3.7 \(\int \frac {\cosh ^{-1}(a x)}{x^2} \, dx\)

Optimal. Leaf size=32 \[ a \tan ^{-1}\left (\sqrt {a x-1} \sqrt {a x+1}\right )-\frac {\cosh ^{-1}(a x)}{x} \]

[Out]

-arccosh(a*x)/x+a*arctan((a*x-1)^(1/2)*(a*x+1)^(1/2))

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Rubi [A]  time = 0.02, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {5662, 92, 205} \[ a \tan ^{-1}\left (\sqrt {a x-1} \sqrt {a x+1}\right )-\frac {\cosh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcCosh[a*x]/x^2,x]

[Out]

-(ArcCosh[a*x]/x) + a*ArcTan[Sqrt[-1 + a*x]*Sqrt[1 + a*x]]

Rule 92

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 5662

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcC
osh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcCosh[c*x])^(n - 1))/(Sqr
t[-1 + c*x]*Sqrt[1 + c*x]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\cosh ^{-1}(a x)}{x^2} \, dx &=-\frac {\cosh ^{-1}(a x)}{x}+a \int \frac {1}{x \sqrt {-1+a x} \sqrt {1+a x}} \, dx\\ &=-\frac {\cosh ^{-1}(a x)}{x}+a^2 \operatorname {Subst}\left (\int \frac {1}{a+a x^2} \, dx,x,\sqrt {-1+a x} \sqrt {1+a x}\right )\\ &=-\frac {\cosh ^{-1}(a x)}{x}+a \tan ^{-1}\left (\sqrt {-1+a x} \sqrt {1+a x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 57, normalized size = 1.78 \[ \frac {a \sqrt {a^2 x^2-1} \tan ^{-1}\left (\sqrt {a^2 x^2-1}\right )}{\sqrt {a x-1} \sqrt {a x+1}}-\frac {\cosh ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCosh[a*x]/x^2,x]

[Out]

-(ArcCosh[a*x]/x) + (a*Sqrt[-1 + a^2*x^2]*ArcTan[Sqrt[-1 + a^2*x^2]])/(Sqrt[-1 + a*x]*Sqrt[1 + a*x])

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fricas [B]  time = 0.50, size = 65, normalized size = 2.03 \[ \frac {2 \, a x \arctan \left (-a x + \sqrt {a^{2} x^{2} - 1}\right ) + {\left (x - 1\right )} \log \left (a x + \sqrt {a^{2} x^{2} - 1}\right ) + x \log \left (-a x + \sqrt {a^{2} x^{2} - 1}\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/x^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.43, size = 36, normalized size = 1.12 \[ a \arctan \left (\sqrt {a^{2} x^{2} - 1}\right ) - \frac {\log \left (a x + \sqrt {a^{2} x^{2} - 1}\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/x^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.02, size = 51, normalized size = 1.59 \[ -\frac {\mathrm {arccosh}\left (a x \right )}{x}-\frac {a \sqrt {a x -1}\, \sqrt {a x +1}\, \arctan \left (\frac {1}{\sqrt {a^{2} x^{2}-1}}\right )}{\sqrt {a^{2} x^{2}-1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccosh(a*x)/x^2,x)

[Out]

-arccosh(a*x)/x-a*(a*x-1)^(1/2)*(a*x+1)^(1/2)/(a^2*x^2-1)^(1/2)*arctan(1/(a^2*x^2-1)^(1/2))

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maxima [A]  time = 0.87, size = 22, normalized size = 0.69 \[ -a \arcsin \left (\frac {1}{a {\left | x \right |}}\right ) - \frac {\operatorname {arcosh}\left (a x\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/x^2,x, algorithm="maxima")

[Out]

-a*arcsin(1/(a*abs(x))) - arccosh(a*x)/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\mathrm {acosh}\left (a\,x\right )}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh(a*x)/x^2,x)

[Out]

int(acosh(a*x)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acosh(a*x)/x**2,x)

[Out]

Integral(acosh(a*x)/x**2, x)

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