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

Optimal. Leaf size=410 \[ \frac{f^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c \sqrt{d-c^2 d x^2}}-\frac{2 f g \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 \sqrt{d-c^2 d x^2}}+\frac{4 b f g x \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c \sqrt{d-c^2 d x^2}}+\frac{g^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c^3 \sqrt{d-c^2 d x^2}}-\frac{g^2 x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 \sqrt{d-c^2 d x^2}}+\frac{b g^2 x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c \sqrt{d-c^2 d x^2}}+\frac{4 b^2 f g \left (1-c^2 x^2\right )}{c^2 \sqrt{d-c^2 d x^2}}+\frac{b^2 g^2 x \left (1-c^2 x^2\right )}{4 c^2 \sqrt{d-c^2 d x^2}}-\frac{b^2 g^2 \sqrt{1-c^2 x^2} \sin ^{-1}(c x)}{4 c^3 \sqrt{d-c^2 d x^2}} \]

[Out]

(4*b^2*f*g*(1 - c^2*x^2))/(c^2*Sqrt[d - c^2*d*x^2]) + (b^2*g^2*x*(1 - c^2*x^2))/(4*c^2*Sqrt[d - c^2*d*x^2]) -
(b^2*g^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(4*c^3*Sqrt[d - c^2*d*x^2]) + (4*b*f*g*x*Sqrt[1 - c^2*x^2]*(a + b*ArcS
in[c*x]))/(c*Sqrt[d - c^2*d*x^2]) + (b*g^2*x^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(2*c*Sqrt[d - c^2*d*x^2]
) - (2*f*g*(1 - c^2*x^2)*(a + b*ArcSin[c*x])^2)/(c^2*Sqrt[d - c^2*d*x^2]) - (g^2*x*(1 - c^2*x^2)*(a + b*ArcSin
[c*x])^2)/(2*c^2*Sqrt[d - c^2*d*x^2]) + (f^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^3)/(3*b*c*Sqrt[d - c^2*d*x^
2]) + (g^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^3)/(6*b*c^3*Sqrt[d - c^2*d*x^2])

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Rubi [A]  time = 0.553056, antiderivative size = 410, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4777, 4773, 3317, 3296, 2638, 3311, 32, 2635, 8} \[ \frac{f^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c \sqrt{d-c^2 d x^2}}-\frac{2 f g \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 \sqrt{d-c^2 d x^2}}+\frac{4 b f g x \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c \sqrt{d-c^2 d x^2}}+\frac{g^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c^3 \sqrt{d-c^2 d x^2}}-\frac{g^2 x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 \sqrt{d-c^2 d x^2}}+\frac{b g^2 x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c \sqrt{d-c^2 d x^2}}+\frac{4 b^2 f g \left (1-c^2 x^2\right )}{c^2 \sqrt{d-c^2 d x^2}}+\frac{b^2 g^2 x \left (1-c^2 x^2\right )}{4 c^2 \sqrt{d-c^2 d x^2}}-\frac{b^2 g^2 \sqrt{1-c^2 x^2} \sin ^{-1}(c x)}{4 c^3 \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b^2*f*g*(1 - c^2*x^2))/(c^2*Sqrt[d - c^2*d*x^2]) + (b^2*g^2*x*(1 - c^2*x^2))/(4*c^2*Sqrt[d - c^2*d*x^2]) -
(b^2*g^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(4*c^3*Sqrt[d - c^2*d*x^2]) + (4*b*f*g*x*Sqrt[1 - c^2*x^2]*(a + b*ArcS
in[c*x]))/(c*Sqrt[d - c^2*d*x^2]) + (b*g^2*x^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(2*c*Sqrt[d - c^2*d*x^2]
) - (2*f*g*(1 - c^2*x^2)*(a + b*ArcSin[c*x])^2)/(c^2*Sqrt[d - c^2*d*x^2]) - (g^2*x*(1 - c^2*x^2)*(a + b*ArcSin
[c*x])^2)/(2*c^2*Sqrt[d - c^2*d*x^2]) + (f^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^3)/(3*b*c*Sqrt[d - c^2*d*x^
2]) + (g^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^3)/(6*b*c^3*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 3317

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] ||
IGtQ[m, 0] || NeQ[a^2 - b^2, 0])

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{(f+g x)^2 \left (a+b \sin ^{-1}(c x)\right )^2}{\sqrt{d-c^2 d x^2}} \, dx &=\frac{\sqrt{1-c^2 x^2} \int \frac{(f+g x)^2 \left (a+b \sin ^{-1}(c x)\right )^2}{\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 (a+b x)^2 (c f+g \sin (x))^2 \, dx,x,\sin ^{-1}(c x)\right )}{c^3 \sqrt{d-c^2 d x^2}}\\ &=\frac{\sqrt{1-c^2 x^2} \operatorname{Subst}\left (\int \left (c^2 f^2 (a+b x)^2+2 c f g (a+b x)^2 \sin (x)+g^2 (a+b x)^2 \sin ^2(x)\right ) \, dx,x,\sin ^{-1}(c x)\right )}{c^3 \sqrt{d-c^2 d x^2}}\\ &=\frac{f^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c \sqrt{d-c^2 d x^2}}+\frac{\left (2 f g \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x)^2 \sin (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^2 \sqrt{d-c^2 d x^2}}+\frac{\left (g^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x)^2 \sin ^2(x) \, dx,x,\sin ^{-1}(c x)\right )}{c^3 \sqrt{d-c^2 d x^2}}\\ &=\frac{b g^2 x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c \sqrt{d-c^2 d x^2}}-\frac{2 f g \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 \sqrt{d-c^2 d x^2}}-\frac{g^2 x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 \sqrt{d-c^2 d x^2}}+\frac{f^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c \sqrt{d-c^2 d x^2}}+\frac{\left (4 b f g \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \cos (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^2 \sqrt{d-c^2 d x^2}}+\frac{\left (g^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x)^2 \, dx,x,\sin ^{-1}(c x)\right )}{2 c^3 \sqrt{d-c^2 d x^2}}-\frac{\left (b^2 g^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \sin ^2(x) \, dx,x,\sin ^{-1}(c x)\right )}{2 c^3 \sqrt{d-c^2 d x^2}}\\ &=\frac{b^2 g^2 x \left (1-c^2 x^2\right )}{4 c^2 \sqrt{d-c^2 d x^2}}+\frac{4 b f g x \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c \sqrt{d-c^2 d x^2}}+\frac{b g^2 x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c \sqrt{d-c^2 d x^2}}-\frac{2 f g \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 \sqrt{d-c^2 d x^2}}-\frac{g^2 x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 \sqrt{d-c^2 d x^2}}+\frac{f^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c \sqrt{d-c^2 d x^2}}+\frac{g^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c^3 \sqrt{d-c^2 d x^2}}-\frac{\left (4 b^2 f g \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \sin (x) \, dx,x,\sin ^{-1}(c x)\right )}{c^2 \sqrt{d-c^2 d x^2}}-\frac{\left (b^2 g^2 \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int 1 \, dx,x,\sin ^{-1}(c x)\right )}{4 c^3 \sqrt{d-c^2 d x^2}}\\ &=\frac{4 b^2 f g \left (1-c^2 x^2\right )}{c^2 \sqrt{d-c^2 d x^2}}+\frac{b^2 g^2 x \left (1-c^2 x^2\right )}{4 c^2 \sqrt{d-c^2 d x^2}}-\frac{b^2 g^2 \sqrt{1-c^2 x^2} \sin ^{-1}(c x)}{4 c^3 \sqrt{d-c^2 d x^2}}+\frac{4 b f g x \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c \sqrt{d-c^2 d x^2}}+\frac{b g^2 x^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 c \sqrt{d-c^2 d x^2}}-\frac{2 f g \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c^2 \sqrt{d-c^2 d x^2}}-\frac{g^2 x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 c^2 \sqrt{d-c^2 d x^2}}+\frac{f^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{3 b c \sqrt{d-c^2 d x^2}}+\frac{g^2 \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{6 b c^3 \sqrt{d-c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 1.37078, size = 400, normalized size = 0.98 \[ \frac{3 \sqrt{d} g \left (c^2 x^2-1\right ) \left (4 c \left (a^2 \sqrt{1-c^2 x^2} (4 f+g x)-8 a b c f x-8 b^2 f \sqrt{1-c^2 x^2}\right )+2 a b g \cos \left (2 \sin ^{-1}(c x)\right )+b^2 (-g) \sin \left (2 \sin ^{-1}(c x)\right )\right )-12 a^2 \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2} \left (2 c^2 f^2+g^2\right ) \tan ^{-1}\left (\frac{c x \sqrt{d-c^2 d x^2}}{\sqrt{d} \left (c^2 x^2-1\right )}\right )+6 b \sqrt{d} \left (c^2 x^2-1\right ) \sin ^{-1}(c x)^2 \left (-2 a \left (2 c^2 f^2+g^2\right )+8 b c f g \sqrt{1-c^2 x^2}+b g^2 \sin \left (2 \sin ^{-1}(c x)\right )\right )+6 b \sqrt{d} g \left (c^2 x^2-1\right ) \sin ^{-1}(c x) \left (16 c f \left (a \sqrt{1-c^2 x^2}-b c x\right )+2 a g \sin \left (2 \sin ^{-1}(c x)\right )+b g \cos \left (2 \sin ^{-1}(c x)\right )\right )-4 b^2 \sqrt{d} \left (c^2 x^2-1\right ) \left (2 c^2 f^2+g^2\right ) \sin ^{-1}(c x)^3}{24 c^3 \sqrt{d} \sqrt{1-c^2 x^2} \sqrt{d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b^2*Sqrt[d]*(2*c^2*f^2 + g^2)*(-1 + c^2*x^2)*ArcSin[c*x]^3 - 12*a^2*(2*c^2*f^2 + g^2)*Sqrt[1 - c^2*x^2]*Sq
rt[d - c^2*d*x^2]*ArcTan[(c*x*Sqrt[d - c^2*d*x^2])/(Sqrt[d]*(-1 + c^2*x^2))] + 6*b*Sqrt[d]*g*(-1 + c^2*x^2)*Ar
cSin[c*x]*(16*c*f*(-(b*c*x) + a*Sqrt[1 - c^2*x^2]) + b*g*Cos[2*ArcSin[c*x]] + 2*a*g*Sin[2*ArcSin[c*x]]) + 3*Sq
rt[d]*g*(-1 + c^2*x^2)*(4*c*(-8*a*b*c*f*x - 8*b^2*f*Sqrt[1 - c^2*x^2] + a^2*(4*f + g*x)*Sqrt[1 - c^2*x^2]) + 2
*a*b*g*Cos[2*ArcSin[c*x]] - b^2*g*Sin[2*ArcSin[c*x]]) + 6*b*Sqrt[d]*(-1 + c^2*x^2)*ArcSin[c*x]^2*(-2*a*(2*c^2*
f^2 + g^2) + 8*b*c*f*g*Sqrt[1 - c^2*x^2] + b*g^2*Sin[2*ArcSin[c*x]]))/(24*c^3*Sqrt[d]*Sqrt[1 - c^2*x^2]*Sqrt[d
- c^2*d*x^2])

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Maple [B]  time = 0.387, size = 1181, normalized size = 2.9 \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)^2*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^(1/2),x)

[Out]

-4*a*b*(-d*(c^2*x^2-1))^(1/2)*f*g/d/(c^2*x^2-1)*arcsin(c*x)*x^2-a*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/
c/d/(c^2*x^2-1)*arcsin(c*x)^2*f^2-1/2*a*b*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^3/d/(c^2*x^2-1)*arcsin(c
*x)^2*g^2+1/2*a^2*g^2/c^2/(c^2*d)^(1/2)*arctan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))+a^2*f^2/(c^2*d)^(1/2)*arc
tan((c^2*d)^(1/2)*x/(-c^2*d*x^2+d)^(1/2))-4*a*b*(-d*(c^2*x^2-1))^(1/2)*f*g/c/d/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*
x-4*b^2*(-d*(c^2*x^2-1))^(1/2)*f*g/c/d/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x-1/2*b^2*(-d*(c^2*x^2-1))^(
1/2)*g^2/d/(c^2*x^2-1)*arcsin(c*x)^2*x^3+4*b^2*(-d*(c^2*x^2-1))^(1/2)*f*g/d/(c^2*x^2-1)*x^2-4*b^2*(-d*(c^2*x^2
-1))^(1/2)*f*g/c^2/d/(c^2*x^2-1)-1/4*b^2*(-d*(c^2*x^2-1))^(1/2)*g^2/c^2/d/(c^2*x^2-1)*x+1/4*a*b*(-d*(c^2*x^2-1
))^(1/2)*g^2/c^3/d/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)-a*b*(-d*(c^2*x^2-1))^(1/2)*g^2/d/(c^2*x^2-1)*arcsin(c*x)*x^3
+1/2*b^2*(-d*(c^2*x^2-1))^(1/2)*g^2/c^2/d/(c^2*x^2-1)*arcsin(c*x)^2*x+2*b^2*(-d*(c^2*x^2-1))^(1/2)*f*g/c^2/d/(
c^2*x^2-1)*arcsin(c*x)^2+1/4*b^2*(-d*(c^2*x^2-1))^(1/2)*g^2/c^3/d/(c^2*x^2-1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)-2
*b^2*(-d*(c^2*x^2-1))^(1/2)*f*g/d/(c^2*x^2-1)*arcsin(c*x)^2*x^2-1/3*b^2*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1
/2)/c/d/(c^2*x^2-1)*arcsin(c*x)^3*f^2-1/6*b^2*(-d*(c^2*x^2-1))^(1/2)*(-c^2*x^2+1)^(1/2)/c^3/d/(c^2*x^2-1)*arcs
in(c*x)^3*g^2+4*a*b*(-d*(c^2*x^2-1))^(1/2)*f*g/c^2/d/(c^2*x^2-1)*arcsin(c*x)-1/2*a*b*(-d*(c^2*x^2-1))^(1/2)*g^
2/c/d/(c^2*x^2-1)*(-c^2*x^2+1)^(1/2)*x^2-1/2*a^2*g^2*x/c^2/d*(-c^2*d*x^2+d)^(1/2)-2*a^2*f*g/c^2/d*(-c^2*d*x^2+
d)^(1/2)+1/4*b^2*(-d*(c^2*x^2-1))^(1/2)*g^2/d/(c^2*x^2-1)*x^3-1/2*b^2*(-d*(c^2*x^2-1))^(1/2)*g^2/c/d/(c^2*x^2-
1)*arcsin(c*x)*(-c^2*x^2+1)^(1/2)*x^2+a*b*(-d*(c^2*x^2-1))^(1/2)*g^2/c^2/d/(c^2*x^2-1)*arcsin(c*x)*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((g*x+f)^2*(a+b*arcsin(c*x))^2/(-c^2*d*x^2+d)^(1/2),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 (a^{2} g^{2} x^{2} + 2 \, a^{2} f g x + a^{2} f^{2} +{\left (b^{2} g^{2} x^{2} + 2 \, b^{2} f g x + b^{2} f^{2}\right )} \arcsin \left (c x\right )^{2} + 2 \,{\left (a b g^{2} x^{2} + 2 \, a b f g x + a b f^{2}\right )} \arcsin \left (c x\right )\right )} \sqrt{-c^{2} d x^{2} + d}}{c^{2} d x^{2} - d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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