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

Optimal. Leaf size=410 \[ -\frac{i b^2 f \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{c d \sqrt{d-c^2 d x^2}}-\frac{2 i b^2 g \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{c^2 d \sqrt{d-c^2 d x^2}}+\frac{2 i b^2 g \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{c^2 d \sqrt{d-c^2 d x^2}}+\frac{f x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt{d-c^2 d x^2}}-\frac{i f \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c d \sqrt{d-c^2 d x^2}}+\frac{2 b f \sqrt{1-c^2 x^2} \log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d \sqrt{d-c^2 d x^2}}+\frac{g \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d \sqrt{d-c^2 d x^2}}+\frac{4 i b g \sqrt{1-c^2 x^2} \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt{d-c^2 d x^2}} \]

[Out]

(g*(a + b*ArcSin[c*x])^2)/(c^2*d*Sqrt[d - c^2*d*x^2]) + (f*x*(a + b*ArcSin[c*x])^2)/(d*Sqrt[d - c^2*d*x^2]) -
(I*f*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(c*d*Sqrt[d - c^2*d*x^2]) + ((4*I)*b*g*Sqrt[1 - c^2*x^2]*(a + b*
ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/(c^2*d*Sqrt[d - c^2*d*x^2]) + (2*b*f*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c
*x])*Log[1 + E^((2*I)*ArcSin[c*x])])/(c*d*Sqrt[d - c^2*d*x^2]) - ((2*I)*b^2*g*Sqrt[1 - c^2*x^2]*PolyLog[2, (-I
)*E^(I*ArcSin[c*x])])/(c^2*d*Sqrt[d - c^2*d*x^2]) + ((2*I)*b^2*g*Sqrt[1 - c^2*x^2]*PolyLog[2, I*E^(I*ArcSin[c*
x])])/(c^2*d*Sqrt[d - c^2*d*x^2]) - (I*b^2*f*Sqrt[1 - c^2*x^2]*PolyLog[2, -E^((2*I)*ArcSin[c*x])])/(c*d*Sqrt[d
- c^2*d*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.572884, antiderivative size = 410, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 11, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.355, Rules used = {4777, 4763, 4651, 4675, 3719, 2190, 2279, 2391, 4677, 4657, 4181} \[ -\frac{i b^2 f \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{c d \sqrt{d-c^2 d x^2}}-\frac{2 i b^2 g \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )}{c^2 d \sqrt{d-c^2 d x^2}}+\frac{2 i b^2 g \sqrt{1-c^2 x^2} \text{PolyLog}\left (2,i e^{i \sin ^{-1}(c x)}\right )}{c^2 d \sqrt{d-c^2 d x^2}}+\frac{f x \left (a+b \sin ^{-1}(c x)\right )^2}{d \sqrt{d-c^2 d x^2}}-\frac{i f \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^2}{c d \sqrt{d-c^2 d x^2}}+\frac{2 b f \sqrt{1-c^2 x^2} \log \left (1+e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c d \sqrt{d-c^2 d x^2}}+\frac{g \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 d \sqrt{d-c^2 d x^2}}+\frac{4 i b g \sqrt{1-c^2 x^2} \tan ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{c^2 d \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(g*(a + b*ArcSin[c*x])^2)/(c^2*d*Sqrt[d - c^2*d*x^2]) + (f*x*(a + b*ArcSin[c*x])^2)/(d*Sqrt[d - c^2*d*x^2]) -
(I*f*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^2)/(c*d*Sqrt[d - c^2*d*x^2]) + ((4*I)*b*g*Sqrt[1 - c^2*x^2]*(a + b*
ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])])/(c^2*d*Sqrt[d - c^2*d*x^2]) + (2*b*f*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c
*x])*Log[1 + E^((2*I)*ArcSin[c*x])])/(c*d*Sqrt[d - c^2*d*x^2]) - ((2*I)*b^2*g*Sqrt[1 - c^2*x^2]*PolyLog[2, (-I
)*E^(I*ArcSin[c*x])])/(c^2*d*Sqrt[d - c^2*d*x^2]) + ((2*I)*b^2*g*Sqrt[1 - c^2*x^2]*PolyLog[2, I*E^(I*ArcSin[c*
x])])/(c^2*d*Sqrt[d - c^2*d*x^2]) - (I*b^2*f*Sqrt[1 - c^2*x^2]*PolyLog[2, -E^((2*I)*ArcSin[c*x])])/(c*d*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 4763

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))

Rule 4651

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)/Sqrt[d], Int[(x*(a + b*ArcSin[c*x])^(n - 1))/(d + e*x^2), x], x
] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[d, 0]

Rule 4675

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 3719

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 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 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]

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 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]

Rubi steps

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

Mathematica [A]  time = 1.36067, size = 237, normalized size = 0.58 \[ \frac{\sqrt{1-c^2 x^2} \left ((c f-g) \left (-\cot \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right ) \left (a+b \sin ^{-1}(c x)\right )^2+i \left (4 b^2 \text{PolyLog}\left (2,-i e^{-i \sin ^{-1}(c x)}\right )+\left (a+b \sin ^{-1}(c x)\right ) \left (a+b \sin ^{-1}(c x)-4 i b \log \left (1+i e^{-i \sin ^{-1}(c x)}\right )\right )\right )\right )-(c f+g) \left (-\tan \left (\frac{1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right ) \left (a+b \sin ^{-1}(c x)\right )^2+i \left (4 b^2 \text{PolyLog}\left (2,-i e^{i \sin ^{-1}(c x)}\right )+\left (a+b \sin ^{-1}(c x)\right ) \left (a+b \sin ^{-1}(c x)+4 i b \log \left (1+i e^{i \sin ^{-1}(c x)}\right )\right )\right )\right )\right )}{2 c^2 d \sqrt{d-c^2 d x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.312, size = 1047, normalized size = 2.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*b*c*f*sqrt(1/(c^4*d))*log(x^2 - 1/c^2)/d + 2*a*b*f*x*arcsin(c*x)/(sqrt(-c^2*d*x^2 + d)*d) + a^2*f*x/(sqrt(-
c^2*d*x^2 + d)*d) - sqrt(d)*integrate((2*a*b*g*x*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)) + (b^2*g*x + b^2*f
)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2)/((c^2*d^2*x^2 - d^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)), x) + a^2*g
/(sqrt(-c^2*d*x^2 + d)*c^2*d)

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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 (a^{2} g x + a^{2} f +{\left (b^{2} g x + b^{2} f\right )} \arcsin \left (c x\right )^{2} + 2 \,{\left (a b g x + a b f\right )} \arcsin \left (c x\right )\right )}}{c^{4} d^{2} x^{4} - 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-c^2*d*x^2 + d)*(a^2*g*x + a^2*f + (b^2*g*x + b^2*f)*arcsin(c*x)^2 + 2*(a*b*g*x + a*b*f)*arcsin(
c*x))/(c^4*d^2*x^4 - 2*c^2*d^2*x^2 + d^2), 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} \left (f + g x\right )}{\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((g*x+f)*(a+b*asin(c*x))**2/(-c**2*d*x**2+d)**(3/2),x)

[Out]

Integral((a + b*asin(c*x))**2*(f + g*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 (g x + f\right )}{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (-c^{2} d x^{2} + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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