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

Optimal. Leaf size=507 \[ \frac {g \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right ) (f+g x)}-\frac {i c^2 f \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )^{3/2}}+\frac {i c^2 f \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{\sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )^{3/2}}-\frac {b c^2 f \sqrt {1-c^2 x^2} \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )^{3/2}}+\frac {b c^2 f \sqrt {1-c^2 x^2} \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )^{3/2}}-\frac {b c \sqrt {1-c^2 x^2} \log (f+g x)}{\sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.71, antiderivative size = 507, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {4777, 4773, 3324, 3323, 2264, 2190, 2279, 2391, 2668, 31} \[ -\frac {b c^2 f \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )^{3/2}}+\frac {b c^2 f \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{\sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )^{3/2}}+\frac {g \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right ) (f+g x)}-\frac {i c^2 f \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{c f-\sqrt {c^2 f^2-g^2}}\right )}{\sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )^{3/2}}+\frac {i c^2 f \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )}{\sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )^{3/2}}-\frac {b c \sqrt {1-c^2 x^2} \log (f+g x)}{\sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 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 3324

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*(c + d*x)^m*Cos[
e + f*x])/(f*(a^2 - b^2)*(a + b*Sin[e + f*x])), x] + (Dist[a/(a^2 - b^2), Int[(c + d*x)^m/(a + b*Sin[e + f*x])
, x], x] - Dist[(b*d*m)/(f*(a^2 - b^2)), Int[((c + d*x)^(m - 1)*Cos[e + f*x])/(a + b*Sin[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)^2 \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)^2 \sqrt {1-c^2 x^2}} \, dx}{\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 f+g \sin (x))^2} \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}}\\ &=\frac {g \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 f^2-g^2\right ) (f+g x) \sqrt {d-c^2 d x^2}}+\frac {\left (c^2 f \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 )}{\left (c^2 f^2-g^2\right ) \sqrt {d-c^2 d x^2}}-\frac {\left (b c g \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\cos (x)}{c f+g \sin (x)} \, dx,x,\sin ^{-1}(c x)\right )}{\left (c^2 f^2-g^2\right ) \sqrt {d-c^2 d x^2}}\\ &=\frac {g \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 f^2-g^2\right ) (f+g x) \sqrt {d-c^2 d x^2}}-\frac {\left (b c \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{c f+x} \, dx,x,c g x\right )}{\left (c^2 f^2-g^2\right ) \sqrt {d-c^2 d x^2}}+\frac {\left (2 c^2 f \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 )}{\left (c^2 f^2-g^2\right ) \sqrt {d-c^2 d x^2}}\\ &=\frac {g \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 f^2-g^2\right ) (f+g x) \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {1-c^2 x^2} \log (f+g x)}{\left (c^2 f^2-g^2\right ) \sqrt {d-c^2 d x^2}}-\frac {\left (2 i c^2 f 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 )}{\left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}+\frac {\left (2 i c^2 f 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 )}{\left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}\\ &=\frac {g \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 f^2-g^2\right ) (f+g x) \sqrt {d-c^2 d x^2}}-\frac {i c^2 f \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 )}{\left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}+\frac {i c^2 f \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 )}{\left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {1-c^2 x^2} \log (f+g x)}{\left (c^2 f^2-g^2\right ) \sqrt {d-c^2 d x^2}}+\frac {\left (i b c^2 f \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 )}{\left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}-\frac {\left (i b c^2 f \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 )}{\left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}\\ &=\frac {g \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 f^2-g^2\right ) (f+g x) \sqrt {d-c^2 d x^2}}-\frac {i c^2 f \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 )}{\left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}+\frac {i c^2 f \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 )}{\left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {1-c^2 x^2} \log (f+g x)}{\left (c^2 f^2-g^2\right ) \sqrt {d-c^2 d x^2}}+\frac {\left (b c^2 f \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 )}{\left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}-\frac {\left (b c^2 f \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 )}{\left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}\\ &=\frac {g \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 f^2-g^2\right ) (f+g x) \sqrt {d-c^2 d x^2}}-\frac {i c^2 f \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 )}{\left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}+\frac {i c^2 f \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 )}{\left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}-\frac {b c \sqrt {1-c^2 x^2} \log (f+g x)}{\left (c^2 f^2-g^2\right ) \sqrt {d-c^2 d x^2}}-\frac {b c^2 f \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 )}{\left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}+\frac {b c^2 f \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 )}{\left (c^2 f^2-g^2\right )^{3/2} \sqrt {d-c^2 d x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 295, normalized size = 0.58 \[ \frac {c \sqrt {1-c^2 x^2} \left (\frac {c f \left (-i \left (a+b \sin ^{-1}(c x)\right ) \left (\log \left (1+\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}-c f}\right )-\log \left (1-\frac {i g e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 f^2-g^2}+c f}\right )\right )-b \text {Li}_2\left (-\frac {i e^{i \sin ^{-1}(c x)} g}{\sqrt {c^2 f^2-g^2}-c f}\right )+b \text {Li}_2\left (\frac {i e^{i \sin ^{-1}(c x)} g}{c f+\sqrt {c^2 f^2-g^2}}\right )\right )}{\sqrt {c^2 f^2-g^2}}+\frac {g \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{c f+c g x}-b \log (f+g x)\right )}{\sqrt {d-c^2 d x^2} \left (c^2 f^2-g^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.82, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{2} d g^{2} x^{4} + 2 \, c^{2} d f g x^{3} - 2 \, d f g x - d f^{2} + {\left (c^{2} d f^{2} - d g^{2}\right )} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))/(g*x+f)^2/(-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^2*x^4 + 2*c^2*d*f*g*x^3 - 2*d*f*g*x - d*f^2 + (c^2
*d*f^2 - d*g^2)*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.95, size = 1678, normalized size = 3.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/d/(c^2*f^2-g^2)/(x+f/g)*(-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/g*c^2*f/(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))+b*(-d*(c^2*x^2-1))^(1/2)*arcsin(c
*x)/d/(c^2*x^2-1)/(c^2*f^2-g^2)/(g*x+f)*(-c^2*x^2+1)*x*c^2*f+b*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)/d/(c^2*x^2-1
)/(c^2*f^2-g^2)/(g*x+f)*x^3*c^4*f+I*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d/(c^2*x^2-1)/(c^2*f^2-g^2)^2*
c^2*(-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)))*f+b*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)/d/(c^2*x^2-1)/(c^2*f^2-g^2)/(g*x+f)*x^2*c^2*g+I*b*(-d*(c^2*x^2
-1))^(1/2)*arcsin(c*x)/d/(c^2*x^2-1)/(c^2*f^2-g^2)/(g*x+f)*(-c^2*x^2+1)^(1/2)*c*f-b*(-d*(c^2*x^2-1))^(1/2)*arc
sin(c*x)/d/(c^2*x^2-1)/(c^2*f^2-g^2)/(g*x+f)*x*c^2*f-b*(-d*(c^2*x^2-1))^(1/2)*arcsin(c*x)/d/(c^2*x^2-1)/(c^2*f
^2-g^2)/(g*x+f)*g+b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d/(c^2*x^2-1)/(c^2*f^2-g^2)^2*c^2*(-c^2*f^2+g^2)
^(1/2)*f*arcsin(c*x)*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
)))-b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d/(c^2*x^2-1)/(c^2*f^2-g^2)^2*c^2*(-c^2*f^2+g^2)^(1/2)*f*arcsi
n(c*x)*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)))+I*b*(-d*(c^2
*x^2-1))^(1/2)*arcsin(c*x)/d/(c^2*x^2-1)/(c^2*f^2-g^2)/(g*x+f)*(-c^2*x^2+1)^(1/2)*x*c*g-I*b*(-c^2*x^2+1)^(1/2)
*(-d*(c^2*x^2-1))^(1/2)/d/(c^2*x^2-1)/(c^2*f^2-g^2)^2*c^2*(-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)))*f+b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)
/d/(c^2*x^2-1)/(c^2*f^2-g^2)^2*c^3*ln((I*c*x+(-c^2*x^2+1)^(1/2))^2*g+2*I*c*f*(I*c*x+(-c^2*x^2+1)^(1/2))-g)*f^2
-2*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d/(c^2*x^2-1)/(c^2*f^2-g^2)^2*c^3*ln(I*c*x+(-c^2*x^2+1)^(1/2))*
f^2-b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d/(c^2*x^2-1)/(c^2*f^2-g^2)^2*c*ln((I*c*x+(-c^2*x^2+1)^(1/2))^
2*g+2*I*c*f*(I*c*x+(-c^2*x^2+1)^(1/2))-g)*g^2+2*b*(-c^2*x^2+1)^(1/2)*(-d*(c^2*x^2-1))^(1/2)/d/(c^2*x^2-1)/(c^2
*f^2-g^2)^2*c*ln(I*c*x+(-c^2*x^2+1)^(1/2))*g^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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