3.40 \(\int \frac{\cosh ^{-1}(a x)^4}{x^3} \, dx\)

Optimal. Leaf size=115 \[ -6 a^2 \cosh ^{-1}(a x) \text{PolyLog}\left (2,-e^{2 \cosh ^{-1}(a x)}\right )+3 a^2 \text{PolyLog}\left (3,-e^{2 \cosh ^{-1}(a x)}\right )+2 a^2 \cosh ^{-1}(a x)^3-6 a^2 \cosh ^{-1}(a x)^2 \log \left (e^{2 \cosh ^{-1}(a x)}+1\right )-\frac{\cosh ^{-1}(a x)^4}{2 x^2}+\frac{2 a \sqrt{a x-1} \sqrt{a x+1} \cosh ^{-1}(a x)^3}{x} \]

[Out]

2*a^2*ArcCosh[a*x]^3 + (2*a*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*ArcCosh[a*x]^3)/x - ArcCosh[a*x]^4/(2*x^2) - 6*a^2*Ar
cCosh[a*x]^2*Log[1 + E^(2*ArcCosh[a*x])] - 6*a^2*ArcCosh[a*x]*PolyLog[2, -E^(2*ArcCosh[a*x])] + 3*a^2*PolyLog[
3, -E^(2*ArcCosh[a*x])]

________________________________________________________________________________________

Rubi [A]  time = 0.358169, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {5662, 5724, 5660, 3718, 2190, 2531, 2282, 6589} \[ -6 a^2 \cosh ^{-1}(a x) \text{PolyLog}\left (2,-e^{2 \cosh ^{-1}(a x)}\right )+3 a^2 \text{PolyLog}\left (3,-e^{2 \cosh ^{-1}(a x)}\right )+2 a^2 \cosh ^{-1}(a x)^3-6 a^2 \cosh ^{-1}(a x)^2 \log \left (e^{2 \cosh ^{-1}(a x)}+1\right )-\frac{\cosh ^{-1}(a x)^4}{2 x^2}+\frac{2 a \sqrt{a x-1} \sqrt{a x+1} \cosh ^{-1}(a x)^3}{x} \]

Antiderivative was successfully verified.

[In]

Int[ArcCosh[a*x]^4/x^3,x]

[Out]

2*a^2*ArcCosh[a*x]^3 + (2*a*Sqrt[-1 + a*x]*Sqrt[1 + a*x]*ArcCosh[a*x]^3)/x - ArcCosh[a*x]^4/(2*x^2) - 6*a^2*Ar
cCosh[a*x]^2*Log[1 + E^(2*ArcCosh[a*x])] - 6*a^2*ArcCosh[a*x]*PolyLog[2, -E^(2*ArcCosh[a*x])] + 3*a^2*PolyLog[
3, -E^(2*ArcCosh[a*x])]

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 5724

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d1_) + (e1_.)*(x_))^(p_.)*((d2_) + (e2_.)*(x
_))^(p_.), x_Symbol] :> Simp[((f*x)^(m + 1)*(d1 + e1*x)^(p + 1)*(d2 + e2*x)^(p + 1)*(a + b*ArcCosh[c*x])^n)/(d
1*d2*f*(m + 1)), x] + Dist[(b*c*n*(-(d1*d2))^IntPart[p]*(d1 + e1*x)^FracPart[p]*(d2 + e2*x)^FracPart[p])/(f*(m
+ 1)*(1 + c*x)^FracPart[p]*(-1 + c*x)^FracPart[p]), Int[(f*x)^(m + 1)*(-1 + c^2*x^2)^(p + 1/2)*(a + b*ArcCosh
[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d1, e1, d2, e2, f, m, p}, x] && EqQ[e1 - c*d1, 0] && EqQ[e2 + c*d2,
0] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1] && IntegerQ[p + 1/2]

Rule 5660

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Coth[x], x], x, ArcCosh
[c*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin{align*} \int \frac{\cosh ^{-1}(a x)^4}{x^3} \, dx &=-\frac{\cosh ^{-1}(a x)^4}{2 x^2}+(2 a) \int \frac{\cosh ^{-1}(a x)^3}{x^2 \sqrt{-1+a x} \sqrt{1+a x}} \, dx\\ &=\frac{2 a \sqrt{-1+a x} \sqrt{1+a x} \cosh ^{-1}(a x)^3}{x}-\frac{\cosh ^{-1}(a x)^4}{2 x^2}-\left (6 a^2\right ) \int \frac{\cosh ^{-1}(a x)^2}{x} \, dx\\ &=\frac{2 a \sqrt{-1+a x} \sqrt{1+a x} \cosh ^{-1}(a x)^3}{x}-\frac{\cosh ^{-1}(a x)^4}{2 x^2}-\left (6 a^2\right ) \operatorname{Subst}\left (\int x^2 \tanh (x) \, dx,x,\cosh ^{-1}(a x)\right )\\ &=2 a^2 \cosh ^{-1}(a x)^3+\frac{2 a \sqrt{-1+a x} \sqrt{1+a x} \cosh ^{-1}(a x)^3}{x}-\frac{\cosh ^{-1}(a x)^4}{2 x^2}-\left (12 a^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 x} x^2}{1+e^{2 x}} \, dx,x,\cosh ^{-1}(a x)\right )\\ &=2 a^2 \cosh ^{-1}(a x)^3+\frac{2 a \sqrt{-1+a x} \sqrt{1+a x} \cosh ^{-1}(a x)^3}{x}-\frac{\cosh ^{-1}(a x)^4}{2 x^2}-6 a^2 \cosh ^{-1}(a x)^2 \log \left (1+e^{2 \cosh ^{-1}(a x)}\right )+\left (12 a^2\right ) \operatorname{Subst}\left (\int x \log \left (1+e^{2 x}\right ) \, dx,x,\cosh ^{-1}(a x)\right )\\ &=2 a^2 \cosh ^{-1}(a x)^3+\frac{2 a \sqrt{-1+a x} \sqrt{1+a x} \cosh ^{-1}(a x)^3}{x}-\frac{\cosh ^{-1}(a x)^4}{2 x^2}-6 a^2 \cosh ^{-1}(a x)^2 \log \left (1+e^{2 \cosh ^{-1}(a x)}\right )-6 a^2 \cosh ^{-1}(a x) \text{Li}_2\left (-e^{2 \cosh ^{-1}(a x)}\right )+\left (6 a^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 x}\right ) \, dx,x,\cosh ^{-1}(a x)\right )\\ &=2 a^2 \cosh ^{-1}(a x)^3+\frac{2 a \sqrt{-1+a x} \sqrt{1+a x} \cosh ^{-1}(a x)^3}{x}-\frac{\cosh ^{-1}(a x)^4}{2 x^2}-6 a^2 \cosh ^{-1}(a x)^2 \log \left (1+e^{2 \cosh ^{-1}(a x)}\right )-6 a^2 \cosh ^{-1}(a x) \text{Li}_2\left (-e^{2 \cosh ^{-1}(a x)}\right )+\left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 \cosh ^{-1}(a x)}\right )\\ &=2 a^2 \cosh ^{-1}(a x)^3+\frac{2 a \sqrt{-1+a x} \sqrt{1+a x} \cosh ^{-1}(a x)^3}{x}-\frac{\cosh ^{-1}(a x)^4}{2 x^2}-6 a^2 \cosh ^{-1}(a x)^2 \log \left (1+e^{2 \cosh ^{-1}(a x)}\right )-6 a^2 \cosh ^{-1}(a x) \text{Li}_2\left (-e^{2 \cosh ^{-1}(a x)}\right )+3 a^2 \text{Li}_3\left (-e^{2 \cosh ^{-1}(a x)}\right )\\ \end{align*}

Mathematica [A]  time = 1.07254, size = 112, normalized size = 0.97 \[ a^2 \left (6 \cosh ^{-1}(a x) \text{PolyLog}\left (2,-e^{-2 \cosh ^{-1}(a x)}\right )+3 \text{PolyLog}\left (3,-e^{-2 \cosh ^{-1}(a x)}\right )+2 \cosh ^{-1}(a x)^2 \left (\frac{\sqrt{\frac{a x-1}{a x+1}} (a x+1) \cosh ^{-1}(a x)}{a x}-\cosh ^{-1}(a x)-3 \log \left (e^{-2 \cosh ^{-1}(a x)}+1\right )\right )\right )-\frac{\cosh ^{-1}(a x)^4}{2 x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCosh[a*x]^4/x^3,x]

[Out]

-ArcCosh[a*x]^4/(2*x^2) + a^2*(2*ArcCosh[a*x]^2*(-ArcCosh[a*x] + (Sqrt[(-1 + a*x)/(1 + a*x)]*(1 + a*x)*ArcCosh
[a*x])/(a*x) - 3*Log[1 + E^(-2*ArcCosh[a*x])]) + 6*ArcCosh[a*x]*PolyLog[2, -E^(-2*ArcCosh[a*x])] + 3*PolyLog[3
, -E^(-2*ArcCosh[a*x])])

________________________________________________________________________________________

Maple [A]  time = 0.071, size = 149, normalized size = 1.3 \begin{align*} 2\,{a}^{2} \left ({\rm arccosh} \left (ax\right ) \right ) ^{3}-{\frac{ \left ({\rm arccosh} \left (ax\right ) \right ) ^{4}}{2\,{x}^{2}}}-6\,{a}^{2} \left ({\rm arccosh} \left (ax\right ) \right ) ^{2}\ln \left ( 1+ \left ( ax+\sqrt{ax-1}\sqrt{ax+1} \right ) ^{2} \right ) -6\,{a}^{2}{\rm arccosh} \left (ax\right ){\it polylog} \left ( 2,- \left ( ax+\sqrt{ax-1}\sqrt{ax+1} \right ) ^{2} \right ) +3\,{a}^{2}{\it polylog} \left ( 3,- \left ( ax+\sqrt{ax-1}\sqrt{ax+1} \right ) ^{2} \right ) +2\,{\frac{a \left ({\rm arccosh} \left (ax\right ) \right ) ^{3}\sqrt{ax-1}\sqrt{ax+1}}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccosh(a*x)^4/x^3,x)

[Out]

2*a^2*arccosh(a*x)^3-1/2*arccosh(a*x)^4/x^2-6*a^2*arccosh(a*x)^2*ln(1+(a*x+(a*x-1)^(1/2)*(a*x+1)^(1/2))^2)-6*a
^2*arccosh(a*x)*polylog(2,-(a*x+(a*x-1)^(1/2)*(a*x+1)^(1/2))^2)+3*a^2*polylog(3,-(a*x+(a*x-1)^(1/2)*(a*x+1)^(1
/2))^2)+2*a*arccosh(a*x)^3*(a*x-1)^(1/2)*(a*x+1)^(1/2)/x

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{\log \left (a x + \sqrt{a x + 1} \sqrt{a x - 1}\right )^{4}}{2 \, x^{2}} + \int \frac{2 \,{\left (a^{3} x^{2} + \sqrt{a x + 1} \sqrt{a x - 1} a^{2} x - a\right )} \log \left (a x + \sqrt{a x + 1} \sqrt{a x - 1}\right )^{3}}{a^{3} x^{5} - a x^{3} +{\left (a^{2} x^{4} - x^{2}\right )} \sqrt{a x + 1} \sqrt{a x - 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)^4/x^3,x, algorithm="maxima")

[Out]

-1/2*log(a*x + sqrt(a*x + 1)*sqrt(a*x - 1))^4/x^2 + integrate(2*(a^3*x^2 + sqrt(a*x + 1)*sqrt(a*x - 1)*a^2*x -
a)*log(a*x + sqrt(a*x + 1)*sqrt(a*x - 1))^3/(a^3*x^5 - a*x^3 + (a^2*x^4 - x^2)*sqrt(a*x + 1)*sqrt(a*x - 1)),
x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\operatorname{arcosh}\left (a x\right )^{4}}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)^4/x^3,x, algorithm="fricas")

[Out]

integral(arccosh(a*x)^4/x^3, x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acosh}^{4}{\left (a x \right )}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acosh(a*x)**4/x**3,x)

[Out]

Integral(acosh(a*x)**4/x**3, x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{arcosh}\left (a x\right )^{4}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccosh(a*x)^4/x^3,x, algorithm="giac")

[Out]

integrate(arccosh(a*x)^4/x^3, x)