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

Optimal. Leaf size=337 \[ \frac {x \sin ^{-1}(a x)^3}{2 c^2 \left (1-a^2 x^2\right )}-\frac {3 \sin ^{-1}(a x)^2}{2 a c^2 \sqrt {1-a^2 x^2}}+\frac {3 i \sin ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \sin ^{-1}(a x)}\right )}{2 a c^2}-\frac {3 i \sin ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \sin ^{-1}(a x)}\right )}{2 a c^2}-\frac {3 \sin ^{-1}(a x) \text {Li}_3\left (-i e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 \sin ^{-1}(a x) \text {Li}_3\left (i e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \text {Li}_2\left (-i e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \text {Li}_2\left (i e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \text {Li}_4\left (-i e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \text {Li}_4\left (i e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {i \sin ^{-1}(a x)^3 \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {6 i \sin ^{-1}(a x) \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c^2} \]

[Out]

1/2*x*arcsin(a*x)^3/c^2/(-a^2*x^2+1)-6*I*arcsin(a*x)*arctan(I*a*x+(-a^2*x^2+1)^(1/2))/a/c^2-I*arcsin(a*x)^3*ar
ctan(I*a*x+(-a^2*x^2+1)^(1/2))/a/c^2+3*I*polylog(2,-I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^2+3/2*I*arcsin(a*x)^2*po
lylog(2,-I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^2-3*I*polylog(2,I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^2-3/2*I*arcsin(a*
x)^2*polylog(2,I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^2-3*arcsin(a*x)*polylog(3,-I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^
2+3*arcsin(a*x)*polylog(3,I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^2-3*I*polylog(4,-I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c
^2+3*I*polylog(4,I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^2-3/2*arcsin(a*x)^2/a/c^2/(-a^2*x^2+1)^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4655, 4657, 4181, 2531, 6609, 2282, 6589, 4677, 2279, 2391} \[ \frac {3 i \sin ^{-1}(a x)^2 \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(a x)}\right )}{2 a c^2}-\frac {3 i \sin ^{-1}(a x)^2 \text {PolyLog}\left (2,i e^{i \sin ^{-1}(a x)}\right )}{2 a c^2}-\frac {3 \sin ^{-1}(a x) \text {PolyLog}\left (3,-i e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 \sin ^{-1}(a x) \text {PolyLog}\left (3,i e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \text {PolyLog}\left (2,-i e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \text {PolyLog}\left (2,i e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \text {PolyLog}\left (4,-i e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \text {PolyLog}\left (4,i e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {x \sin ^{-1}(a x)^3}{2 c^2 \left (1-a^2 x^2\right )}-\frac {3 \sin ^{-1}(a x)^2}{2 a c^2 \sqrt {1-a^2 x^2}}-\frac {i \sin ^{-1}(a x)^3 \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {6 i \sin ^{-1}(a x) \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcSin[a*x]^2)/(2*a*c^2*Sqrt[1 - a^2*x^2]) + (x*ArcSin[a*x]^3)/(2*c^2*(1 - a^2*x^2)) - ((6*I)*ArcSin[a*x]*
ArcTan[E^(I*ArcSin[a*x])])/(a*c^2) - (I*ArcSin[a*x]^3*ArcTan[E^(I*ArcSin[a*x])])/(a*c^2) + ((3*I)*PolyLog[2, (
-I)*E^(I*ArcSin[a*x])])/(a*c^2) + (((3*I)/2)*ArcSin[a*x]^2*PolyLog[2, (-I)*E^(I*ArcSin[a*x])])/(a*c^2) - ((3*I
)*PolyLog[2, I*E^(I*ArcSin[a*x])])/(a*c^2) - (((3*I)/2)*ArcSin[a*x]^2*PolyLog[2, I*E^(I*ArcSin[a*x])])/(a*c^2)
- (3*ArcSin[a*x]*PolyLog[3, (-I)*E^(I*ArcSin[a*x])])/(a*c^2) + (3*ArcSin[a*x]*PolyLog[3, I*E^(I*ArcSin[a*x])]
)/(a*c^2) - ((3*I)*PolyLog[4, (-I)*E^(I*ArcSin[a*x])])/(a*c^2) + ((3*I)*PolyLog[4, I*E^(I*ArcSin[a*x])])/(a*c^
2)

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 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 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 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 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4655

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p
+ 1)*(a + b*ArcSin[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a + b*
ArcSin[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 - c^2*x^2)^FracPart[p
]), Int[x*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2
*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rule 4657

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/(c*d), Subst[Int[(a +
b*x)^n*Sec[x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]

Rule 4677

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^
(p + 1)*(a + b*ArcSin[c*x])^n)/(2*e*(p + 1)), x] + Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p + 1
)*(1 - c^2*x^2)^FracPart[p]), Int[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b,
c, d, e, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && NeQ[p, -1]

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]

Rule 6609

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

Rubi steps

\begin {align*} \int \frac {\sin ^{-1}(a x)^3}{\left (c-a^2 c x^2\right )^2} \, dx &=\frac {x \sin ^{-1}(a x)^3}{2 c^2 \left (1-a^2 x^2\right )}-\frac {(3 a) \int \frac {x \sin ^{-1}(a x)^2}{\left (1-a^2 x^2\right )^{3/2}} \, dx}{2 c^2}+\frac {\int \frac {\sin ^{-1}(a x)^3}{c-a^2 c x^2} \, dx}{2 c}\\ &=-\frac {3 \sin ^{-1}(a x)^2}{2 a c^2 \sqrt {1-a^2 x^2}}+\frac {x \sin ^{-1}(a x)^3}{2 c^2 \left (1-a^2 x^2\right )}+\frac {3 \int \frac {\sin ^{-1}(a x)}{1-a^2 x^2} \, dx}{c^2}+\frac {\operatorname {Subst}\left (\int x^3 \sec (x) \, dx,x,\sin ^{-1}(a x)\right )}{2 a c^2}\\ &=-\frac {3 \sin ^{-1}(a x)^2}{2 a c^2 \sqrt {1-a^2 x^2}}+\frac {x \sin ^{-1}(a x)^3}{2 c^2 \left (1-a^2 x^2\right )}-\frac {i \sin ^{-1}(a x)^3 \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {3 \operatorname {Subst}\left (\int x^2 \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{2 a c^2}+\frac {3 \operatorname {Subst}\left (\int x^2 \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{2 a c^2}+\frac {3 \operatorname {Subst}\left (\int x \sec (x) \, dx,x,\sin ^{-1}(a x)\right )}{a c^2}\\ &=-\frac {3 \sin ^{-1}(a x)^2}{2 a c^2 \sqrt {1-a^2 x^2}}+\frac {x \sin ^{-1}(a x)^3}{2 c^2 \left (1-a^2 x^2\right )}-\frac {6 i \sin ^{-1}(a x) \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {i \sin ^{-1}(a x)^3 \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \sin ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \sin ^{-1}(a x)}\right )}{2 a c^2}-\frac {3 i \sin ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \sin ^{-1}(a x)}\right )}{2 a c^2}-\frac {(3 i) \operatorname {Subst}\left (\int x \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{a c^2}+\frac {(3 i) \operatorname {Subst}\left (\int x \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{a c^2}-\frac {3 \operatorname {Subst}\left (\int \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{a c^2}+\frac {3 \operatorname {Subst}\left (\int \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{a c^2}\\ &=-\frac {3 \sin ^{-1}(a x)^2}{2 a c^2 \sqrt {1-a^2 x^2}}+\frac {x \sin ^{-1}(a x)^3}{2 c^2 \left (1-a^2 x^2\right )}-\frac {6 i \sin ^{-1}(a x) \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {i \sin ^{-1}(a x)^3 \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \sin ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \sin ^{-1}(a x)}\right )}{2 a c^2}-\frac {3 i \sin ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \sin ^{-1}(a x)}\right )}{2 a c^2}-\frac {3 \sin ^{-1}(a x) \text {Li}_3\left (-i e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 \sin ^{-1}(a x) \text {Li}_3\left (i e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {(3 i) \operatorname {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 \operatorname {Subst}\left (\int \text {Li}_3\left (-i e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{a c^2}-\frac {3 \operatorname {Subst}\left (\int \text {Li}_3\left (i e^{i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{a c^2}\\ &=-\frac {3 \sin ^{-1}(a x)^2}{2 a c^2 \sqrt {1-a^2 x^2}}+\frac {x \sin ^{-1}(a x)^3}{2 c^2 \left (1-a^2 x^2\right )}-\frac {6 i \sin ^{-1}(a x) \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {i \sin ^{-1}(a x)^3 \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \text {Li}_2\left (-i e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \sin ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \sin ^{-1}(a x)}\right )}{2 a c^2}-\frac {3 i \text {Li}_2\left (i e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \sin ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \sin ^{-1}(a x)}\right )}{2 a c^2}-\frac {3 \sin ^{-1}(a x) \text {Li}_3\left (-i e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 \sin ^{-1}(a x) \text {Li}_3\left (i e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-i x)}{x} \, dx,x,e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {(3 i) \operatorname {Subst}\left (\int \frac {\text {Li}_3(i x)}{x} \, dx,x,e^{i \sin ^{-1}(a x)}\right )}{a c^2}\\ &=-\frac {3 \sin ^{-1}(a x)^2}{2 a c^2 \sqrt {1-a^2 x^2}}+\frac {x \sin ^{-1}(a x)^3}{2 c^2 \left (1-a^2 x^2\right )}-\frac {6 i \sin ^{-1}(a x) \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {i \sin ^{-1}(a x)^3 \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \text {Li}_2\left (-i e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \sin ^{-1}(a x)^2 \text {Li}_2\left (-i e^{i \sin ^{-1}(a x)}\right )}{2 a c^2}-\frac {3 i \text {Li}_2\left (i e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \sin ^{-1}(a x)^2 \text {Li}_2\left (i e^{i \sin ^{-1}(a x)}\right )}{2 a c^2}-\frac {3 \sin ^{-1}(a x) \text {Li}_3\left (-i e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 \sin ^{-1}(a x) \text {Li}_3\left (i e^{i \sin ^{-1}(a x)}\right )}{a c^2}-\frac {3 i \text {Li}_4\left (-i e^{i \sin ^{-1}(a x)}\right )}{a c^2}+\frac {3 i \text {Li}_4\left (i e^{i \sin ^{-1}(a x)}\right )}{a c^2}\\ \end {align*}

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Mathematica [A]  time = 0.52, size = 234, normalized size = 0.69 \[ \frac {\frac {a x \sin ^{-1}(a x)^3}{1-a^2 x^2}-\frac {3 \sin ^{-1}(a x)^2}{\sqrt {1-a^2 x^2}}-6 \sin ^{-1}(a x) \text {Li}_3\left (-i e^{i \sin ^{-1}(a x)}\right )+6 \sin ^{-1}(a x) \text {Li}_3\left (i e^{i \sin ^{-1}(a x)}\right )+3 i \left (\sin ^{-1}(a x)^2+2\right ) \text {Li}_2\left (-i e^{i \sin ^{-1}(a x)}\right )-3 i \left (\sin ^{-1}(a x)^2+2\right ) \text {Li}_2\left (i e^{i \sin ^{-1}(a x)}\right )-6 i \text {Li}_4\left (-i e^{i \sin ^{-1}(a x)}\right )+6 i \text {Li}_4\left (i e^{i \sin ^{-1}(a x)}\right )-2 i \sin ^{-1}(a x)^3 \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )-12 i \sin ^{-1}(a x) \tan ^{-1}\left (e^{i \sin ^{-1}(a x)}\right )}{2 a c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*ArcSin[a*x]^2)/Sqrt[1 - a^2*x^2] + (a*x*ArcSin[a*x]^3)/(1 - a^2*x^2) - (12*I)*ArcSin[a*x]*ArcTan[E^(I*Arc
Sin[a*x])] - (2*I)*ArcSin[a*x]^3*ArcTan[E^(I*ArcSin[a*x])] + (3*I)*(2 + ArcSin[a*x]^2)*PolyLog[2, (-I)*E^(I*Ar
cSin[a*x])] - (3*I)*(2 + ArcSin[a*x]^2)*PolyLog[2, I*E^(I*ArcSin[a*x])] - 6*ArcSin[a*x]*PolyLog[3, (-I)*E^(I*A
rcSin[a*x])] + 6*ArcSin[a*x]*PolyLog[3, I*E^(I*ArcSin[a*x])] - (6*I)*PolyLog[4, (-I)*E^(I*ArcSin[a*x])] + (6*I
)*PolyLog[4, I*E^(I*ArcSin[a*x])])/(2*a*c^2)

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\arcsin \left (a x\right )^{3}}{a^{4} c^{2} x^{4} - 2 \, a^{2} c^{2} x^{2} + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsin(a*x)^3/(a^4*c^2*x^4 - 2*a^2*c^2*x^2 + c^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsin(a*x)^3/(a^2*c*x^2 - c)^2, x)

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maple [A]  time = 0.25, size = 486, normalized size = 1.44 \[ -\frac {\arcsin \left (a x \right )^{3} x}{2 \left (a^{2} x^{2}-1\right ) c^{2}}+\frac {3 \arcsin \left (a x \right )^{2} \sqrt {-a^{2} x^{2}+1}}{2 a \left (a^{2} x^{2}-1\right ) c^{2}}+\frac {\arcsin \left (a x \right )^{3} \ln \left (1-i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{2 a \,c^{2}}-\frac {3 i \arcsin \left (a x \right )^{2} \polylog \left (2, i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{2 a \,c^{2}}+\frac {3 \arcsin \left (a x \right ) \polylog \left (3, i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{a \,c^{2}}+\frac {3 i \polylog \left (4, i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{a \,c^{2}}-\frac {\arcsin \left (a x \right )^{3} \ln \left (1+i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{2 a \,c^{2}}+\frac {3 i \arcsin \left (a x \right )^{2} \polylog \left (2, -i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{2 a \,c^{2}}-\frac {3 \arcsin \left (a x \right ) \polylog \left (3, -i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{a \,c^{2}}-\frac {3 i \polylog \left (4, -i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{a \,c^{2}}-\frac {3 \arcsin \left (a x \right ) \ln \left (1+i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{a \,c^{2}}+\frac {3 \arcsin \left (a x \right ) \ln \left (1-i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{a \,c^{2}}+\frac {3 i \dilog \left (1+i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{a \,c^{2}}-\frac {3 i \dilog \left (1-i \left (i a x +\sqrt {-a^{2} x^{2}+1}\right )\right )}{a \,c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/(a^2*x^2-1)*arcsin(a*x)^3/c^2*x+3/2/a/(a^2*x^2-1)*arcsin(a*x)^2/c^2*(-a^2*x^2+1)^(1/2)+1/2/a/c^2*arcsin(a
*x)^3*ln(1-I*(I*a*x+(-a^2*x^2+1)^(1/2)))-3/2*I*arcsin(a*x)^2*polylog(2,I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^2+3*a
rcsin(a*x)*polylog(3,I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^2+3*I*polylog(4,I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^2-1/2
/a/c^2*arcsin(a*x)^3*ln(1+I*(I*a*x+(-a^2*x^2+1)^(1/2)))+3/2*I*arcsin(a*x)^2*polylog(2,-I*(I*a*x+(-a^2*x^2+1)^(
1/2)))/a/c^2-3*arcsin(a*x)*polylog(3,-I*(I*a*x+(-a^2*x^2+1)^(1/2)))/a/c^2-3*I*polylog(4,-I*(I*a*x+(-a^2*x^2+1)
^(1/2)))/a/c^2-3/a/c^2*arcsin(a*x)*ln(1+I*(I*a*x+(-a^2*x^2+1)^(1/2)))+3/a/c^2*arcsin(a*x)*ln(1-I*(I*a*x+(-a^2*
x^2+1)^(1/2)))+3*I/a/c^2*dilog(1+I*(I*a*x+(-a^2*x^2+1)^(1/2)))-3*I/a/c^2*dilog(1-I*(I*a*x+(-a^2*x^2+1)^(1/2)))

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maxima [A]  time = 0.84, size = 57, normalized size = 0.17 \[ -\frac {1}{4} \, {\left (\frac {2 \, x}{a^{2} c^{2} x^{2} - c^{2}} - \frac {\log \left (a x + 1\right )}{a c^{2}} + \frac {\log \left (a x - 1\right )}{a c^{2}}\right )} \arcsin \left (a x\right )^{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*x/(a^2*c^2*x^2 - c^2) - log(a*x + 1)/(a*c^2) + log(a*x - 1)/(a*c^2))*arcsin(a*x)^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a*x)^3/(c - a^2*c*x^2)^2,x)

[Out]

int(asin(a*x)^3/(c - a^2*c*x^2)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\operatorname {asin}^{3}{\left (a x \right )}}{a^{4} x^{4} - 2 a^{2} x^{2} + 1}\, dx}{c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asin(a*x)**3/(a**4*x**4 - 2*a**2*x**2 + 1), x)/c**2

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