3.10 \(\int \frac{\cosh ^{-1}(a x)}{x^5} \, dx\)

Optimal. Leaf size=66 \[ \frac{a^3 \sqrt{a x-1} \sqrt{a x+1}}{6 x}+\frac{a \sqrt{a x-1} \sqrt{a x+1}}{12 x^3}-\frac{\cosh ^{-1}(a x)}{4 x^4} \]

[Out]

(a*Sqrt[-1 + a*x]*Sqrt[1 + a*x])/(12*x^3) + (a^3*Sqrt[-1 + a*x]*Sqrt[1 + a*x])/(6*x) - ArcCosh[a*x]/(4*x^4)

________________________________________________________________________________________

Rubi [A]  time = 0.0244926, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5662, 103, 12, 95} \[ \frac{a^3 \sqrt{a x-1} \sqrt{a x+1}}{6 x}+\frac{a \sqrt{a x-1} \sqrt{a x+1}}{12 x^3}-\frac{\cosh ^{-1}(a x)}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[ArcCosh[a*x]/x^5,x]

[Out]

(a*Sqrt[-1 + a*x]*Sqrt[1 + a*x])/(12*x^3) + (a^3*Sqrt[-1 + a*x]*Sqrt[1 + a*x])/(6*x) - ArcCosh[a*x]/(4*x^4)

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 95

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] /; FreeQ[{a, b, c, d,
e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && EqQ[a*d*f*(m + 1) + b*c*f*(n + 1) + b*d*e*(p + 1), 0
] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{\cosh ^{-1}(a x)}{x^5} \, dx &=-\frac{\cosh ^{-1}(a x)}{4 x^4}+\frac{1}{4} a \int \frac{1}{x^4 \sqrt{-1+a x} \sqrt{1+a x}} \, dx\\ &=\frac{a \sqrt{-1+a x} \sqrt{1+a x}}{12 x^3}-\frac{\cosh ^{-1}(a x)}{4 x^4}+\frac{1}{12} a \int \frac{2 a^2}{x^2 \sqrt{-1+a x} \sqrt{1+a x}} \, dx\\ &=\frac{a \sqrt{-1+a x} \sqrt{1+a x}}{12 x^3}-\frac{\cosh ^{-1}(a x)}{4 x^4}+\frac{1}{6} a^3 \int \frac{1}{x^2 \sqrt{-1+a x} \sqrt{1+a x}} \, dx\\ &=\frac{a \sqrt{-1+a x} \sqrt{1+a x}}{12 x^3}+\frac{a^3 \sqrt{-1+a x} \sqrt{1+a x}}{6 x}-\frac{\cosh ^{-1}(a x)}{4 x^4}\\ \end{align*}

Mathematica [A]  time = 0.0229603, size = 45, normalized size = 0.68 \[ \frac{a x \sqrt{a x-1} \sqrt{a x+1} \left (2 a^2 x^2+1\right )-3 \cosh ^{-1}(a x)}{12 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCosh[a*x]/x^5,x]

[Out]

(a*x*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*(1 + 2*a^2*x^2) - 3*ArcCosh[a*x])/(12*x^4)

________________________________________________________________________________________

Maple [A]  time = 0.012, size = 50, normalized size = 0.8 \begin{align*}{a}^{4} \left ( -{\frac{{\rm arccosh} \left (ax\right )}{4\,{x}^{4}{a}^{4}}}+{\frac{2\,{a}^{2}{x}^{2}+1}{12\,{x}^{3}{a}^{3}}\sqrt{ax-1}\sqrt{ax+1}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccosh(a*x)/x^5,x)

[Out]

a^4*(-1/4*arccosh(a*x)/a^4/x^4+1/12*(a*x-1)^(1/2)*(a*x+1)^(1/2)*(2*a^2*x^2+1)/a^3/x^3)

________________________________________________________________________________________

Maxima [A]  time = 1.77233, size = 65, normalized size = 0.98 \begin{align*} \frac{1}{12} \,{\left (\frac{2 \, \sqrt{a^{2} x^{2} - 1} a^{2}}{x} + \frac{\sqrt{a^{2} x^{2} - 1}}{x^{3}}\right )} a - \frac{\operatorname{arcosh}\left (a x\right )}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/x^5,x, algorithm="maxima")

[Out]

1/12*(2*sqrt(a^2*x^2 - 1)*a^2/x + sqrt(a^2*x^2 - 1)/x^3)*a - 1/4*arccosh(a*x)/x^4

________________________________________________________________________________________

Fricas [A]  time = 2.46552, size = 109, normalized size = 1.65 \begin{align*} \frac{{\left (2 \, a^{3} x^{3} + a x\right )} \sqrt{a^{2} x^{2} - 1} - 3 \, \log \left (a x + \sqrt{a^{2} x^{2} - 1}\right )}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/x^5,x, algorithm="fricas")

[Out]

1/12*((2*a^3*x^3 + a*x)*sqrt(a^2*x^2 - 1) - 3*log(a*x + sqrt(a^2*x^2 - 1)))/x^4

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acosh}{\left (a x \right )}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acosh(a*x)/x**5,x)

[Out]

Integral(acosh(a*x)/x**5, x)

________________________________________________________________________________________

Giac [A]  time = 1.34505, size = 104, normalized size = 1.58 \begin{align*} \frac{{\left (3 \,{\left (x{\left | a \right |} - \sqrt{a^{2} x^{2} - 1}\right )}^{2} + 1\right )} a^{3}{\left | a \right |}}{3 \,{\left ({\left (x{\left | a \right |} - \sqrt{a^{2} x^{2} - 1}\right )}^{2} + 1\right )}^{3}} - \frac{\log \left (a x + \sqrt{a^{2} x^{2} - 1}\right )}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)/x^5,x, algorithm="giac")

[Out]

1/3*(3*(x*abs(a) - sqrt(a^2*x^2 - 1))^2 + 1)*a^3*abs(a)/((x*abs(a) - sqrt(a^2*x^2 - 1))^2 + 1)^3 - 1/4*log(a*x
+ sqrt(a^2*x^2 - 1))/x^4