3.287 \(\int \frac {\sinh ^{-1}(a x)^2}{x \sqrt {1+a^2 x^2}} \, dx\)

Optimal. Leaf size=68 \[ -2 \sinh ^{-1}(a x) \text {Li}_2\left (-e^{\sinh ^{-1}(a x)}\right )+2 \sinh ^{-1}(a x) \text {Li}_2\left (e^{\sinh ^{-1}(a x)}\right )+2 \text {Li}_3\left (-e^{\sinh ^{-1}(a x)}\right )-2 \text {Li}_3\left (e^{\sinh ^{-1}(a x)}\right )-2 \sinh ^{-1}(a x)^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right ) \]

[Out]

-2*arcsinh(a*x)^2*arctanh(a*x+(a^2*x^2+1)^(1/2))-2*arcsinh(a*x)*polylog(2,-a*x-(a^2*x^2+1)^(1/2))+2*arcsinh(a*
x)*polylog(2,a*x+(a^2*x^2+1)^(1/2))+2*polylog(3,-a*x-(a^2*x^2+1)^(1/2))-2*polylog(3,a*x+(a^2*x^2+1)^(1/2))

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {5760, 4182, 2531, 2282, 6589} \[ -2 \sinh ^{-1}(a x) \text {PolyLog}\left (2,-e^{\sinh ^{-1}(a x)}\right )+2 \sinh ^{-1}(a x) \text {PolyLog}\left (2,e^{\sinh ^{-1}(a x)}\right )+2 \text {PolyLog}\left (3,-e^{\sinh ^{-1}(a x)}\right )-2 \text {PolyLog}\left (3,e^{\sinh ^{-1}(a x)}\right )-2 \sinh ^{-1}(a x)^2 \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x]^2/(x*Sqrt[1 + a^2*x^2]),x]

[Out]

-2*ArcSinh[a*x]^2*ArcTanh[E^ArcSinh[a*x]] - 2*ArcSinh[a*x]*PolyLog[2, -E^ArcSinh[a*x]] + 2*ArcSinh[a*x]*PolyLo
g[2, E^ArcSinh[a*x]] + 2*PolyLog[3, -E^ArcSinh[a*x]] - 2*PolyLog[3, E^ArcSinh[a*x]]

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 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 4182

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 5760

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.11, size = 100, normalized size = 1.47 \[ 2 \sinh ^{-1}(a x) \text {Li}_2\left (-e^{-\sinh ^{-1}(a x)}\right )-2 \sinh ^{-1}(a x) \text {Li}_2\left (e^{-\sinh ^{-1}(a x)}\right )+2 \text {Li}_3\left (-e^{-\sinh ^{-1}(a x)}\right )-2 \text {Li}_3\left (e^{-\sinh ^{-1}(a x)}\right )+\sinh ^{-1}(a x)^2 \log \left (1-e^{-\sinh ^{-1}(a x)}\right )-\sinh ^{-1}(a x)^2 \log \left (e^{-\sinh ^{-1}(a x)}+1\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSinh[a*x]^2/(x*Sqrt[1 + a^2*x^2]),x]

[Out]

ArcSinh[a*x]^2*Log[1 - E^(-ArcSinh[a*x])] - ArcSinh[a*x]^2*Log[1 + E^(-ArcSinh[a*x])] + 2*ArcSinh[a*x]*PolyLog
[2, -E^(-ArcSinh[a*x])] - 2*ArcSinh[a*x]*PolyLog[2, E^(-ArcSinh[a*x])] + 2*PolyLog[3, -E^(-ArcSinh[a*x])] - 2*
PolyLog[3, E^(-ArcSinh[a*x])]

________________________________________________________________________________________

fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a^{2} x^{2} + 1} \operatorname {arsinh}\left (a x\right )^{2}}{a^{2} x^{3} + x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^2/x/(a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(a^2*x^2 + 1)*arcsinh(a*x)^2/(a^2*x^3 + x), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arsinh}\left (a x\right )^{2}}{\sqrt {a^{2} x^{2} + 1} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^2/x/(a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

integrate(arcsinh(a*x)^2/(sqrt(a^2*x^2 + 1)*x), x)

________________________________________________________________________________________

maple [A]  time = 0.09, size = 144, normalized size = 2.12 \[ -\arcsinh \left (a x \right )^{2} \ln \left (1+a x +\sqrt {a^{2} x^{2}+1}\right )-2 \arcsinh \left (a x \right ) \polylog \left (2, -a x -\sqrt {a^{2} x^{2}+1}\right )+2 \polylog \left (3, -a x -\sqrt {a^{2} x^{2}+1}\right )+\arcsinh \left (a x \right )^{2} \ln \left (1-a x -\sqrt {a^{2} x^{2}+1}\right )+2 \arcsinh \left (a x \right ) \polylog \left (2, a x +\sqrt {a^{2} x^{2}+1}\right )-2 \polylog \left (3, a x +\sqrt {a^{2} x^{2}+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(a*x)^2/x/(a^2*x^2+1)^(1/2),x)

[Out]

-arcsinh(a*x)^2*ln(1+a*x+(a^2*x^2+1)^(1/2))-2*arcsinh(a*x)*polylog(2,-a*x-(a^2*x^2+1)^(1/2))+2*polylog(3,-a*x-
(a^2*x^2+1)^(1/2))+arcsinh(a*x)^2*ln(1-a*x-(a^2*x^2+1)^(1/2))+2*arcsinh(a*x)*polylog(2,a*x+(a^2*x^2+1)^(1/2))-
2*polylog(3,a*x+(a^2*x^2+1)^(1/2))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)^2/x/(a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(arcsinh(a*x)^2/(sqrt(a^2*x^2 + 1)*x), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {asinh}\left (a\,x\right )}^2}{x\,\sqrt {a^2\,x^2+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a*x)^2/(x*(a^2*x^2 + 1)^(1/2)),x)

[Out]

int(asinh(a*x)^2/(x*(a^2*x^2 + 1)^(1/2)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}^{2}{\left (a x \right )}}{x \sqrt {a^{2} x^{2} + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a*x)**2/x/(a**2*x**2+1)**(1/2),x)

[Out]

Integral(asinh(a*x)**2/(x*sqrt(a**2*x**2 + 1)), x)

________________________________________________________________________________________