3.298 \(\int \frac{\sinh ^{-1}(\sqrt{x})}{x^4} \, dx\)

Optimal. Leaf size=62 \[ \frac{4 \sqrt{x+1}}{45 x^{3/2}}-\frac{\sqrt{x+1}}{15 x^{5/2}}-\frac{\sinh ^{-1}\left (\sqrt{x}\right )}{3 x^3}-\frac{8 \sqrt{x+1}}{45 \sqrt{x}} \]

[Out]

-Sqrt[1 + x]/(15*x^(5/2)) + (4*Sqrt[1 + x])/(45*x^(3/2)) - (8*Sqrt[1 + x])/(45*Sqrt[x]) - ArcSinh[Sqrt[x]]/(3*
x^3)

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Rubi [A]  time = 0.0215487, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {5902, 12, 45, 37} \[ \frac{4 \sqrt{x+1}}{45 x^{3/2}}-\frac{\sqrt{x+1}}{15 x^{5/2}}-\frac{\sinh ^{-1}\left (\sqrt{x}\right )}{3 x^3}-\frac{8 \sqrt{x+1}}{45 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[Sqrt[x]]/x^4,x]

[Out]

-Sqrt[1 + x]/(15*x^(5/2)) + (4*Sqrt[1 + x])/(45*x^(3/2)) - (8*Sqrt[1 + x])/(45*Sqrt[x]) - ArcSinh[Sqrt[x]]/(3*
x^3)

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]

Rule 12

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

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 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]

Rubi steps

\begin{align*} \int \frac{\sinh ^{-1}\left (\sqrt{x}\right )}{x^4} \, dx &=-\frac{\sinh ^{-1}\left (\sqrt{x}\right )}{3 x^3}+\frac{1}{3} \int \frac{1}{2 x^{7/2} \sqrt{1+x}} \, dx\\ &=-\frac{\sinh ^{-1}\left (\sqrt{x}\right )}{3 x^3}+\frac{1}{6} \int \frac{1}{x^{7/2} \sqrt{1+x}} \, dx\\ &=-\frac{\sqrt{1+x}}{15 x^{5/2}}-\frac{\sinh ^{-1}\left (\sqrt{x}\right )}{3 x^3}-\frac{2}{15} \int \frac{1}{x^{5/2} \sqrt{1+x}} \, dx\\ &=-\frac{\sqrt{1+x}}{15 x^{5/2}}+\frac{4 \sqrt{1+x}}{45 x^{3/2}}-\frac{\sinh ^{-1}\left (\sqrt{x}\right )}{3 x^3}+\frac{4}{45} \int \frac{1}{x^{3/2} \sqrt{1+x}} \, dx\\ &=-\frac{\sqrt{1+x}}{15 x^{5/2}}+\frac{4 \sqrt{1+x}}{45 x^{3/2}}-\frac{8 \sqrt{1+x}}{45 \sqrt{x}}-\frac{\sinh ^{-1}\left (\sqrt{x}\right )}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0149408, size = 39, normalized size = 0.63 \[ \frac{\sqrt{x} \sqrt{x+1} \left (-8 x^2+4 x-3\right )-15 \sinh ^{-1}\left (\sqrt{x}\right )}{45 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[Sqrt[x]]/x^4,x]

[Out]

(Sqrt[x]*Sqrt[1 + x]*(-3 + 4*x - 8*x^2) - 15*ArcSinh[Sqrt[x]])/(45*x^3)

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Maple [A]  time = 0.006, size = 41, normalized size = 0.7 \begin{align*} -{\frac{1}{3\,{x}^{3}}{\it Arcsinh} \left ( \sqrt{x} \right ) }-{\frac{1}{15}\sqrt{1+x}{x}^{-{\frac{5}{2}}}}+{\frac{4}{45}\sqrt{1+x}{x}^{-{\frac{3}{2}}}}-{\frac{8}{45}\sqrt{1+x}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(x^(1/2))/x^4,x)

[Out]

-1/3*arcsinh(x^(1/2))/x^3-1/15*(1+x)^(1/2)/x^(5/2)+4/45*(1+x)^(1/2)/x^(3/2)-8/45*(1+x)^(1/2)/x^(1/2)

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Maxima [A]  time = 1.66074, size = 54, normalized size = 0.87 \begin{align*} -\frac{8 \, \sqrt{x + 1}}{45 \, \sqrt{x}} + \frac{4 \, \sqrt{x + 1}}{45 \, x^{\frac{3}{2}}} - \frac{\sqrt{x + 1}}{15 \, x^{\frac{5}{2}}} - \frac{\operatorname{arsinh}\left (\sqrt{x}\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(x^(1/2))/x^4,x, algorithm="maxima")

[Out]

-8/45*sqrt(x + 1)/sqrt(x) + 4/45*sqrt(x + 1)/x^(3/2) - 1/15*sqrt(x + 1)/x^(5/2) - 1/3*arcsinh(sqrt(x))/x^3

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Fricas [A]  time = 3.05551, size = 112, normalized size = 1.81 \begin{align*} -\frac{{\left (8 \, x^{2} - 4 \, x + 3\right )} \sqrt{x + 1} \sqrt{x} + 15 \, \log \left (\sqrt{x + 1} + \sqrt{x}\right )}{45 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(x^(1/2))/x^4,x, algorithm="fricas")

[Out]

-1/45*((8*x^2 - 4*x + 3)*sqrt(x + 1)*sqrt(x) + 15*log(sqrt(x + 1) + sqrt(x)))/x^3

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asinh}{\left (\sqrt{x} \right )}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(x**(1/2))/x**4,x)

[Out]

Integral(asinh(sqrt(x))/x**4, x)

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Giac [A]  time = 1.32342, size = 90, normalized size = 1.45 \begin{align*} -\frac{\log \left (\sqrt{x + 1} + \sqrt{x}\right )}{3 \, x^{3}} + \frac{16 \,{\left (10 \,{\left (\sqrt{x + 1} - \sqrt{x}\right )}^{4} - 5 \,{\left (\sqrt{x + 1} - \sqrt{x}\right )}^{2} + 1\right )}}{45 \,{\left ({\left (\sqrt{x + 1} - \sqrt{x}\right )}^{2} - 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(x^(1/2))/x^4,x, algorithm="giac")

[Out]

-1/3*log(sqrt(x + 1) + sqrt(x))/x^3 + 16/45*(10*(sqrt(x + 1) - sqrt(x))^4 - 5*(sqrt(x + 1) - sqrt(x))^2 + 1)/(
(sqrt(x + 1) - sqrt(x))^2 - 1)^5