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

Optimal. Leaf size=401 \[ -\frac{b^2 c^3 d \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac{b^2 c^3 d \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac{b c \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac{i b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac{i b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac{b^2 c^2 \log (d+e x)}{e \left (c^2 d^2-e^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.635537, antiderivative size = 401, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {4743, 4773, 3324, 3323, 2264, 2190, 2279, 2391, 2668, 31} \[ -\frac{b^2 c^3 d \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac{b^2 c^3 d \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac{b c \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac{i b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}+\frac{i b c^3 d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )}{e \left (c^2 d^2-e^2\right )^{3/2}}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{2 e (d+e x)^2}-\frac{b^2 c^2 \log (d+e x)}{e \left (c^2 d^2-e^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4743

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(m_.), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*ArcSin[c*x])^n)/(e*(m + 1)), x] - Dist[(b*c*n)/(e*(m + 1)), Int[((d + e*x)^(m + 1)*(a + b*ArcSin[c*x])^(
n - 1))/Sqrt[1 - c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4773

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol]
:> Dist[1/(c^(m + 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*(c*f + g*Sin[x])^m, x], x, ArcSin[c*x]], x] /; FreeQ[{a,
b, c, d, e, f, g, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && GtQ[d, 0] && (GtQ[m, 0] || IGtQ[n, 0])

Rule 3324

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

Rule 3323

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

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 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 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 1.01079, size = 315, normalized size = 0.79 \[ \frac{\frac{2 b c^3 d \left (-b \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{c d-\sqrt{c^2 d^2-e^2}}\right )+b \text{PolyLog}\left (2,\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )-i \left (a+b \sin ^{-1}(c x)\right ) \left (\log \left (1+\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}-c d}\right )-\log \left (1-\frac{i e e^{i \sin ^{-1}(c x)}}{\sqrt{c^2 d^2-e^2}+c d}\right )\right )\right )}{\left (c^2 d^2-e^2\right )^{3/2}}+\frac{2 b c e \sqrt{1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{\left (c^2 d^2-e^2\right ) (d+e x)}-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{(d+e x)^2}-\frac{2 b^2 c^2 \log (d+e x)}{c^2 d^2-e^2}}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*b*c*e*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/((c^2*d^2 - e^2)*(d + e*x)) - (a + b*ArcSin[c*x])^2/(d + e*x)
^2 - (2*b^2*c^2*Log[d + e*x])/(c^2*d^2 - e^2) + (2*b*c^3*d*((-I)*(a + b*ArcSin[c*x])*(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])]) - b*Po
lyLog[2, (I*e*E^(I*ArcSin[c*x]))/(c*d - Sqrt[c^2*d^2 - e^2])] + b*PolyLog[2, (I*e*E^(I*ArcSin[c*x]))/(c*d + Sq
rt[c^2*d^2 - e^2])]))/(c^2*d^2 - e^2)^(3/2))/(2*e)

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Maple [B]  time = 0.868, size = 1173, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(c*x))^2/(e*x+d)^3,x)

[Out]

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

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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((a+b*arcsin(c*x))^2/(e*x+d)^3,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{b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^2/(e*x+d)^3,x, algorithm="fricas")

[Out]

integral((b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)/(e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3), 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 )^{2}}{\left (d + e x\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(c*x))**2/(e*x+d)**3,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{{\left (e x + d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(c*x))^2/(e*x+d)^3,x, algorithm="giac")

[Out]

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