3.432 \(\int \frac{(a+b \sin ^{-1}(\frac{\sqrt{1-c x}}{\sqrt{1+c x}}))^3}{1-c^2 x^2} \, dx\)

Optimal. Leaf size=275 \[ -\frac{3 b^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )}{2 c}+\frac{3 i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^2}{2 c}-\frac{3 i b^3 \text{PolyLog}\left (4,e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right )}{4 c}+\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^4}{4 b c}-\frac{\log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^3}{c} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.223743, antiderivative size = 275, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6681, 4625, 3717, 2190, 2531, 6609, 2282, 6589} \[ -\frac{3 b^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )}{2 c}+\frac{3 i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^2}{2 c}-\frac{3 i b^3 \text{PolyLog}\left (4,e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right )}{4 c}+\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^4}{4 b c}-\frac{\log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^3}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6681

Int[((a_.) + (b_.)*(F_)[((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.)*(x_)]])^(n_.)/((A_.) + (C_.)*(x_)^
2), x_Symbol] :> Dist[(2*e*g)/(C*(e*f - d*g)), Subst[Int[(a + b*F[c*x])^n/x, x], x, Sqrt[d + e*x]/Sqrt[f + g*x
]], x] /; FreeQ[{a, b, c, d, e, f, g, A, C, F}, x] && EqQ[C*d*f - A*e*g, 0] && EqQ[e*f + d*g, 0] && IGtQ[n, 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]

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 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 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 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]

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 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{\left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^3}{x} \, dx,x,\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{c}\\ &=-\frac{\operatorname{Subst}\left (\int (a+b x)^3 \cot (x) \, dx,x,\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )}{c}\\ &=\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^4}{4 b c}+\frac{(2 i) \operatorname{Subst}\left (\int \frac{e^{2 i x} (a+b x)^3}{1-e^{2 i x}} \, dx,x,\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )}{c}\\ &=\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^4}{4 b c}-\frac{\left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^3 \log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}+\frac{(3 b) \operatorname{Subst}\left (\int (a+b x)^2 \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )}{c}\\ &=\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^4}{4 b c}-\frac{\left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^3 \log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}+\frac{3 i b \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2 \text{Li}_2\left (e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{2 c}-\frac{\left (3 i b^2\right ) \operatorname{Subst}\left (\int (a+b x) \text{Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )}{c}\\ &=\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^4}{4 b c}-\frac{\left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^3 \log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}+\frac{3 i b \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2 \text{Li}_2\left (e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{2 c}-\frac{3 b^2 \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \text{Li}_3\left (e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{2 c}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \text{Li}_3\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )}{2 c}\\ &=\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^4}{4 b c}-\frac{\left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^3 \log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}+\frac{3 i b \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2 \text{Li}_2\left (e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{2 c}-\frac{3 b^2 \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \text{Li}_3\left (e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{2 c}-\frac{\left (3 i b^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{4 c}\\ &=\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^4}{4 b c}-\frac{\left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^3 \log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}+\frac{3 i b \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2 \text{Li}_2\left (e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{2 c}-\frac{3 b^2 \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \text{Li}_3\left (e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{2 c}-\frac{3 i b^3 \text{Li}_4\left (e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{4 c}\\ \end{align*}

Mathematica [F]  time = 0.297938, size = 0, normalized size = 0. \[ \int \frac{\left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^3}{1-c^2 x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [B]  time = 0.799, size = 1232, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^3/c*ln(c*x+1)-1/2*a^3/c*ln(c*x-1)+1/4*I*b^3/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^4-b^3/c*arcsin((-c*x+
1)^(1/2)/(c*x+1)^(1/2))^3*ln(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))+3*I*b^3/c*arcsin((-c
*x+1)^(1/2)/(c*x+1)^(1/2))^2*polylog(2,I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))-6*b^3/c*arcs
in((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(3,-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))-6*I*b^3
/c*polylog(4,-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))-b^3/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(
1/2))^3*ln(1-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))+3*I*a^2*b/c*polylog(2,I*(-c*x+1)^(1/2)
/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))-6*b^3/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(3,I*(-c*x+1)^(
1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))+6*I*a*b^2/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(2,I*(-
c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))+3*I*b^3/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*polylo
g(2,-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))-3*a*b^2/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))
^2*ln(1-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))+6*I*a*b^2/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(
1/2))*polylog(2,-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))-6*a*b^2/c*polylog(3,I*(-c*x+1)^(1/
2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))-3*a*b^2/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*ln(1+I*(-c*x+1)^
(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))-6*I*b^3/c*polylog(4,I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)
/(c*x+1))^(1/2))-6*a*b^2/c*polylog(3,-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))+I*a*b^2/c*arc
sin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^3-3*a^2*b/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))*ln(1+I*(-c*x+1)^(1/2)/(c*x+
1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))+3/2*I*a^2*b/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2-3*a^2*b/c*arcsin((-c
*x+1)^(1/2)/(c*x+1)^(1/2))*ln(1-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))+3*I*a^2*b/c*polylog
(2,-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a^{3}{\left (\frac{\log \left (c x + 1\right )}{c} - \frac{\log \left (c x - 1\right )}{c}\right )} - \int \frac{b^{3} \arctan \left (\sqrt{-c x + 1}, \sqrt{2} \sqrt{c} \sqrt{x}\right )^{3} + 3 \, a b^{2} \arctan \left (\sqrt{-c x + 1}, \sqrt{2} \sqrt{c} \sqrt{x}\right )^{2} + 3 \, a^{2} b \arctan \left (\sqrt{-c x + 1}, \sqrt{2} \sqrt{c} \sqrt{x}\right )}{c^{2} x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^3*(log(c*x + 1)/c - log(c*x - 1)/c) - integrate((b^3*arctan2(sqrt(-c*x + 1), sqrt(2)*sqrt(c)*sqrt(x))^3
+ 3*a*b^2*arctan2(sqrt(-c*x + 1), sqrt(2)*sqrt(c)*sqrt(x))^2 + 3*a^2*b*arctan2(sqrt(-c*x + 1), sqrt(2)*sqrt(c)
*sqrt(x)))/(c^2*x^2 - 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b^{3} \arcsin \left (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right )^{3} + 3 \, a b^{2} \arcsin \left (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right )^{2} + 3 \, a^{2} b \arcsin \left (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right ) + a^{3}}{c^{2} x^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError