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

Optimal. Leaf size=548 \[ \frac{2 i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d e \sqrt{c d x+d} \sqrt{e-c e x}}-\frac{2 i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d e \sqrt{c d x+d} \sqrt{e-c e x}}-\frac{2 i b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{d e \sqrt{c d x+d} \sqrt{e-c e x}}+\frac{2 i b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{d e \sqrt{c d x+d} \sqrt{e-c e x}}-\frac{2 b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )}{d e \sqrt{c d x+d} \sqrt{e-c e x}}+\frac{2 b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )}{d e \sqrt{c d x+d} \sqrt{e-c e x}}+\frac{4 i b \sqrt{1-c^2 x^2} \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d e \sqrt{c d x+d} \sqrt{e-c e x}}-\frac{2 \sqrt{1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d e \sqrt{c d x+d} \sqrt{e-c e x}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d e \sqrt{c d x+d} \sqrt{e-c e x}} \]

[Out]

(a + b*ArcSin[c*x])^2/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + ((4*I)*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*A
rcTan[E^(I*ArcSin[c*x])])/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) - (2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2*A
rcTanh[E^(I*ArcSin[c*x])])/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + ((2*I)*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*
x])*PolyLog[2, -E^(I*ArcSin[c*x])])/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) - ((2*I)*b^2*Sqrt[1 - c^2*x^2]*PolyL
og[2, (-I)*E^(I*ArcSin[c*x])])/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + ((2*I)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2,
I*E^(I*ArcSin[c*x])])/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) - ((2*I)*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*
PolyLog[2, E^(I*ArcSin[c*x])])/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) - (2*b^2*Sqrt[1 - c^2*x^2]*PolyLog[3, -E^
(I*ArcSin[c*x])])/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + (2*b^2*Sqrt[1 - c^2*x^2]*PolyLog[3, E^(I*ArcSin[c*x]
)])/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])

________________________________________________________________________________________

Rubi [A]  time = 0.853429, antiderivative size = 548, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 11, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.314, Rules used = {4739, 4705, 4709, 4183, 2531, 2282, 6589, 4657, 4181, 2279, 2391} \[ \frac{2 i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d e \sqrt{c d x+d} \sqrt{e-c e x}}-\frac{2 i b \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d e \sqrt{c d x+d} \sqrt{e-c e x}}-\frac{2 i b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{d e \sqrt{c d x+d} \sqrt{e-c e x}}+\frac{2 i b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{d e \sqrt{c d x+d} \sqrt{e-c e x}}-\frac{2 b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,-e^{i \sin ^{-1}(c x)}\right )}{d e \sqrt{c d x+d} \sqrt{e-c e x}}+\frac{2 b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,e^{i \sin ^{-1}(c x)}\right )}{d e \sqrt{c d x+d} \sqrt{e-c e x}}+\frac{4 i b \sqrt{1-c^2 x^2} \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d e \sqrt{c d x+d} \sqrt{e-c e x}}-\frac{2 \sqrt{1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d e \sqrt{c d x+d} \sqrt{e-c e x}}+\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{d e \sqrt{c d x+d} \sqrt{e-c e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*ArcSin[c*x])^2/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + ((4*I)*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*A
rcTan[E^(I*ArcSin[c*x])])/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) - (2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2*A
rcTanh[E^(I*ArcSin[c*x])])/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + ((2*I)*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*
x])*PolyLog[2, -E^(I*ArcSin[c*x])])/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) - ((2*I)*b^2*Sqrt[1 - c^2*x^2]*PolyL
og[2, (-I)*E^(I*ArcSin[c*x])])/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + ((2*I)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2,
I*E^(I*ArcSin[c*x])])/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) - ((2*I)*b*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*
PolyLog[2, E^(I*ArcSin[c*x])])/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) - (2*b^2*Sqrt[1 - c^2*x^2]*PolyLog[3, -E^
(I*ArcSin[c*x])])/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + (2*b^2*Sqrt[1 - c^2*x^2]*PolyLog[3, E^(I*ArcSin[c*x]
)])/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])

Rule 4739

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((h_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(
q_), x_Symbol] :> Dist[((-((d^2*g)/e))^IntPart[q]*(d + e*x)^FracPart[q]*(f + g*x)^FracPart[q])/(1 - c^2*x^2)^F
racPart[q], Int[(h*x)^m*(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q] && GeQ[p - q, 0]

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

Mathematica [A]  time = 5.74661, size = 877, normalized size = 1.6 \[ \frac{\sqrt{d} \sqrt{e} \log (c x) a^2-\sqrt{d} \sqrt{e} \log \left (d e+\sqrt{d} \sqrt{c x d+d} \sqrt{e-c e x} \sqrt{e}\right ) a^2-\frac{\sqrt{c x d+d} \sqrt{e-c e x} a^2}{c^2 x^2-1}+\frac{2 b d e \left (\sqrt{1-c^2 x^2} \log \left (1-e^{i \sin ^{-1}(c x)}\right ) \sin ^{-1}(c x)-\sqrt{1-c^2 x^2} \log \left (1+e^{i \sin ^{-1}(c x)}\right ) \sin ^{-1}(c x)+\sin ^{-1}(c x)+\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 (\cos \left (\frac{1}{2} \sin ^{-1}(c x)\right )+\sin \left (\frac{1}{2} \sin ^{-1}(c x)\right )\right )+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 )\right ) a}{\sqrt{c x d+d} \sqrt{e-c e x}}+\frac{b^2 d e \left (\sqrt{1-c^2 x^2} \log \left (1-e^{i \sin ^{-1}(c x)}\right ) \sin ^{-1}(c x)^2-\sqrt{1-c^2 x^2} \log \left (1+e^{i \sin ^{-1}(c x)}\right ) \sin ^{-1}(c x)^2+\sin ^{-1}(c x)^2-2 \sqrt{1-c^2 x^2} \log \left (1-i e^{i \sin ^{-1}(c x)}\right ) \sin ^{-1}(c x)+2 \sqrt{1-c^2 x^2} \log \left (1+i e^{i \sin ^{-1}(c x)}\right ) \sin ^{-1}(c x)+2 i \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right ) \sin ^{-1}(c x)-2 i \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right ) \sin ^{-1}(c x)+i \pi \sqrt{1-c^2 x^2} \sin ^{-1}(c x)-\pi \sqrt{1-c^2 x^2} \log \left (1-i e^{i \sin ^{-1}(c x)}\right )-\pi \sqrt{1-c^2 x^2} \log \left (1+i e^{i \sin ^{-1}(c x)}\right )+\pi \sqrt{1-c^2 x^2} \log \left (-\cos \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )+\pi \sqrt{1-c^2 x^2} \log \left (\sin \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\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 )\right )}{\sqrt{c x d+d} \sqrt{e-c e x}}}{d^2 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])/(-1 + c^2*x^2)) + a^2*Sqrt[d]*Sqrt[e]*Log[c*x] - a^2*Sqrt[d]*Sqrt[e]*
Log[d*e + Sqrt[d]*Sqrt[e]*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]] + (2*a*b*d*e*(ArcSin[c*x] + Sqrt[1 - c^2*x^2]*ArcSi
n[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]] - Sqrt[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])]))/
(Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + (b^2*d*e*(I*Pi*Sqrt[1 - c^2*x^2]*ArcSin[c*x] + ArcSin[c*x]^2 + Sqrt[1 - c^
2*x^2]*ArcSin[c*x]^2*Log[1 - E^(I*ArcSin[c*x])] - Pi*Sqrt[1 - c^2*x^2]*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])] - Pi*Sqrt[1 - c^2*x^2]*Log[1 + I*E^(I*ArcSin[c*x])] + 2*S
qrt[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*ArcSi
n[c*x])] + Pi*Sqrt[1 - c^2*x^2]*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] + Pi*Sqrt[1 - c^2*x^2]*Log[Sin[(Pi + 2*ArcSi
n[c*x])/4]] + (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]*Pol
yLog[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])]))/(Sqrt[d + c*d*x]*Sqrt[e - c*e*x]))/(d^2*e^2)

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Maple [F]  time = 0.284, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\arcsin \left ( cx \right ) \right ) ^{2}}{x} \left ( cdx+d \right ) ^{-{\frac{3}{2}}} \left ( -cex+e \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*d*x+d)^(3/2)/(-c*e*x+e)^(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{{\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )} \sqrt{c d x + d} \sqrt{-c e x + e}}{c^{4} d^{2} e^{2} x^{5} - 2 \, c^{2} d^{2} e^{2} x^{3} + d^{2} e^{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*d*x+d)^(3/2)/(-c*e*x+e)^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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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 d x + d\right )}^{\frac{3}{2}}{\left (-c e x + e\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*d*x+d)^(3/2)/(-c*e*x+e)^(3/2),x, algorithm="giac")

[Out]

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