3.9 \(\int \frac{\cosh ^{-1}(a x)}{x^4} \, dx\)

Optimal. Leaf size=65 \[ \frac{1}{6} a^3 \tan ^{-1}\left (\sqrt{a x-1} \sqrt{a x+1}\right )+\frac{a \sqrt{a x-1} \sqrt{a x+1}}{6 x^2}-\frac{\cosh ^{-1}(a x)}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.027259, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {5662, 103, 12, 92, 205} \[ \frac{1}{6} a^3 \tan ^{-1}\left (\sqrt{a x-1} \sqrt{a x+1}\right )+\frac{a \sqrt{a x-1} \sqrt{a x+1}}{6 x^2}-\frac{\cosh ^{-1}(a x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcCosh[a*x]/x^4,x]

[Out]

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

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]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 12

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

Rubi steps

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

Mathematica [A]  time = 0.0752107, size = 78, normalized size = 1.2 \[ \frac{\frac{a x \left (a^2 x^2+a^2 x^2 \sqrt{a^2 x^2-1} \tan ^{-1}\left (\sqrt{a^2 x^2-1}\right )-1\right )}{\sqrt{a x-1} \sqrt{a x+1}}-2 \cosh ^{-1}(a x)}{6 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCosh[a*x]/x^4,x]

[Out]

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

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Maple [A]  time = 0.013, size = 73, normalized size = 1.1 \begin{align*} -{\frac{{\rm arccosh} \left (ax\right )}{3\,{x}^{3}}}-{\frac{{a}^{3}}{6}\sqrt{ax-1}\sqrt{ax+1}\arctan \left ({\frac{1}{\sqrt{{a}^{2}{x}^{2}-1}}} \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}-1}}}}+{\frac{a}{6\,{x}^{2}}\sqrt{ax-1}\sqrt{ax+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccosh(a*x)/x^4,x)

[Out]

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

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Maxima [A]  time = 1.82707, size = 61, normalized size = 0.94 \begin{align*} -\frac{1}{6} \,{\left (a^{2} \arcsin \left (\frac{1}{\sqrt{a^{2}}{\left | x \right |}}\right ) - \frac{\sqrt{a^{2} x^{2} - 1}}{x^{2}}\right )} a - \frac{\operatorname{arcosh}\left (a x\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/x^4,x, algorithm="maxima")

[Out]

-1/6*(a^2*arcsin(1/(sqrt(a^2)*abs(x))) - sqrt(a^2*x^2 - 1)/x^2)*a - 1/3*arccosh(a*x)/x^3

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Fricas [A]  time = 2.45052, size = 215, normalized size = 3.31 \begin{align*} \frac{2 \, a^{3} x^{3} \arctan \left (-a x + \sqrt{a^{2} x^{2} - 1}\right ) + 2 \, x^{3} \log \left (-a x + \sqrt{a^{2} x^{2} - 1}\right ) + \sqrt{a^{2} x^{2} - 1} a x + 2 \,{\left (x^{3} - 1\right )} \log \left (a x + \sqrt{a^{2} x^{2} - 1}\right )}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/x^4,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acosh(a*x)/x**4,x)

[Out]

Integral(acosh(a*x)/x**4, x)

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Giac [A]  time = 1.33463, size = 78, normalized size = 1.2 \begin{align*} \frac{1}{6} \, a^{3}{\left (\frac{\sqrt{a^{2} x^{2} - 1}}{a^{2} x^{2}} + \arctan \left (\sqrt{a^{2} x^{2} - 1}\right )\right )} - \frac{\log \left (a x + \sqrt{a^{2} x^{2} - 1}\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/x^4,x, algorithm="giac")

[Out]

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