3.3.88 \(\int \frac {\sinh ^{-1}(a x)^2}{x^2 \sqrt {1+a^2 x^2}} \, dx\) [288]

Optimal. Leaf size=66 \[ -a \sinh ^{-1}(a x)^2-\frac {\sqrt {1+a^2 x^2} \sinh ^{-1}(a x)^2}{x}+2 a \sinh ^{-1}(a x) \log \left (1-e^{2 \sinh ^{-1}(a x)}\right )+a \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(a x)}\right ) \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {5800, 5775, 3797, 2221, 2317, 2438} \begin {gather*} -\frac {\sqrt {a^2 x^2+1} \sinh ^{-1}(a x)^2}{x}+a \text {Li}_2\left (e^{2 \sinh ^{-1}(a x)}\right )-a \sinh ^{-1}(a x)^2+2 a \sinh ^{-1}(a x) \log \left (1-e^{2 \sinh ^{-1}(a x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

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 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> Simp[(-I)*((
c + d*x)^(m + 1)/(d*(m + 1))), x] + Dist[2*I, Int[((c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5775

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Dist[1/b, Subst[Int[x^n*Coth[-a/b + x/b], x],
x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 5800

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[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, m, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.23, size = 65, normalized size = 0.98 \begin {gather*} a \left (\sinh ^{-1}(a x) \left (\sinh ^{-1}(a x)-\frac {\sqrt {1+a^2 x^2} \sinh ^{-1}(a x)}{a x}+2 \log \left (1-e^{-2 \sinh ^{-1}(a x)}\right )\right )-\text {PolyLog}\left (2,e^{-2 \sinh ^{-1}(a x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a*(ArcSinh[a*x]*(ArcSinh[a*x] - (Sqrt[1 + a^2*x^2]*ArcSinh[a*x])/(a*x) + 2*Log[1 - E^(-2*ArcSinh[a*x])]) - Pol
yLog[2, E^(-2*ArcSinh[a*x])])

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Maple [A]
time = 4.24, size = 132, normalized size = 2.00

method	result	size
default	\(\frac {\left (a x -\sqrt {a^{2} x^{2}+1}\right ) \arcsinh \left (a x \right )^{2}}{x}-2 a \arcsinh \left (a x \right )^{2}+2 a \arcsinh \left (a x \right ) \ln \left (1-a x -\sqrt {a^{2} x^{2}+1}\right )+2 a \polylog \left (2, a x +\sqrt {a^{2} x^{2}+1}\right )+2 a \arcsinh \left (a x \right ) \ln \left (1+a x +\sqrt {a^{2} x^{2}+1}\right )+2 a \polylog \left (2, -a x -\sqrt {a^{2} x^{2}+1}\right )\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*x-(a^2*x^2+1)^(1/2))/x*arcsinh(a*x)^2-2*a*arcsinh(a*x)^2+2*a*arcsinh(a*x)*ln(1-a*x-(a^2*x^2+1)^(1/2))+2*a*p
olylog(2,a*x+(a^2*x^2+1)^(1/2))+2*a*arcsinh(a*x)*ln(1+a*x+(a^2*x^2+1)^(1/2))+2*a*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/(a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

-sqrt(a^2*x^2 + 1)*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))/(sqrt(a^2*x^2 + 1)*a*x^2 + (a^2*x^2 + 1)*x), 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/(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^4 + 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} \sqrt {a^{2} x^{2} + 1}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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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\,\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)^2/(x^2*(a^2*x^2 + 1)^(1/2)),x)

[Out]

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

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