3.2.19 \(\int \frac {\sinh ^{-1}(a x)}{x^3 \sqrt {1+a^2 x^2}} \, dx\) [119]

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

[Out]

-1/2*a/x+a^2*arcsinh(a*x)*arctanh(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*pol
ylog(2,a*x+(a^2*x^2+1)^(1/2))-1/2*arcsinh(a*x)*(a^2*x^2+1)^(1/2)/x^2

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Rubi [A]
time = 0.11, antiderivative size = 80, normalized size of antiderivative = 1.00, 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 = {5809, 5816, 4267, 2317, 2438, 30} \begin {gather*} \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 )-\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*a/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 30

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

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 5809

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/(f*(m + 1)))*Simp[(d +
e*x^2)^p/(1 + c^2*x^2)^p], Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /;
FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && ILtQ[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)}{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 \text {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 \text {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(a x)\right )-\frac {1}{2} a^2 \text {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 \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(a x)}\right )-\frac {1}{2} a^2 \text {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*}

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Mathematica [A]
time = 0.46, size = 126, normalized size = 1.58 \begin {gather*} \frac {1}{8} a^2 \left (-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 )-4 \sinh ^{-1}(a x) \log \left (1-e^{-\sinh ^{-1}(a x)}\right )+4 \sinh ^{-1}(a x) \log \left (1+e^{-\sinh ^{-1}(a x)}\right )-4 \text {PolyLog}\left (2,-e^{-\sinh ^{-1}(a x)}\right )+4 \text {PolyLog}\left (2,e^{-\sinh ^{-1}(a x)}\right )-\sinh ^{-1}(a x) \text {sech}^2\left (\frac {1}{2} \sinh ^{-1}(a x)\right )+2 \tanh \left (\frac {1}{2} \sinh ^{-1}(a x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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 = 4.94, size = 150, normalized size = 1.88

method	result	size
default	\(-\frac {\arcsinh \left (a x \right ) a^{2} x^{2}+a x \sqrt {a^{2} x^{2}+1}+\arcsinh \left (a x \right )}{2 \sqrt {a^{2} x^{2}+1}\, x^{2}}-\frac {a^{2} \arcsinh \left (a x \right ) \ln \left (1-a x -\sqrt {a^{2} x^{2}+1}\right )}{2}-\frac {a^{2} \polylog \left (2, a x +\sqrt {a^{2} x^{2}+1}\right )}{2}+\frac {a^{2} \arcsinh \left (a x \right ) \ln \left (1+a x +\sqrt {a^{2} x^{2}+1}\right )}{2}+\frac {a^{2} \polylog \left (2, -a x -\sqrt {a^{2} x^{2}+1}\right )}{2}\)	\(150\)

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,method=_RETURNVERBOSE)

[Out]

-1/2/(a^2*x^2+1)^(1/2)*(arcsinh(a*x)*a^2*x^2+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 = 0.00, 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)/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 = 0.00, 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)/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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asinh}{\left (a x \right )}}{x^{3} \sqrt {a^{2} x^{2} + 1}}\, dx \end {gather*}

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 = 0.00, 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)/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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {asinh}\left (a\,x\right )}{x^3\,\sqrt {a^2\,x^2+1}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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