3.271 \(\int \frac {\sin ^{-1}(a x)^2}{x^3 \sqrt {1-a^2 x^2}} \, dx\)

Optimal. Leaf size=163 \[ i a^2 \sin ^{-1}(a x) \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-i a^2 \sin ^{-1}(a x) \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-a^2 \text {Li}_3\left (-e^{i \sin ^{-1}(a x)}\right )+a^2 \text {Li}_3\left (e^{i \sin ^{-1}(a x)}\right )-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{2 x^2}-a^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+a^2 \left (-\sin ^{-1}(a x)^2\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-\frac {a \sin ^{-1}(a x)}{x} \]

[Out]

-a*arcsin(a*x)/x-a^2*arcsin(a*x)^2*arctanh(I*a*x+(-a^2*x^2+1)^(1/2))-a^2*arctanh((-a^2*x^2+1)^(1/2))+I*a^2*arc
sin(a*x)*polylog(2,-I*a*x-(-a^2*x^2+1)^(1/2))-I*a^2*arcsin(a*x)*polylog(2,I*a*x+(-a^2*x^2+1)^(1/2))-a^2*polylo
g(3,-I*a*x-(-a^2*x^2+1)^(1/2))+a^2*polylog(3,I*a*x+(-a^2*x^2+1)^(1/2))-1/2*arcsin(a*x)^2*(-a^2*x^2+1)^(1/2)/x^
2

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Rubi [A]  time = 0.25, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {4701, 4709, 4183, 2531, 2282, 6589, 4627, 266, 63, 208} \[ i a^2 \sin ^{-1}(a x) \text {PolyLog}\left (2,-e^{i \sin ^{-1}(a x)}\right )-i a^2 \sin ^{-1}(a x) \text {PolyLog}\left (2,e^{i \sin ^{-1}(a x)}\right )-a^2 \text {PolyLog}\left (3,-e^{i \sin ^{-1}(a x)}\right )+a^2 \text {PolyLog}\left (3,e^{i \sin ^{-1}(a x)}\right )-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{2 x^2}-a^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+a^2 \left (-\sin ^{-1}(a x)^2\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-\frac {a \sin ^{-1}(a x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*ArcSin[a*x])/x) - (Sqrt[1 - a^2*x^2]*ArcSin[a*x]^2)/(2*x^2) - a^2*ArcSin[a*x]^2*ArcTanh[E^(I*ArcSin[a*x])
] - a^2*ArcTanh[Sqrt[1 - a^2*x^2]] + I*a^2*ArcSin[a*x]*PolyLog[2, -E^(I*ArcSin[a*x])] - I*a^2*ArcSin[a*x]*Poly
Log[2, E^(I*ArcSin[a*x])] - a^2*PolyLog[3, -E^(I*ArcSin[a*x])] + a^2*PolyLog[3, E^(I*ArcSin[a*x])]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

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 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4701

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*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(f*(m + 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, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1] && Inte
gerQ[m]

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 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {\sin ^{-1}(a x)^2}{x^3 \sqrt {1-a^2 x^2}} \, dx &=-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{2 x^2}+a \int \frac {\sin ^{-1}(a x)}{x^2} \, dx+\frac {1}{2} a^2 \int \frac {\sin ^{-1}(a x)^2}{x \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {a \sin ^{-1}(a x)}{x}-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{2 x^2}+\frac {1}{2} a^2 \operatorname {Subst}\left (\int x^2 \csc (x) \, dx,x,\sin ^{-1}(a x)\right )+a^2 \int \frac {1}{x \sqrt {1-a^2 x^2}} \, dx\\ &=-\frac {a \sin ^{-1}(a x)}{x}-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{2 x^2}-a^2 \sin ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )+\frac {1}{2} a^2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-a^2 x}} \, dx,x,x^2\right )-a^2 \operatorname {Subst}\left (\int x \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )+a^2 \operatorname {Subst}\left (\int x \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )\\ &=-\frac {a \sin ^{-1}(a x)}{x}-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{2 x^2}-a^2 \sin ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )+i a^2 \sin ^{-1}(a x) \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-i a^2 \sin ^{-1}(a x) \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-\left (i a^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )+\left (i a^2\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )-\operatorname {Subst}\left (\int \frac {1}{\frac {1}{a^2}-\frac {x^2}{a^2}} \, dx,x,\sqrt {1-a^2 x^2}\right )\\ &=-\frac {a \sin ^{-1}(a x)}{x}-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{2 x^2}-a^2 \sin ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-a^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+i a^2 \sin ^{-1}(a x) \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-i a^2 \sin ^{-1}(a x) \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-a^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i \sin ^{-1}(a x)}\right )+a^2 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i \sin ^{-1}(a x)}\right )\\ &=-\frac {a \sin ^{-1}(a x)}{x}-\frac {\sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2}{2 x^2}-a^2 \sin ^{-1}(a x)^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-a^2 \tanh ^{-1}\left (\sqrt {1-a^2 x^2}\right )+i a^2 \sin ^{-1}(a x) \text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-i a^2 \sin ^{-1}(a x) \text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )-a^2 \text {Li}_3\left (-e^{i \sin ^{-1}(a x)}\right )+a^2 \text {Li}_3\left (e^{i \sin ^{-1}(a x)}\right )\\ \end {align*}

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Mathematica [A]  time = 1.57, size = 194, normalized size = 1.19 \[ \frac {1}{8} a^2 \left (8 i \sin ^{-1}(a x) \left (\text {Li}_2\left (-e^{i \sin ^{-1}(a x)}\right )-\text {Li}_2\left (e^{i \sin ^{-1}(a x)}\right )\right )+8 \left (\text {Li}_3\left (e^{i \sin ^{-1}(a x)}\right )-\text {Li}_3\left (-e^{i \sin ^{-1}(a x)}\right )\right )+4 \sin ^{-1}(a x)^2 \left (\log \left (1-e^{i \sin ^{-1}(a x)}\right )-\log \left (1+e^{i \sin ^{-1}(a x)}\right )\right )-4 \sin ^{-1}(a x) \tan \left (\frac {1}{2} \sin ^{-1}(a x)\right )-4 \sin ^{-1}(a x) \cot \left (\frac {1}{2} \sin ^{-1}(a x)\right )+\sin ^{-1}(a x)^2 \left (-\csc ^2\left (\frac {1}{2} \sin ^{-1}(a x)\right )\right )+\sin ^{-1}(a x)^2 \sec ^2\left (\frac {1}{2} \sin ^{-1}(a x)\right )+8 \log \left (\tan \left (\frac {1}{2} \sin ^{-1}(a x)\right )\right )\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*(-4*ArcSin[a*x]*Cot[ArcSin[a*x]/2] - ArcSin[a*x]^2*Csc[ArcSin[a*x]/2]^2 + 4*ArcSin[a*x]^2*(Log[1 - E^(I*A
rcSin[a*x])] - Log[1 + E^(I*ArcSin[a*x])]) + 8*Log[Tan[ArcSin[a*x]/2]] + (8*I)*ArcSin[a*x]*(PolyLog[2, -E^(I*A
rcSin[a*x])] - PolyLog[2, E^(I*ArcSin[a*x])]) + 8*(-PolyLog[3, -E^(I*ArcSin[a*x])] + PolyLog[3, E^(I*ArcSin[a*
x])]) + ArcSin[a*x]^2*Sec[ArcSin[a*x]/2]^2 - 4*ArcSin[a*x]*Tan[ArcSin[a*x]/2]))/8

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-a^{2} x^{2} + 1} \arcsin \left (a x\right )^{2}}{a^{2} x^{5} - x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.33, size = 269, normalized size = 1.65 \[ -\frac {\sqrt {-a^{2} x^{2}+1}\, \arcsin \left (a x \right ) \left (a^{2} x^{2} \arcsin \left (a x \right )-2 a x \sqrt {-a^{2} x^{2}+1}-\arcsin \left (a x \right )\right )}{2 \left (a^{2} x^{2}-1\right ) x^{2}}+\frac {a^{2} \arcsin \left (a x \right )^{2} \ln \left (1-i a x -\sqrt {-a^{2} x^{2}+1}\right )}{2}-i a^{2} \arcsin \left (a x \right ) \polylog \left (2, i a x +\sqrt {-a^{2} x^{2}+1}\right )-\frac {a^{2} \arcsin \left (a x \right )^{2} \ln \left (1+i a x +\sqrt {-a^{2} x^{2}+1}\right )}{2}+i a^{2} \arcsin \left (a x \right ) \polylog \left (2, -i a x -\sqrt {-a^{2} x^{2}+1}\right )+a^{2} \polylog \left (3, i a x +\sqrt {-a^{2} x^{2}+1}\right )-a^{2} \polylog \left (3, -i a x -\sqrt {-a^{2} x^{2}+1}\right )-2 \arctanh \left (i a x +\sqrt {-a^{2} x^{2}+1}\right ) a^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-a^2*x^2+1)^(1/2)/(a^2*x^2-1)/x^2*arcsin(a*x)*(a^2*x^2*arcsin(a*x)-2*a*x*(-a^2*x^2+1)^(1/2)-arcsin(a*x))
+1/2*a^2*arcsin(a*x)^2*ln(1-I*a*x-(-a^2*x^2+1)^(1/2))-I*a^2*arcsin(a*x)*polylog(2,I*a*x+(-a^2*x^2+1)^(1/2))-1/
2*a^2*arcsin(a*x)^2*ln(1+I*a*x+(-a^2*x^2+1)^(1/2))+I*a^2*arcsin(a*x)*polylog(2,-I*a*x-(-a^2*x^2+1)^(1/2))+a^2*
polylog(3,I*a*x+(-a^2*x^2+1)^(1/2))-a^2*polylog(3,-I*a*x-(-a^2*x^2+1)^(1/2))-2*arctanh(I*a*x+(-a^2*x^2+1)^(1/2
))*a^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asin}^{2}{\left (a x \right )}}{x^{3} \sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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