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

Optimal. Leaf size=202 \[ \frac{3 b^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e}+\frac{3 i b^3 \text{PolyLog}\left (4,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e}-\frac{2 i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e}-\frac{3 b^4 \text{PolyLog}\left (5,e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}+\frac{\log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^4}{d e} \]

[Out]

((-I/5)*(a + b*ArcSin[c + d*x])^5)/(b*d*e) + ((a + b*ArcSin[c + d*x])^4*Log[1 - E^((2*I)*ArcSin[c + d*x])])/(d
*e) - ((2*I)*b*(a + b*ArcSin[c + d*x])^3*PolyLog[2, E^((2*I)*ArcSin[c + d*x])])/(d*e) + (3*b^2*(a + b*ArcSin[c
+ d*x])^2*PolyLog[3, E^((2*I)*ArcSin[c + d*x])])/(d*e) + ((3*I)*b^3*(a + b*ArcSin[c + d*x])*PolyLog[4, E^((2*
I)*ArcSin[c + d*x])])/(d*e) - (3*b^4*PolyLog[5, E^((2*I)*ArcSin[c + d*x])])/(2*d*e)

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Rubi [A]  time = 0.234107, antiderivative size = 202, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {4805, 12, 4625, 3717, 2190, 2531, 6609, 2282, 6589} \[ \frac{3 b^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e}+\frac{3 i b^3 \text{PolyLog}\left (4,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e}-\frac{2 i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e}-\frac{3 b^4 \text{PolyLog}\left (5,e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}+\frac{\log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^4}{d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/5)*(a + b*ArcSin[c + d*x])^5)/(b*d*e) + ((a + b*ArcSin[c + d*x])^4*Log[1 - E^((2*I)*ArcSin[c + d*x])])/(d
*e) - ((2*I)*b*(a + b*ArcSin[c + d*x])^3*PolyLog[2, E^((2*I)*ArcSin[c + d*x])])/(d*e) + (3*b^2*(a + b*ArcSin[c
+ d*x])^2*PolyLog[3, E^((2*I)*ArcSin[c + d*x])])/(d*e) + ((3*I)*b^3*(a + b*ArcSin[c + d*x])*PolyLog[4, E^((2*
I)*ArcSin[c + d*x])])/(d*e) - (3*b^4*PolyLog[5, E^((2*I)*ArcSin[c + d*x])])/(2*d*e)

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 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 6609

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, 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} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^4}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^4}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac{\operatorname{Subst}\left (\int (a+b x)^4 \cot (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}-\frac{(2 i) \operatorname{Subst}\left (\int \frac{e^{2 i x} (a+b x)^4}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}+\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{(4 b) \operatorname{Subst}\left (\int (a+b x)^3 \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}+\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3 \text{Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac{\left (6 i b^2\right ) \operatorname{Subst}\left (\int (a+b x)^2 \text{Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}+\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3 \text{Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac{3 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text{Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{\left (6 b^3\right ) \operatorname{Subst}\left (\int (a+b x) \text{Li}_3\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}+\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3 \text{Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac{3 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text{Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac{3 i b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_4\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{\left (3 i b^4\right ) \operatorname{Subst}\left (\int \text{Li}_4\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}+\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3 \text{Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac{3 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text{Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac{3 i b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_4\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{\left (3 b^4\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_4(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}\\ &=-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )^5}{5 b d e}+\frac{\left (a+b \sin ^{-1}(c+d x)\right )^4 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{2 i b \left (a+b \sin ^{-1}(c+d x)\right )^3 \text{Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac{3 b^2 \left (a+b \sin ^{-1}(c+d x)\right )^2 \text{Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}+\frac{3 i b^3 \left (a+b \sin ^{-1}(c+d x)\right ) \text{Li}_4\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac{3 b^4 \text{Li}_5\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}\\ \end{align*}

Mathematica [B]  time = 0.377343, size = 439, normalized size = 2.17 \[ \frac{4 a^2 b^2 \left (24 i \sin ^{-1}(c+d x) \text{PolyLog}\left (2,e^{-2 i \sin ^{-1}(c+d x)}\right )+12 \text{PolyLog}\left (3,e^{-2 i \sin ^{-1}(c+d x)}\right )+8 i \sin ^{-1}(c+d x)^3+24 \sin ^{-1}(c+d x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c+d x)}\right )-i \pi ^3\right )+64 a^3 b \left (\sin ^{-1}(c+d x) \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )-\frac{1}{2} i \left (\sin ^{-1}(c+d x)^2+\text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right )\right )\right )-i a b^3 \left (-96 \sin ^{-1}(c+d x)^2 \text{PolyLog}\left (2,e^{-2 i \sin ^{-1}(c+d x)}\right )+96 i \sin ^{-1}(c+d x) \text{PolyLog}\left (3,e^{-2 i \sin ^{-1}(c+d x)}\right )+48 \text{PolyLog}\left (4,e^{-2 i \sin ^{-1}(c+d x)}\right )-16 \sin ^{-1}(c+d x)^4+64 i \sin ^{-1}(c+d x)^3 \log \left (1-e^{-2 i \sin ^{-1}(c+d x)}\right )+\pi ^4\right )+16 b^4 \left (2 i \sin ^{-1}(c+d x)^3 \text{PolyLog}\left (2,e^{-2 i \sin ^{-1}(c+d x)}\right )+3 \sin ^{-1}(c+d x)^2 \text{PolyLog}\left (3,e^{-2 i \sin ^{-1}(c+d x)}\right )-3 i \sin ^{-1}(c+d x) \text{PolyLog}\left (4,e^{-2 i \sin ^{-1}(c+d x)}\right )-\frac{3}{2} \text{PolyLog}\left (5,e^{-2 i \sin ^{-1}(c+d x)}\right )+\frac{1}{5} i \sin ^{-1}(c+d x)^5+\sin ^{-1}(c+d x)^4 \log \left (1-e^{-2 i \sin ^{-1}(c+d x)}\right )-\frac{i \pi ^5}{160}\right )+16 a^4 \log (c+d x)}{16 d e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(16*a^4*Log[c + d*x] + 64*a^3*b*(ArcSin[c + d*x]*Log[1 - E^((2*I)*ArcSin[c + d*x])] - (I/2)*(ArcSin[c + d*x]^2
+ PolyLog[2, E^((2*I)*ArcSin[c + d*x])])) + 4*a^2*b^2*((-I)*Pi^3 + (8*I)*ArcSin[c + d*x]^3 + 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])]) - I*a*b^3*(Pi^4 - 16*ArcSin[c + d*x]^4 + (64*I)*ArcSin[c + d*x]^3*Log[
1 - E^((-2*I)*ArcSin[c + d*x])] - 96*ArcSin[c + d*x]^2*PolyLog[2, E^((-2*I)*ArcSin[c + d*x])] + (96*I)*ArcSin[
c + d*x]*PolyLog[3, E^((-2*I)*ArcSin[c + d*x])] + 48*PolyLog[4, E^((-2*I)*ArcSin[c + d*x])]) + 16*b^4*((-I/160
)*Pi^5 + (I/5)*ArcSin[c + d*x]^5 + ArcSin[c + d*x]^4*Log[1 - E^((-2*I)*ArcSin[c + d*x])] + (2*I)*ArcSin[c + d*
x]^3*PolyLog[2, E^((-2*I)*ArcSin[c + d*x])] + 3*ArcSin[c + d*x]^2*PolyLog[3, E^((-2*I)*ArcSin[c + d*x])] - (3*
I)*ArcSin[c + d*x]*PolyLog[4, E^((-2*I)*ArcSin[c + d*x])] - (3*PolyLog[5, E^((-2*I)*ArcSin[c + d*x])])/2))/(16
*d*e)

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Maple [B]  time = 0.045, size = 1295, normalized size = 6.4 \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),x)

[Out]

-4*I/d*a^3*b/e*polylog(2,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))-2*I/d*a^2*b^2/e*arcsin(d*x+c)^3-2*I/d*a^3*b/e*arcsin(
d*x+c)^2-4*I/d*a^3*b/e*polylog(2,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))-12*I/d*a^2*b^2/e*arcsin(d*x+c)*polylog(2,I*(d*
x+c)+(1-(d*x+c)^2)^(1/2))+12/d*a^2*b^2/e*polylog(3,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))+12/d*b^4/e*arcsin(d*x+c)^2*p
olylog(3,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))+12/d*b^4/e*arcsin(d*x+c)^2*polylog(3,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))+1
/d*b^4/e*arcsin(d*x+c)^4*ln(1-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))+1/d*b^4/e*arcsin(d*x+c)^4*ln(1+I*(d*x+c)+(1-(d*x+
c)^2)^(1/2))+12/d*a^2*b^2/e*polylog(3,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))-1/5*I/d*b^4/e*arcsin(d*x+c)^5-4*I/d*b^4/
e*arcsin(d*x+c)^3*polylog(2,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))-4*I/d*b^4/e*arcsin(d*x+c)^3*polylog(2,-I*(d*x+c)-(1
-(d*x+c)^2)^(1/2))+24*I/d*b^4/e*arcsin(d*x+c)*polylog(4,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))+24*I/d*b^4/e*arcsin(d*x
+c)*polylog(4,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))+24/d*a*b^3/e*arcsin(d*x+c)*polylog(3,-I*(d*x+c)-(1-(d*x+c)^2)^(1
/2))+6/d*a^2*b^2/e*arcsin(d*x+c)^2*ln(1-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))+4/d*a*b^3/e*arcsin(d*x+c)^3*ln(1+I*(d*x
+c)+(1-(d*x+c)^2)^(1/2))+4/d*a^3*b/e*arcsin(d*x+c)*ln(1+I*(d*x+c)+(1-(d*x+c)^2)^(1/2))+4/d*a^3*b/e*arcsin(d*x+
c)*ln(1-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))+6/d*a^2*b^2/e*arcsin(d*x+c)^2*ln(1+I*(d*x+c)+(1-(d*x+c)^2)^(1/2))+24/d*
a*b^3/e*arcsin(d*x+c)*polylog(3,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))+4/d*a*b^3/e*arcsin(d*x+c)^3*ln(1-I*(d*x+c)-(1-(
d*x+c)^2)^(1/2))+24*I/d*a*b^3/e*polylog(4,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))+24*I/d*a*b^3/e*polylog(4,I*(d*x+c)+(
1-(d*x+c)^2)^(1/2))-I/d*a*b^3/e*arcsin(d*x+c)^4-24/d*b^4/e*polylog(5,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))+1/d*a^4/e
*ln(d*x+c)-24/d*b^4/e*polylog(5,I*(d*x+c)+(1-(d*x+c)^2)^(1/2))-12*I/d*a*b^3/e*arcsin(d*x+c)^2*polylog(2,I*(d*x
+c)+(1-(d*x+c)^2)^(1/2))-12*I/d*a*b^3/e*arcsin(d*x+c)^2*polylog(2,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))-12*I/d*a^2*b
^2/e*arcsin(d*x+c)*polylog(2,-I*(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),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 e x + c e}, 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),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*e*x + c*e), x)

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

[Out]

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

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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}}{d e x + c e}\,{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),x, algorithm="giac")

[Out]

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