3.119 \(\int \frac{\sinh ^{-1}(a x)}{x^3 \sqrt{1+a^2 x^2}} \, dx\)

Optimal. Leaf size=80 \[ \frac{1}{2} a^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(a x)}\right )-\frac{1}{2} a^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(a x)}\right )-\frac{\sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{2 x^2}+a^2 \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-\frac{a}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.148335, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {5747, 5760, 4182, 2279, 2391, 30} \[ \frac{1}{2} a^2 \text{PolyLog}\left (2,-e^{\sinh ^{-1}(a x)}\right )-\frac{1}{2} a^2 \text{PolyLog}\left (2,e^{\sinh ^{-1}(a x)}\right )-\frac{\sqrt{a^2 x^2+1} \sinh ^{-1}(a x)}{2 x^2}+a^2 \sinh ^{-1}(a x) \tanh ^{-1}\left (e^{\sinh ^{-1}(a x)}\right )-\frac{a}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a*x]/(x^3*Sqrt[1 + a^2*x^2]),x]

[Out]

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

Rule 5747

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
((f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(d*f*(m + 1)), x] + (-Dist[(c^2*(m + 2*p + 3))/(f^2
*(m + 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^
2)^FracPart[p])/(f*(m + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSin
h[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[m, -1] && Int
egerQ[m]

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 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 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.671114, size = 126, normalized size = 1.58 \[ \frac{1}{8} a^2 \left (-4 \text{PolyLog}\left (2,-e^{-\sinh ^{-1}(a x)}\right )+4 \text{PolyLog}\left (2,e^{-\sinh ^{-1}(a x)}\right )-4 \sinh ^{-1}(a x) \log \left (1-e^{-\sinh ^{-1}(a x)}\right )+4 \sinh ^{-1}(a x) \log \left (e^{-\sinh ^{-1}(a x)}+1\right )+2 \tanh \left (\frac{1}{2} \sinh ^{-1}(a x)\right )-2 \coth \left (\frac{1}{2} \sinh ^{-1}(a x)\right )-\sinh ^{-1}(a x) \text{csch}^2\left (\frac{1}{2} \sinh ^{-1}(a x)\right )-\sinh ^{-1}(a x) \text{sech}^2\left (\frac{1}{2} \sinh ^{-1}(a x)\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSinh[a*x]/(x^3*Sqrt[1 + a^2*x^2]),x]

[Out]

(a^2*(-2*Coth[ArcSinh[a*x]/2] - ArcSinh[a*x]*Csch[ArcSinh[a*x]/2]^2 - 4*ArcSinh[a*x]*Log[1 - E^(-ArcSinh[a*x])
] + 4*ArcSinh[a*x]*Log[1 + E^(-ArcSinh[a*x])] - 4*PolyLog[2, -E^(-ArcSinh[a*x])] + 4*PolyLog[2, E^(-ArcSinh[a*
x])] - ArcSinh[a*x]*Sech[ArcSinh[a*x]/2]^2 + 2*Tanh[ArcSinh[a*x]/2]))/8

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Maple [A]  time = 0.052, size = 150, normalized size = 1.9 \begin{align*} -{\frac{1}{2\,{x}^{2}} \left ({a}^{2}{x}^{2}{\it Arcsinh} \left ( ax \right ) +ax\sqrt{{a}^{2}{x}^{2}+1}+{\it Arcsinh} \left ( ax \right ) \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+{\frac{{a}^{2}{\it Arcsinh} \left ( ax \right ) }{2}\ln \left ( 1+ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) }+{\frac{{a}^{2}}{2}{\it polylog} \left ( 2,-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) }-{\frac{{a}^{2}{\it Arcsinh} \left ( ax \right ) }{2}\ln \left ( 1-ax-\sqrt{{a}^{2}{x}^{2}+1} \right ) }-{\frac{{a}^{2}}{2}{\it polylog} \left ( 2,ax+\sqrt{{a}^{2}{x}^{2}+1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)/x^3/(a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a^{2} x^{2} + 1} \operatorname{arsinh}\left (a x\right )}{a^{2} x^{5} + x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)/x^3/(a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(a^2*x^2 + 1)*arcsinh(a*x)/(a^2*x^5 + x^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asinh}{\left (a x \right )}}{x^{3} \sqrt{a^{2} x^{2} + 1}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(a*x)/x**3/(a**2*x**2+1)**(1/2),x)

[Out]

Integral(asinh(a*x)/(x**3*sqrt(a**2*x**2 + 1)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(a*x)/x^3/(a^2*x^2+1)^(1/2),x, algorithm="giac")

[Out]

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