3.2.8 \(\int \frac {\text {ArcSin}(a x)}{x^3 \sqrt {1-a^2 x^2}} \, dx\) [108]

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

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 98, normalized size of antiderivative = 1.00, 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 = {4789, 4803, 4268, 2317, 2438, 30} \begin {gather*} \frac {1}{2} i a^2 \text {Li}_2\left (-e^{i \text {ArcSin}(a x)}\right )-\frac {1}{2} i a^2 \text {Li}_2\left (e^{i \text {ArcSin}(a x)}\right )-\frac {\sqrt {1-a^2 x^2} \text {ArcSin}(a x)}{2 x^2}+a^2 (-\text {ArcSin}(a x)) \tanh ^{-1}\left (e^{i \text {ArcSin}(a x)}\right )-\frac {a}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*a/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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 4789

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/(f*(m + 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]) /; Free
Q[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && ILtQ[m, -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}(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 \text {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 \text {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )+\frac {1}{2} a^2 \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.56, size = 137, normalized size = 1.40 \begin {gather*} \frac {1}{8} a^2 \left (-2 \cot \left (\frac {1}{2} \text {ArcSin}(a x)\right )-\text {ArcSin}(a x) \csc ^2\left (\frac {1}{2} \text {ArcSin}(a x)\right )+4 \text {ArcSin}(a x) \log \left (1-e^{i \text {ArcSin}(a x)}\right )-4 \text {ArcSin}(a x) \log \left (1+e^{i \text {ArcSin}(a x)}\right )+4 i \text {PolyLog}\left (2,-e^{i \text {ArcSin}(a x)}\right )-4 i \text {PolyLog}\left (2,e^{i \text {ArcSin}(a x)}\right )+\text {ArcSin}(a x) \sec ^2\left (\frac {1}{2} \text {ArcSin}(a x)\right )-2 \tan \left (\frac {1}{2} \text {ArcSin}(a x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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 = 0.39, size = 171, normalized size = 1.74

method	result	size
default	\(-\frac {\sqrt {-a^{2} x^{2}+1}\, \left (a^{2} x^{2} \arcsin \left (a x \right )-a x \sqrt {-a^{2} x^{2}+1}-\arcsin \left (a x \right )\right )}{2 \left (a^{2} x^{2}-1\right ) x^{2}}-\frac {i a^{2} \left (i \arcsin \left (a x \right ) \ln \left (1-i a x -\sqrt {-a^{2} x^{2}+1}\right )-i \arcsin \left (a x \right ) \ln \left (1+i a x +\sqrt {-a^{2} x^{2}+1}\right )+\polylog \left (2, i a x +\sqrt {-a^{2} x^{2}+1}\right )-\polylog \left (2, -i a x -\sqrt {-a^{2} x^{2}+1}\right )\right )}{2}\)	\(171\)

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,method=_RETURNVERBOSE)

[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*I*a^2*(I*
arcsin(a*x)*ln(1-I*a*x-(-a^2*x^2+1)^(1/2))-I*arcsin(a*x)*ln(1+I*a*x+(-a^2*x^2+1)^(1/2))+polylog(2,I*a*x+(-a^2*
x^2+1)^(1/2))-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, 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(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asin}{\left (a x \right )}}{x^{3} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \end {gather*}

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 = 0.00, 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(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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\mathrm {asin}\left (a\,x\right )}{x^3\,\sqrt {1-a^2\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a*x)/(x^3*(1 - a^2*x^2)^(1/2)),x)

[Out]

int(asin(a*x)/(x^3*(1 - a^2*x^2)^(1/2)), x)

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