3.665 \(\int \frac{\sin ^{-1}(x)}{x (1-x^2)^{3/2}} \, dx\)

Optimal. Leaf size=62 \[ i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(x)}\right )-i \text{PolyLog}\left (2,e^{i \sin ^{-1}(x)}\right )+\frac{\sin ^{-1}(x)}{\sqrt{1-x^2}}-\tanh ^{-1}(x)-2 \sin ^{-1}(x) \tanh ^{-1}\left (e^{i \sin ^{-1}(x)}\right ) \]

[Out]

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

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Rubi [A]  time = 0.11596, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {4705, 4709, 4183, 2279, 2391, 206} \[ i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(x)}\right )-i \text{PolyLog}\left (2,e^{i \sin ^{-1}(x)}\right )+\frac{\sin ^{-1}(x)}{\sqrt{1-x^2}}-\tanh ^{-1}(x)-2 \sin ^{-1}(x) \tanh ^{-1}\left (e^{i \sin ^{-1}(x)}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[x]/(x*(1 - x^2)^(3/2)),x]

[Out]

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

Rule 4705

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)/(2*d*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d*(p +
1)), Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^Frac
Part[p])/(2*f*(p + 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, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ
[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 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]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.180873, size = 112, normalized size = 1.81 \[ i \text{PolyLog}\left (2,-e^{i \sin ^{-1}(x)}\right )-i \text{PolyLog}\left (2,e^{i \sin ^{-1}(x)}\right )+\frac{\sin ^{-1}(x)}{\sqrt{1-x^2}}+\sin ^{-1}(x) \log \left (1-e^{i \sin ^{-1}(x)}\right )-\sin ^{-1}(x) \log \left (1+e^{i \sin ^{-1}(x)}\right )+\log \left (\cos \left (\frac{1}{2} \sin ^{-1}(x)\right )-\sin \left (\frac{1}{2} \sin ^{-1}(x)\right )\right )-\log \left (\sin \left (\frac{1}{2} \sin ^{-1}(x)\right )+\cos \left (\frac{1}{2} \sin ^{-1}(x)\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

ArcSin[x]/Sqrt[1 - x^2] + ArcSin[x]*Log[1 - E^(I*ArcSin[x])] - ArcSin[x]*Log[1 + E^(I*ArcSin[x])] + Log[Cos[Ar
cSin[x]/2] - Sin[ArcSin[x]/2]] - Log[Cos[ArcSin[x]/2] + Sin[ArcSin[x]/2]] + I*PolyLog[2, -E^(I*ArcSin[x])] - I
*PolyLog[2, E^(I*ArcSin[x])]

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Maple [A]  time = 0.159, size = 97, normalized size = 1.6 \begin{align*} -{\frac{\arcsin \left ( x \right ) }{{x}^{2}-1}\sqrt{-{x}^{2}+1}}+2\,i\arctan \left ( ix+\sqrt{-{x}^{2}+1} \right ) +i{\it dilog} \left ( ix+\sqrt{-{x}^{2}+1}+1 \right ) -\arcsin \left ( x \right ) \ln \left ( ix+\sqrt{-{x}^{2}+1}+1 \right ) +i{\it dilog} \left ( ix+\sqrt{-{x}^{2}+1} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(x)/x/(-x^2+1)^(3/2),x)

[Out]

-(-x^2+1)^(1/2)/(x^2-1)*arcsin(x)+2*I*arctan(I*x+(-x^2+1)^(1/2))+I*dilog(I*x+(-x^2+1)^(1/2)+1)-arcsin(x)*ln(I*
x+(-x^2+1)^(1/2)+1)+I*dilog(I*x+(-x^2+1)^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arcsin \left (x\right )}{{\left (-x^{2} + 1\right )}^{\frac{3}{2}} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-x^{2} + 1} \arcsin \left (x\right )}{x^{5} - 2 \, x^{3} + x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-x^2 + 1)*arcsin(x)/(x^5 - 2*x^3 + x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asin}{\left (x \right )}}{x \left (- \left (x - 1\right ) \left (x + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(x)/x/(-x**2+1)**(3/2),x)

[Out]

Integral(asin(x)/(x*(-(x - 1)*(x + 1))**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arcsin \left (x\right )}{{\left (-x^{2} + 1\right )}^{\frac{3}{2}} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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