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

Optimal. Leaf size=488 \[ -\frac {\left (a+b \sin ^{-1}(c x)\right ) \left (d^2 h-d e g+e^2 f\right )}{2 e^3 (d+e x)^2}-\frac {(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac {h \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )}{e^3}+\frac {b c \sqrt {1-c^2 x^2} \left (d^2 h-d e g+e^2 f\right )}{2 e^2 \left (c^2 d^2-e^2\right ) (d+e x)}-\frac {b c \tan ^{-1}\left (\frac {c^2 d x+e}{\sqrt {1-c^2 x^2} \sqrt {c^2 d^2-e^2}}\right ) \left (2 e^2 (e g-2 d h)-c^2 d \left (-3 d^2 h+d e g+e^2 f\right )\right )}{2 e^3 \left (c^2 d^2-e^2\right )^{3/2}}-\frac {i b h \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^3}-\frac {i b h \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e^3}+\frac {b h \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^3}+\frac {b h \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e^3}-\frac {b h \sin ^{-1}(c x) \log (d+e x)}{e^3}-\frac {i b h \sin ^{-1}(c x)^2}{2 e^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.26, antiderivative size = 488, 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, 807, 725, 204, 216, 2404, 4741, 4519, 2190, 2279, 2391} \[ -\frac {i b h \text {PolyLog}\left (2,\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^3}-\frac {i b h \text {PolyLog}\left (2,\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e^3}-\frac {\left (a+b \sin ^{-1}(c x)\right ) \left (d^2 h-d e g+e^2 f\right )}{2 e^3 (d+e x)^2}-\frac {(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac {h \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )}{e^3}+\frac {b c \sqrt {1-c^2 x^2} \left (d^2 h-d e g+e^2 f\right )}{2 e^2 \left (c^2 d^2-e^2\right ) (d+e x)}-\frac {b c \tan ^{-1}\left (\frac {c^2 d x+e}{\sqrt {1-c^2 x^2} \sqrt {c^2 d^2-e^2}}\right ) \left (2 e^2 (e g-2 d h)-c^2 d \left (-3 d^2 h+d e g+e^2 f\right )\right )}{2 e^3 \left (c^2 d^2-e^2\right )^{3/2}}+\frac {b h \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^3}+\frac {b h \sin ^{-1}(c x) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e^3}-\frac {b h \sin ^{-1}(c x) \log (d+e x)}{e^3}-\frac {i b h \sin ^{-1}(c x)^2}{2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*(e^2*f - d*e*g + d^2*h)*Sqrt[1 - c^2*x^2])/(2*e^2*(c^2*d^2 - e^2)*(d + e*x)) - ((I/2)*b*h*ArcSin[c*x]^2)/
e^3 - ((e^2*f - d*e*g + d^2*h)*(a + b*ArcSin[c*x]))/(2*e^3*(d + e*x)^2) - ((e*g - 2*d*h)*(a + b*ArcSin[c*x]))/
(e^3*(d + e*x)) - (b*c*(2*e^2*(e*g - 2*d*h) - c^2*d*(e^2*f + d*e*g - 3*d^2*h))*ArcTan[(e + c^2*d*x)/(Sqrt[c^2*
d^2 - e^2]*Sqrt[1 - c^2*x^2])])/(2*e^3*(c^2*d^2 - e^2)^(3/2)) + (b*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*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*h*ArcSin[c*x]*Log[d + e*x])/e^3 + (h*(a + b*ArcSin[c*x])*Log[d + e*x])/e^3 - (I*b*h*PolyLog[2
, (I*e*E^(I*ArcSin[c*x]))/(c*d - Sqrt[c^2*d^2 - e^2])])/e^3 - (I*b*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 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 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

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)^3} \, dx &=-\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 e^3 (d+e x)^2}-\frac {(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac {h \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-(b c) \int \frac {3 d^2 h-e^2 (f+2 g x)-d e (g-4 h x)+2 h (d+e x)^2 \log (d+e x)}{2 e^3 (d+e x)^2 \sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 e^3 (d+e x)^2}-\frac {(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac {h \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-\frac {(b c) \int \frac {3 d^2 h-e^2 (f+2 g x)-d e (g-4 h x)+2 h (d+e x)^2 \log (d+e x)}{(d+e x)^2 \sqrt {1-c^2 x^2}} \, dx}{2 e^3}\\ &=-\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 e^3 (d+e x)^2}-\frac {(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac {h \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-\frac {(b c) \int \left (\frac {-e^2 f-d e g+3 d^2 h-2 e (e g-2 d h) x}{(d+e x)^2 \sqrt {1-c^2 x^2}}+\frac {2 h \log (d+e x)}{\sqrt {1-c^2 x^2}}\right ) \, dx}{2 e^3}\\ &=-\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 e^3 (d+e x)^2}-\frac {(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}+\frac {h \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-\frac {(b c) \int \frac {-e^2 f-d e g+3 d^2 h-2 e (e g-2 d h) x}{(d+e x)^2 \sqrt {1-c^2 x^2}} \, dx}{2 e^3}-\frac {(b c h) \int \frac {\log (d+e x)}{\sqrt {1-c^2 x^2}} \, dx}{e^3}\\ &=\frac {b c \left (e^2 f-d e g+d^2 h\right ) \sqrt {1-c^2 x^2}}{2 e^2 \left (c^2 d^2-e^2\right ) (d+e x)}-\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 e^3 (d+e x)^2}-\frac {(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}-\frac {b h \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac {h \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}+\frac {(b c h) \int \frac {\sin ^{-1}(c x)}{c d+c e x} \, dx}{e^2}-\frac {\left (b c \left (2 e^2 (e g-2 d h)-c^2 d \left (e^2 f+d e g-3 d^2 h\right )\right )\right ) \int \frac {1}{(d+e x) \sqrt {1-c^2 x^2}} \, dx}{2 e^3 \left (c^2 d^2-e^2\right )}\\ &=\frac {b c \left (e^2 f-d e g+d^2 h\right ) \sqrt {1-c^2 x^2}}{2 e^2 \left (c^2 d^2-e^2\right ) (d+e x)}-\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 e^3 (d+e x)^2}-\frac {(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}-\frac {b h \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac {h \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}+\frac {(b c 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 {\left (b c \left (2 e^2 (e g-2 d h)-c^2 d \left (e^2 f+d e g-3 d^2 h\right )\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 )}{2 e^3 \left (c^2 d^2-e^2\right )}\\ &=\frac {b c \left (e^2 f-d e g+d^2 h\right ) \sqrt {1-c^2 x^2}}{2 e^2 \left (c^2 d^2-e^2\right ) (d+e x)}-\frac {i b h \sin ^{-1}(c x)^2}{2 e^3}-\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 e^3 (d+e x)^2}-\frac {(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}-\frac {b c \left (2 e^2 (e g-2 d h)-c^2 d \left (e^2 f+d e g-3 d^2 h\right )\right ) \tan ^{-1}\left (\frac {e+c^2 d x}{\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}}\right )}{2 e^3 \left (c^2 d^2-e^2\right )^{3/2}}-\frac {b h \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac {h \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}+\frac {(b c 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 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 \left (e^2 f-d e g+d^2 h\right ) \sqrt {1-c^2 x^2}}{2 e^2 \left (c^2 d^2-e^2\right ) (d+e x)}-\frac {i b h \sin ^{-1}(c x)^2}{2 e^3}-\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 e^3 (d+e x)^2}-\frac {(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}-\frac {b c \left (2 e^2 (e g-2 d h)-c^2 d \left (e^2 f+d e g-3 d^2 h\right )\right ) \tan ^{-1}\left (\frac {e+c^2 d x}{\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}}\right )}{2 e^3 \left (c^2 d^2-e^2\right )^{3/2}}+\frac {b 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 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 h \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac {h \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-\frac {(b 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) \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 c \left (e^2 f-d e g+d^2 h\right ) \sqrt {1-c^2 x^2}}{2 e^2 \left (c^2 d^2-e^2\right ) (d+e x)}-\frac {i b h \sin ^{-1}(c x)^2}{2 e^3}-\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 e^3 (d+e x)^2}-\frac {(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}-\frac {b c \left (2 e^2 (e g-2 d h)-c^2 d \left (e^2 f+d e g-3 d^2 h\right )\right ) \tan ^{-1}\left (\frac {e+c^2 d x}{\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}}\right )}{2 e^3 \left (c^2 d^2-e^2\right )^{3/2}}+\frac {b 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 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 h \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac {h \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}+\frac {(i b 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 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 c \left (e^2 f-d e g+d^2 h\right ) \sqrt {1-c^2 x^2}}{2 e^2 \left (c^2 d^2-e^2\right ) (d+e x)}-\frac {i b h \sin ^{-1}(c x)^2}{2 e^3}-\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right )}{2 e^3 (d+e x)^2}-\frac {(e g-2 d h) \left (a+b \sin ^{-1}(c x)\right )}{e^3 (d+e x)}-\frac {b c \left (2 e^2 (e g-2 d h)-c^2 d \left (e^2 f+d e g-3 d^2 h\right )\right ) \tan ^{-1}\left (\frac {e+c^2 d x}{\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}}\right )}{2 e^3 \left (c^2 d^2-e^2\right )^{3/2}}+\frac {b 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 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 h \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac {h \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-\frac {i b 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 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 [C]  time = 6.69, size = 996, normalized size = 2.04 \[ \frac {2 a d h-a e g}{e^3 (d+e x)}+b f \left (-\frac {c \sqrt {\frac {-d-\sqrt {\frac {1}{c^2}} e}{d+e x}+1} \sqrt {\frac {\sqrt {\frac {1}{c^2}} e-d}{d+e x}+1} F_1\left (2;\frac {1}{2},\frac {1}{2};3;-\frac {\sqrt {\frac {1}{c^2}} e-d}{d+e x},-\frac {-d-\sqrt {\frac {1}{c^2}} e}{d+e x}\right )}{4 e^2 (d+e x) \sqrt {1-c^2 x^2}}-\frac {\sin ^{-1}(c x)}{2 e (d+e x)^2}\right )+\frac {a h \log (d+e x)}{e^3}+b g \left (\frac {\frac {c \tan ^{-1}\left (\frac {d x c^2+e}{\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}}\right )}{\sqrt {c^2 d^2-e^2}}-\frac {\sin ^{-1}(c x)}{d+e x}}{e^2}-\frac {d \left (-\frac {i d \left (\log \left (\frac {e^2 \sqrt {c^2 d^2-e^2} \left (i d x c^2+i e+\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}\right )}{c^3 d (d+e x)}\right )+\log (4)\right ) c^3}{(c d-e) e (c d+e) \sqrt {c^2 d^2-e^2}}+\frac {\sqrt {1-c^2 x^2} c}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac {\sin ^{-1}(c x)}{e (d+e x)^2}\right )}{2 e}\right )+b h \left (\frac {\left (-\frac {i d \left (\log \left (\frac {e^2 \sqrt {c^2 d^2-e^2} \left (i d x c^2+i e+\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}\right )}{c^3 d (d+e x)}\right )+\log (4)\right ) c^3}{(c d-e) e (c d+e) \sqrt {c^2 d^2-e^2}}+\frac {\sqrt {1-c^2 x^2} c}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac {\sin ^{-1}(c x)}{e (d+e x)^2}\right ) d^2}{2 e^2}-\frac {2 \left (\frac {c \tan ^{-1}\left (\frac {d x c^2+e}{\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}}\right )}{\sqrt {c^2 d^2-e^2}}-\frac {\sin ^{-1}(c x)}{d+e x}\right ) d}{e^3}+\frac {-\frac {i \sin ^{-1}(c x)^2}{2 e}+\frac {\log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right ) \sin ^{-1}(c x)}{e}+\frac {\log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right ) \sin ^{-1}(c x)}{e}-\frac {i \text {Li}_2\left (-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}-c d}\right )}{e}-\frac {i \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e}}{e^2}\right )+\frac {-a h d^2+a e g d-a e^2 f}{2 e^3 (d+e x)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(a*e^2*f) + a*d*e*g - a*d^2*h)/(2*e^3*(d + e*x)^2) + (-(a*e*g) + 2*a*d*h)/(e^3*(d + e*x)) + b*f*(-1/4*(c*Sqr
t[1 + (-d - Sqrt[c^(-2)]*e)/(d + e*x)]*Sqrt[1 + (-d + Sqrt[c^(-2)]*e)/(d + e*x)]*AppellF1[2, 1/2, 1/2, 3, -((-
d + Sqrt[c^(-2)]*e)/(d + e*x)), -((-d - Sqrt[c^(-2)]*e)/(d + e*x))])/(e^2*(d + e*x)*Sqrt[1 - c^2*x^2]) - ArcSi
n[c*x]/(2*e*(d + e*x)^2)) + (a*h*Log[d + e*x])/e^3 + b*g*((-(ArcSin[c*x]/(d + e*x)) + (c*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])/e^2 - (d*((c*Sqrt[1 - c^2*x^2])/((c^2*d^2 - e^2
)*(d + e*x)) - ArcSin[c*x]/(e*(d + e*x)^2) - (I*c^3*d*(Log[4] + Log[(e^2*Sqrt[c^2*d^2 - e^2]*(I*e + I*c^2*d*x
+ Sqrt[c^2*d^2 - e^2]*Sqrt[1 - c^2*x^2]))/(c^3*d*(d + e*x))]))/((c*d - e)*e*(c*d + e)*Sqrt[c^2*d^2 - e^2])))/(
2*e)) + b*h*((-2*d*(-(ArcSin[c*x]/(d + e*x)) + (c*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]))/e^3 + (d^2*((c*Sqrt[1 - c^2*x^2])/((c^2*d^2 - e^2)*(d + e*x)) - ArcSin[c*x]/(e*(d + e
*x)^2) - (I*c^3*d*(Log[4] + Log[(e^2*Sqrt[c^2*d^2 - e^2]*(I*e + I*c^2*d*x + Sqrt[c^2*d^2 - e^2]*Sqrt[1 - c^2*x
^2]))/(c^3*d*(d + e*x))]))/((c*d - e)*e*(c*d + e)*Sqrt[c^2*d^2 - e^2])))/(2*e^2) + (((-1/2*I)*ArcSin[c*x]^2)/e
+ (ArcSin[c*x]*Log[1 - (I*e*E^(I*ArcSin[c*x]))/(c*d - Sqrt[c^2*d^2 - e^2])])/e + (ArcSin[c*x]*Log[1 - (I*e*E^
(I*ArcSin[c*x]))/(c*d + Sqrt[c^2*d^2 - e^2])])/e - (I*PolyLog[2, ((-I)*e*E^(I*ArcSin[c*x]))/(-(c*d) + Sqrt[c^2
*d^2 - e^2])])/e - (I*PolyLog[2, (I*e*E^(I*ArcSin[c*x]))/(c*d + Sqrt[c^2*d^2 - e^2])])/e)/e^2)

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fricas [F]  time = 2.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a h x^{2} + a g x + a f + {\left (b h x^{2} + b g x + b f\right )} \arcsin \left (c x\right )}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \]

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)^3,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^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^
3), x)

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

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)^3,x, algorithm="giac")

[Out]

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

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maple [B]  time = 2.55, size = 2697, normalized size = 5.53 \[ \text {result too large to display} \]

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)^3,x)

[Out]

-2*c^2*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2*arcsin(c*x)*x*d*h+1/2*c^3*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2*e*(-c^2*x^2+1)^(1
/2)*x*f+c^2*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2*e*arcsin(c*x)*x*g-2*c^2*b/e/(c^2*d^2-e^2)^2*h*arcsin(c*x)*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^2-2*c^2*b/e/(c^2*d^2-e^2)
^2*d^2*h*arcsin(c*x)*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))
)-1/2*I*c^4*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2*e*x^2*f-I*c^4*b/e^3/(c^2*d^2-e^2)^2*d^4*h*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*c^2*b/e/(c^2*d^2-e^2)^2*h*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-I*c^2*b/e^3/(c^2*d^2-e^2
)*d^2*h*arcsin(c*x)^2-1/2*I*c^4*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2/e^3*d^4*h+1/2*I*c^4*b/(c^2*d^2-e^2)/(c*e*x+c*d)^
2/e^2*d^3*g-1/2*I*c^4*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2/e*d^2*f+2*I*c^2*b/e/(c^2*d^2-e^2)^2*h*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-I*c^4*b/e^3/(c^2*d^2-e^2)^2*d^4*
h*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)))+1/2*I*c^4*b/(c
^2*d^2-e^2)/(c*e*x+c*d)^2*x^2*d*g-I*c^4*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2*x*d*f+3/2*c^4*b/(c^2*d^2-e^2)/(c*e*x+c*d
)^2/e^3*arcsin(c*x)*d^4*h-1/2*c^4*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2/e^2*arcsin(c*x)*d^3*g-1/2*c^4*b/(c^2*d^2-e^2)/
(c*e*x+c*d)^2/e*arcsin(c*x)*d^2*f+1/2*c^3*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2/e^2*(-c^2*x^2+1)^(1/2)*d^3*h-1/2*c^3*b
/(c^2*d^2-e^2)/(c*e*x+c*d)^2/e*(-c^2*x^2+1)^(1/2)*d^2*g-3/2*c^2*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2/e*arcsin(c*x)*d^
2*h+c^4*b/e^3/(c^2*d^2-e^2)^2*d^4*h*arcsin(c*x)*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^4*b/e^3/(c^2*d^2-e^2)^2*d^4*h*arcsin(c*x)*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)))-1/2*c^3*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2*(-c^2*x^2+1)^(1/
2)*x*d*g+a*h/e^3*ln(c*e*x+c*d)+1/2*c^2*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2*arcsin(c*x)*g*d-3*c^3*b/e^3/(c^2*d^2-e^2)
^(3/2)*d^3*h*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))-1/2*c^2*a/e^3/(c*e*x+c*d
)^2*d^2*h+2*c*a/e^3/(c*e*x+c*d)*d*h+b*e/(c^2*d^2-e^2)^2*h*arcsin(c*x)*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)))+b*e/(c^2*d^2-e^2)^2*h*arcsin(c*x)*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)))+I*b/e/(c^2*d^2-e^2)*h*arcsin(c*x)^2-I*b*e/
(c^2*d^2-e^2)^2*h*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))
)-I*b*e/(c^2*d^2-e^2)^2*h*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/(c^2*d^2-e^2)^(3/2)*g*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))
+1/2*c^2*a/e^2/(c*e*x+c*d)^2*d*g+1/2*I*b*arcsin(c*x)^2*h/e^3+1/2*c^3*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2/e*(-c^2*x^2
+1)^(1/2)*x*d^2*h-c^4*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2/e*arcsin(c*x)*x*d^2*g+2*c^4*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2/
e^2*arcsin(c*x)*x*d^3*h+I*c^4*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2/e*g*d^2*x-1/2*I*c^4*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2/
e*x^2*d^2*h-I*c^4*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2/e^2*x*d^3*h+4*c*b/e/(c^2*d^2-e^2)^(3/2)*d*h*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))+c^3*b/e/(c^2*d^2-e^2)^(3/2)*d*f*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))+c^3*b/e^2/(c^2*d^2-e^2)^(3/2)*d^2*g*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))+1/2*c^2*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2*e*arcsin(c*x)*f+1/2*c^3
*b/(c^2*d^2-e^2)/(c*e*x+c*d)^2*(-c^2*x^2+1)^(1/2)*d*f-1/2*c^2*a/e/(c*e*x+c*d)^2*f-c*a*g/e^2/(c*e*x+c*d)

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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)^3,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 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\left (h\,x^2+g\,x+f\right )}{{\left (d+e\,x\right )}^3} \,d x \]

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)^3,x)

[Out]

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

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

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)**3,x)

[Out]

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

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