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

Optimal. Leaf size=358 \[ -\frac{i b g \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e^2}-\frac{i b g \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e^2}-\frac{(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{e^2 (d+e x)}+\frac{g \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac{b c (e f-d g) \tan ^{-1}\left (\frac{c^2 d x+e}{\sqrt{1-c^2 x^2} \sqrt{c^2 d^2-e^2}}\right )}{e^2 \sqrt{c^2 d^2-e^2}}+\frac{b g \sin ^{-1}(c x) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e^2}+\frac{b g \sin ^{-1}(c x) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e^2}-\frac{b g \sin ^{-1}(c x) \log (d+e x)}{e^2}-\frac{i b g \sin ^{-1}(c x)^2}{2 e^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.952115, antiderivative size = 358, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 13, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.619, Rules used = {43, 4753, 12, 6742, 725, 204, 216, 2404, 4741, 4519, 2190, 2279, 2391} \[ -\frac{i b g \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e^2}-\frac{i b g \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e^2}-\frac{(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{e^2 (d+e x)}+\frac{g \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )}{e^2}+\frac{b c (e f-d g) \tan ^{-1}\left (\frac{c^2 d x+e}{\sqrt{1-c^2 x^2} \sqrt{c^2 d^2-e^2}}\right )}{e^2 \sqrt{c^2 d^2-e^2}}+\frac{b g \sin ^{-1}(c x) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e^2}+\frac{b g \sin ^{-1}(c x) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e^2}-\frac{b g \sin ^{-1}(c x) \log (d+e x)}{e^2}-\frac{i b g \sin ^{-1}(c x)^2}{2 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 12

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

Rule 6742

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

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 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 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 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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.46929, size = 322, normalized size = 0.9 \[ \frac{-\frac{1}{2} i b g \left (2 \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )+2 \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )+\sin ^{-1}(c x) \left (\sin ^{-1}(c x)+2 i \left (\log \left (1+\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}-c d}\right )+\log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )\right )\right )\right )-\frac{(e f-d g) \left (a+b \sin ^{-1}(c x)\right )}{d+e x}+g \log (d+e x) \left (a+b \sin ^{-1}(c x)\right )+\frac{b c (e f-d g) \tan ^{-1}\left (\frac{c^2 d x+e}{\sqrt{1-c^2 x^2} \sqrt{c^2 d^2-e^2}}\right )}{\sqrt{c^2 d^2-e^2}}-b g \sin ^{-1}(c x) \log (d+e x)}{e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.703, size = 982, normalized size = 2.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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*a/e/(c*e*x+c*d)*f+a*g/e^2*ln(c*e*x+c*d)-1/2*I*b*g*arcsin(c*x)^2/e^2+c*b*arcsin(c*x)/
e^2/(c*e*x+c*d)*d*g-c*b*arcsin(c*x)/e/(c*e*x+c*d)*f+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))+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-b*arcsin(c*x)*g/(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*arcsin
(c*x)*g/(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*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*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*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))-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-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., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{a g x + a f +{\left (b g x + b f\right )} \arcsin \left (c x\right )}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{asin}{\left (c x \right )}\right ) \left (f + g x\right )}{\left (d + e x\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (g x + f\right )}{\left (b \arcsin \left (c x\right ) + a\right )}}{{\left (e x + d\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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