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

Optimal. Leaf size=380 \[ -\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}+\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}-\frac {b \sqrt {1-c^2 x^2} \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}+\frac {b \sqrt {1-c^2 x^2} \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}} \]

[Out]

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

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Rubi [A]  time = 0.61, antiderivative size = 380, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4777, 4773, 3323, 2264, 2190, 2279, 2391} \[ -\frac {b \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}+\frac {b \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}-\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}}+\frac {i \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*Log[1 - (I*E^(I*ArcSin[c*x])*g)/(c*f - Sqrt[c^2*f^2 - g^2])])/(Sqr
t[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])*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]) - (b*Sqrt[1 - c^2*x^2]*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]) + (b*Sqrt[1 - c
^2*x^2]*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])

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 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 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 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 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 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]

Rubi steps

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

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Mathematica [A]  time = 0.19, size = 232, normalized size = 0.61 \[ \frac {\sqrt {1-c^2 x^2} \left (-i \left (a+b \sin ^{-1}(c x)\right ) \left (\log \left (1+\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}-c f}\right )-\log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )\right )-b \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(c x)} g}{\sqrt {c^2 f^2-g^2}-c f}\right )+b \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )\right )}{\sqrt {d-c^2 d x^2} \sqrt {c^2 f^2-g^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{2} d g x^{3} + c^{2} d f x^{2} - d g x - d f}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep^4-1)]index.cc index_m i_lex_is_greater Error: Bad Argument Value

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maple [A]  time = 0.39, size = 502, normalized size = 1.32 \[ -\frac {a \ln \left (\frac {-\frac {2 d \left (c^{2} f^{2}-g^{2}\right )}{g^{2}}+\frac {2 c^{2} d f \left (x +\frac {f}{g}\right )}{g}+2 \sqrt {-\frac {d \left (c^{2} f^{2}-g^{2}\right )}{g^{2}}}\, \sqrt {-c^{2} d \left (x +\frac {f}{g}\right )^{2}+\frac {2 c^{2} d f \left (x +\frac {f}{g}\right )}{g}-\frac {d \left (c^{2} f^{2}-g^{2}\right )}{g^{2}}}}{x +\frac {f}{g}}\right )}{g \sqrt {-\frac {d \left (c^{2} f^{2}-g^{2}\right )}{g^{2}}}}-\frac {i b \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \sqrt {-c^{2} f^{2}+g^{2}}\, \sqrt {-c^{2} x^{2}+1}\, \left (i \ln \left (\frac {-i c f -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) g +\sqrt {-c^{2} f^{2}+g^{2}}}{-i c f +\sqrt {-c^{2} f^{2}+g^{2}}}\right ) \arcsin \left (c x \right )-i \ln \left (\frac {i c f +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) g +\sqrt {-c^{2} f^{2}+g^{2}}}{i c f +\sqrt {-c^{2} f^{2}+g^{2}}}\right ) \arcsin \left (c x \right )+\dilog \left (\frac {-i c f -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) g +\sqrt {-c^{2} f^{2}+g^{2}}}{-i c f +\sqrt {-c^{2} f^{2}+g^{2}}}\right )-\dilog \left (\frac {i c f +\left (i c x +\sqrt {-c^{2} x^{2}+1}\right ) g +\sqrt {-c^{2} f^{2}+g^{2}}}{i c f +\sqrt {-c^{2} f^{2}+g^{2}}}\right )\right )}{d \left (c^{2} x^{2}-1\right ) \left (c^{2} f^{2}-g^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \arcsin \left (c x\right ) + a}{\sqrt {-c^{2} d x^{2} + d} {\left (g x + f\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{\left (f+g\,x\right )\,\sqrt {d-c^2\,d\,x^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asin}{\left (c x \right )}}{\sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )} \left (f + g x\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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