3.7.65 \(\int \frac {\sin ^{-1}(x)}{x (1-x^2)^{3/2}} \, dx\) [665]

Optimal. Leaf size=62 \[ \frac {\sin ^{-1}(x)}{\sqrt {1-x^2}}-2 \sin ^{-1}(x) \tanh ^{-1}\left (e^{i \sin ^{-1}(x)}\right )-\tanh ^{-1}(x)+i \text {Li}_2\left (-e^{i \sin ^{-1}(x)}\right )-i \text {Li}_2\left (e^{i \sin ^{-1}(x)}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 62, normalized size of antiderivative = 1.00, 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 = {4793, 4803, 4268, 2317, 2438, 212} \begin {gather*} i \text {PolyLog}\left (2,-e^{i \text {ArcSin}(x)}\right )-i \text {PolyLog}\left (2,e^{i \text {ArcSin}(x)}\right )+\frac {\text {ArcSin}(x)}{\sqrt {1-x^2}}-2 \text {ArcSin}(x) \tanh ^{-1}\left (e^{i \text {ArcSin}(x)}\right )-\tanh ^{-1}(x) \end {gather*}

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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 4268

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 4793

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/(2*f*(p + 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*ArcSin[c*x])^(n - 1), x], x]) /; Fre
eQ[{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 4803

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[(1/c^(m
+ 1))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]], Subst[Int[(a + b*x)^n*Sin[x]^m, x], x, ArcSin[c*x]], x] /; Free
Q[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0] && IntegerQ[m]

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)+\text {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)-\text {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(x)\right )+\text {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 \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(x)}\right )-i \text {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 = 0.13, size = 112, normalized size = 1.81 \begin {gather*} \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 (\cos \left (\frac {1}{2} \sin ^{-1}(x)\right )+\sin \left (\frac {1}{2} \sin ^{-1}(x)\right )\right )+i \text {Li}_2\left (-e^{i \sin ^{-1}(x)}\right )-i \text {Li}_2\left (e^{i \sin ^{-1}(x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[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])]

________________________________________________________________________________________

Maple [A]
time = 0.43, size = 97, normalized size = 1.56

method	result	size
default	\(-\frac {\sqrt {-x^{2}+1}\, \arcsin \left (x \right )}{x^{2}-1}+2 i \arctan \left (i x +\sqrt {-x^{2}+1}\right )+i \dilog \left (i x +\sqrt {-x^{2}+1}+1\right )-\arcsin \left (x \right ) \ln \left (i x +\sqrt {-x^{2}+1}+1\right )+i \dilog \left (i x +\sqrt {-x^{2}+1}\right )\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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))

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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(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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asin}{\left (x \right )}}{x \left (- \left (x - 1\right ) \left (x + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

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)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\mathrm {asin}\left (x\right )}{x\,{\left (1-x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(x)/(x*(1 - x^2)^(3/2)),x)

[Out]

int(asin(x)/(x*(1 - x^2)^(3/2)), x)

________________________________________________________________________________________