3.6.67 \(\int \frac {(a+b \text {ArcSin}(c x))^2}{\sqrt {d+c d x} (e-c e x)^{3/2}} \, dx\) [567]

Optimal. Leaf size=454 \[ \frac {d \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {d x \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {i d \left (1-c^2 x^2\right )^{3/2} (a+b \text {ArcSin}(c x))^2}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {4 i b d \left (1-c^2 x^2\right )^{3/2} (a+b \text {ArcSin}(c x)) \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {2 b d \left (1-c^2 x^2\right )^{3/2} (a+b \text {ArcSin}(c x)) \log \left (1+e^{2 i \text {ArcSin}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {2 i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}+\frac {2 i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {PolyLog}\left (2,i e^{i \text {ArcSin}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}}-\frac {i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {PolyLog}\left (2,-e^{2 i \text {ArcSin}(c x)}\right )}{c (d+c d x)^{3/2} (e-c e x)^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.46, antiderivative size = 454, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.344, Rules used = {4763, 4847, 4745, 4765, 3800, 2221, 2317, 2438, 4767, 4749, 4266} \begin {gather*} \frac {4 i b d \left (1-c^2 x^2\right )^{3/2} \text {ArcTan}\left (e^{i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}-\frac {i d \left (1-c^2 x^2\right )^{3/2} (a+b \text {ArcSin}(c x))^2}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {d \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))^2}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {d x \left (1-c^2 x^2\right ) (a+b \text {ArcSin}(c x))^2}{(c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {2 b d \left (1-c^2 x^2\right )^{3/2} \log \left (1+e^{2 i \text {ArcSin}(c x)}\right ) (a+b \text {ArcSin}(c x))}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}-\frac {2 i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {Li}_2\left (-i e^{i \text {ArcSin}(c x)}\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}+\frac {2 i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {Li}_2\left (i e^{i \text {ArcSin}(c x)}\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}}-\frac {i b^2 d \left (1-c^2 x^2\right )^{3/2} \text {Li}_2\left (-e^{2 i \text {ArcSin}(c x)}\right )}{c (c d x+d)^{3/2} (e-c e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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*(e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] &&
IGtQ[m, 0]

Rule 4266

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 4745

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

Rule 4749

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 4763

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> D
ist[(d + e*x)^q*((f + g*x)^q/(1 - c^2*x^2)^q), Int[(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, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q]
&& GeQ[p - q, 0]

Rule 4765

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

Rule 4767

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^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 4847

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

Rubi steps

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

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Mathematica [A]
time = 1.07, size = 221, normalized size = 0.49 \begin {gather*} -\frac {\sqrt {d+c d x} \sqrt {e-c e x} \left (a \left (a+a c x+4 b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )\right )-4 i b^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-i e^{i \text {ArcSin}(c x)}\right )+b^2 \sqrt {1-c^2 x^2} \text {ArcSin}(c x)^2 \left (-i+\tan \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )+2 b \sqrt {1-c^2 x^2} \text {ArcSin}(c x) \left (2 b \log \left (1+i e^{i \text {ArcSin}(c x)}\right )+a \tan \left (\frac {1}{4} (\pi +2 \text {ArcSin}(c x))\right )\right )\right )}{c d e^2 (-1+c x) (1+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(a*(a + a*c*x + 4*b*Sqrt[1 - c^2*x^2]*Log[Cos[(Pi + 2*ArcSin[c*x])/4]]) - (
4*I)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] + b^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x]^2*(-I + Tan[
(Pi + 2*ArcSin[c*x])/4]) + 2*b*Sqrt[1 - c^2*x^2]*ArcSin[c*x]*(2*b*Log[1 + I*E^(I*ArcSin[c*x])] + a*Tan[(Pi + 2
*ArcSin[c*x])/4])))/(c*d*e^2*(-1 + c*x)*(1 + c*x)))

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Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2}}{\sqrt {c d x +d}\, \left (-c e x +e \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\sqrt {d \left (c x + 1\right )} \left (- e \left (c x - 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{\sqrt {d+c\,d\,x}\,{\left (e-c\,e\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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