3.618 \(\int \frac{\sqrt{x}}{(2+b x)^{3/2}} \, dx\)

Optimal. Leaf size=44 \[ \frac{2 \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{3/2}}-\frac{2 \sqrt{x}}{b \sqrt{b x+2}} \]

[Out]

(-2*Sqrt[x])/(b*Sqrt[2 + b*x]) + (2*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(3/2)

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Rubi [A]  time = 0.0093868, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {47, 54, 215} \[ \frac{2 \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{3/2}}-\frac{2 \sqrt{x}}{b \sqrt{b x+2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(2 + b*x)^(3/2),x]

[Out]

(-2*Sqrt[x])/(b*Sqrt[2 + b*x]) + (2*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(3/2)

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

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

Mathematica [A]  time = 0.0289237, size = 44, normalized size = 1. \[ \frac{2 \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{2}}\right )}{b^{3/2}}-\frac{2 \sqrt{x}}{b \sqrt{b x+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(2 + b*x)^(3/2),x]

[Out]

(-2*Sqrt[x])/(b*Sqrt[2 + b*x]) + (2*ArcSinh[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(3/2)

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Maple [A]  time = 0.034, size = 48, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{{b}^{3/2}\sqrt{\pi }} \left ( -1/2\,{\frac{\sqrt{\pi }\sqrt{x}\sqrt{2}\sqrt{b}}{\sqrt{1/2\,bx+1}}}+\sqrt{\pi }{\it Arcsinh} \left ( 1/2\,\sqrt{b}\sqrt{x}\sqrt{2} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/b^(3/2)/Pi^(1/2)*(-1/2*Pi^(1/2)*x^(1/2)*2^(1/2)*b^(1/2)/(1/2*b*x+1)^(1/2)+Pi^(1/2)*arcsinh(1/2*b^(1/2)*x^(1/
2)*2^(1/2)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.86869, size = 302, normalized size = 6.86 \begin{align*} \left [\frac{{\left (b x + 2\right )} \sqrt{b} \log \left (b x + \sqrt{b x + 2} \sqrt{b} \sqrt{x} + 1\right ) - 2 \, \sqrt{b x + 2} b \sqrt{x}}{b^{3} x + 2 \, b^{2}}, -\frac{2 \,{\left ({\left (b x + 2\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{b x + 2} \sqrt{-b}}{b \sqrt{x}}\right ) + \sqrt{b x + 2} b \sqrt{x}\right )}}{b^{3} x + 2 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[((b*x + 2)*sqrt(b)*log(b*x + sqrt(b*x + 2)*sqrt(b)*sqrt(x) + 1) - 2*sqrt(b*x + 2)*b*sqrt(x))/(b^3*x + 2*b^2),
-2*((b*x + 2)*sqrt(-b)*arctan(sqrt(b*x + 2)*sqrt(-b)/(b*sqrt(x))) + sqrt(b*x + 2)*b*sqrt(x))/(b^3*x + 2*b^2)]

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Sympy [A]  time = 1.75628, size = 41, normalized size = 0.93 \begin{align*} - \frac{2 \sqrt{x}}{b \sqrt{b x + 2}} + \frac{2 \operatorname{asinh}{\left (\frac{\sqrt{2} \sqrt{b} \sqrt{x}}{2} \right )}}{b^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(x)/(b*sqrt(b*x + 2)) + 2*asinh(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(3/2)

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Giac [B]  time = 24.9303, size = 111, normalized size = 2.52 \begin{align*} -\frac{{\left (\frac{\log \left ({\left (\sqrt{b x + 2} \sqrt{b} - \sqrt{{\left (b x + 2\right )} b - 2 \, b}\right )}^{2}\right )}{\sqrt{b}} + \frac{8 \, \sqrt{b}}{{\left (\sqrt{b x + 2} \sqrt{b} - \sqrt{{\left (b x + 2\right )} b - 2 \, b}\right )}^{2} + 2 \, b}\right )}{\left | b \right |}}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log((sqrt(b*x + 2)*sqrt(b) - sqrt((b*x + 2)*b - 2*b))^2)/sqrt(b) + 8*sqrt(b)/((sqrt(b*x + 2)*sqrt(b) - sqrt(
(b*x + 2)*b - 2*b))^2 + 2*b))*abs(b)/b^2