3.108 \(\int \frac{\sin ^{-1}(a x)}{x^3 \sqrt{1-a^2 x^2}} \, dx\)

Optimal. Leaf size=98 \[ \frac{1}{2} i a^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-\frac{1}{2} i a^2 \text{PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-\frac{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)}{2 x^2}+a^2 \left (-\sin ^{-1}(a x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-\frac{a}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.146777, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4701, 4709, 4183, 2279, 2391, 30} \[ \frac{1}{2} i a^2 \text{PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-\frac{1}{2} i a^2 \text{PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-\frac{\sqrt{1-a^2 x^2} \sin ^{-1}(a x)}{2 x^2}+a^2 \left (-\sin ^{-1}(a x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-\frac{a}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a*x]/(x^3*Sqrt[1 - a^2*x^2]),x]

[Out]

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

Rule 4701

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[(c^2*(m + 2*p + 3))/(f^2*(m
+ 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[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[c*x
])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1] && Inte
gerQ[m]

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]

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

Mathematica [A]  time = 0.838323, size = 137, normalized size = 1.4 \[ \frac{1}{8} a^2 \left (4 i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-4 i \text{PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )+4 \sin ^{-1}(a x) \log \left (1-e^{i \sin ^{-1}(a x)}\right )-4 \sin ^{-1}(a x) \log \left (1+e^{i \sin ^{-1}(a x)}\right )-2 \tan \left (\frac{1}{2} \sin ^{-1}(a x)\right )-2 \cot \left (\frac{1}{2} \sin ^{-1}(a x)\right )-\sin ^{-1}(a x) \csc ^2\left (\frac{1}{2} \sin ^{-1}(a x)\right )+\sin ^{-1}(a x) \sec ^2\left (\frac{1}{2} \sin ^{-1}(a x)\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSin[a*x]/(x^3*Sqrt[1 - a^2*x^2]),x]

[Out]

(a^2*(-2*Cot[ArcSin[a*x]/2] - ArcSin[a*x]*Csc[ArcSin[a*x]/2]^2 + 4*ArcSin[a*x]*Log[1 - E^(I*ArcSin[a*x])] - 4*
ArcSin[a*x]*Log[1 + E^(I*ArcSin[a*x])] + (4*I)*PolyLog[2, -E^(I*ArcSin[a*x])] - (4*I)*PolyLog[2, E^(I*ArcSin[a
*x])] + ArcSin[a*x]*Sec[ArcSin[a*x]/2]^2 - 2*Tan[ArcSin[a*x]/2]))/8

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Maple [A]  time = 0.172, size = 178, normalized size = 1.8 \begin{align*} -{\frac{1}{ \left ( 2\,{a}^{2}{x}^{2}-2 \right ){x}^{2}}\sqrt{-{a}^{2}{x}^{2}+1} \left ({a}^{2}{x}^{2}\arcsin \left ( ax \right ) -ax\sqrt{-{a}^{2}{x}^{2}+1}-\arcsin \left ( ax \right ) \right ) }-{\frac{{a}^{2}\arcsin \left ( ax \right ) }{2}\ln \left ( 1+iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) }+{\frac{{a}^{2}\arcsin \left ( ax \right ) }{2}\ln \left ( 1-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) }+{\frac{i}{2}}{a}^{2}{\it polylog} \left ( 2,-iax-\sqrt{-{a}^{2}{x}^{2}+1} \right ) -{\frac{i}{2}}{a}^{2}{\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)/x^3/(-a^2*x^2+1)^(1/2),x)

[Out]

-1/2*(-a^2*x^2+1)^(1/2)/(a^2*x^2-1)/x^2*(a^2*x^2*arcsin(a*x)-a*x*(-a^2*x^2+1)^(1/2)-arcsin(a*x))-1/2*a^2*arcsi
n(a*x)*ln(1+I*a*x+(-a^2*x^2+1)^(1/2))+1/2*a^2*arcsin(a*x)*ln(1-I*a*x-(-a^2*x^2+1)^(1/2))+1/2*I*a^2*polylog(2,-
I*a*x-(-a^2*x^2+1)^(1/2))-1/2*I*a^2*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*} \int \frac{\arcsin \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(arcsin(a*x)/x^3/(-a^2*x^2+1)^(1/2),x, algorithm="maxima")

[Out]

integrate(arcsin(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} \arcsin \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(arcsin(a*x)/x^3/(-a^2*x^2+1)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-a^2*x^2 + 1)*arcsin(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{asin}{\left (a x \right )}}{x^{3} \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)/x**3/(-a**2*x**2+1)**(1/2),x)

[Out]

Integral(asin(a*x)/(x**3*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 )}{\sqrt{-a^{2} x^{2} + 1} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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