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*x^(1/2))^2

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.02, size = 16, normalized size = 1.00 \[ \frac {x}{a \left (a+b \sqrt {x}\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sqrt[x])^(-3),x]

[Out]

x/(a*(a + b*Sqrt[x])^2)

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fricas [B]  time = 0.63, size = 47, normalized size = 2.94 \[ -\frac {2 \, b^{3} x^{\frac {3}{2}} - 3 \, a b^{2} x + a^{3}}{b^{6} x^{2} - 2 \, a^{2} b^{4} x + a^{4} b^{2}} \]

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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giac [A]  time = 0.19, size = 22, normalized size = 1.38 \[ -\frac {2 \, b \sqrt {x} + a}{{\left (b \sqrt {x} + a\right )}^{2} b^{2}} \]

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)

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maple [B]  time = 0.02, size = 131, normalized size = 8.19 \[ -3 \left (-\frac {a^{2}}{2 \left (b^{2} x -a^{2}\right )^{2} b^{4}}-\frac {1}{\left (b^{2} x -a^{2}\right ) b^{4}}\right ) a \,b^{2}+\frac {a^{3}}{2 \left (b^{2} x -a^{2}\right )^{2} b^{2}}+\frac {a}{2 \left (b \sqrt {x}+a \right )^{2} b^{2}}-\frac {a}{2 \left (b \sqrt {x}-a \right )^{2} b^{2}}-\frac {1}{\left (b \sqrt {x}+a \right ) b^{2}}-\frac {1}{\left (b \sqrt {x}-a \right ) b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/b^2/(b*x^(1/2)+a)+1/2/b^2*a/(b*x^(1/2)+a)^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*a*b^2*(-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 = 0.76, size = 29, normalized size = 1.81 \[ -\frac {2}{{\left (b \sqrt {x} + a\right )} b^{2}} + \frac {a}{{\left (b \sqrt {x} + a\right )}^{2} b^{2}} \]

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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mupad [B]  time = 1.12, size = 34, normalized size = 2.12 \[ -\frac {\frac {a}{b^2}+\frac {2\,\sqrt {x}}{b}}{b^2\,x+a^2+2\,a\,b\,\sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*x^(1/2))^3,x)

[Out]

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

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sympy [A]  time = 1.15, size = 63, normalized size = 3.94 \[ \begin {cases} - \frac {a}{a^{2} b^{2} + 2 a b^{3} \sqrt {x} + b^{4} x} - \frac {2 b \sqrt {x}}{a^{2} b^{2} + 2 a b^{3} \sqrt {x} + b^{4} x} & \text {for}\: b \neq 0 \\\frac {x}{a^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a/(a**2*b**2 + 2*a*b**3*sqrt(x) + b**4*x) - 2*b*sqrt(x)/(a**2*b**2 + 2*a*b**3*sqrt(x) + b**4*x), N
e(b, 0)), (x/a**3, True))

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