \, dx &=2 \operatorname{Subst}\left (\int \frac{x}{(a+b x)^3} \, dx,x,\sqrt{x}\right )\\ &=\frac{x}{a \left (a+b \sqrt{x}\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.0039473, size = 16, normalized size = 1. \[ \frac{x}{a \left (a+b \sqrt{x}\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[x])^(-3),x]

[Out]

x/(a*(a + b*Sqrt[x])^2)

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Maple [B]  time = 0.024, size = 131, normalized size = 8.2 \begin{align*} -{\frac{1}{{b}^{2}} \left ( a+b\sqrt{x} \right ) ^{-1}}+{\frac{a}{2\,{b}^{2}} \left ( a+b\sqrt{x} \right ) ^{-2}}-{\frac{1}{{b}^{2}} \left ( b\sqrt{x}-a \right ) ^{-1}}-{\frac{a}{2\,{b}^{2}} \left ( b\sqrt{x}-a \right ) ^{-2}}+{\frac{{a}^{3}}{2\, \left ({b}^{2}x-{a}^{2} \right ) ^{2}{b}^{2}}}-3\,{b}^{2}a \left ( -1/2\,{\frac{{a}^{2}}{{b}^{4} \left ({b}^{2}x-{a}^{2} \right ) ^{2}}}-{\frac{1}{{b}^{4} \left ({b}^{2}x-{a}^{2} \right ) }} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 0.937931, size = 39, normalized size = 2.44 \begin{align*} -\frac{2}{{\left (b \sqrt{x} + a\right )} b^{2}} + \frac{a}{{\left (b \sqrt{x} + a\right )}^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.2579, size = 95, normalized size = 5.94 \begin{align*} -\frac{2 \, b^{3} x^{\frac{3}{2}} - 3 \, a b^{2} x + a^{3}}{b^{6} x^{2} - 2 \, a^{2} b^{4} x + a^{4} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*b^3*x^(3/2) - 3*a*b^2*x + a^3)/(b^6*x^2 - 2*a^2*b^4*x + a^4*b^2)

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Sympy [A]  time = 0.941293, size = 34, normalized size = 2.12 \begin{align*} \begin{cases} \frac{x}{a^{3} + 2 a^{2} b \sqrt{x} + a b^{2} x} & \text{for}\: a \neq 0 \\- \frac{2}{b^{3} \sqrt{x}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/(a**3 + 2*a**2*b*sqrt(x) + a*b**2*x), Ne(a, 0)), (-2/(b**3*sqrt(x)), True))

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Giac [A]  time = 1.1001, size = 30, normalized size = 1.88 \begin{align*} -\frac{2 \, b \sqrt{x} + a}{{\left (b \sqrt{x} + a\right )}^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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