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

Optimal. Leaf size=467 \[ \frac{2 i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt{d-c^2 d x^2}}-\frac{2 i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt{d-c^2 d x^2}}-\frac{2 i b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{d \sqrt{d-c^2 d x^2}}+\frac{2 i b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{d \sqrt{d-c^2 d x^2}}-\frac{2 b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )}{d \sqrt{d-c^2 d x^2}}+\frac{2 b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )}{d \sqrt{d-c^2 d x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt{d-c^2 d x^2}}+\frac{4 i b \sqrt{1-c^2 x^2} \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt{d-c^2 d x^2}}-\frac{2 \sqrt{1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt{d-c^2 d x^2}} \]

[Out]

(a + b*ArcSin[c*x])^2/(d*Sqrt[d - c^2*d*x^2]) + ((4*I)*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*ArcTan[E^(I*Arc
Sin[c*x])])/(d*Sqrt[d - c^2*d*x^2]) - (2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2*ArcTanh[E^(I*ArcSin[c*x])])/(
d*Sqrt[d - c^2*d*x^2]) + ((2*I)*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*PolyLog[2, -E^(I*ArcSin[c*x])])/(d*Sqr
t[d - c^2*d*x^2]) - ((2*I)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(d*Sqrt[d - c^2*d*x^2]) +
((2*I)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, I*E^(I*ArcSin[c*x])])/(d*Sqrt[d - c^2*d*x^2]) - ((2*I)*b*Sqrt[1 - c^2
*x^2]*(a + b*ArcSin[c*x])*PolyLog[2, E^(I*ArcSin[c*x])])/(d*Sqrt[d - c^2*d*x^2]) - (2*b^2*Sqrt[1 - c^2*x^2]*Po
lyLog[3, -E^(I*ArcSin[c*x])])/(d*Sqrt[d - c^2*d*x^2]) + (2*b^2*Sqrt[1 - c^2*x^2]*PolyLog[3, E^(I*ArcSin[c*x])]
)/(d*Sqrt[d - c^2*d*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.573021, antiderivative size = 467, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 11, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.379, Rules used = {4705, 4713, 4709, 4183, 2531, 2282, 6589, 4657, 4181, 2279, 2391} \[ \frac{2 i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt{d-c^2 d x^2}}-\frac{2 i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt{d-c^2 d x^2}}-\frac{2 i b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{d \sqrt{d-c^2 d x^2}}+\frac{2 i b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{d \sqrt{d-c^2 d x^2}}-\frac{2 b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )}{d \sqrt{d-c^2 d x^2}}+\frac{2 b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )}{d \sqrt{d-c^2 d x^2}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt{d-c^2 d x^2}}+\frac{4 i b \sqrt{1-c^2 x^2} \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d \sqrt{d-c^2 d x^2}}-\frac{2 \sqrt{1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*ArcSin[c*x])^2/(d*Sqrt[d - c^2*d*x^2]) + ((4*I)*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*ArcTan[E^(I*Arc
Sin[c*x])])/(d*Sqrt[d - c^2*d*x^2]) - (2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2*ArcTanh[E^(I*ArcSin[c*x])])/(
d*Sqrt[d - c^2*d*x^2]) + ((2*I)*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*PolyLog[2, -E^(I*ArcSin[c*x])])/(d*Sqr
t[d - c^2*d*x^2]) - ((2*I)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, (-I)*E^(I*ArcSin[c*x])])/(d*Sqrt[d - c^2*d*x^2]) +
((2*I)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, I*E^(I*ArcSin[c*x])])/(d*Sqrt[d - c^2*d*x^2]) - ((2*I)*b*Sqrt[1 - c^2
*x^2]*(a + b*ArcSin[c*x])*PolyLog[2, E^(I*ArcSin[c*x])])/(d*Sqrt[d - c^2*d*x^2]) - (2*b^2*Sqrt[1 - c^2*x^2]*Po
lyLog[3, -E^(I*ArcSin[c*x])])/(d*Sqrt[d - c^2*d*x^2]) + (2*b^2*Sqrt[1 - c^2*x^2]*PolyLog[3, E^(I*ArcSin[c*x])]
)/(d*Sqrt[d - c^2*d*x^2])

Rule 4705

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)/(2*d*f*(p + 1)), x] + (Dist[(m + 2*p + 3)/(2*d*(p +
1)), Int[(f*x)^m*(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)^Frac
Part[p])/(2*f*(p + 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, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ
[m, 1] && (IntegerQ[m] || IntegerQ[p] || EqQ[n, 1])

Rule 4713

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[
Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2], Int[((f*x)^m*(a + b*ArcSin[c*x])^n)/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a,
b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] &&  !GtQ[d, 0] && (IntegerQ[m] || EqQ[n, 1])

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 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 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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 1.94327, size = 667, normalized size = 1.43 \[ \frac{2 a b d \left (i \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )-i \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )+\sqrt{1-c^2 x^2} \sin ^{-1}(c x) \log \left (1-e^{i \sin ^{-1}(c x)}\right )-\sqrt{1-c^2 x^2} \sin ^{-1}(c x) \log \left (1+e^{i \sin ^{-1}(c x)}\right )+\sqrt{1-c^2 x^2} \log \left (\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )-\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )-\sqrt{1-c^2 x^2} \log \left (\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )+\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )+\sin ^{-1}(c x)\right )+b^2 d \left (2 i \sqrt{1-c^2 x^2} \sin ^{-1}(c x) \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )-2 i \sqrt{1-c^2 x^2} \sin ^{-1}(c x) \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )-2 i \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )+2 i \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )-2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )+2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )+\sqrt{1-c^2 x^2} \sin ^{-1}(c x)^2 \log \left (1-e^{i \sin ^{-1}(c x)}\right )-\sqrt{1-c^2 x^2} \sin ^{-1}(c x)^2 \log \left (1+e^{i \sin ^{-1}(c x)}\right )-2 \sqrt{1-c^2 x^2} \sin ^{-1}(c x) \log \left (1-i e^{i \sin ^{-1}(c x)}\right )+2 \sqrt{1-c^2 x^2} \sin ^{-1}(c x) \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+\sin ^{-1}(c x)^2\right )+a^2 \sqrt{d} \sqrt{d-c^2 d x^2} \log (c x)-a^2 \sqrt{d} \sqrt{d-c^2 d x^2} \log \left (\sqrt{d} \sqrt{d-c^2 d x^2}+d\right )+a^2 d}{d^2 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*d + a^2*Sqrt[d]*Sqrt[d - c^2*d*x^2]*Log[c*x] - a^2*Sqrt[d]*Sqrt[d - c^2*d*x^2]*Log[d + Sqrt[d]*Sqrt[d - c
^2*d*x^2]] + 2*a*b*d*(ArcSin[c*x] + Sqrt[1 - c^2*x^2]*ArcSin[c*x]*Log[1 - E^(I*ArcSin[c*x])] - Sqrt[1 - c^2*x^
2]*ArcSin[c*x]*Log[1 + E^(I*ArcSin[c*x])] + Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] - S
qrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]] + I*Sqrt[1 - c^2*x^2]*PolyLog[2, -E^(I*ArcSin[c*
x])] - I*Sqrt[1 - c^2*x^2]*PolyLog[2, E^(I*ArcSin[c*x])]) + b^2*d*(ArcSin[c*x]^2 + Sqrt[1 - c^2*x^2]*ArcSin[c*
x]^2*Log[1 - E^(I*ArcSin[c*x])] - 2*Sqrt[1 - c^2*x^2]*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] + 2*Sqrt[1 - c^
2*x^2]*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] - Sqrt[1 - c^2*x^2]*ArcSin[c*x]^2*Log[1 + E^(I*ArcSin[c*x])] +
(2*I)*Sqrt[1 - c^2*x^2]*ArcSin[c*x]*PolyLog[2, -E^(I*ArcSin[c*x])] - (2*I)*Sqrt[1 - c^2*x^2]*PolyLog[2, (-I)*
E^(I*ArcSin[c*x])] + (2*I)*Sqrt[1 - c^2*x^2]*PolyLog[2, I*E^(I*ArcSin[c*x])] - (2*I)*Sqrt[1 - c^2*x^2]*ArcSin[
c*x]*PolyLog[2, E^(I*ArcSin[c*x])] - 2*Sqrt[1 - c^2*x^2]*PolyLog[3, -E^(I*ArcSin[c*x])] + 2*Sqrt[1 - c^2*x^2]*
PolyLog[3, E^(I*ArcSin[c*x])]))/(d^2*Sqrt[d - c^2*d*x^2])

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Maple [B]  time = 0.244, size = 1096, normalized size = 2.4 \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/x/(-c^2*d*x^2+d)^(3/2),x)

[Out]

a^2/d/(-c^2*d*x^2+d)^(1/2)-a^2/d^(3/2)*ln((2*d+2*d^(1/2)*(-c^2*d*x^2+d)^(1/2))/x)-b^2*(-d*(c^2*x^2-1))^(1/2)/d
^2/(c^2*x^2-1)*arcsin(c*x)^2+b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)^2*ln(1+
I*c*x+(-c^2*x^2+1)^(1/2))-b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)^2*ln(1-I*c
*x-(-c^2*x^2+1)^(1/2))-2*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)*ln(1+I*(I*c
*x+(-c^2*x^2+1)^(1/2)))+2*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)*ln(1-I*(I*
c*x+(-c^2*x^2+1)^(1/2)))+2*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*polylog(3,-I*c*x-(-c^
2*x^2+1)^(1/2))-2*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*polylog(3,I*c*x+(-c^2*x^2+1)^(
1/2))+2*I*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)*polylog(2,I*c*x+(-c^2*x^2+
1)^(1/2))+2*I*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*dilog(1+I*(I*c*x+(-c^2*x^2+1)^(1/2
)))-2*I*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*dilog(1+I*c*x+(-c^2*x^2+1)^(1/2))-4*I*a*
b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arctan(I*c*x+(-c^2*x^2+1)^(1/2))-2*a*b*(-d*(c^2*x^
2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)-2*I*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*dilo
g(1-I*(I*c*x+(-c^2*x^2+1)^(1/2)))-2*I*b^2*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x
)*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))-2*I*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*dilog
(I*c*x+(-c^2*x^2+1)^(1/2))+2*a*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d^2/(c^2*x^2-1)*arcsin(c*x)*ln(1+I*
c*x+(-c^2*x^2+1)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-c^{2} d x^{2} + d}{\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )}}{c^{4} d^{2} x^{5} - 2 \, c^{2} d^{2} x^{3} + d^{2} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-c^2*d*x^2 + d)*(b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)/(c^4*d^2*x^5 - 2*c^2*d^2*x^3 + d^2
*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}^{\frac{3}{2}} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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