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

Optimal. Leaf size=332 \[ \frac{3 i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3}-\frac{3 i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3}-\frac{3 b^2 \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac{3 b^2 \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}-\frac{3 b \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3 \sqrt{1-c^2 x^2}}-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{3 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{3 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{4 c d^3}+\frac{b^2 x}{12 d^3 \left (1-c^2 x^2\right )}+\frac{5 b^2 \tanh ^{-1}(c x)}{6 c d^3} \]

[Out]

(b^2*x)/(12*d^3*(1 - c^2*x^2)) - (b*(a + b*ArcSin[c*x]))/(6*c*d^3*(1 - c^2*x^2)^(3/2)) - (3*b*(a + b*ArcSin[c*
x]))/(4*c*d^3*Sqrt[1 - c^2*x^2]) + (x*(a + b*ArcSin[c*x])^2)/(4*d^3*(1 - c^2*x^2)^2) + (3*x*(a + b*ArcSin[c*x]
)^2)/(8*d^3*(1 - c^2*x^2)) - (((3*I)/4)*(a + b*ArcSin[c*x])^2*ArcTan[E^(I*ArcSin[c*x])])/(c*d^3) + (5*b^2*ArcT
anh[c*x])/(6*c*d^3) + (((3*I)/4)*b*(a + b*ArcSin[c*x])*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(c*d^3) - (((3*I)/4
)*b*(a + b*ArcSin[c*x])*PolyLog[2, I*E^(I*ArcSin[c*x])])/(c*d^3) - (3*b^2*PolyLog[3, (-I)*E^(I*ArcSin[c*x])])/
(4*c*d^3) + (3*b^2*PolyLog[3, I*E^(I*ArcSin[c*x])])/(4*c*d^3)

________________________________________________________________________________________

Rubi [A]  time = 0.350826, antiderivative size = 332, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {4655, 4657, 4181, 2531, 2282, 6589, 4677, 206, 199} \[ \frac{3 i b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3}-\frac{3 i b \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3}-\frac{3 b^2 \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac{3 b^2 \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}-\frac{3 b \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3 \sqrt{1-c^2 x^2}}-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{3 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{3 i \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{4 c d^3}+\frac{b^2 x}{12 d^3 \left (1-c^2 x^2\right )}+\frac{5 b^2 \tanh ^{-1}(c x)}{6 c d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*x)/(12*d^3*(1 - c^2*x^2)) - (b*(a + b*ArcSin[c*x]))/(6*c*d^3*(1 - c^2*x^2)^(3/2)) - (3*b*(a + b*ArcSin[c*
x]))/(4*c*d^3*Sqrt[1 - c^2*x^2]) + (x*(a + b*ArcSin[c*x])^2)/(4*d^3*(1 - c^2*x^2)^2) + (3*x*(a + b*ArcSin[c*x]
)^2)/(8*d^3*(1 - c^2*x^2)) - (((3*I)/4)*(a + b*ArcSin[c*x])^2*ArcTan[E^(I*ArcSin[c*x])])/(c*d^3) + (5*b^2*ArcT
anh[c*x])/(6*c*d^3) + (((3*I)/4)*b*(a + b*ArcSin[c*x])*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(c*d^3) - (((3*I)/4
)*b*(a + b*ArcSin[c*x])*PolyLog[2, I*E^(I*ArcSin[c*x])])/(c*d^3) - (3*b^2*PolyLog[3, (-I)*E^(I*ArcSin[c*x])])/
(4*c*d^3) + (3*b^2*PolyLog[3, I*E^(I*ArcSin[c*x])])/(4*c*d^3)

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 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 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 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]

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rubi steps

\begin{align*} \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^3} \, dx &=\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac{(b c) \int \frac{x \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{2 d^3}+\frac{3 \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{\left (d-c^2 d x^2\right )^2} \, dx}{4 d}\\ &=-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{3 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac{b^2 \int \frac{1}{\left (1-c^2 x^2\right )^2} \, dx}{6 d^3}-\frac{(3 b c) \int \frac{x \left (a+b \sin ^{-1}(c x)\right )}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{4 d^3}+\frac{3 \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d-c^2 d x^2} \, dx}{8 d^2}\\ &=\frac{b^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{3 b \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3 \sqrt{1-c^2 x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{3 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}+\frac{b^2 \int \frac{1}{1-c^2 x^2} \, dx}{12 d^3}+\frac{\left (3 b^2\right ) \int \frac{1}{1-c^2 x^2} \, dx}{4 d^3}+\frac{3 \operatorname{Subst}\left (\int (a+b x)^2 \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{8 c d^3}\\ &=\frac{b^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{3 b \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3 \sqrt{1-c^2 x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{3 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac{3 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac{5 b^2 \tanh ^{-1}(c x)}{6 c d^3}-\frac{(3 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 c d^3}+\frac{(3 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1+i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 c d^3}\\ &=\frac{b^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{3 b \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3 \sqrt{1-c^2 x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{3 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac{3 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac{5 b^2 \tanh ^{-1}(c x)}{6 c d^3}+\frac{3 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}-\frac{3 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}-\frac{\left (3 i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 c d^3}+\frac{\left (3 i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (i e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 c d^3}\\ &=\frac{b^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{3 b \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3 \sqrt{1-c^2 x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{3 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac{3 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac{5 b^2 \tanh ^{-1}(c x)}{6 c d^3}+\frac{3 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}-\frac{3 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}-\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(-i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}\\ &=\frac{b^2 x}{12 d^3 \left (1-c^2 x^2\right )}-\frac{b \left (a+b \sin ^{-1}(c x)\right )}{6 c d^3 \left (1-c^2 x^2\right )^{3/2}}-\frac{3 b \left (a+b \sin ^{-1}(c x)\right )}{4 c d^3 \sqrt{1-c^2 x^2}}+\frac{x \left (a+b \sin ^{-1}(c x)\right )^2}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac{3 x \left (a+b \sin ^{-1}(c x)\right )^2}{8 d^3 \left (1-c^2 x^2\right )}-\frac{3 i \left (a+b \sin ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac{5 b^2 \tanh ^{-1}(c x)}{6 c d^3}+\frac{3 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}-\frac{3 i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}-\frac{3 b^2 \text{Li}_3\left (-i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}+\frac{3 b^2 \text{Li}_3\left (i e^{i \sin ^{-1}(c x)}\right )}{4 c d^3}\\ \end{align*}

Mathematica [A]  time = 5.80358, size = 556, normalized size = 1.67 \[ \frac{\frac{a b \left (-72 i \left (c^2 x^2-1\right )^2 \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )-70 \sqrt{1-c^2 x^2}+40 \cos \left (2 \sin ^{-1}(c x)\right )-18 \cos \left (3 \sin ^{-1}(c x)\right )+10 \cos \left (4 \sin ^{-1}(c x)\right )+3 \sin ^{-1}(c x) \left (22 c x+6 \sin \left (3 \sin ^{-1}(c x)\right )+9 \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-9 \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+12 \left (\log \left (1-i e^{i \sin ^{-1}(c x)}\right )-\log \left (1+i e^{i \sin ^{-1}(c x)}\right )\right ) \cos \left (2 \sin ^{-1}(c x)\right )+3 \left (\log \left (1-i e^{i \sin ^{-1}(c x)}\right )-\log \left (1+i e^{i \sin ^{-1}(c x)}\right )\right ) \cos \left (4 \sin ^{-1}(c x)\right )\right )+30\right )}{c \left (c^2 x^2-1\right )^2}+\frac{72 i a b \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{c}-\frac{4 b^2 \left (-18 i \sin ^{-1}(c x) \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )+18 i \sin ^{-1}(c x) \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )+18 \text{PolyLog}\left (3,-i e^{i \sin ^{-1}(c x)}\right )-18 \text{PolyLog}\left (3,i e^{i \sin ^{-1}(c x)}\right )+\frac{2 c x}{c^2 x^2-1}+\frac{9 c x \sin ^{-1}(c x)^2}{c^2 x^2-1}-\frac{6 c x \sin ^{-1}(c x)^2}{\left (c^2 x^2-1\right )^2}+\frac{18 \sin ^{-1}(c x)}{\sqrt{1-c^2 x^2}}+\frac{4 \sin ^{-1}(c x)}{\left (1-c^2 x^2\right )^{3/2}}-20 \tanh ^{-1}(c x)+18 i \sin ^{-1}(c x)^2 \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )\right )}{c}-\frac{36 a^2 x}{c^2 x^2-1}+\frac{24 a^2 x}{\left (c^2 x^2-1\right )^2}-\frac{18 a^2 \log (1-c x)}{c}+\frac{18 a^2 \log (c x+1)}{c}}{96 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((24*a^2*x)/(-1 + c^2*x^2)^2 - (36*a^2*x)/(-1 + c^2*x^2) - (18*a^2*Log[1 - c*x])/c + (18*a^2*Log[1 + c*x])/c +
((72*I)*a*b*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/c - (4*b^2*((2*c*x)/(-1 + c^2*x^2) + (4*ArcSin[c*x])/(1 - c^2
*x^2)^(3/2) + (18*ArcSin[c*x])/Sqrt[1 - c^2*x^2] - (6*c*x*ArcSin[c*x]^2)/(-1 + c^2*x^2)^2 + (9*c*x*ArcSin[c*x]
^2)/(-1 + c^2*x^2) + (18*I)*ArcSin[c*x]^2*ArcTan[E^(I*ArcSin[c*x])] - 20*ArcTanh[c*x] - (18*I)*ArcSin[c*x]*Pol
yLog[2, (-I)*E^(I*ArcSin[c*x])] + (18*I)*ArcSin[c*x]*PolyLog[2, I*E^(I*ArcSin[c*x])] + 18*PolyLog[3, (-I)*E^(I
*ArcSin[c*x])] - 18*PolyLog[3, I*E^(I*ArcSin[c*x])]))/c + (a*b*(30 - 70*Sqrt[1 - c^2*x^2] + 40*Cos[2*ArcSin[c*
x]] - 18*Cos[3*ArcSin[c*x]] + 10*Cos[4*ArcSin[c*x]] - (72*I)*(-1 + c^2*x^2)^2*PolyLog[2, I*E^(I*ArcSin[c*x])]
+ 3*ArcSin[c*x]*(22*c*x + 9*Log[1 - I*E^(I*ArcSin[c*x])] + 12*Cos[2*ArcSin[c*x]]*(Log[1 - I*E^(I*ArcSin[c*x])]
- Log[1 + I*E^(I*ArcSin[c*x])]) + 3*Cos[4*ArcSin[c*x]]*(Log[1 - I*E^(I*ArcSin[c*x])] - Log[1 + I*E^(I*ArcSin[
c*x])]) - 9*Log[1 + I*E^(I*ArcSin[c*x])] + 6*Sin[3*ArcSin[c*x]])))/(c*(-1 + c^2*x^2)^2))/(96*d^3)

________________________________________________________________________________________

Maple [B]  time = 0.233, size = 890, normalized size = 2.7 \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))^2/(-c^2*d*x^2+d)^3,x)

[Out]

1/16/c*a^2/d^3/(c*x-1)^2-3/16/c*a^2/d^3/(c*x-1)-1/16/c*a^2/d^3/(c*x+1)^2-3/16/c*a^2/d^3/(c*x+1)-3/16/c*a^2/d^3
*ln(c*x-1)+3/16/c*a^2/d^3*ln(c*x+1)+1/12*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*x-3/8/c*b^2/d^3*arcsin(c*x)^2*ln(1+I*(I
*c*x+(-c^2*x^2+1)^(1/2)))+3/8/c*b^2/d^3*arcsin(c*x)^2*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))+5/8*b^2/d^3/(c^4*x^4-
2*c^2*x^2+1)*arcsin(c*x)^2*x-5/3*I/c*b^2/d^3*arctan(I*c*x+(-c^2*x^2+1)^(1/2))-1/12*c^2*b^2/d^3/(c^4*x^4-2*c^2*
x^2+1)*x^3-3/4*b^2*polylog(3,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))/c/d^3+3/4*b^2*polylog(3,I*(I*c*x+(-c^2*x^2+1)^(1/2
)))/c/d^3+3/4*c*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x^2-3/4*c^2*a*b/d^3/(c^4*x^4-2*c^
2*x^2+1)*arcsin(c*x)*x^3+3/4*c*a*b/d^3/(c^4*x^4-2*c^2*x^2+1)*x^2*(-c^2*x^2+1)^(1/2)-3/4*I/c*a*b/d^3*dilog(1-I*
(I*c*x+(-c^2*x^2+1)^(1/2)))+3/4*I/c*b^2/d^3*arcsin(c*x)*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-3/4*I/c*b^2/d
^3*arcsin(c*x)*polylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))+3/4*I/c*a*b/d^3*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2)))+5
/4*a*b/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*x-11/12/c*b^2/d^3/(c^4*x^4-2*c^2*x^2+1)*arcsin(c*x)*(-c^2*x^2+1)^
(1/2)-11/12/c*a*b/d^3/(c^4*x^4-2*c^2*x^2+1)*(-c^2*x^2+1)^(1/2)-3/4/c*a*b/d^3*arcsin(c*x)*ln(1+I*(I*c*x+(-c^2*x
^2+1)^(1/2)))+3/4/c*a*b/d^3*arcsin(c*x)*ln(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-3/8*c^2*b^2/d^3/(c^4*x^4-2*c^2*x^2+
1)*arcsin(c*x)^2*x^3

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{16} \, a^{2}{\left (\frac{2 \,{\left (3 \, c^{2} x^{3} - 5 \, x\right )}}{c^{4} d^{3} x^{4} - 2 \, c^{2} d^{3} x^{2} + d^{3}} - \frac{3 \, \log \left (c x + 1\right )}{c d^{3}} + \frac{3 \, \log \left (c x - 1\right )}{c d^{3}}\right )} + \frac{3 \,{\left (b^{2} c^{4} x^{4} - 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} \log \left (c x + 1\right ) - 3 \,{\left (b^{2} c^{4} x^{4} - 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} \log \left (-c x + 1\right ) - 2 \,{\left (3 \, b^{2} c^{3} x^{3} - 5 \, b^{2} c x\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )^{2} - 2 \,{\left (c^{5} d^{3} x^{4} - 2 \, c^{3} d^{3} x^{2} + c d^{3}\right )} \int \frac{16 \, a b \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) -{\left (3 \,{\left (b^{2} c^{4} x^{4} - 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (c x + 1\right ) - 3 \,{\left (b^{2} c^{4} x^{4} - 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right ) \log \left (-c x + 1\right ) - 2 \,{\left (3 \, b^{2} c^{3} x^{3} - 5 \, b^{2} c x\right )} \arctan \left (c x, \sqrt{c x + 1} \sqrt{-c x + 1}\right )\right )} \sqrt{c x + 1} \sqrt{-c x + 1}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}\,{d x}}{16 \,{\left (c^{5} d^{3} x^{4} - 2 \, c^{3} d^{3} x^{2} + c d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*a^2*(2*(3*c^2*x^3 - 5*x)/(c^4*d^3*x^4 - 2*c^2*d^3*x^2 + d^3) - 3*log(c*x + 1)/(c*d^3) + 3*log(c*x - 1)/(
c*d^3)) + 1/16*(3*(b^2*c^4*x^4 - 2*b^2*c^2*x^2 + b^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2*log(c*x + 1
) - 3*(b^2*c^4*x^4 - 2*b^2*c^2*x^2 + b^2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2*log(-c*x + 1) - 2*(3*b^
2*c^3*x^3 - 5*b^2*c*x)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 16*(c^5*d^3*x^4 - 2*c^3*d^3*x^2 + c*d^3)
*integrate(-1/8*(16*a*b*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) - (3*(b^2*c^4*x^4 - 2*b^2*c^2*x^2 + b^2)*ar
ctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))*log(c*x + 1) - 3*(b^2*c^4*x^4 - 2*b^2*c^2*x^2 + b^2)*arctan2(c*x, sqr
t(c*x + 1)*sqrt(-c*x + 1))*log(-c*x + 1) - 2*(3*b^2*c^3*x^3 - 5*b^2*c*x)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x
+ 1)))*sqrt(c*x + 1)*sqrt(-c*x + 1))/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x))/(c^5*d^3*x^4 - 2
*c^3*d^3*x^2 + c*d^3)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}{c^{6} d^{3} x^{6} - 3 \, c^{4} d^{3} x^{4} + 3 \, c^{2} d^{3} x^{2} - d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)/(c^6*d^3*x^6 - 3*c^4*d^3*x^4 + 3*c^2*d^3*x^2 - d^3), x
)

________________________________________________________________________________________

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))**2/(-c**2*d*x**2+d)**3,x)

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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