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]

-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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Rubi [A]  time = 0.02, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5902, 12, 45, 37} \[ \frac {4 \sqrt {x+1}}{45 x^{3/2}}-\frac {\sqrt {x+1}}{15 x^{5/2}}-\frac {\sinh ^{-1}\left (\sqrt {x}\right )}{3 x^3}-\frac {8 \sqrt {x+1}}{45 \sqrt {x}} \]

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 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^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*}

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Mathematica [A]  time = 0.02, 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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fricas [A]  time = 0.61, size = 37, normalized size = 0.60 \[ -\frac {{\left (8 \, x^{2} - 4 \, x + 3\right )} \sqrt {x + 1} \sqrt {x} + 15 \, \log \left (\sqrt {x + 1} + \sqrt {x}\right )}{45 \, x^{3}} \]

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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giac [A]  time = 0.37, size = 67, normalized size = 1.08 \[ -\frac {\log \left (\sqrt {x + 1} + \sqrt {x}\right )}{3 \, x^{3}} + \frac {16 \, {\left (10 \, {\left (\sqrt {x + 1} - \sqrt {x}\right )}^{4} - 5 \, {\left (\sqrt {x + 1} - \sqrt {x}\right )}^{2} + 1\right )}}{45 \, {\left ({\left (\sqrt {x + 1} - \sqrt {x}\right )}^{2} - 1\right )}^{5}} \]

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

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maple [A]  time = 0.00, size = 41, normalized size = 0.66 \[ -\frac {\arcsinh \left (\sqrt {x}\right )}{3 x^{3}}-\frac {\sqrt {1+x}}{15 x^{\frac {5}{2}}}+\frac {4 \sqrt {1+x}}{45 x^{\frac {3}{2}}}-\frac {8 \sqrt {1+x}}{45 \sqrt {x}} \]

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 = 0.89, size = 40, normalized size = 0.65 \[ -\frac {8 \, \sqrt {x + 1}}{45 \, \sqrt {x}} + \frac {4 \, \sqrt {x + 1}}{45 \, x^{\frac {3}{2}}} - \frac {\sqrt {x + 1}}{15 \, x^{\frac {5}{2}}} - \frac {\operatorname {arsinh}\left (\sqrt {x}\right )}{3 \, x^{3}} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {asinh}\left (\sqrt {x}\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(asinh(x^(1/2))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}{\left (\sqrt {x} \right )}}{x^{4}}\, dx \]

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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