3.73 \(\int \frac{(a+b \sin ^{-1}(c x))^2}{(f+g x) \sqrt{d-c^2 d x^2}} \, dx\)

Optimal. Leaf size=589 \[ -\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text{PolyLog}\left (2,\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{d-c^2 d x^2} \sqrt{c^2 f^2-g^2}}+\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text{PolyLog}\left (2,\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{\sqrt{d-c^2 d x^2} \sqrt{c^2 f^2-g^2}}-\frac{2 i b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{d-c^2 d x^2} \sqrt{c^2 f^2-g^2}}+\frac{2 i b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{\sqrt{d-c^2 d x^2} \sqrt{c^2 f^2-g^2}}-\frac{i \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{d-c^2 d x^2} \sqrt{c^2 f^2-g^2}}+\frac{i \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{\sqrt{d-c^2 d x^2} \sqrt{c^2 f^2-g^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.01781, antiderivative size = 589, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {4777, 4773, 3323, 2264, 2190, 2531, 2282, 6589} \[ -\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text{PolyLog}\left (2,\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{d-c^2 d x^2} \sqrt{c^2 f^2-g^2}}+\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text{PolyLog}\left (2,\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{\sqrt{d-c^2 d x^2} \sqrt{c^2 f^2-g^2}}-\frac{2 i b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{d-c^2 d x^2} \sqrt{c^2 f^2-g^2}}+\frac{2 i b^2 \sqrt{1-c^2 x^2} \text{PolyLog}\left (3,\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{\sqrt{d-c^2 d x^2} \sqrt{c^2 f^2-g^2}}-\frac{i \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{d-c^2 d x^2} \sqrt{c^2 f^2-g^2}}+\frac{i \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )}{\sqrt{d-c^2 d x^2} \sqrt{c^2 f^2-g^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4777

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

Rule 4773

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

Rule 3323

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c + d*x)^m*E
^(I*(e + f*x)))/(I*b + 2*a*E^(I*(e + f*x)) - I*b*E^(2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && 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 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}(c x)\right )^2}{(f+g x) \sqrt{d-c^2 d x^2}} \, dx &=\frac{\sqrt{1-c^2 x^2} \int \frac{\left (a+b \sin ^{-1}(c x)\right )^2}{(f+g x) \sqrt{1-c^2 x^2}} \, dx}{\sqrt{d-c^2 d x^2}}\\ &=\frac{\sqrt{1-c^2 x^2} \operatorname{Subst}\left (\int \frac{(a+b x)^2}{c f+g \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}}\\ &=\frac{\left (2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{e^{i x} (a+b x)^2}{2 c e^{i x} f+i g-i e^{2 i x} g} \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt{d-c^2 d x^2}}\\ &=-\frac{\left (2 i g \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{e^{i x} (a+b x)^2}{2 c f-2 i e^{i x} g-2 \sqrt{c^2 f^2-g^2}} \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}+\frac{\left (2 i g \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{e^{i x} (a+b x)^2}{2 c f-2 i e^{i x} g+2 \sqrt{c^2 f^2-g^2}} \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}\\ &=-\frac{i \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}+\frac{i \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}+\frac{\left (2 i b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1-\frac{2 i e^{i x} g}{2 c f-2 \sqrt{c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}-\frac{\left (2 i b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1-\frac{2 i e^{i x} g}{2 c f+2 \sqrt{c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}\\ &=-\frac{i \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}+\frac{i \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}-\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}+\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}+\frac{\left (2 b^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (\frac{2 i e^{i x} g}{2 c f-2 \sqrt{c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}-\frac{\left (2 b^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (\frac{2 i e^{i x} g}{2 c f+2 \sqrt{c^2 f^2-g^2}}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}\\ &=-\frac{i \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}+\frac{i \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}-\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}+\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{c^2 f^2-g^2} \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{\text{Li}_2\left (\frac{i g x}{c f-\sqrt{c^2 f^2-g^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{\sqrt{c^2 f^2-g^2} \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{\text{Li}_2\left (\frac{i g x}{c f+\sqrt{c^2 f^2-g^2}}\right )}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}\\ &=-\frac{i \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}+\frac{i \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}-\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}+\frac{2 b \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}-\frac{2 i b^2 \sqrt{1-c^2 x^2} \text{Li}_3\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}+\frac{2 i b^2 \sqrt{1-c^2 x^2} \text{Li}_3\left (\frac{i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt{c^2 f^2-g^2}}\right )}{\sqrt{c^2 f^2-g^2} \sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 0.391118, size = 357, normalized size = 0.61 \[ -\frac{i \sqrt{1-c^2 x^2} \left (-2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{PolyLog}\left (2,\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )+2 i b \left (a+b \sin ^{-1}(c x)\right ) \text{PolyLog}\left (2,\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )+2 b^2 \text{PolyLog}\left (3,\frac{i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt{c^2 f^2-g^2}}\right )-2 b^2 \text{PolyLog}\left (3,\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )+\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1+\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}-c f}\right )-\left (a+b \sin ^{-1}(c x)\right )^2 \log \left (1-\frac{i g e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 f^2-g^2}+c f}\right )\right )}{\sqrt{d-c^2 d x^2} \sqrt{c^2 f^2-g^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.589, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b\arcsin \left ( cx \right ) \right ) ^{2}}{gx+f}{\frac{1}{\sqrt{-{c}^{2}d{x}^{2}+d}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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^{2} d g x^{3} + c^{2} d f x^{2} - d g x - d f}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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