3.3.12 \(\int \frac {(a+b \text {ArcSin}(c+d x))^4}{(c e+d e x)^3} \, dx\) [212]

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

[Out]

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

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Rubi [A]
time = 0.22, antiderivative size = 198, normalized size of antiderivative = 1.00, 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 = {4889, 12, 4723, 4771, 4721, 3798, 2221, 2611, 2320, 6724} \begin {gather*} -\frac {6 i b^3 \text {Li}_2\left (e^{2 i \text {ArcSin}(c+d x)}\right ) (a+b \text {ArcSin}(c+d x))}{d e^3}+\frac {6 b^2 \log \left (1-e^{2 i \text {ArcSin}(c+d x)}\right ) (a+b \text {ArcSin}(c+d x))^2}{d e^3}-\frac {2 b \sqrt {1-(c+d x)^2} (a+b \text {ArcSin}(c+d x))^3}{d e^3 (c+d x)}-\frac {2 i b (a+b \text {ArcSin}(c+d x))^3}{d e^3}-\frac {(a+b \text {ArcSin}(c+d x))^4}{2 d e^3 (c+d x)^2}+\frac {3 b^4 \text {Li}_3\left (e^{2 i \text {ArcSin}(c+d x)}\right )}{d e^3} \end {gather*}

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 12

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320

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 2611

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 3798

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 4721

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

Rule 4723

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 4771

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/(f*(m + 1)))*Simp[(d
+ e*x^2)^p/(1 - c^2*x^2)^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, m, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]

Rule 4889

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 6724

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 {\text {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^4}{e^3 x^3} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {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) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.85, size = 385, normalized size = 1.94 \begin {gather*} \frac {-\frac {2 a^4}{(c+d x)^2}-\frac {8 a^3 b \sqrt {1-(c+d x)^2}}{c+d x}-\frac {8 a^3 b \text {ArcSin}(c+d x)}{(c+d x)^2}-\frac {2 b^4 \text {ArcSin}(c+d x)^4}{(c+d x)^2}+24 a^2 b^2 \left (-\frac {\sqrt {1-(c+d x)^2} \text {ArcSin}(c+d x)}{c+d x}-\frac {\text {ArcSin}(c+d x)^2}{2 (c+d x)^2}+\log (c+d x)\right )+8 a b^3 \left (-\frac {3 \sqrt {1-(c+d x)^2} \text {ArcSin}(c+d x)^2}{c+d x}-\frac {\text {ArcSin}(c+d x)^3}{(c+d x)^2}+6 \text {ArcSin}(c+d x) \log \left (1-e^{2 i \text {ArcSin}(c+d x)}\right )-3 i \left (\text {ArcSin}(c+d x)^2+\text {PolyLog}\left (2,e^{2 i \text {ArcSin}(c+d x)}\right )\right )\right )+b^4 \left (-i \pi ^3+8 i \text {ArcSin}(c+d x)^3-\frac {8 \sqrt {1-(c+d x)^2} \text {ArcSin}(c+d x)^3}{c+d x}+24 \text {ArcSin}(c+d x)^2 \log \left (1-e^{-2 i \text {ArcSin}(c+d x)}\right )+24 i \text {ArcSin}(c+d x) \text {PolyLog}\left (2,e^{-2 i \text {ArcSin}(c+d x)}\right )+12 \text {PolyLog}\left (3,e^{-2 i \text {ArcSin}(c+d x)}\right )\right )}{4 d e^3} \end {gather*}

Antiderivative was successfully verified.

[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] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 680 vs. \(2 (228 ) = 456\).
time = 0.38, size = 681, normalized size = 3.44

method	result	size
derivativedivides	\(\frac {-\frac {a^{4}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {12 i a \,b^{3} \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {2 b^{4} \arcsin \left (d x +c \right )^{3} \sqrt {1-\left (d x +c \right )^{2}}}{e^{3} \left (d x +c \right )}-\frac {b^{4} \arcsin \left (d x +c \right )^{4}}{2 e^{3} \left (d x +c \right )^{2}}+\frac {6 b^{4} \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 )}{e^{3}}-\frac {12 i a \,b^{3} \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {12 b^{4} \polylog \left (3, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {6 b^{4} \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 )}{e^{3}}-\frac {6 i a \,b^{3} \arcsin \left (d x +c \right )^{2}}{e^{3}}+\frac {12 b^{4} \polylog \left (3, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {12 i b^{4} \arcsin \left (d x +c \right ) \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {6 a \,b^{3} \arcsin \left (d x +c \right )^{2} \sqrt {1-\left (d x +c \right )^{2}}}{e^{3} \left (d x +c \right )}-\frac {2 a \,b^{3} \arcsin \left (d x +c \right )^{3}}{e^{3} \left (d x +c \right )^{2}}+\frac {12 a \,b^{3} \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {12 a \,b^{3} \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {2 i b^{4} \arcsin \left (d x +c \right )^{3}}{e^{3}}-\frac {12 i b^{4} \arcsin \left (d x +c \right ) \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {3 a^{2} b^{2} \arcsin \left (d x +c \right )^{2}}{e^{3} \left (d x +c \right )^{2}}-\frac {6 a^{2} b^{2} \arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{e^{3} \left (d x +c \right )}+\frac {6 a^{2} b^{2} \ln \left (d x +c \right )}{e^{3}}+\frac {4 a^{3} b \left (-\frac {\arcsin \left (d x +c \right )}{2 \left (d x +c \right )^{2}}-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{2 \left (d x +c \right )}\right )}{e^{3}}}{d}\)	\(681\)
default	\(\frac {-\frac {a^{4}}{2 e^{3} \left (d x +c \right )^{2}}-\frac {12 i a \,b^{3} \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {2 b^{4} \arcsin \left (d x +c \right )^{3} \sqrt {1-\left (d x +c \right )^{2}}}{e^{3} \left (d x +c \right )}-\frac {b^{4} \arcsin \left (d x +c \right )^{4}}{2 e^{3} \left (d x +c \right )^{2}}+\frac {6 b^{4} \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 )}{e^{3}}-\frac {12 i a \,b^{3} \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {12 b^{4} \polylog \left (3, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {6 b^{4} \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 )}{e^{3}}-\frac {6 i a \,b^{3} \arcsin \left (d x +c \right )^{2}}{e^{3}}+\frac {12 b^{4} \polylog \left (3, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {12 i b^{4} \arcsin \left (d x +c \right ) \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {6 a \,b^{3} \arcsin \left (d x +c \right )^{2} \sqrt {1-\left (d x +c \right )^{2}}}{e^{3} \left (d x +c \right )}-\frac {2 a \,b^{3} \arcsin \left (d x +c \right )^{3}}{e^{3} \left (d x +c \right )^{2}}+\frac {12 a \,b^{3} \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}+\frac {12 a \,b^{3} \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {2 i b^{4} \arcsin \left (d x +c \right )^{3}}{e^{3}}-\frac {12 i b^{4} \arcsin \left (d x +c \right ) \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{e^{3}}-\frac {3 a^{2} b^{2} \arcsin \left (d x +c \right )^{2}}{e^{3} \left (d x +c \right )^{2}}-\frac {6 a^{2} b^{2} \arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{e^{3} \left (d x +c \right )}+\frac {6 a^{2} b^{2} \ln \left (d x +c \right )}{e^{3}}+\frac {4 a^{3} b \left (-\frac {\arcsin \left (d x +c \right )}{2 \left (d x +c \right )^{2}}-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{2 \left (d x +c \right )}\right )}{e^{3}}}{d}\)	\(681\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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]

Timed out

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)*e^(-3)/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^4}{{\left (c\,e+d\,e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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