3.237 \(\int \frac {\cosh ^{-1}(\sqrt {x})}{x^3} \, dx\)

Optimal. Leaf size=76 \[ \frac {\sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1}}{6 x^{3/2}}-\frac {\cosh ^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {\sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1}}{3 \sqrt {x}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 76, 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 = {5903, 12, 272, 265} \[ \frac {\sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1}}{6 x^{3/2}}-\frac {\cosh ^{-1}\left (\sqrt {x}\right )}{2 x^2}+\frac {\sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1}}{3 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]])/(6*x^(3/2)) + (Sqrt[-1 + Sqrt[x]]*Sqrt[1 + Sqrt[x]])/(3*Sqrt[x]) - ArcC
osh[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 265

Int[((c_.)*(x_))^(m_.)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*
x)^(m + 1)*(a1 + b1*x^n)^(p + 1)*(a2 + b2*x^n)^(p + 1))/(a1*a2*c*(m + 1)), x] /; FreeQ[{a1, b1, a2, b2, c, m,
n, p}, x] && EqQ[a2*b1 + a1*b2, 0] && EqQ[(m + 1)/(2*n) + p + 1, 0] && NeQ[m, -1]

Rule 272

Int[(x_)^(m_)*((a1_) + (b1_.)*(x_)^(n_))^(p_)*((a2_) + (b2_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a
1 + b1*x^n)^(p + 1)*(a2 + b2*x^n)^(p + 1))/(a1*a2*(m + 1)), x] - Dist[(b1*b2*(m + 2*n*(p + 1) + 1))/(a1*a2*(m
+ 1)), Int[x^(m + 2*n)*(a1 + b1*x^n)^p*(a2 + b2*x^n)^p, x], x] /; FreeQ[{a1, b1, a2, b2, m, n, p}, x] && EqQ[a
2*b1 + a1*b2, 0] && ILtQ[Simplify[(m + 1)/(2*n) + p + 1], 0] && NeQ[m, -1]

Rule 5903

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 49, normalized size = 0.64 \[ \frac {\sqrt {\sqrt {x}-1} \sqrt {\sqrt {x}+1} \sqrt {x} (2 x+1)-3 \cosh ^{-1}\left (\sqrt {x}\right )}{6 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.62, size = 32, normalized size = 0.42 \[ \frac {{\left (2 \, x + 1\right )} \sqrt {x - 1} \sqrt {x} - 3 \, \log \left (\sqrt {x - 1} + \sqrt {x}\right )}{6 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(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 = 1.53, size = 62, normalized size = 0.82 \[ -\frac {\log \left (\sqrt {\sqrt {x} + 1} \sqrt {\sqrt {x} - 1} + \sqrt {x}\right )}{2 \, x^{2}} + \frac {2 \, {\left (3 \, {\left (\sqrt {x - 1} - \sqrt {x}\right )}^{2} + 1\right )}}{3 \, {\left ({\left (\sqrt {x - 1} - \sqrt {x}\right )}^{2} + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(sqrt(sqrt(x) + 1)*sqrt(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 = 35, normalized size = 0.46 \[ -\frac {\mathrm {arccosh}\left (\sqrt {x}\right )}{2 x^{2}}+\frac {\sqrt {-1+\sqrt {x}}\, \sqrt {1+\sqrt {x}}\, \left (1+2 x \right )}{6 x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.81, size = 30, normalized size = 0.39 \[ \frac {\sqrt {x - 1}}{3 \, \sqrt {x}} + \frac {\sqrt {x - 1}}{6 \, x^{\frac {3}{2}}} - \frac {\operatorname {arcosh}\left (\sqrt {x}\right )}{2 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(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*arccosh(sqrt(x))/x^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {acosh}\left (\sqrt {x}\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acosh(x^(1/2))/x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acosh(sqrt(x))/x**3, x)

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