3.297 \(\int \frac {\sinh ^{-1}(\sqrt {x})}{x^3} \, dx\)

Optimal. Leaf size=46 \[ -\frac {\sqrt {x+1}}{6 x^{3/2}}-\frac {\sinh ^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {\sqrt {x+1}}{3 \sqrt {x}} \]

[Out]

-1/2*arcsinh(x^(1/2))/x^2-1/6*(1+x)^(1/2)/x^(3/2)+1/3*(1+x)^(1/2)/x^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5902, 12, 45, 37} \[ -\frac {\sqrt {x+1}}{6 x^{3/2}}-\frac {\sinh ^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {\sqrt {x+1}}{3 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[Sqrt[x]]/x^3,x]

[Out]

-Sqrt[1 + x]/(6*x^(3/2)) + Sqrt[1 + x]/(3*Sqrt[x]) - ArcSinh[Sqrt[x]]/(2*x^2)

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 45

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] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 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^3} \, dx &=-\frac {\sinh ^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {1}{2} \int \frac {1}{2 x^{5/2} \sqrt {1+x}} \, dx\\ &=-\frac {\sinh ^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {1}{4} \int \frac {1}{x^{5/2} \sqrt {1+x}} \, dx\\ &=-\frac {\sqrt {1+x}}{6 x^{3/2}}-\frac {\sinh ^{-1}\left (\sqrt {x}\right )}{2 x^2}-\frac {1}{6} \int \frac {1}{x^{3/2} \sqrt {1+x}} \, dx\\ &=-\frac {\sqrt {1+x}}{6 x^{3/2}}+\frac {\sqrt {1+x}}{3 \sqrt {x}}-\frac {\sinh ^{-1}\left (\sqrt {x}\right )}{2 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 34, normalized size = 0.74 \[ \frac {\sqrt {x} \sqrt {x+1} (2 x-1)-3 \sinh ^{-1}\left (\sqrt {x}\right )}{6 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[Sqrt[x]]/x^3,x]

[Out]

(Sqrt[x]*Sqrt[1 + x]*(-1 + 2*x) - 3*ArcSinh[Sqrt[x]])/(6*x^2)

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fricas [A]  time = 0.57, size = 32, normalized size = 0.70 \[ \frac {{\left (2 \, x - 1\right )} \sqrt {x + 1} \sqrt {x} - 3 \, \log \left (\sqrt {x + 1} + \sqrt {x}\right )}{6 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(x^(1/2))/x^3,x, algorithm="fricas")

[Out]

1/6*((2*x - 1)*sqrt(x + 1)*sqrt(x) - 3*log(sqrt(x + 1) + sqrt(x)))/x^2

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giac [A]  time = 0.32, size = 52, normalized size = 1.13 \[ -\frac {\log \left (\sqrt {x + 1} + \sqrt {x}\right )}{2 \, x^{2}} + \frac {2 \, {\left (3 \, {\left (\sqrt {x + 1} - \sqrt {x}\right )}^{2} - 1\right )}}{3 \, {\left ({\left (\sqrt {x + 1} - \sqrt {x}\right )}^{2} - 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(x^(1/2))/x^3,x, algorithm="giac")

[Out]

-1/2*log(sqrt(x + 1) + sqrt(x))/x^2 + 2/3*(3*(sqrt(x + 1) - sqrt(x))^2 - 1)/((sqrt(x + 1) - sqrt(x))^2 - 1)^3

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maple [A]  time = 0.00, size = 31, normalized size = 0.67 \[ -\frac {\arcsinh \left (\sqrt {x}\right )}{2 x^{2}}-\frac {\sqrt {1+x}}{6 x^{\frac {3}{2}}}+\frac {\sqrt {1+x}}{3 \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(x^(1/2))/x^3,x)

[Out]

-1/2*arcsinh(x^(1/2))/x^2-1/6*(1+x)^(1/2)/x^(3/2)+1/3*(1+x)^(1/2)/x^(1/2)

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maxima [A]  time = 0.84, size = 30, normalized size = 0.65 \[ \frac {\sqrt {x + 1}}{3 \, \sqrt {x}} - \frac {\sqrt {x + 1}}{6 \, x^{\frac {3}{2}}} - \frac {\operatorname {arsinh}\left (\sqrt {x}\right )}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(x^(1/2))/x^3,x, algorithm="maxima")

[Out]

1/3*sqrt(x + 1)/sqrt(x) - 1/6*sqrt(x + 1)/x^(3/2) - 1/2*arcsinh(sqrt(x))/x^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {asinh}\left (\sqrt {x}\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(x^(1/2))/x^3,x)

[Out]

int(asinh(x^(1/2))/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}{\left (\sqrt {x} \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(x**(1/2))/x**3,x)

[Out]

Integral(asinh(sqrt(x))/x**3, x)

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