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

Optimal. Leaf size=460 \[ -\frac {\left (a+b \sin ^{-1}(c x)\right ) \left (d^2 h-d e g+e^2 f\right )}{e^3 (d+e x)}+\frac {(e g-2 d h) \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )}{e^3}+\frac {h x \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {b c \tan ^{-1}\left (\frac {c^2 d x+e}{\sqrt {1-c^2 x^2} \sqrt {c^2 d^2-e^2}}\right ) \left (d^2 h-d e g+e^2 f\right )}{e^3 \sqrt {c^2 d^2-e^2}}-\frac {i b (e g-2 d h) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^3}-\frac {i b (e g-2 d h) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e^3}+\frac {b \sin ^{-1}(c x) (e g-2 d h) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^3}+\frac {b \sin ^{-1}(c x) (e g-2 d h) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e^3}+\frac {b h \sqrt {1-c^2 x^2}}{c e^2}-\frac {i b \sin ^{-1}(c x)^2 (e g-2 d h)}{2 e^3}-\frac {b \sin ^{-1}(c x) (e g-2 d h) \log (d+e x)}{e^3} \]

[Out]

-1/2*I*b*(-2*d*h+e*g)*arcsin(c*x)^2/e^3+h*x*(a+b*arcsin(c*x))/e^2-(d^2*h-d*e*g+e^2*f)*(a+b*arcsin(c*x))/e^3/(e
*x+d)-b*(-2*d*h+e*g)*arcsin(c*x)*ln(e*x+d)/e^3+(-2*d*h+e*g)*(a+b*arcsin(c*x))*ln(e*x+d)/e^3+b*(-2*d*h+e*g)*arc
sin(c*x)*ln(1-I*e*(I*c*x+(-c^2*x^2+1)^(1/2))/(c*d-(c^2*d^2-e^2)^(1/2)))/e^3+b*(-2*d*h+e*g)*arcsin(c*x)*ln(1-I*
e*(I*c*x+(-c^2*x^2+1)^(1/2))/(c*d+(c^2*d^2-e^2)^(1/2)))/e^3-I*b*(-2*d*h+e*g)*polylog(2,I*e*(I*c*x+(-c^2*x^2+1)
^(1/2))/(c*d-(c^2*d^2-e^2)^(1/2)))/e^3-I*b*(-2*d*h+e*g)*polylog(2,I*e*(I*c*x+(-c^2*x^2+1)^(1/2))/(c*d+(c^2*d^2
-e^2)^(1/2)))/e^3+b*c*(d^2*h-d*e*g+e^2*f)*arctan((c^2*d*x+e)/(c^2*d^2-e^2)^(1/2)/(-c^2*x^2+1)^(1/2))/e^3/(c^2*
d^2-e^2)^(1/2)+b*h*(-c^2*x^2+1)^(1/2)/c/e^2

________________________________________________________________________________________

Rubi [A]  time = 0.85, antiderivative size = 460, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 14, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {698, 4753, 12, 6742, 261, 725, 204, 216, 2404, 4741, 4519, 2190, 2279, 2391} \[ -\frac {i b (e g-2 d h) \text {PolyLog}\left (2,\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^3}-\frac {i b (e g-2 d h) \text {PolyLog}\left (2,\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e^3}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \left (d^2 h-d e g+e^2 f\right )}{e^3 (d+e x)}+\frac {(e g-2 d h) \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )}{e^3}+\frac {h x \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {b c \tan ^{-1}\left (\frac {c^2 d x+e}{\sqrt {1-c^2 x^2} \sqrt {c^2 d^2-e^2}}\right ) \left (d^2 h-d e g+e^2 f\right )}{e^3 \sqrt {c^2 d^2-e^2}}+\frac {b \sin ^{-1}(c x) (e g-2 d h) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^3}+\frac {b \sin ^{-1}(c x) (e g-2 d h) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e^3}+\frac {b h \sqrt {1-c^2 x^2}}{c e^2}-\frac {i b \sin ^{-1}(c x)^2 (e g-2 d h)}{2 e^3}-\frac {b \sin ^{-1}(c x) (e g-2 d h) \log (d+e x)}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*h*Sqrt[1 - c^2*x^2])/(c*e^2) - ((I/2)*b*(e*g - 2*d*h)*ArcSin[c*x]^2)/e^3 + (h*x*(a + b*ArcSin[c*x]))/e^2 -
((e^2*f - d*e*g + d^2*h)*(a + b*ArcSin[c*x]))/(e^3*(d + e*x)) + (b*c*(e^2*f - d*e*g + d^2*h)*ArcTan[(e + c^2*d
*x)/(Sqrt[c^2*d^2 - e^2]*Sqrt[1 - c^2*x^2])])/(e^3*Sqrt[c^2*d^2 - e^2]) + (b*(e*g - 2*d*h)*ArcSin[c*x]*Log[1 -
(I*e*E^(I*ArcSin[c*x]))/(c*d - Sqrt[c^2*d^2 - e^2])])/e^3 + (b*(e*g - 2*d*h)*ArcSin[c*x]*Log[1 - (I*e*E^(I*Ar
cSin[c*x]))/(c*d + Sqrt[c^2*d^2 - e^2])])/e^3 - (b*(e*g - 2*d*h)*ArcSin[c*x]*Log[d + e*x])/e^3 + ((e*g - 2*d*h
)*(a + b*ArcSin[c*x])*Log[d + e*x])/e^3 - (I*b*(e*g - 2*d*h)*PolyLog[2, (I*e*E^(I*ArcSin[c*x]))/(c*d - Sqrt[c^
2*d^2 - e^2])])/e^3 - (I*b*(e*g - 2*d*h)*PolyLog[2, (I*e*E^(I*ArcSin[c*x]))/(c*d + Sqrt[c^2*d^2 - e^2])])/e^3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, 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 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 2404

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/Sqrt[(f_) + (g_.)*(x_)^2], x_Symbol] :> With[{u = Int
Hide[1/Sqrt[f + g*x^2], x]}, Simp[u*(a + b*Log[c*(d + e*x)^n]), x] - Dist[b*e*n, Int[SimplifyIntegrand[u/(d +
e*x), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && GtQ[f, 0]

Rule 4519

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2, 2] - I*b
*E^(I*(c + d*x))), x] + Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x))), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

Rule 4741

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Subst[Int[((a + b*x)^n*Cos[x])/
(c*d + e*Sin[x]), x], x, ArcSin[c*x]] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[n, 0]

Rule 4753

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*(Px_)*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> With[{u = IntHide[Px*(d
+ e*x)^m, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 - c^2*x^2], x], x], x]
] /; FreeQ[{a, b, c, d, e, m}, x] && PolynomialQ[Px, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.97, size = 392, normalized size = 0.85 \[ \frac {-\frac {\left (a+b \sin ^{-1}(c x)\right ) \left (d^2 h-d e g+e^2 f\right )}{d+e x}+(e g-2 d h) \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )+e h x \left (a+b \sin ^{-1}(c x)\right )+\frac {b c \tan ^{-1}\left (\frac {c^2 d x+e}{\sqrt {1-c^2 x^2} \sqrt {c^2 d^2-e^2}}\right ) \left (d^2 h-d e g+e^2 f\right )}{\sqrt {c^2 d^2-e^2}}-\frac {1}{2} i b (e g-2 d h) \left (2 \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )+2 \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )+\sin ^{-1}(c x) \left (\sin ^{-1}(c x)+2 i \left (\log \left (1+\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}-c d}\right )+\log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )\right )\right )\right )+\frac {b e h \sqrt {1-c^2 x^2}}{c}-b \sin ^{-1}(c x) (e g-2 d h) \log (d+e x)}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*e*h*Sqrt[1 - c^2*x^2])/c + e*h*x*(a + b*ArcSin[c*x]) - ((e^2*f - d*e*g + d^2*h)*(a + b*ArcSin[c*x]))/(d +
e*x) + (b*c*(e^2*f - d*e*g + d^2*h)*ArcTan[(e + c^2*d*x)/(Sqrt[c^2*d^2 - e^2]*Sqrt[1 - c^2*x^2])])/Sqrt[c^2*d^
2 - e^2] - b*(e*g - 2*d*h)*ArcSin[c*x]*Log[d + e*x] + (e*g - 2*d*h)*(a + b*ArcSin[c*x])*Log[d + e*x] - (I/2)*b
*(e*g - 2*d*h)*(ArcSin[c*x]*(ArcSin[c*x] + (2*I)*(Log[1 + (I*e*E^(I*ArcSin[c*x]))/(-(c*d) + Sqrt[c^2*d^2 - e^2
])] + Log[1 - (I*e*E^(I*ArcSin[c*x]))/(c*d + Sqrt[c^2*d^2 - e^2])])) + 2*PolyLog[2, (I*e*E^(I*ArcSin[c*x]))/(c
*d - Sqrt[c^2*d^2 - e^2])] + 2*PolyLog[2, (I*e*E^(I*ArcSin[c*x]))/(c*d + Sqrt[c^2*d^2 - e^2])]))/e^3

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a h x^{2} + a g x + a f + {\left (b h x^{2} + b g x + b f\right )} \arcsin \left (c x\right )}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.49, size = 1910, normalized size = 4.15 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*a/e^2/(c*e*x+c*d)*d*g-c*b*arcsin(c*x)/e/(c*e*x+c*d)*f+c*b*arcsin(c*x)/e^2/(c*e*x+c*d)*d*g+b*h*(-c^2*x^2+1)^(
1/2)/c/e^2-1/2*I*b*g*arcsin(c*x)^2/e^2-c*b*arcsin(c*x)/e^3/(c*e*x+c*d)*d^2*h+2*c*b/e^3*d^2*h/(c^2*d^2-e^2)^(1/
2)*arctan(1/2*(2*(I*c*x+(-c^2*x^2+1)^(1/2))*e+2*I*d*c)/(c^2*d^2-e^2)^(1/2))+2*b/e*d*h*arcsin(c*x)/(c^2*d^2-e^2
)*ln((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(I*d*c+(-c^2*d^2+e^2)^(1/2)))+2*b/e*d*h*arcsin(
c*x)/(c^2*d^2-e^2)*ln((-I*d*c-(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(-I*d*c+(-c^2*d^2+e^2)^(1/2))
)-2*I*b/e*d*h/(c^2*d^2-e^2)*dilog((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(I*d*c+(-c^2*d^2+e
^2)^(1/2)))-2*I*b/e*d*h/(c^2*d^2-e^2)*dilog((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e-(-c^2*d^2+e^2)^(1/2))/(I*d*c-(
-c^2*d^2+e^2)^(1/2)))+2*c*b/e*f/(c^2*d^2-e^2)^(1/2)*arctan(1/2*(2*(I*c*x+(-c^2*x^2+1)^(1/2))*e+2*I*d*c)/(c^2*d
^2-e^2)^(1/2))+a*h/e^2*x-2*c*b/e^2*d*g/(c^2*d^2-e^2)^(1/2)*arctan(1/2*(2*(I*c*x+(-c^2*x^2+1)^(1/2))*e+2*I*d*c)
/(c^2*d^2-e^2)^(1/2))-2*c^2*b/e^3*d^3*h*arcsin(c*x)/(c^2*d^2-e^2)*ln((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2
*d^2+e^2)^(1/2))/(I*d*c+(-c^2*d^2+e^2)^(1/2)))-2*c^2*b/e^3*d^3*h*arcsin(c*x)/(c^2*d^2-e^2)*ln((-I*d*c-(I*c*x+(
-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(-I*d*c+(-c^2*d^2+e^2)^(1/2)))+2*I*c^2*b/e^3*d^3*h/(c^2*d^2-e^2)*di
log((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e-(-c^2*d^2+e^2)^(1/2))/(I*d*c-(-c^2*d^2+e^2)^(1/2)))+2*I*c^2*b/e^3*d^3*
h/(c^2*d^2-e^2)*dilog((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(I*d*c+(-c^2*d^2+e^2)^(1/2)))-
c*a/e^3/(c*e*x+c*d)*d^2*h+I*b*arcsin(c*x)^2/e^3*d*h+a*g/e^2*ln(c*e*x+c*d)+I*b*g/(c^2*d^2-e^2)*dilog((I*d*c+(I*
c*x+(-c^2*x^2+1)^(1/2))*e-(-c^2*d^2+e^2)^(1/2))/(I*d*c-(-c^2*d^2+e^2)^(1/2)))-b*ln((-I*d*c-(I*c*x+(-c^2*x^2+1)
^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(-I*d*c+(-c^2*d^2+e^2)^(1/2)))/(c^2*d^2-e^2)*arcsin(c*x)*g-b*ln((I*d*c+(I*c*x+
(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(I*d*c+(-c^2*d^2+e^2)^(1/2)))/(c^2*d^2-e^2)*arcsin(c*x)*g-c*a/e/(c
*e*x+c*d)*f+I*b*g/(c^2*d^2-e^2)*dilog((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(I*d*c+(-c^2*d
^2+e^2)^(1/2)))-2*a/e^3*ln(c*e*x+c*d)*d*h+b*arcsin(c*x)*h/e^2*x-I*c^2*b/e^2*g/(c^2*d^2-e^2)*dilog((I*d*c+(I*c*
x+(-c^2*x^2+1)^(1/2))*e-(-c^2*d^2+e^2)^(1/2))/(I*d*c-(-c^2*d^2+e^2)^(1/2)))*d^2+c^2*b/e^2*g*arcsin(c*x)/(c^2*d
^2-e^2)*ln((-I*d*c-(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(-I*d*c+(-c^2*d^2+e^2)^(1/2)))*d^2+c^2*b
/e^2*g*arcsin(c*x)/(c^2*d^2-e^2)*ln((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/(I*d*c+(-c^2*d^2
+e^2)^(1/2)))*d^2-I*c^2*b/e^2*g/(c^2*d^2-e^2)*dilog((I*d*c+(I*c*x+(-c^2*x^2+1)^(1/2))*e+(-c^2*d^2+e^2)^(1/2))/
(I*d*c+(-c^2*d^2+e^2)^(1/2)))*d^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e-c*d>0)', see `assume?` for m
ore details)Is e-c*d positive, negative or zero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asin(c*x))*(f + g*x + h*x**2)/(d + e*x)**2, x)

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