3.2.1 \(\int \frac {(f+g x+h x^2) (a+b \text {ArcSin}(c x))}{(d+e x)^2} \, dx\) [101]

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

[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 = 0.60, 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 = {712, 4837, 12, 6874, 267, 739, 210, 222, 2451, 4825, 4615, 2221, 2317, 2438} \begin {gather*} -\frac {(a+b \text {ArcSin}(c x)) \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) (a+b \text {ArcSin}(c x))}{e^3}+\frac {h x (a+b \text {ArcSin}(c x))}{e^2}-\frac {i b (e g-2 d h) \text {Li}_2\left (\frac {i e e^{i \text {ArcSin}(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 \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )}{e^3}+\frac {b \text {ArcSin}(c x) (e g-2 d h) \log \left (1-\frac {i e e^{i \text {ArcSin}(c x)}}{c d-\sqrt {c^2 d^2-e^2}}\right )}{e^3}+\frac {b \text {ArcSin}(c x) (e g-2 d h) \log \left (1-\frac {i e e^{i \text {ArcSin}(c x)}}{\sqrt {c^2 d^2-e^2}+c d}\right )}{e^3}-\frac {i b \text {ArcSin}(c x)^2 (e g-2 d h)}{2 e^3}-\frac {b \text {ArcSin}(c x) (e g-2 d h) \log (d+e x)}{e^3}+\frac {b c \text {ArcTan}\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 h \sqrt {1-c^2 x^2}}{c e^2} \end {gather*}

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 210

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

Rule 222

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 267

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 712

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 739

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2317

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 2438

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 2451

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 4615

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 4825

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 4837

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 6874

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 ) \text {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)) \text {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)) \text {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)) \text {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)) \text {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)) \text {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)) \text {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)) \text {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 [C] Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
time = 1.85, size = 770, normalized size = 1.67 \begin {gather*} \frac {2 a e h x-\frac {2 a \left (e^2 f-d e g+d^2 h\right )}{d+e x}-2 b e f \left (\frac {c \sqrt {\frac {e \left (-\sqrt {\frac {1}{c^2}}+x\right )}{d+e x}} \sqrt {\frac {e \left (\sqrt {\frac {1}{c^2}}+x\right )}{d+e x}} F_1\left (1;\frac {1}{2},\frac {1}{2};2;\frac {d-\sqrt {\frac {1}{c^2}} e}{d+e x},\frac {d+\sqrt {\frac {1}{c^2}} e}{d+e x}\right )}{\sqrt {1-c^2 x^2}}+\frac {e \text {ArcSin}(c x)}{d+e x}\right )+2 a (e g-2 d h) \log (d+e x)+b e g \left (\frac {2 d \text {ArcSin}(c x)}{d+e x}-i \text {ArcSin}(c x)^2-\frac {2 c d \text {ArcTan}\left (\frac {e+c^2 d x}{\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}}\right )}{\sqrt {c^2 d^2-e^2}}+2 \text {ArcSin}(c x) \log \left (1+\frac {i e e^{i \text {ArcSin}(c x)}}{-c d+\sqrt {c^2 d^2-e^2}}\right )+2 \text {ArcSin}(c x) \log \left (1-\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )-2 i \text {PolyLog}\left (2,-\frac {i e e^{i \text {ArcSin}(c x)}}{-c d+\sqrt {c^2 d^2-e^2}}\right )-2 i \text {PolyLog}\left (2,\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )\right )+2 b h \left (\frac {e \sqrt {1-c^2 x^2}}{c}+e x \text {ArcSin}(c x)-\frac {d^2 \text {ArcSin}(c x)}{d+e x}+i d \text {ArcSin}(c x)^2+\frac {c d^2 \text {ArcTan}\left (\frac {e+c^2 d x}{\sqrt {c^2 d^2-e^2} \sqrt {1-c^2 x^2}}\right )}{\sqrt {c^2 d^2-e^2}}-2 d \text {ArcSin}(c x) \log \left (1+\frac {i e e^{i \text {ArcSin}(c x)}}{-c d+\sqrt {c^2 d^2-e^2}}\right )-2 d \text {ArcSin}(c x) \log \left (1-\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )+2 i d \text {PolyLog}\left (2,-\frac {i e e^{i \text {ArcSin}(c x)}}{-c d+\sqrt {c^2 d^2-e^2}}\right )+2 i d \text {PolyLog}\left (2,\frac {i e e^{i \text {ArcSin}(c x)}}{c d+\sqrt {c^2 d^2-e^2}}\right )\right )}{2 e^3} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1957 vs. \(2 (467 ) = 934\).
time = 1.72, size = 1958, normalized size = 4.26

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1958\)
default	\(\text {Expression too large to display}\)	\(1958\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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(c*d-%e>0)', see `assume?` for
more details

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))/(x^2*e^2 + 2*d*x*e + d^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

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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