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

Optimal. Leaf size=52 \[ i \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-i \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-2 \sin ^{-1}(a x) \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right ) \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4709, 4183, 2279, 2391} \[ i \text {PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-i \text {PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-2 \sin ^{-1}(a x) \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 71, normalized size = 1.37 \[ i \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-i \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )+\sin ^{-1}(a x) \left (\log \left (1-e^{i \sin ^{-1}(a x)}\right )-\log \left (1+e^{i \sin ^{-1}(a x)}\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [F]  time = 2.03, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} \arcsin \left (a x\right )}{a^{2} x^{3} - x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arcsin \left (a x\right )}{\sqrt {-a^{2} x^{2} + 1} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.08, size = 103, normalized size = 1.98 \[ \arcsin \left (a x \right ) \ln \left (1-i a x -\sqrt {-a^{2} x^{2}+1}\right )-\arcsin \left (a x \right ) \ln \left (1+i a x +\sqrt {-a^{2} x^{2}+1}\right )+i \dilog \left (1+i a x +\sqrt {-a^{2} x^{2}+1}\right )-i \dilog \left (1-i a x -\sqrt {-a^{2} x^{2}+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arcsin \left (a x\right )}{\sqrt {-a^{2} x^{2} + 1} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {asin}\left (a\,x\right )}{x\,\sqrt {1-a^2\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a*x)/(x*(1 - a^2*x^2)^(1/2)),x)

[Out]

int(asin(a*x)/(x*(1 - a^2*x^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asin}{\left (a x \right )}}{x \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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