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

Optimal. Leaf size=459 \[ \frac {\log (d+e x) \left (a+b \sin ^{-1}(c x)\right ) \left (d^2 h-d e g+e^2 f\right )}{e^3}+\frac {x (e g-d h) \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {h x^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}-\frac {i b \left (d^2 h-d e g+e^2 f\right ) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^3}-\frac {i b \left (d^2 h-d e g+e^2 f\right ) \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e^3}+\frac {b \sin ^{-1}(c x) \left (d^2 h-d e g+e^2 f\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^3}+\frac {b \sin ^{-1}(c x) \left (d^2 h-d e g+e^2 f\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e^3}+\frac {b \sqrt {1-c^2 x^2} (4 (e g-d h)+e h x)}{4 c e^2}-\frac {b h \sin ^{-1}(c x)}{4 c^2 e}-\frac {i b \sin ^{-1}(c x)^2 \left (d^2 h-d e g+e^2 f\right )}{2 e^3}-\frac {b \sin ^{-1}(c x) \log (d+e x) \left (d^2 h-d e g+e^2 f\right )}{e^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.79, antiderivative size = 459, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 12, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {698, 4753, 12, 6742, 780, 216, 2404, 4741, 4519, 2190, 2279, 2391} \[ -\frac {i b \left (d^2 h-d e g+e^2 f\right ) \text {PolyLog}\left (2,\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^3}-\frac {i b \left (d^2 h-d e g+e^2 f\right ) \text {PolyLog}\left (2,\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e^3}+\frac {\log (d+e x) \left (a+b \sin ^{-1}(c x)\right ) \left (d^2 h-d e g+e^2 f\right )}{e^3}+\frac {x (e g-d h) \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {h x^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}+\frac {b \sin ^{-1}(c x) \left (d^2 h-d e g+e^2 f\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^3}+\frac {b \sin ^{-1}(c x) \left (d^2 h-d e g+e^2 f\right ) \log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e^3}+\frac {b \sqrt {1-c^2 x^2} (4 (e g-d h)+e h x)}{4 c e^2}-\frac {b h \sin ^{-1}(c x)}{4 c^2 e}-\frac {i b \sin ^{-1}(c x)^2 \left (d^2 h-d e g+e^2 f\right )}{2 e^3}-\frac {b \sin ^{-1}(c x) \log (d+e x) \left (d^2 h-d e g+e^2 f\right )}{e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(4*(e*g - d*h) + e*h*x)*Sqrt[1 - c^2*x^2])/(4*c*e^2) - (b*h*ArcSin[c*x])/(4*c^2*e) - ((I/2)*b*(e^2*f - d*e*
g + d^2*h)*ArcSin[c*x]^2)/e^3 + ((e*g - d*h)*x*(a + b*ArcSin[c*x]))/e^2 + (h*x^2*(a + b*ArcSin[c*x]))/(2*e) +
(b*(e^2*f - d*e*g + d^2*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^2*f - d*e*g + d^2*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^
2*f - d*e*g + d^2*h)*ArcSin[c*x]*Log[d + e*x])/e^3 + ((e^2*f - d*e*g + d^2*h)*(a + b*ArcSin[c*x])*Log[d + e*x]
)/e^3 - (I*b*(e^2*f - d*e*g + d^2*h)*PolyLog[2, (I*e*E^(I*ArcSin[c*x]))/(c*d - Sqrt[c^2*d^2 - e^2])])/e^3 - (I
*b*(e^2*f - d*e*g + d^2*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 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 780

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

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} \, dx &=\frac {(e g-d h) x \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {h x^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}+\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-(b c) \int \frac {e x (2 e g-2 d h+e h x)+2 \left (e^2 f-d e g+d^2 h\right ) \log (d+e x)}{2 e^3 \sqrt {1-c^2 x^2}} \, dx\\ &=\frac {(e g-d h) x \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {h x^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}+\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-\frac {(b c) \int \frac {e x (2 e g-2 d h+e h x)+2 \left (e^2 f-d e g+d^2 h\right ) \log (d+e x)}{\sqrt {1-c^2 x^2}} \, dx}{2 e^3}\\ &=\frac {(e g-d h) x \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {h x^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}+\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-\frac {(b c) \int \left (\frac {e x (2 e g-2 d h+e h x)}{\sqrt {1-c^2 x^2}}+\frac {2 \left (e^2 f-d e g+d^2 h\right ) \log (d+e x)}{\sqrt {1-c^2 x^2}}\right ) \, dx}{2 e^3}\\ &=\frac {(e g-d h) x \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {h x^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}+\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-\frac {(b c) \int \frac {x (2 e g-2 d h+e h x)}{\sqrt {1-c^2 x^2}} \, dx}{2 e^2}-\frac {\left (b c \left (e^2 f-d e g+d^2 h\right )\right ) \int \frac {\log (d+e x)}{\sqrt {1-c^2 x^2}} \, dx}{e^3}\\ &=\frac {b (4 (e g-d h)+e h x) \sqrt {1-c^2 x^2}}{4 c e^2}+\frac {(e g-d h) x \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {h x^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}-\frac {b \left (e^2 f-d e g+d^2 h\right ) \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-\frac {(b h) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{4 c e}+\frac {\left (b c \left (e^2 f-d e g+d^2 h\right )\right ) \int \frac {\sin ^{-1}(c x)}{c d+c e x} \, dx}{e^2}\\ &=\frac {b (4 (e g-d h)+e h x) \sqrt {1-c^2 x^2}}{4 c e^2}-\frac {b h \sin ^{-1}(c x)}{4 c^2 e}+\frac {(e g-d h) x \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {h x^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}-\frac {b \left (e^2 f-d e g+d^2 h\right ) \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}+\frac {\left (b c \left (e^2 f-d e g+d^2 h\right )\right ) \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 (4 (e g-d h)+e h x) \sqrt {1-c^2 x^2}}{4 c e^2}-\frac {b h \sin ^{-1}(c x)}{4 c^2 e}-\frac {i b \left (e^2 f-d e g+d^2 h\right ) \sin ^{-1}(c x)^2}{2 e^3}+\frac {(e g-d h) x \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {h x^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}-\frac {b \left (e^2 f-d e g+d^2 h\right ) \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}+\frac {\left (b c \left (e^2 f-d e g+d^2 h\right )\right ) \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 {\left (b c \left (e^2 f-d e g+d^2 h\right )\right ) \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 (4 (e g-d h)+e h x) \sqrt {1-c^2 x^2}}{4 c e^2}-\frac {b h \sin ^{-1}(c x)}{4 c^2 e}-\frac {i b \left (e^2 f-d e g+d^2 h\right ) \sin ^{-1}(c x)^2}{2 e^3}+\frac {(e g-d h) x \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {h x^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}+\frac {b \left (e^2 f-d e g+d^2 h\right ) \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 \left (e^2 f-d e g+d^2 h\right ) \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 \left (e^2 f-d e g+d^2 h\right ) \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-\frac {\left (b \left (e^2 f-d e g+d^2 h\right )\right ) \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 {\left (b \left (e^2 f-d e g+d^2 h\right )\right ) \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 (4 (e g-d h)+e h x) \sqrt {1-c^2 x^2}}{4 c e^2}-\frac {b h \sin ^{-1}(c x)}{4 c^2 e}-\frac {i b \left (e^2 f-d e g+d^2 h\right ) \sin ^{-1}(c x)^2}{2 e^3}+\frac {(e g-d h) x \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {h x^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}+\frac {b \left (e^2 f-d e g+d^2 h\right ) \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 \left (e^2 f-d e g+d^2 h\right ) \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 \left (e^2 f-d e g+d^2 h\right ) \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}+\frac {\left (i b \left (e^2 f-d e g+d^2 h\right )\right ) \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 {\left (i b \left (e^2 f-d e g+d^2 h\right )\right ) \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 (4 (e g-d h)+e h x) \sqrt {1-c^2 x^2}}{4 c e^2}-\frac {b h \sin ^{-1}(c x)}{4 c^2 e}-\frac {i b \left (e^2 f-d e g+d^2 h\right ) \sin ^{-1}(c x)^2}{2 e^3}+\frac {(e g-d h) x \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac {h x^2 \left (a+b \sin ^{-1}(c x)\right )}{2 e}+\frac {b \left (e^2 f-d e g+d^2 h\right ) \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 \left (e^2 f-d e g+d^2 h\right ) \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 \left (e^2 f-d e g+d^2 h\right ) \sin ^{-1}(c x) \log (d+e x)}{e^3}+\frac {\left (e^2 f-d e g+d^2 h\right ) \left (a+b \sin ^{-1}(c x)\right ) \log (d+e x)}{e^3}-\frac {i b \left (e^2 f-d e g+d^2 h\right ) \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 \left (e^2 f-d e g+d^2 h\right ) \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 = 0.65, size = 381, normalized size = 0.83 \[ \frac {2 \log (d+e x) \left (a+b \sin ^{-1}(c x)\right ) \left (d^2 h-d e g+e^2 f\right )+2 e x (e g-d h) \left (a+b \sin ^{-1}(c x)\right )+e^2 h x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {b \left (-2 i c^2 \left (d^2 h-d e g+e^2 f\right ) \left (2 \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )+2 \text {Li}_2\left (\frac {i e e^{i \sin ^{-1}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )+\sin ^{-1}(c x) \left (\sin ^{-1}(c x)+2 i \left (\log \left (1+\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}-c d}\right )+\log \left (1-\frac {i e e^{i \sin ^{-1}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )\right )\right )\right )-4 c^2 \sin ^{-1}(c x) \log (d+e x) \left (d^2 h-d e g+e^2 f\right )+4 c e \sqrt {1-c^2 x^2} (e g-d h)+c e^2 h x \sqrt {1-c^2 x^2}-e^2 h \sin ^{-1}(c x)\right )}{2 c^2}}{2 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*(e*g - d*h)*x*(a + b*ArcSin[c*x]) + e^2*h*x^2*(a + b*ArcSin[c*x]) + 2*(e^2*f - d*e*g + d^2*h)*(a + b*ArcS
in[c*x])*Log[d + e*x] + (b*(4*c*e*(e*g - d*h)*Sqrt[1 - c^2*x^2] + c*e^2*h*x*Sqrt[1 - c^2*x^2] - e^2*h*ArcSin[c
*x] - 4*c^2*(e^2*f - d*e*g + d^2*h)*ArcSin[c*x]*Log[d + e*x] - (2*I)*c^2*(e^2*f - d*e*g + d^2*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*ArcS
in[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])])))/(2*c^2))/(2*e^3)

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fricas [F]  time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a h x^{2} + a g x + a f + {\left (b h x^{2} + b g x + b f\right )} \arcsin \left (c x\right )}{e x + d}, x\right ) \]

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),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*x + d), x)

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

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),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.74, size = 2446, normalized size = 5.33 \[ \text {result too large to display} \]

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),x)

[Out]

a*g/e*x+a/e*ln(c*e*x+c*d)*f+b*g*(-c^2*x^2+1)^(1/2)/c/e+b*arcsin(c*x)*g/e*x-1/2*I*b*arcsin(c*x)^2/e*f-c^2*b/e^2
*d^3*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)))+c^2*b/e*f*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*f/(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+I*c^2*b/e^2*d^3*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)))-I*c^2*b/e*f/(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*f*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*d^3*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)))+I*c^2*b/e^2*d^3*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)))+I*b/e*d^2*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)))+I*b/e*d^2*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)))-b/e*d^2*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)))-b/e*d^2*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)))+a/e^3*ln(c*e*x+c*d)*d
^2*h+1/8/c^2*b/e*h*sin(2*arcsin(c*x))-a/e^2*x*d*h+b*d*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)))+b*d*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)))-b*e*f*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)))+1/2*I*b*arcsin(c*
x)^2/e^2*d*g+I*b*e*f/(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)))+I*b*e*f/(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)))-I*b*d*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*e*f*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)))-I*b*d*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)))-1/4/c^2*b/e*h*arcsin(c*x)*cos(2*arcsin(c*x))-1/c*b/e^2*(-c^2
*x^2+1)^(1/2)*d*h-b*arcsin(c*x)/e^2*x*d*h-1/2*I*b*arcsin(c*x)^2/e^3*d^2*h+1/2*a/e*h*x^2+c^2*b/e^3*d^4*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)))
+c^2*b/e^3*d^4*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)))-I*c^2*b/e^3*d^4*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)))-I*c^2*b/e^3*d^4*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)))-a/e^2*ln(c*e*x+c*d)*d*g

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a g {\left (\frac {x}{e} - \frac {d \log \left (e x + d\right )}{e^{2}}\right )} + \frac {1}{2} \, a h {\left (\frac {2 \, d^{2} \log \left (e x + d\right )}{e^{3}} + \frac {e x^{2} - 2 \, d x}{e^{2}}\right )} + \frac {a f \log \left (e x + d\right )}{e} + \int \frac {{\left (b h x^{2} + b g x + b f\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{e x + d}\,{d x} \]

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),x, algorithm="maxima")

[Out]

a*g*(x/e - d*log(e*x + d)/e^2) + 1/2*a*h*(2*d^2*log(e*x + d)/e^3 + (e*x^2 - 2*d*x)/e^2) + a*f*log(e*x + d)/e +
integrate((b*h*x^2 + b*g*x + b*f)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/(e*x + d), x)

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

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),x)

[Out]

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

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

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),x)

[Out]

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

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