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

Optimal. Leaf size=198 \[ -\frac{6 i b^3 \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^3}+\frac{3 b^4 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}+\frac{6 b^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^3}-\frac{2 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2} \]

[Out]

((-2*I)*b*(a + b*ArcSin[c + d*x])^3)/(d*e^3) - (2*b*Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x])^3)/(d*e^3*(c
+ d*x)) - (a + b*ArcSin[c + d*x])^4/(2*d*e^3*(c + d*x)^2) + (6*b^2*(a + b*ArcSin[c + d*x])^2*Log[1 - E^((2*I)
*ArcSin[c + d*x])])/(d*e^3) - ((6*I)*b^3*(a + b*ArcSin[c + d*x])*PolyLog[2, E^((2*I)*ArcSin[c + d*x])])/(d*e^3
) + (3*b^4*PolyLog[3, E^((2*I)*ArcSin[c + d*x])])/(d*e^3)

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Rubi [A]  time = 0.319608, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.435, Rules used = {4805, 12, 4627, 4681, 4625, 3717, 2190, 2531, 2282, 6589} \[ -\frac{6 i b^3 \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^3}+\frac{3 b^4 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}+\frac{6 b^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e^3}-\frac{2 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*b*(a + b*ArcSin[c + d*x])^3)/(d*e^3) - (2*b*Sqrt[1 - (c + d*x)^2]*(a + b*ArcSin[c + d*x])^3)/(d*e^3*(c
+ d*x)) - (a + b*ArcSin[c + d*x])^4/(2*d*e^3*(c + d*x)^2) + (6*b^2*(a + b*ArcSin[c + d*x])^2*Log[1 - E^((2*I)
*ArcSin[c + d*x])])/(d*e^3) - ((6*I)*b^3*(a + b*ArcSin[c + d*x])*PolyLog[2, E^((2*I)*ArcSin[c + d*x])])/(d*e^3
) + (3*b^4*PolyLog[3, E^((2*I)*ArcSin[c + d*x])])/(d*e^3)

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 12

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

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 4681

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[(b*c*n*d^IntPart[p]*(d + e*x
^2)^FracPart[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*ArcSi
n[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p
+ 3, 0] && NeQ[m, -1]

Rule 4625

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tan[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && 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 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 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 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 )^4}{(c e+d e x)^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^4}{e^3 x^3} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^4}{x^3} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^3}{x^2 \sqrt{1-x^2}} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac{2 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^2}{x} \, dx,x,c+d x\right )}{d e^3}\\ &=-\frac{2 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int (a+b x)^2 \cot (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac{2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3}-\frac{2 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}-\frac{\left (12 i b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 i x} (a+b x)^2}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac{2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3}-\frac{2 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac{6 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}-\frac{\left (12 b^3\right ) \operatorname{Subst}\left (\int (a+b x) \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac{2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3}-\frac{2 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac{6 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}-\frac{6 i b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}+\frac{\left (6 i b^4\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e^3}\\ &=-\frac{2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3}-\frac{2 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac{6 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}-\frac{6 i b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}+\frac{\left (3 b^4\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}\\ &=-\frac{2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3}-\frac{2 b \sqrt{1-(c+d x)^2} \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e^3 (c+d x)}-\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4}{2 d e^3 (c+d x)^2}+\frac{6 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}-\frac{6 i b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}+\frac{3 b^4 \text{Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e^3}\\ \end{align*}

Mathematica [A]  time = 1.20757, size = 385, normalized size = 1.94 \[ \frac{8 a b^3 \left (-3 i \left (\sin ^{-1}(c+d x)^2+\text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right )\right )-\frac{\sin ^{-1}(c+d x)^3}{(c+d x)^2}-\frac{3 \sqrt{1-(c+d x)^2} \sin ^{-1}(c+d x)^2}{c+d x}+6 \sin ^{-1}(c+d x) \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )\right )+b^4 \left (24 i \sin ^{-1}(c+d x) \text{PolyLog}\left (2,e^{-2 i \sin ^{-1}(c+d x)}\right )+12 \text{PolyLog}\left (3,e^{-2 i \sin ^{-1}(c+d x)}\right )-\frac{8 \sqrt{1-(c+d x)^2} \sin ^{-1}(c+d x)^3}{c+d x}+8 i \sin ^{-1}(c+d x)^3+24 \sin ^{-1}(c+d x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c+d x)}\right )-i \pi ^3\right )+24 a^2 b^2 \left (\log (c+d x)-\frac{\sin ^{-1}(c+d x)^2}{2 (c+d x)^2}-\frac{\sqrt{1-(c+d x)^2} \sin ^{-1}(c+d x)}{c+d x}\right )-\frac{8 a^3 b \sqrt{1-(c+d x)^2}}{c+d x}-\frac{8 a^3 b \sin ^{-1}(c+d x)}{(c+d x)^2}-\frac{2 a^4}{(c+d x)^2}-\frac{2 b^4 \sin ^{-1}(c+d x)^4}{(c+d x)^2}}{4 d e^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-2*a^4)/(c + d*x)^2 - (8*a^3*b*Sqrt[1 - (c + d*x)^2])/(c + d*x) - (8*a^3*b*ArcSin[c + d*x])/(c + d*x)^2 - (2
*b^4*ArcSin[c + d*x]^4)/(c + d*x)^2 + 24*a^2*b^2*(-((Sqrt[1 - (c + d*x)^2]*ArcSin[c + d*x])/(c + d*x)) - ArcSi
n[c + d*x]^2/(2*(c + d*x)^2) + Log[c + d*x]) + 8*a*b^3*((-3*Sqrt[1 - (c + d*x)^2]*ArcSin[c + d*x]^2)/(c + d*x)
- ArcSin[c + d*x]^3/(c + d*x)^2 + 6*ArcSin[c + d*x]*Log[1 - E^((2*I)*ArcSin[c + d*x])] - (3*I)*(ArcSin[c + d*
x]^2 + PolyLog[2, E^((2*I)*ArcSin[c + d*x])])) + b^4*((-I)*Pi^3 + (8*I)*ArcSin[c + d*x]^3 - (8*Sqrt[1 - (c + d
*x)^2]*ArcSin[c + d*x]^3)/(c + d*x) + 24*ArcSin[c + d*x]^2*Log[1 - E^((-2*I)*ArcSin[c + d*x])] + (24*I)*ArcSin
[c + d*x]*PolyLog[2, E^((-2*I)*ArcSin[c + d*x])] + 12*PolyLog[3, E^((-2*I)*ArcSin[c + d*x])]))/(4*d*e^3)

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Maple [B]  time = 0.085, size = 747, normalized size = 3.8 \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(d*x+c))^4/(d*e*x+c*e)^3,x)

[Out]

-1/2/d*a^4/e^3/(d*x+c)^2-12*I/d*a*b^3/e^3*polylog(2,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))-2/d*b^4/e^3*arcsin(d*x+c)^
3/(d*x+c)*(1-(d*x+c)^2)^(1/2)-1/2/d*b^4/e^3*arcsin(d*x+c)^4/(d*x+c)^2+6/d*b^4/e^3*arcsin(d*x+c)^2*ln(1+I*(d*x+
c)+(1-(d*x+c)^2)^(1/2))-12*I/d*a*b^3/e^3*polylog(2,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))+12/d*b^4/e^3*polylog(3,-I*(d
*x+c)-(1-(d*x+c)^2)^(1/2))+6/d*b^4/e^3*arcsin(d*x+c)^2*ln(1-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))-2*I/d*b^4/e^3*arcsi
n(d*x+c)^3+12/d*b^4/e^3*polylog(3,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))-12*I/d*b^4/e^3*arcsin(d*x+c)*polylog(2,I*(d*x
+c)+(1-(d*x+c)^2)^(1/2))-6/d*a*b^3/e^3*arcsin(d*x+c)^2/(d*x+c)*(1-(d*x+c)^2)^(1/2)-2/d*a*b^3/e^3*arcsin(d*x+c)
^3/(d*x+c)^2+12/d*a*b^3/e^3*arcsin(d*x+c)*ln(1+I*(d*x+c)+(1-(d*x+c)^2)^(1/2))+12/d*a*b^3/e^3*arcsin(d*x+c)*ln(
1-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))-6*I/d*a*b^3/e^3*arcsin(d*x+c)^2-12*I/d*b^4/e^3*arcsin(d*x+c)*polylog(2,-I*(d*
x+c)-(1-(d*x+c)^2)^(1/2))-3/d*a^2*b^2/e^3*arcsin(d*x+c)^2/(d*x+c)^2-6/d*a^2*b^2/e^3*arcsin(d*x+c)/(d*x+c)*(1-(
d*x+c)^2)^(1/2)+6/d*a^2*b^2/e^3*ln(d*x+c)-2/d*a^3*b/e^3/(d*x+c)^2*arcsin(d*x+c)-2/d*a^3*b/e^3/(d*x+c)*(1-(d*x+
c)^2)^(1/2)

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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((a+b*arcsin(d*x+c))^4/(d*e*x+c*e)^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^{4} \arcsin \left (d x + c\right )^{4} + 4 \, a b^{3} \arcsin \left (d x + c\right )^{3} + 6 \, a^{2} b^{2} \arcsin \left (d x + c\right )^{2} + 4 \, a^{3} b \arcsin \left (d x + c\right ) + a^{4}}{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arcsin \left (d x + c\right ) + a\right )}^{4}}{{\left (d e x + c e\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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