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

Optimal. Leaf size=654 \[ \frac {i g^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{d \sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )^{3/2}}-\frac {i g^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{d \sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )^{3/2}}+\frac {\sqrt {1-c^2 x^2} \tan \left (\frac {1}{2} \sin ^{-1}(c x)+\frac {\pi }{4}\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 d \sqrt {d-c^2 d x^2} (c f+g)}-\frac {\sqrt {1-c^2 x^2} \cot \left (\frac {1}{2} \sin ^{-1}(c x)+\frac {\pi }{4}\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 d \sqrt {d-c^2 d x^2} (c f-g)}+\frac {b g^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{d \sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )^{3/2}}-\frac {b g^2 \sqrt {1-c^2 x^2} \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{d \sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )^{3/2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)+\frac {\pi }{4}\right )\right )}{d \sqrt {d-c^2 d x^2} (c f-g)}+\frac {b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)+\frac {\pi }{4}\right )\right )}{d \sqrt {d-c^2 d x^2} (c f+g)} \]

[Out]

-1/2*(a+b*arcsin(c*x))*cot(1/4*Pi+1/2*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/d/(c*f-g)/(-c^2*d*x^2+d)^(1/2)+b*ln(cos(
1/4*Pi+1/2*arcsin(c*x)))*(-c^2*x^2+1)^(1/2)/d/(c*f+g)/(-c^2*d*x^2+d)^(1/2)+b*ln(sin(1/4*Pi+1/2*arcsin(c*x)))*(
-c^2*x^2+1)^(1/2)/d/(c*f-g)/(-c^2*d*x^2+d)^(1/2)+I*g^2*(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)/d/(c^2*f^2-g^2)^(3/2)/(-c^2*d*x^2+d)^(1/2)-I*g^2*(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)/d/(c^2*f^2-g^2)^(3/2)/(-c^
2*d*x^2+d)^(1/2)+b*g^2*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)/
d/(c^2*f^2-g^2)^(3/2)/(-c^2*d*x^2+d)^(1/2)-b*g^2*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)/d/(c^2*f^2-g^2)^(3/2)/(-c^2*d*x^2+d)^(1/2)+1/2*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)*
tan(1/4*Pi+1/2*arcsin(c*x))/d/(c*f+g)/(-c^2*d*x^2+d)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 1.16, antiderivative size = 654, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 11, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {4777, 4775, 4773, 3318, 4184, 3475, 3323, 2264, 2190, 2279, 2391} \[ \frac {b g^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{d \sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )^{3/2}}-\frac {b g^2 \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{d \sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )^{3/2}}+\frac {i g^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{d \sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )^{3/2}}-\frac {i g^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{d \sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )^{3/2}}+\frac {\sqrt {1-c^2 x^2} \tan \left (\frac {1}{2} \sin ^{-1}(c x)+\frac {\pi }{4}\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 d \sqrt {d-c^2 d x^2} (c f+g)}-\frac {\sqrt {1-c^2 x^2} \cot \left (\frac {1}{2} \sin ^{-1}(c x)+\frac {\pi }{4}\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 d \sqrt {d-c^2 d x^2} (c f-g)}+\frac {b \sqrt {1-c^2 x^2} \log \left (\sin \left (\frac {1}{2} \sin ^{-1}(c x)+\frac {\pi }{4}\right )\right )}{d \sqrt {d-c^2 d x^2} (c f-g)}+\frac {b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \sin ^{-1}(c x)+\frac {\pi }{4}\right )\right )}{d \sqrt {d-c^2 d x^2} (c f+g)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])*Cot[Pi/4 + ArcSin[c*x]/2])/(2*d*(c*f - g)*Sqrt[d - c^2*d*x^2]) + (I*g^
2*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])])/(d*(c^2*
f^2 - g^2)^(3/2)*Sqrt[d - c^2*d*x^2]) - (I*g^2*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])])/(d*(c^2*f^2 - g^2)^(3/2)*Sqrt[d - c^2*d*x^2]) + (b*Sqrt[1 - c^2*x^2]*Log[
Cos[Pi/4 + ArcSin[c*x]/2]])/(d*(c*f + g)*Sqrt[d - c^2*d*x^2]) + (b*Sqrt[1 - c^2*x^2]*Log[Sin[Pi/4 + ArcSin[c*x
]/2]])/(d*(c*f - g)*Sqrt[d - c^2*d*x^2]) + (b*g^2*Sqrt[1 - c^2*x^2]*PolyLog[2, (I*E^(I*ArcSin[c*x])*g)/(c*f -
Sqrt[c^2*f^2 - g^2])])/(d*(c^2*f^2 - g^2)^(3/2)*Sqrt[d - c^2*d*x^2]) - (b*g^2*Sqrt[1 - c^2*x^2]*PolyLog[2, (I*
E^(I*ArcSin[c*x])*g)/(c*f + Sqrt[c^2*f^2 - g^2])])/(d*(c^2*f^2 - g^2)^(3/2)*Sqrt[d - c^2*d*x^2]) + (Sqrt[1 - c
^2*x^2]*(a + b*ArcSin[c*x])*Tan[Pi/4 + ArcSin[c*x]/2])/(2*d*(c*f + g)*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 3318

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1*(e + (Pi*a)/(2*b)))/2 + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 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 3475

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

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[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 4775

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :
> Int[ExpandIntegrand[(a + b*ArcSin[c*x])^n/Sqrt[d + e*x^2], (f + g*x)^m*(d + e*x^2)^(p + 1/2), x], x] /; Free
Q[{a, b, c, d, e, f, g}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && ILtQ[p + 1/2, 0] && GtQ[d, 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) \left (d-c^2 d x^2\right )^{3/2}} \, dx &=\frac {\sqrt {1-c^2 x^2} \int \frac {a+b \sin ^{-1}(c x)}{(f+g x) \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 {c \left (a+b \sin ^{-1}(c x)\right )}{2 (c f+g) (-1+c x) \sqrt {1-c^2 x^2}}+\frac {c \left (a+b \sin ^{-1}(c x)\right )}{2 (c f-g) (1+c x) \sqrt {1-c^2 x^2}}+\frac {g^2 \left (a+b \sin ^{-1}(c x)\right )}{(-c f+g) (c f+g) (f+g x) \sqrt {1-c^2 x^2}}\right ) \, dx}{d \sqrt {d-c^2 d x^2}}\\ &=\frac {\left (c \sqrt {1-c^2 x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{(1+c x) \sqrt {1-c^2 x^2}} \, dx}{2 d (c f-g) \sqrt {d-c^2 d x^2}}-\frac {\left (c \sqrt {1-c^2 x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{(-1+c x) \sqrt {1-c^2 x^2}} \, dx}{2 d (c f+g) \sqrt {d-c^2 d x^2}}+\frac {\left (g^2 \sqrt {1-c^2 x^2}\right ) \int \frac {a+b \sin ^{-1}(c x)}{(f+g x) \sqrt {1-c^2 x^2}} \, dx}{d (-c f+g) (c f+g) \sqrt {d-c^2 d x^2}}\\ &=\frac {\left (c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {a+b x}{c+c \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )}{2 d (c f-g) \sqrt {d-c^2 d x^2}}-\frac {\left (c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {a+b x}{-c+c \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )}{2 d (c f+g) \sqrt {d-c^2 d x^2}}+\frac {\left (g^2 \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {a+b x}{c f+g \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )}{d (-c f+g) (c f+g) \sqrt {d-c^2 d x^2}}\\ &=\frac {\sqrt {1-c^2 x^2} \operatorname {Subst}\left (\int (a+b x) \csc ^2\left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 d (c f-g) \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} \operatorname {Subst}\left (\int (a+b x) \csc ^2\left (\frac {\pi }{4}-\frac {x}{2}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{4 d (c f+g) \sqrt {d-c^2 d x^2}}+\frac {\left (2 g^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 )}{d (-c f+g) (c f+g) \sqrt {d-c^2 d x^2}}\\ &=-\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \cot \left (\frac {\pi }{4}+\frac {1}{2} \sin ^{-1}(c x)\right )}{2 d (c f-g) \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} \sin ^{-1}(c x)\right )}{2 d (c f+g) \sqrt {d-c^2 d x^2}}+\frac {\left (b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \cot \left (\frac {\pi }{4}+\frac {x}{2}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{2 d (c f-g) \sqrt {d-c^2 d x^2}}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \cot \left (\frac {\pi }{4}-\frac {x}{2}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{2 d (c f+g) \sqrt {d-c^2 d x^2}}-\frac {\left (2 i g^3 \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 )}{d (-c f+g) (c f+g) \sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}+\frac {\left (2 i g^3 \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 )}{d (-c f+g) (c f+g) \sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}\\ &=-\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \cot \left (\frac {\pi }{4}+\frac {1}{2} \sin ^{-1}(c x)\right )}{2 d (c f-g) \sqrt {d-c^2 d x^2}}+\frac {i g^2 \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 )}{d \left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}-\frac {i g^2 \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 )}{d \left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {\pi }{4}+\frac {1}{2} \sin ^{-1}(c x)\right )\right )}{d (c f+g) \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (\sin \left (\frac {\pi }{4}+\frac {1}{2} \sin ^{-1}(c x)\right )\right )}{d (c f-g) \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} \sin ^{-1}(c x)\right )}{2 d (c f+g) \sqrt {d-c^2 d x^2}}+\frac {\left (i b g^2 \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 )}{d (-c f+g) (c f+g) \sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}-\frac {\left (i b g^2 \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 )}{d (-c f+g) (c f+g) \sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}\\ &=-\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \cot \left (\frac {\pi }{4}+\frac {1}{2} \sin ^{-1}(c x)\right )}{2 d (c f-g) \sqrt {d-c^2 d x^2}}+\frac {i g^2 \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 )}{d \left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}-\frac {i g^2 \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 )}{d \left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {\pi }{4}+\frac {1}{2} \sin ^{-1}(c x)\right )\right )}{d (c f+g) \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (\sin \left (\frac {\pi }{4}+\frac {1}{2} \sin ^{-1}(c x)\right )\right )}{d (c f-g) \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} \sin ^{-1}(c x)\right )}{2 d (c f+g) \sqrt {d-c^2 d x^2}}+\frac {\left (b g^2 \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 )}{d (-c f+g) (c f+g) \sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}-\frac {\left (b g^2 \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 )}{d (-c f+g) (c f+g) \sqrt {c^2 f^2-g^2} \sqrt {d-c^2 d x^2}}\\ &=-\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \cot \left (\frac {\pi }{4}+\frac {1}{2} \sin ^{-1}(c x)\right )}{2 d (c f-g) \sqrt {d-c^2 d x^2}}+\frac {i g^2 \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 )}{d \left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}-\frac {i g^2 \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 )}{d \left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {\pi }{4}+\frac {1}{2} \sin ^{-1}(c x)\right )\right )}{d (c f+g) \sqrt {d-c^2 d x^2}}+\frac {b \sqrt {1-c^2 x^2} \log \left (\sin \left (\frac {\pi }{4}+\frac {1}{2} \sin ^{-1}(c x)\right )\right )}{d (c f-g) \sqrt {d-c^2 d x^2}}+\frac {b g^2 \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 )}{d \left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}-\frac {b g^2 \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 )}{d \left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}+\frac {\sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} \sin ^{-1}(c x)\right )}{2 d (c f+g) \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 2.04, size = 359, normalized size = 0.55 \[ \frac {\sqrt {1-c^2 x^2} \left (\frac {2 g^2 \left (i \left (a+b \sin ^{-1}(c x)\right ) \left (\log \left (1+\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}-c f}\right )-\log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )\right )+b \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(c x)} g}{\sqrt {c^2 f^2-g^2}-c f}\right )-b \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )\right )}{(c f-g) (c f+g) \sqrt {c^2 f^2-g^2}}+\frac {2 b \log \left (\sin \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )-\cot \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right ) \left (a+b \sin ^{-1}(c x)\right )}{c f-g}+\frac {\tan \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right ) \left (a+b \sin ^{-1}(c x)\right )+2 b \log \left (\cos \left (\frac {1}{4} \left (2 \sin ^{-1}(c x)+\pi \right )\right )\right )}{c f+g}\right )}{2 d \sqrt {d-c^2 d x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{4} d^{2} g x^{5} + c^{4} d^{2} f x^{4} - 2 \, c^{2} d^{2} g x^{3} - 2 \, c^{2} d^{2} f x^{2} + d^{2} g x + d^{2} f}, x\right ) \]

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)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(-c^2*d*x^2 + d)*(b*arcsin(c*x) + a)/(c^4*d^2*g*x^5 + c^4*d^2*f*x^4 - 2*c^2*d^2*g*x^3 - 2*c^2*d^2
*f*x^2 + d^2*g*x + d^2*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)^(3/2),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 [B]  time = 0.82, size = 1519, normalized size = 2.32 \[ \text {result too large to display} \]

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)^(3/2),x)

[Out]

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

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

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)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{\left (f+g\,x\right )\,{\left (d-c^2\,d\,x^2\right )}^{3/2}} \,d x \]

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)^(3/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {asin}{\left (c x \right )}}{\left (- d \left (c x - 1\right ) \left (c x + 1\right )\right )^{\frac {3}{2}} \left (f + g x\right )}\, dx \]

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)**(3/2),x)

[Out]

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

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