3.18 \(\int \frac{\sin ^{-1}(a x)^2}{x^2} \, dx\)

Optimal. Leaf size=66 \[ 2 i a \text{PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-2 i a \text{PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-\frac{\sin ^{-1}(a x)^2}{x}-4 a \sin ^{-1}(a x) \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.102522, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4627, 4709, 4183, 2279, 2391} \[ 2 i a \text{PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-2 i a \text{PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-\frac{\sin ^{-1}(a x)^2}{x}-4 a \sin ^{-1}(a x) \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[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 4709

Int[(((a_.) + ArcSin[(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*Sin[x]^m, x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2
*d + e, 0] && GtQ[d, 0] && IGtQ[n, 0] && IntegerQ[m]

Rule 4183

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

Rubi steps

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

Mathematica [A]  time = 0.161989, size = 87, normalized size = 1.32 \[ a \left (2 i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-2 i \text{PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-\sin ^{-1}(a x) \left (\frac{\sin ^{-1}(a x)}{a x}-2 \log \left (1-e^{i \sin ^{-1}(a x)}\right )+2 \log \left (1+e^{i \sin ^{-1}(a x)}\right )\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

a*(-(ArcSin[a*x]*(ArcSin[a*x]/(a*x) - 2*Log[1 - E^(I*ArcSin[a*x])] + 2*Log[1 + E^(I*ArcSin[a*x])])) + (2*I)*Po
lyLog[2, -E^(I*ArcSin[a*x])] - (2*I)*PolyLog[2, E^(I*ArcSin[a*x])])

________________________________________________________________________________________

Maple [A]  time = 0.092, size = 119, normalized size = 1.8 \begin{align*} -{\frac{ \left ( \arcsin \left ( ax \right ) \right ) ^{2}}{x}}-2\,a\arcsin \left ( ax \right ) \ln \left ( 1+iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) +2\,a\arcsin \left ( ax \right ) \ln \left ( 1-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) +2\,ia{\it polylog} \left ( 2,-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) -2\,ia{\it polylog} \left ( 2,iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a*x*integrate(sqrt(a*x + 1)*sqrt(-a*x + 1)*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))/(a^2*x^3 - x), x) +
arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^2)/x

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\arcsin \left (a x\right )^{2}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsin(a*x)^2/x^2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asin(a*x)**2/x**2, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsin(a*x)^2/x^2, x)