3.6 \(\int \frac {\sin ^{-1}(a x)}{x} \, dx\)

Optimal. Leaf size=51 \[ -\frac {1}{2} i \text {Li}_2\left (e^{2 i \sin ^{-1}(a x)}\right )-\frac {1}{2} i \sin ^{-1}(a x)^2+\sin ^{-1}(a x) \log \left (1-e^{2 i \sin ^{-1}(a x)}\right ) \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {4625, 3717, 2190, 2279, 2391} \[ -\frac {1}{2} i \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(a x)}\right )-\frac {1}{2} i \sin ^{-1}(a x)^2+\sin ^{-1}(a x) \log \left (1-e^{2 i \sin ^{-1}(a x)}\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a*x]/x,x]

[Out]

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

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]

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 46, normalized size = 0.90 \[ \sin ^{-1}(a x) \log \left (1-e^{2 i \sin ^{-1}(a x)}\right )-\frac {1}{2} i \left (\sin ^{-1}(a x)^2+\text {Li}_2\left (e^{2 i \sin ^{-1}(a x)}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a*x]/x,x]

[Out]

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

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arcsin \left (a x\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)/x,x, algorithm="fricas")

[Out]

integral(arcsin(a*x)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)/x,x, algorithm="giac")

[Out]

integrate(arcsin(a*x)/x, x)

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maple [A]  time = 0.40, size = 111, normalized size = 2.18 \[ -\frac {i \arcsin \left (a x \right )^{2}}{2}+\arcsin \left (a x \right ) \ln \left (1+i a x +\sqrt {-a^{2} x^{2}+1}\right )+\arcsin \left (a x \right ) \ln \left (1-i a x -\sqrt {-a^{2} x^{2}+1}\right )-i \polylog \left (2, -i a x -\sqrt {-a^{2} x^{2}+1}\right )-i \polylog \left (2, i a x +\sqrt {-a^{2} x^{2}+1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(a*x)/x,x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)/x,x, algorithm="maxima")

[Out]

integrate(arcsin(a*x)/x, x)

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mupad [B]  time = 0.08, size = 41, normalized size = 0.80 \[ \ln \left (1-{\mathrm {e}}^{\mathrm {asin}\left (a\,x\right )\,2{}\mathrm {i}}\right )\,\mathrm {asin}\left (a\,x\right )-\frac {\mathrm {polylog}\left (2,{\mathrm {e}}^{\mathrm {asin}\left (a\,x\right )\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{2}-\frac {{\mathrm {asin}\left (a\,x\right )}^2\,1{}\mathrm {i}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a*x)/x,x)

[Out]

log(1 - exp(asin(a*x)*2i))*asin(a*x) - (polylog(2, exp(asin(a*x)*2i))*1i)/2 - (asin(a*x)^2*1i)/2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(a*x)/x,x)

[Out]

Integral(asin(a*x)/x, x)

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