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

Optimal. Leaf size=60 \[ -2 i a \text {Li}_2\left (-i e^{\cosh ^{-1}(a x)}\right )+2 i a \text {Li}_2\left (i e^{\cosh ^{-1}(a x)}\right )-\frac {\cosh ^{-1}(a x)^2}{x}+4 a \cosh ^{-1}(a x) \tan ^{-1}\left (e^{\cosh ^{-1}(a x)}\right ) \]

[Out]

-arccosh(a*x)^2/x+4*a*arccosh(a*x)*arctan(a*x+(a*x-1)^(1/2)*(a*x+1)^(1/2))-2*I*a*polylog(2,-I*(a*x+(a*x-1)^(1/
2)*(a*x+1)^(1/2)))+2*I*a*polylog(2,I*(a*x+(a*x-1)^(1/2)*(a*x+1)^(1/2)))

________________________________________________________________________________________

Rubi [A]  time = 0.21, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5662, 5761, 4180, 2279, 2391} \[ -2 i a \text {PolyLog}\left (2,-i e^{\cosh ^{-1}(a x)}\right )+2 i a \text {PolyLog}\left (2,i e^{\cosh ^{-1}(a x)}\right )-\frac {\cosh ^{-1}(a x)^2}{x}+4 a \cosh ^{-1}(a x) \tan ^{-1}\left (e^{\cosh ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcCosh[a*x]^2/x) + 4*a*ArcCosh[a*x]*ArcTan[E^ArcCosh[a*x]] - (2*I)*a*PolyLog[2, (-I)*E^ArcCosh[a*x]] + (2*I
)*a*PolyLog[2, I*E^ArcCosh[a*x]]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
 e, n}, x] && EqQ[c*d, 1]

Rule 4180

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

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 5761

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.27, size = 92, normalized size = 1.53 \[ -i a \left (2 \text {Li}_2\left (-i e^{-\cosh ^{-1}(a x)}\right )-2 \text {Li}_2\left (i e^{-\cosh ^{-1}(a x)}\right )+\cosh ^{-1}(a x) \left (-\frac {i \cosh ^{-1}(a x)}{a x}+2 \log \left (1-i e^{-\cosh ^{-1}(a x)}\right )-2 \log \left (1+i e^{-\cosh ^{-1}(a x)}\right )\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-I)*a*(ArcCosh[a*x]*(((-I)*ArcCosh[a*x])/(a*x) + 2*Log[1 - I/E^ArcCosh[a*x]] - 2*Log[1 + I/E^ArcCosh[a*x]]) +
 2*PolyLog[2, (-I)/E^ArcCosh[a*x]] - 2*PolyLog[2, I/E^ArcCosh[a*x]])

________________________________________________________________________________________

fricas [F]  time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcosh}\left (a x\right )^{2}}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arccosh(a*x)^2/x^2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccosh(a*x)^2/x^2, x)

________________________________________________________________________________________

maple [A]  time = 0.16, size = 137, normalized size = 2.28 \[ -\frac {\mathrm {arccosh}\left (a x \right )^{2}}{x}-2 i a \,\mathrm {arccosh}\left (a x \right ) \ln \left (1+i \left (a x +\sqrt {a x -1}\, \sqrt {a x +1}\right )\right )+2 i a \,\mathrm {arccosh}\left (a x \right ) \ln \left (1-i \left (a x +\sqrt {a x -1}\, \sqrt {a x +1}\right )\right )-2 i a \dilog \left (1+i \left (a x +\sqrt {a x -1}\, \sqrt {a x +1}\right )\right )+2 i a \dilog \left (1-i \left (a x +\sqrt {a x -1}\, \sqrt {a x +1}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-arccosh(a*x)^2/x-2*I*a*arccosh(a*x)*ln(1+I*(a*x+(a*x-1)^(1/2)*(a*x+1)^(1/2)))+2*I*a*arccosh(a*x)*ln(1-I*(a*x+
(a*x-1)^(1/2)*(a*x+1)^(1/2)))-2*I*a*dilog(1+I*(a*x+(a*x-1)^(1/2)*(a*x+1)^(1/2)))+2*I*a*dilog(1-I*(a*x+(a*x-1)^
(1/2)*(a*x+1)^(1/2)))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {\log \left (a x + \sqrt {a x + 1} \sqrt {a x - 1}\right )^{2}}{x} + \int \frac {2 \, {\left (a^{3} x^{2} + \sqrt {a x + 1} \sqrt {a x - 1} a^{2} x - a\right )} \log \left (a x + \sqrt {a x + 1} \sqrt {a x - 1}\right )}{a^{3} x^{4} - a x^{2} + {\left (a^{2} x^{3} - x\right )} \sqrt {a x + 1} \sqrt {a x - 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\mathrm {acosh}\left (a\,x\right )}^2}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(acosh(a*x)^2/x^2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(acosh(a*x)**2/x**2, x)

________________________________________________________________________________________