3.205 \(\int \frac {(a+b \sin ^{-1}(c+d x))^3}{(c e+d e x)^4} \, dx\)

Optimal. Leaf size=291 \[ \frac {i b^2 \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4}-\frac {i b^2 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4}-\frac {b^2 \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^4}-\frac {b^3 \text {Li}_3\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {b^3 \text {Li}_3\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^3 \tanh ^{-1}\left (\sqrt {1-(c+d x)^2}\right )}{d e^4} \]

[Out]

-b^2*(a+b*arcsin(d*x+c))/d/e^4/(d*x+c)-1/3*(a+b*arcsin(d*x+c))^3/d/e^4/(d*x+c)^3-b*(a+b*arcsin(d*x+c))^2*arcta
nh(I*(d*x+c)+(1-(d*x+c)^2)^(1/2))/d/e^4-b^3*arctanh((1-(d*x+c)^2)^(1/2))/d/e^4+I*b^2*(a+b*arcsin(d*x+c))*polyl
og(2,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))/d/e^4-I*b^2*(a+b*arcsin(d*x+c))*polylog(2,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))/
d/e^4-b^3*polylog(3,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))/d/e^4+b^3*polylog(3,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))/d/e^4-1
/2*b*(a+b*arcsin(d*x+c))^2*(1-(d*x+c)^2)^(1/2)/d/e^4/(d*x+c)^2

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Rubi [A]  time = 0.39, antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 12, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {4805, 12, 4627, 4701, 4709, 4183, 2531, 2282, 6589, 266, 63, 206} \[ \frac {i b^2 \text {PolyLog}\left (2,-e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4}-\frac {i b^2 \text {PolyLog}\left (2,e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4}-\frac {b^3 \text {PolyLog}\left (3,-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {b^3 \text {PolyLog}\left (3,e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^2 \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^4}-\frac {b^3 \tanh ^{-1}\left (\sqrt {1-(c+d x)^2}\right )}{d e^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c + d*x])^3/(c*e + d*e*x)^4,x]

[Out]

-((b^2*(a + b*ArcSin[c + d*x]))/(d*e^4*(c + d*x))) - (b*Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x])^2)/(2*d*
e^4*(c + d*x)^2) - (a + b*ArcSin[c + d*x])^3/(3*d*e^4*(c + d*x)^3) - (b*(a + b*ArcSin[c + d*x])^2*ArcTanh[E^(I
*ArcSin[c + d*x])])/(d*e^4) - (b^3*ArcTanh[Sqrt[1 - (c + d*x)^2]])/(d*e^4) + (I*b^2*(a + b*ArcSin[c + d*x])*Po
lyLog[2, -E^(I*ArcSin[c + d*x])])/(d*e^4) - (I*b^2*(a + b*ArcSin[c + d*x])*PolyLog[2, E^(I*ArcSin[c + d*x])])/
(d*e^4) - (b^3*PolyLog[3, -E^(I*ArcSin[c + d*x])])/(d*e^4) + (b^3*PolyLog[3, E^(I*ArcSin[c + d*x])])/(d*e^4)

Rule 12

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 4627

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

Rule 4701

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(
(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n)/(d*f*(m + 1)), x] + (Dist[(c^2*(m + 2*p + 3))/(f^2*(m
+ 1)), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSin[c*x])^n, x], x] - Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(f*(m + 1)*(1 - c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x
])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1] && Inte
gerQ[m]

Rule 4709

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

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{(c e+d e x)^4} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^3}{e^4 x^4} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^3}{x^4} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{x^3 \sqrt {1-x^2}} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{x \sqrt {1-x^2}} \, dx,x,c+d x\right )}{2 d e^4}+\frac {b^2 \operatorname {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{x^2} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b^2 \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \operatorname {Subst}\left (\int (a+b x)^2 \csc (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{2 d e^4}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1-x^2}} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b^2 \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \left (a+b \sin ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^2 \operatorname {Subst}\left (\int (a+b x) \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}+\frac {b^2 \operatorname {Subst}\left (\int (a+b x) \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}+\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x} x} \, dx,x,(c+d x)^2\right )}{2 d e^4}\\ &=-\frac {b^2 \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \left (a+b \sin ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {\left (i b^3\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}+\frac {\left (i b^3\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (e^{i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^4}-\frac {b^3 \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt {1-(c+d x)^2}\right )}{d e^4}\\ &=-\frac {b^2 \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \left (a+b \sin ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^3 \tanh ^{-1}\left (\sqrt {1-(c+d x)^2}\right )}{d e^4}+\frac {i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^3 \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {b^3 \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}\\ &=-\frac {b^2 \left (a+b \sin ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \sqrt {1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}-\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}-\frac {b \left (a+b \sin ^{-1}(c+d x)\right )^2 \tanh ^{-1}\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^3 \tanh ^{-1}\left (\sqrt {1-(c+d x)^2}\right )}{d e^4}+\frac {i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {i b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}-\frac {b^3 \text {Li}_3\left (-e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}+\frac {b^3 \text {Li}_3\left (e^{i \sin ^{-1}(c+d x)}\right )}{d e^4}\\ \end {align*}

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Mathematica [B]  time = 8.02, size = 732, normalized size = 2.52 \[ -\frac {a^3}{3 d e^4 (c+d x)^3}-\frac {a^2 b \sqrt {-c^2-2 c d x-d^2 x^2+1}}{2 d e^4 (c+d x)^2}-\frac {a^2 b \log \left (\sqrt {-c^2-2 c d x-d^2 x^2+1}+1\right )}{2 d e^4}+\frac {a^2 b \log (c+d x)}{2 d e^4}-\frac {a^2 b \sin ^{-1}(c+d x)}{d e^4 (c+d x)^3}+\frac {a b^2 \left (8 i \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )-\frac {2 \left (4 i (c+d x)^3 \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )+4 \sin ^{-1}(c+d x)^2+2 \sin ^{-1}(c+d x) \sin \left (2 \sin ^{-1}(c+d x)\right )-3 (c+d x) \sin ^{-1}(c+d x) \log \left (1-e^{i \sin ^{-1}(c+d x)}\right )+3 (c+d x) \sin ^{-1}(c+d x) \log \left (1+e^{i \sin ^{-1}(c+d x)}\right )+\sin ^{-1}(c+d x) \sin \left (3 \sin ^{-1}(c+d x)\right ) \log \left (1-e^{i \sin ^{-1}(c+d x)}\right )-\sin ^{-1}(c+d x) \sin \left (3 \sin ^{-1}(c+d x)\right ) \log \left (1+e^{i \sin ^{-1}(c+d x)}\right )-2 \cos \left (2 \sin ^{-1}(c+d x)\right )+2\right )}{(c+d x)^3}\right )}{8 d e^4}+\frac {b^3 \left (48 i \sin ^{-1}(c+d x) \text {Li}_2\left (-e^{i \sin ^{-1}(c+d x)}\right )-48 i \sin ^{-1}(c+d x) \text {Li}_2\left (e^{i \sin ^{-1}(c+d x)}\right )-48 \text {Li}_3\left (-e^{i \sin ^{-1}(c+d x)}\right )+48 \text {Li}_3\left (e^{i \sin ^{-1}(c+d x)}\right )-\frac {16 \sin ^{-1}(c+d x)^3 \sin ^4\left (\frac {1}{2} \sin ^{-1}(c+d x)\right )}{(c+d x)^3}+24 \sin ^{-1}(c+d x)^2 \log \left (1-e^{i \sin ^{-1}(c+d x)}\right )-24 \sin ^{-1}(c+d x)^2 \log \left (1+e^{i \sin ^{-1}(c+d x)}\right )-4 \sin ^{-1}(c+d x)^3 \tan \left (\frac {1}{2} \sin ^{-1}(c+d x)\right )-24 \sin ^{-1}(c+d x) \tan \left (\frac {1}{2} \sin ^{-1}(c+d x)\right )-4 \sin ^{-1}(c+d x)^3 \cot \left (\frac {1}{2} \sin ^{-1}(c+d x)\right )-24 \sin ^{-1}(c+d x) \cot \left (\frac {1}{2} \sin ^{-1}(c+d x)\right )-\left ((c+d x) \sin ^{-1}(c+d x)^3 \csc ^4\left (\frac {1}{2} \sin ^{-1}(c+d x)\right )\right )-6 \sin ^{-1}(c+d x)^2 \csc ^2\left (\frac {1}{2} \sin ^{-1}(c+d x)\right )+6 \sin ^{-1}(c+d x)^2 \sec ^2\left (\frac {1}{2} \sin ^{-1}(c+d x)\right )+48 \log \left (\tan \left (\frac {1}{2} \sin ^{-1}(c+d x)\right )\right )\right )}{48 d e^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSin[c + d*x])^3/(c*e + d*e*x)^4,x]

[Out]

-1/3*a^3/(d*e^4*(c + d*x)^3) - (a^2*b*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2])/(2*d*e^4*(c + d*x)^2) - (a^2*b*ArcSin
[c + d*x])/(d*e^4*(c + d*x)^3) + (a^2*b*Log[c + d*x])/(2*d*e^4) - (a^2*b*Log[1 + Sqrt[1 - c^2 - 2*c*d*x - d^2*
x^2]])/(2*d*e^4) + (a*b^2*((8*I)*PolyLog[2, -E^(I*ArcSin[c + d*x])] - (2*(2 + 4*ArcSin[c + d*x]^2 - 2*Cos[2*Ar
cSin[c + d*x]] - 3*(c + d*x)*ArcSin[c + d*x]*Log[1 - E^(I*ArcSin[c + d*x])] + 3*(c + d*x)*ArcSin[c + d*x]*Log[
1 + E^(I*ArcSin[c + d*x])] + (4*I)*(c + d*x)^3*PolyLog[2, E^(I*ArcSin[c + d*x])] + 2*ArcSin[c + d*x]*Sin[2*Arc
Sin[c + d*x]] + ArcSin[c + d*x]*Log[1 - E^(I*ArcSin[c + d*x])]*Sin[3*ArcSin[c + d*x]] - ArcSin[c + d*x]*Log[1
+ E^(I*ArcSin[c + d*x])]*Sin[3*ArcSin[c + d*x]]))/(c + d*x)^3))/(8*d*e^4) + (b^3*(-24*ArcSin[c + d*x]*Cot[ArcS
in[c + d*x]/2] - 4*ArcSin[c + d*x]^3*Cot[ArcSin[c + d*x]/2] - 6*ArcSin[c + d*x]^2*Csc[ArcSin[c + d*x]/2]^2 - (
c + d*x)*ArcSin[c + d*x]^3*Csc[ArcSin[c + d*x]/2]^4 + 24*ArcSin[c + d*x]^2*Log[1 - E^(I*ArcSin[c + d*x])] - 24
*ArcSin[c + d*x]^2*Log[1 + E^(I*ArcSin[c + d*x])] + 48*Log[Tan[ArcSin[c + d*x]/2]] + (48*I)*ArcSin[c + d*x]*Po
lyLog[2, -E^(I*ArcSin[c + d*x])] - (48*I)*ArcSin[c + d*x]*PolyLog[2, E^(I*ArcSin[c + d*x])] - 48*PolyLog[3, -E
^(I*ArcSin[c + d*x])] + 48*PolyLog[3, E^(I*ArcSin[c + d*x])] + 6*ArcSin[c + d*x]^2*Sec[ArcSin[c + d*x]/2]^2 -
(16*ArcSin[c + d*x]^3*Sin[ArcSin[c + d*x]/2]^4)/(c + d*x)^3 - 24*ArcSin[c + d*x]*Tan[ArcSin[c + d*x]/2] - 4*Ar
cSin[c + d*x]^3*Tan[ArcSin[c + d*x]/2]))/(48*d*e^4)

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \arcsin \left (d x + c\right )^{3} + 3 \, a b^{2} \arcsin \left (d x + c\right )^{2} + 3 \, a^{2} b \arcsin \left (d x + c\right ) + a^{3}}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))^3/(d*e*x+c*e)^4,x, algorithm="fricas")

[Out]

integral((b^3*arcsin(d*x + c)^3 + 3*a*b^2*arcsin(d*x + c)^2 + 3*a^2*b*arcsin(d*x + c) + a^3)/(d^4*e^4*x^4 + 4*
c*d^3*e^4*x^3 + 6*c^2*d^2*e^4*x^2 + 4*c^3*d*e^4*x + c^4*e^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))^3/(d*e*x+c*e)^4,x, algorithm="giac")

[Out]

integrate((b*arcsin(d*x + c) + a)^3/(d*e*x + c*e)^4, x)

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maple [B]  time = 0.43, size = 716, normalized size = 2.46 \[ -\frac {a^{3}}{3 d \,e^{4} \left (d x +c \right )^{3}}-\frac {b^{3} \arcsin \left (d x +c \right )^{2} \sqrt {1-\left (d x +c \right )^{2}}}{2 d \,e^{4} \left (d x +c \right )^{2}}-\frac {b^{3} \arcsin \left (d x +c \right )^{3}}{3 d \,e^{4} \left (d x +c \right )^{3}}-\frac {b^{3} \arcsin \left (d x +c \right )}{d \,e^{4} \left (d x +c \right )}-\frac {b^{3} \arcsin \left (d x +c \right )^{2} \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{2 d \,e^{4}}-\frac {i a \,b^{2} \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{4}}-\frac {b^{3} \polylog \left (3, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{4}}+\frac {b^{3} \arcsin \left (d x +c \right )^{2} \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{2 d \,e^{4}}+\frac {i a \,b^{2} \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{4}}+\frac {b^{3} \polylog \left (3, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{4}}-\frac {2 b^{3} \arctanh \left (i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{4}}-\frac {a \,b^{2} \arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{d \,e^{4} \left (d x +c \right )^{2}}-\frac {a \,b^{2} \arcsin \left (d x +c \right )^{2}}{d \,e^{4} \left (d x +c \right )^{3}}-\frac {a \,b^{2}}{d \,e^{4} \left (d x +c \right )}-\frac {a \,b^{2} \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{4}}+\frac {i b^{3} \arcsin \left (d x +c \right ) \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{4}}+\frac {a \,b^{2} \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{4}}-\frac {i b^{3} \arcsin \left (d x +c \right ) \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d \,e^{4}}-\frac {a^{2} b \arcsin \left (d x +c \right )}{d \,e^{4} \left (d x +c \right )^{3}}-\frac {a^{2} b \sqrt {1-\left (d x +c \right )^{2}}}{2 d \,e^{4} \left (d x +c \right )^{2}}-\frac {a^{2} b \arctanh \left (\frac {1}{\sqrt {1-\left (d x +c \right )^{2}}}\right )}{2 d \,e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin(d*x+c))^3/(d*e*x+c*e)^4,x)

[Out]

-1/3/d*a^3/e^4/(d*x+c)^3-1/2/d*b^3/e^4/(d*x+c)^2*arcsin(d*x+c)^2*(1-(d*x+c)^2)^(1/2)-1/3/d*b^3/e^4/(d*x+c)^3*a
rcsin(d*x+c)^3-1/d*b^3/e^4/(d*x+c)*arcsin(d*x+c)-1/2/d*b^3/e^4*arcsin(d*x+c)^2*ln(1+I*(d*x+c)+(1-(d*x+c)^2)^(1
/2))-I/d*b^3/e^4*arcsin(d*x+c)*polylog(2,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))-b^3*polylog(3,-I*(d*x+c)-(1-(d*x+c)^2)
^(1/2))/d/e^4+1/2/d*b^3/e^4*arcsin(d*x+c)^2*ln(1-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))+I/d*b^3/e^4*arcsin(d*x+c)*poly
log(2,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))+b^3*polylog(3,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))/d/e^4-2/d*b^3/e^4*arctanh(I
*(d*x+c)+(1-(d*x+c)^2)^(1/2))-1/d*a*b^2/e^4/(d*x+c)^2*arcsin(d*x+c)*(1-(d*x+c)^2)^(1/2)-1/d*a*b^2/e^4/(d*x+c)^
3*arcsin(d*x+c)^2-1/d*a*b^2/e^4/(d*x+c)-1/d*a*b^2/e^4*arcsin(d*x+c)*ln(1+I*(d*x+c)+(1-(d*x+c)^2)^(1/2))+I/d*a*
b^2/e^4*polylog(2,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))+1/d*a*b^2/e^4*arcsin(d*x+c)*ln(1-I*(d*x+c)-(1-(d*x+c)^2)^(1/
2))-I/d*a*b^2/e^4*polylog(2,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))-1/d*a^2*b/e^4/(d*x+c)^3*arcsin(d*x+c)-1/2/d*a^2*b/e
^4/(d*x+c)^2*(1-(d*x+c)^2)^(1/2)-1/2/d*a^2*b/e^4*arctanh(1/(1-(d*x+c)^2)^(1/2))

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin(d*x+c))^3/(d*e*x+c*e)^4,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c + d*x))^3/(c*e + d*e*x)^4,x)

[Out]

int((a + b*asin(c + d*x))^3/(c*e + d*e*x)^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{3}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {3 a^{2} b \operatorname {asin}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin(d*x+c))**3/(d*e*x+c*e)**4,x)

[Out]

(Integral(a**3/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(b**3*asin(c +
d*x)**3/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(3*a*b**2*asin(c + d
*x)**2/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(3*a**2*b*asin(c + d*x
)/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x))/e**4

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