3.294 \(\int \sinh ^{-1}(\sqrt{x}) \, dx\)

Optimal. Leaf size=35 \[ -\frac{1}{2} \sqrt{x} \sqrt{x+1}+x \sinh ^{-1}\left (\sqrt{x}\right )+\frac{1}{2} \sinh ^{-1}\left (\sqrt{x}\right ) \]

[Out]

-(Sqrt[x]*Sqrt[1 + x])/2 + ArcSinh[Sqrt[x]]/2 + x*ArcSinh[Sqrt[x]]

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Rubi [A]  time = 0.0107196, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1., Rules used = {5900, 12, 1958, 50, 54, 215} \[ -\frac{1}{2} \sqrt{x} \sqrt{x+1}+x \sinh ^{-1}\left (\sqrt{x}\right )+\frac{1}{2} \sinh ^{-1}\left (\sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[Sqrt[x]],x]

[Out]

-(Sqrt[x]*Sqrt[1 + x])/2 + ArcSinh[Sqrt[x]]/2 + x*ArcSinh[Sqrt[x]]

Rule 5900

Int[ArcSinh[u_], x_Symbol] :> Simp[x*ArcSinh[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/Sqrt[1 + u^2], x], x]
/; InverseFunctionFreeQ[u, x] &&  !FunctionOfExponentialQ[u, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1958

Int[(u_.)*(((e_.)*((a_.) + (b_.)*(x_)^(n_.)))/((c_) + (d_.)*(x_)^(n_.)))^(p_), x_Symbol] :> Int[(u*(e*(a + b*x
^n))^p)/(c + d*x^n)^p, x] /; FreeQ[{a, b, c, d, e, n, p}, x] && GtQ[b*d*e, 0] && GtQ[c - (a*d)/b, 0]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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 \sinh ^{-1}\left (\sqrt{x}\right ) \, dx &=x \sinh ^{-1}\left (\sqrt{x}\right )-\int \frac{1}{2} \sqrt{\frac{x}{1+x}} \, dx\\ &=x \sinh ^{-1}\left (\sqrt{x}\right )-\frac{1}{2} \int \sqrt{\frac{x}{1+x}} \, dx\\ &=x \sinh ^{-1}\left (\sqrt{x}\right )-\frac{1}{2} \int \frac{\sqrt{x}}{\sqrt{1+x}} \, dx\\ &=-\frac{1}{2} \sqrt{x} \sqrt{1+x}+x \sinh ^{-1}\left (\sqrt{x}\right )+\frac{1}{4} \int \frac{1}{\sqrt{x} \sqrt{1+x}} \, dx\\ &=-\frac{1}{2} \sqrt{x} \sqrt{1+x}+x \sinh ^{-1}\left (\sqrt{x}\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x^2}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{1}{2} \sqrt{x} \sqrt{1+x}+\frac{1}{2} \sinh ^{-1}\left (\sqrt{x}\right )+x \sinh ^{-1}\left (\sqrt{x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0375523, size = 33, normalized size = 0.94 \[ \frac{1}{2} \left ((2 x+1) \sinh ^{-1}\left (\sqrt{x}\right )-\sqrt{\frac{x}{x+1}} (x+1)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[Sqrt[x]],x]

[Out]

(-(Sqrt[x/(1 + x)]*(1 + x)) + (1 + 2*x)*ArcSinh[Sqrt[x]])/2

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Maple [A]  time = 0.002, size = 24, normalized size = 0.7 \begin{align*}{\frac{1}{2}{\it Arcsinh} \left ( \sqrt{x} \right ) }+x{\it Arcsinh} \left ( \sqrt{x} \right ) -{\frac{1}{2}\sqrt{x}\sqrt{1+x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(x^(1/2)),x)

[Out]

1/2*arcsinh(x^(1/2))+x*arcsinh(x^(1/2))-1/2*x^(1/2)*(1+x)^(1/2)

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Maxima [A]  time = 1.75239, size = 31, normalized size = 0.89 \begin{align*} x \operatorname{arsinh}\left (\sqrt{x}\right ) - \frac{1}{2} \, \sqrt{x + 1} \sqrt{x} + \frac{1}{2} \, \operatorname{arsinh}\left (\sqrt{x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(x^(1/2)),x, algorithm="maxima")

[Out]

x*arcsinh(sqrt(x)) - 1/2*sqrt(x + 1)*sqrt(x) + 1/2*arcsinh(sqrt(x))

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Fricas [A]  time = 2.61349, size = 92, normalized size = 2.63 \begin{align*} \frac{1}{2} \,{\left (2 \, x + 1\right )} \log \left (\sqrt{x + 1} + \sqrt{x}\right ) - \frac{1}{2} \, \sqrt{x + 1} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(x^(1/2)),x, algorithm="fricas")

[Out]

1/2*(2*x + 1)*log(sqrt(x + 1) + sqrt(x)) - 1/2*sqrt(x + 1)*sqrt(x)

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Sympy [A]  time = 0.359289, size = 29, normalized size = 0.83 \begin{align*} - \frac{\sqrt{x} \sqrt{x + 1}}{2} + x \operatorname{asinh}{\left (\sqrt{x} \right )} + \frac{\operatorname{asinh}{\left (\sqrt{x} \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(x**(1/2)),x)

[Out]

-sqrt(x)*sqrt(x + 1)/2 + x*asinh(sqrt(x)) + asinh(sqrt(x))/2

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Giac [A]  time = 1.57915, size = 54, normalized size = 1.54 \begin{align*} x \log \left (\sqrt{x + 1} + \sqrt{x}\right ) - \frac{1}{2} \, \sqrt{x^{2} + x} - \frac{1}{4} \, \log \left ({\left | -2 \, x + 2 \, \sqrt{x^{2} + x} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(x^(1/2)),x, algorithm="giac")

[Out]

x*log(sqrt(x + 1) + sqrt(x)) - 1/2*sqrt(x^2 + x) - 1/4*log(abs(-2*x + 2*sqrt(x^2 + x) - 1))