3.296 \(\int \frac {\sinh ^{-1}(\sqrt {x})}{x^2} \, dx\)

Optimal. Leaf size=26 \[ -\frac {\sqrt {x+1}}{\sqrt {x}}-\frac {\sinh ^{-1}\left (\sqrt {x}\right )}{x} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {5902, 12, 37} \[ -\frac {\sqrt {x+1}}{\sqrt {x}}-\frac {\sinh ^{-1}\left (\sqrt {x}\right )}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[Sqrt[x]]/x^2,x]

[Out]

-(Sqrt[1 + x]/Sqrt[x]) - ArcSinh[Sqrt[x]]/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 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 5902

Int[((a_.) + ArcSinh[u_]*(b_.))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m + 1)*(a + b*ArcSin
h[u]))/(d*(m + 1)), x] - Dist[b/(d*(m + 1)), Int[SimplifyIntegrand[((c + d*x)^(m + 1)*D[u, x])/Sqrt[1 + u^2],
x], x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1] && InverseFunctionFreeQ[u, x] &&  !FunctionOfQ[(c + d*x)
^(m + 1), u, x] &&  !FunctionOfExponentialQ[u, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 26, normalized size = 1.00 \[ -\frac {\sqrt {x+1}}{\sqrt {x}}-\frac {\sinh ^{-1}\left (\sqrt {x}\right )}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[Sqrt[x]]/x^2,x]

[Out]

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

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fricas [A]  time = 0.55, size = 25, normalized size = 0.96 \[ -\frac {\sqrt {x + 1} \sqrt {x} + \log \left (\sqrt {x + 1} + \sqrt {x}\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.37, size = 35, normalized size = 1.35 \[ -\frac {\log \left (\sqrt {x + 1} + \sqrt {x}\right )}{x} + \frac {2}{{\left (\sqrt {x + 1} - \sqrt {x}\right )}^{2} - 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 21, normalized size = 0.81 \[ -\frac {\arcsinh \left (\sqrt {x}\right )}{x}-\frac {\sqrt {1+x}}{\sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.86, size = 20, normalized size = 0.77 \[ -\frac {\sqrt {x + 1}}{\sqrt {x}} - \frac {\operatorname {arsinh}\left (\sqrt {x}\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int \frac {\mathrm {asinh}\left (\sqrt {x}\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(asinh(x^(1/2))/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}{\left (\sqrt {x} \right )}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(sqrt(x))/x**2, x)

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