3.1.18 \(\int \frac {\sinh ^{-1}(a x)^2}{x^2} \, dx\) [18]

Optimal. Leaf size=50 \[ -\frac {\sinh ^{-1}(a x)^2}{x}-4 a \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-2 a \text {PolyLog}\left (2,-e^{\sinh ^{-1}(a x)}\right )+2 a \text {PolyLog}\left (2,e^{\sinh ^{-1}(a x)}\right ) \]

[Out]

-arcsinh(a*x)^2/x-4*a*arcsinh(a*x)*arctanh(a*x+(a^2*x^2+1)^(1/2))-2*a*polylog(2,-a*x-(a^2*x^2+1)^(1/2))+2*a*po
lylog(2,a*x+(a^2*x^2+1)^(1/2))

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Rubi [A]
time = 0.07, antiderivative size = 50, 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 = {5776, 5816, 4267, 2317, 2438} \begin {gather*} -2 a \text {Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )+2 a \text {Li}_2\left (e^{\sinh ^{-1}(a x)}\right )-\frac {\sinh ^{-1}(a x)^2}{x}-4 a \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcSinh[a*x]^2/x) - 4*a*ArcSinh[a*x]*ArcTanh[E^ArcSinh[a*x]] - 2*a*PolyLog[2, -E^ArcSinh[a*x]] + 2*a*PolyLog
[2, E^ArcSinh[a*x]]

Rule 2317

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 2438

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 4267

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

Rule 5776

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcS
inh[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[
1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 5816

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[(1/c^(m
+ 1))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]], Subst[Int[(a + b*x)^n*Sinh[x]^m, x], x, ArcSinh[c*x]], x] /; F
reeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]
time = 0.14, size = 75, normalized size = 1.50 \begin {gather*} a \left (-\sinh ^{-1}(a x) \left (\frac {\sinh ^{-1}(a x)}{a x}-2 \log \left (1-e^{-\sinh ^{-1}(a x)}\right )+2 \log \left (1+e^{-\sinh ^{-1}(a x)}\right )\right )+2 \text {PolyLog}\left (2,-e^{-\sinh ^{-1}(a x)}\right )-2 \text {PolyLog}\left (2,e^{-\sinh ^{-1}(a x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 1.84, size = 108, normalized size = 2.16

method	result	size
derivativedivides	\(a \left (-\frac {\arcsinh \left (a x \right )^{2}}{a x}-2 \arcsinh \left (a x \right ) \ln \left (1+a x +\sqrt {a^{2} x^{2}+1}\right )-2 \polylog \left (2, -a x -\sqrt {a^{2} x^{2}+1}\right )+2 \arcsinh \left (a x \right ) \ln \left (1-a x -\sqrt {a^{2} x^{2}+1}\right )+2 \polylog \left (2, a x +\sqrt {a^{2} x^{2}+1}\right )\right )\)	\(108\)
default	\(a \left (-\frac {\arcsinh \left (a x \right )^{2}}{a x}-2 \arcsinh \left (a x \right ) \ln \left (1+a x +\sqrt {a^{2} x^{2}+1}\right )-2 \polylog \left (2, -a x -\sqrt {a^{2} x^{2}+1}\right )+2 \arcsinh \left (a x \right ) \ln \left (1-a x -\sqrt {a^{2} x^{2}+1}\right )+2 \polylog \left (2, a x +\sqrt {a^{2} x^{2}+1}\right )\right )\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x)^2/x^2,x,method=_RETURNVERBOSE)

[Out]

a*(-arcsinh(a*x)^2/a/x-2*arcsinh(a*x)*ln(1+a*x+(a^2*x^2+1)^(1/2))-2*polylog(2,-a*x-(a^2*x^2+1)^(1/2))+2*arcsin
h(a*x)*ln(1-a*x-(a^2*x^2+1)^(1/2))+2*polylog(2,a*x+(a^2*x^2+1)^(1/2)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsinh(a*x)^2/x^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asinh}^{2}{\left (a x \right )}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(a*x)**2/x**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsinh(a*x)^2/x^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\mathrm {asinh}\left (a\,x\right )}^2}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(asinh(a*x)^2/x^2, x)

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