3.62 \(\int \frac{1}{\cosh ^{-1}(a x)^3} \, dx\)

Optimal. Leaf size=55 \[ \frac{\text{Shi}\left (\cosh ^{-1}(a x)\right )}{2 a}-\frac{x}{2 \cosh ^{-1}(a x)}-\frac{\sqrt{a x-1} \sqrt{a x+1}}{2 a \cosh ^{-1}(a x)^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.187885, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {5656, 5775, 5658, 3298} \[ \frac{\text{Shi}\left (\cosh ^{-1}(a x)\right )}{2 a}-\frac{x}{2 \cosh ^{-1}(a x)}-\frac{\sqrt{a x-1} \sqrt{a x+1}}{2 a \cosh ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCosh[a*x]^(-3),x]

[Out]

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

Rule 5656

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Sqrt[-1 + c*x]*Sqrt[1 + c*x]*(a + b*ArcCosh[c
*x])^(n + 1))/(b*c*(n + 1)), x] - Dist[c/(b*(n + 1)), Int[(x*(a + b*ArcCosh[c*x])^(n + 1))/(Sqrt[-1 + c*x]*Sqr
t[1 + c*x]), x], x] /; FreeQ[{a, b, c}, x] && LtQ[n, -1]

Rule 5775

Int[(((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/(Sqrt[(d1_) + (e1_.)*(x_)]*Sqrt[(d2_) + (e2_
.)*(x_)]), x_Symbol] :> Simp[((f*x)^m*(a + b*ArcCosh[c*x])^(n + 1))/(b*c*Sqrt[-(d1*d2)]*(n + 1)), x] - Dist[(f
*m)/(b*c*Sqrt[-(d1*d2)]*(n + 1)), Int[(f*x)^(m - 1)*(a + b*ArcCosh[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d1
, e1, d2, e2, f, m}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2, 0] && LtQ[n, -1] && GtQ[d1, 0] && LtQ[d2, 0]

Rule 5658

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Dist[(b*c)^(-1), Subst[Int[x^n*Sinh[a/b - x/b], x]
, x, a + b*ArcCosh[c*x]], x] /; FreeQ[{a, b, c, n}, x]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rubi steps

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

Mathematica [A]  time = 0.0419904, size = 55, normalized size = 1. \[ \frac{\text{Shi}\left (\cosh ^{-1}(a x)\right )}{2 a}-\frac{x}{2 \cosh ^{-1}(a x)}-\frac{\sqrt{a x-1} \sqrt{a x+1}}{2 a \cosh ^{-1}(a x)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCosh[a*x]^(-3),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.027, size = 45, normalized size = 0.8 \begin{align*}{\frac{1}{a} \left ( -{\frac{1}{2\, \left ({\rm arccosh} \left (ax\right ) \right ) ^{2}}\sqrt{ax-1}\sqrt{ax+1}}-{\frac{ax}{2\,{\rm arccosh} \left (ax\right )}}+{\frac{{\it Shi} \left ({\rm arccosh} \left (ax\right ) \right ) }{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/arccosh(a*x)^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccosh(a*x)^3,x, algorithm="maxima")

[Out]

-1/2*(a^7*x^7 - 3*a^5*x^5 + 3*a^3*x^3 + (a^4*x^4 - a^2*x^2)*(a*x + 1)^(3/2)*(a*x - 1)^(3/2) + (3*a^5*x^5 - 5*a
^3*x^3 + 2*a*x)*(a*x + 1)*(a*x - 1) + (3*a^6*x^6 - 7*a^4*x^4 + 5*a^2*x^2 - 1)*sqrt(a*x + 1)*sqrt(a*x - 1) - a*
x + (a^7*x^7 - 3*a^5*x^5 + 3*a^3*x^3 + (a^4*x^4 - 1)*(a*x + 1)^(3/2)*(a*x - 1)^(3/2) + 3*(a^5*x^5 - a^3*x^3)*(
a*x + 1)*(a*x - 1) + (3*a^6*x^6 - 6*a^4*x^4 + 4*a^2*x^2 - 1)*sqrt(a*x + 1)*sqrt(a*x - 1) - a*x)*log(a*x + sqrt
(a*x + 1)*sqrt(a*x - 1)))/((a^7*x^6 + (a*x + 1)^(3/2)*(a*x - 1)^(3/2)*a^4*x^3 - 3*a^5*x^4 + 3*a^3*x^2 + 3*(a^5
*x^4 - a^3*x^2)*(a*x + 1)*(a*x - 1) + 3*(a^6*x^5 - 2*a^4*x^3 + a^2*x)*sqrt(a*x + 1)*sqrt(a*x - 1) - a)*log(a*x
+ sqrt(a*x + 1)*sqrt(a*x - 1))^2) + integrate(1/2*(a^8*x^8 - 4*a^6*x^6 + 6*a^4*x^4 + (a^4*x^4 + 3)*(a*x + 1)^
2*(a*x - 1)^2 + (4*a^5*x^5 - 4*a^3*x^3 + 3*a*x)*(a*x + 1)^(3/2)*(a*x - 1)^(3/2) - 4*a^2*x^2 + 3*(2*a^6*x^6 - 4
*a^4*x^4 + a^2*x^2 + 1)*(a*x + 1)*(a*x - 1) + (4*a^7*x^7 - 12*a^5*x^5 + 9*a^3*x^3 - a*x)*sqrt(a*x + 1)*sqrt(a*
x - 1) + 1)/((a^8*x^8 + (a*x + 1)^2*(a*x - 1)^2*a^4*x^4 - 4*a^6*x^6 + 6*a^4*x^4 + 4*(a^5*x^5 - a^3*x^3)*(a*x +
1)^(3/2)*(a*x - 1)^(3/2) - 4*a^2*x^2 + 6*(a^6*x^6 - 2*a^4*x^4 + a^2*x^2)*(a*x + 1)*(a*x - 1) + 4*(a^7*x^7 - 3
*a^5*x^5 + 3*a^3*x^3 - a*x)*sqrt(a*x + 1)*sqrt(a*x - 1) + 1)*log(a*x + sqrt(a*x + 1)*sqrt(a*x - 1))), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{\operatorname{arcosh}\left (a x\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/arccosh(a*x)^3,x, algorithm="fricas")

[Out]

integral(arccosh(a*x)^(-3), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\operatorname{acosh}^{3}{\left (a x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acosh(a*x)**(-3), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\operatorname{arcosh}\left (a x\right )^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccosh(a*x)^(-3), x)