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

Optimal. Leaf size=76 \[ -i a \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(a x)}\right )-\frac{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{x}-i a \sin ^{-1}(a x)^2+2 a \sin ^{-1}(a x) \log \left (1-e^{2 i \sin ^{-1}(a x)}\right ) \]

[Out]

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

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Rubi [A]  time = 0.14353, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4681, 4625, 3717, 2190, 2279, 2391} \[ -i a \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(a x)}\right )-\frac{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{x}-i a \sin ^{-1}(a x)^2+2 a \sin ^{-1}(a x) \log \left (1-e^{2 i \sin ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a*x]^2/(x^2*Sqrt[1 - a^2*x^2]),x]

[Out]

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

Rule 4681

Int[((a_.) + ArcSin[(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*ArcSin[c*x])^n)/(d*f*(m + 1)), 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*ArcSi
n[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

Rule 4625

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

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 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 \sqrt{1-a^2 x^2}} \, dx &=-\frac{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{x}+(2 a) \int \frac{\sin ^{-1}(a x)}{x} \, dx\\ &=-\frac{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{x}+(2 a) \operatorname{Subst}\left (\int x \cot (x) \, dx,x,\sin ^{-1}(a x)\right )\\ &=-i a \sin ^{-1}(a x)^2-\frac{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{x}-(4 i a) \operatorname{Subst}\left (\int \frac{e^{2 i x} x}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(a x)\right )\\ &=-i a \sin ^{-1}(a x)^2-\frac{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{x}+2 a \sin ^{-1}(a x) \log \left (1-e^{2 i \sin ^{-1}(a x)}\right )-(2 a) \operatorname{Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )\\ &=-i a \sin ^{-1}(a x)^2-\frac{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{x}+2 a \sin ^{-1}(a x) \log \left (1-e^{2 i \sin ^{-1}(a x)}\right )+(i a) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(a x)}\right )\\ &=-i a \sin ^{-1}(a x)^2-\frac{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{x}+2 a \sin ^{-1}(a x) \log \left (1-e^{2 i \sin ^{-1}(a x)}\right )-i a \text{Li}_2\left (e^{2 i \sin ^{-1}(a x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.260923, size = 72, normalized size = 0.95 \[ \sin ^{-1}(a x) \left (2 a \log \left (1-e^{2 i \sin ^{-1}(a x)}\right )-\frac{\left (\sqrt{1-a^2 x^2}+i a x\right ) \sin ^{-1}(a x)}{x}\right )-i a \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a*x]^2/(x^2*Sqrt[1 - a^2*x^2]),x]

[Out]

ArcSin[a*x]*(-(((I*a*x + Sqrt[1 - a^2*x^2])*ArcSin[a*x])/x) + 2*a*Log[1 - E^((2*I)*ArcSin[a*x])]) - I*a*PolyLo
g[2, E^((2*I)*ArcSin[a*x])]

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Maple [A]  time = 0.107, size = 148, normalized size = 2. \begin{align*}{\frac{ \left ( \arcsin \left ( ax \right ) \right ) ^{2}}{x} \left ( 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\,a\arcsin \left ( ax \right ) \ln \left ( 1-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) -2\,i \left ( \arcsin \left ( ax \right ) \right ) ^{2}a-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/(-a^2*x^2+1)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(sqrt(a*x + 1)*sqrt(-a*x + 1)*arctan2(a*x, sqrt(a*x + 1)*sqrt(-a*x + 1))^2 - 2*a*x*integrate(arctan2(a*x, sqr
t(a*x + 1)*sqrt(-a*x + 1))/x, x))/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} \arcsin \left (a x\right )^{2}}{a^{2} x^{4} - x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-a^2*x^2 + 1)*arcsin(a*x)^2/(a^2*x^4 - x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(a*x)**2/x**2/(-a**2*x**2+1)**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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