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]

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

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Rubi [A]  time = 0.01, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, 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 12

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

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]

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 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]

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*}

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Mathematica [A]  time = 0.04, 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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fricas [A]  time = 0.77, size = 28, normalized size = 0.80 \[ \frac {1}{2} \, {\left (2 \, x + 1\right )} \log \left (\sqrt {x + 1} + \sqrt {x}\right ) - \frac {1}{2} \, \sqrt {x + 1} \sqrt {x} \]

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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giac [A]  time = 0.52, size = 40, normalized size = 1.14 \[ x \log \left (\sqrt {x + 1} + \sqrt {x}\right ) - \frac {1}{2} \, \sqrt {x^{2} + x} - \frac {1}{4} \, \log \left ({\left | -2 \, x + 2 \, \sqrt {x^{2} + x} - 1 \right |}\right ) \]

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))

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maple [A]  time = 0.00, size = 24, normalized size = 0.69 \[ \frac {\arcsinh \left (\sqrt {x}\right )}{2}+x \arcsinh \left (\sqrt {x}\right )-\frac {\sqrt {x}\, \sqrt {1+x}}{2} \]

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 = 0.85, size = 23, normalized size = 0.66 \[ x \operatorname {arsinh}\left (\sqrt {x}\right ) - \frac {1}{2} \, \sqrt {x + 1} \sqrt {x} + \frac {1}{2} \, \operatorname {arsinh}\left (\sqrt {x}\right ) \]

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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mupad [B]  time = 0.92, size = 31, normalized size = 0.89 \[ \mathrm {atanh}\left (\frac {\sqrt {x}}{\sqrt {x+1}-1}\right )+x\,\mathrm {asinh}\left (\sqrt {x}\right )-\frac {\sqrt {x}\,\sqrt {x+1}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atanh(x^(1/2)/((x + 1)^(1/2) - 1)) + x*asinh(x^(1/2)) - (x^(1/2)*(x + 1)^(1/2))/2

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sympy [A]  time = 0.30, size = 29, normalized size = 0.83 \[ - \frac {\sqrt {x} \sqrt {x + 1}}{2} + x \operatorname {asinh}{\left (\sqrt {x} \right )} + \frac {\operatorname {asinh}{\left (\sqrt {x} \right )}}{2} \]

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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