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

Optimal. Leaf size=169 \[ \frac {3 b^2 \text {Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{2 d e}-\frac {3 i b \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e}-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 b d e}+\frac {\log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e}+\frac {3 i b^3 \text {Li}_4\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{4 d e} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 9, 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} \[ \frac {3 b^2 \text {PolyLog}\left (3,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{2 d e}-\frac {3 i b \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^2}{2 d e}+\frac {3 i b^3 \text {PolyLog}\left (4,e^{2 i \sin ^{-1}(c+d x)}\right )}{4 d e}-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 b d e}+\frac {\log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right ) \left (a+b \sin ^{-1}(c+d x)\right )^3}{d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

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 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 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 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 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]

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]

Rubi steps

\begin {align*} \int \frac {\left (a+b \sin ^{-1}(c+d x)\right )^3}{c e+d e x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^3}{e x} \, dx,x,c+d x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^3}{x} \, dx,x,c+d x\right )}{d e}\\ &=\frac {\operatorname {Subst}\left (\int (a+b x)^3 \cot (x) \, dx,x,\sin ^{-1}(c+d x)\right )}{d e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 b d e}-\frac {(2 i) \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)^3}{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 )^4}{4 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {(3 b) \operatorname {Subst}\left (\int (a+b x)^2 \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 )^4}{4 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {3 i b \left (a+b \sin ^{-1}(c+d x)\right )^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}+\frac {\left (3 i b^2\right ) \operatorname {Subst}\left (\int (a+b x) \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 )^4}{4 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {3 i b \left (a+b \sin ^{-1}(c+d x)\right )^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}+\frac {3 b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}-\frac {\left (3 b^3\right ) \operatorname {Subst}\left (\int \text {Li}_3\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{2 d e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {3 i b \left (a+b \sin ^{-1}(c+d x)\right )^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}+\frac {3 b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}+\frac {\left (3 i b^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c+d x)}\right )}{4 d e}\\ &=-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )^4}{4 b d e}+\frac {\left (a+b \sin ^{-1}(c+d x)\right )^3 \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )}{d e}-\frac {3 i b \left (a+b \sin ^{-1}(c+d x)\right )^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}+\frac {3 b^2 \left (a+b \sin ^{-1}(c+d x)\right ) \text {Li}_3\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{2 d e}+\frac {3 i b^3 \text {Li}_4\left (e^{2 i \sin ^{-1}(c+d x)}\right )}{4 d e}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 304, normalized size = 1.80 \[ -\frac {i \left (64 i a^3 \log (c+d x)+96 a^2 b \text {Li}_2\left (e^{2 i \sin ^{-1}(c+d x)}\right )+96 a^2 b \sin ^{-1}(c+d x)^2+192 i a^2 b \sin ^{-1}(c+d x) \log \left (1-e^{2 i \sin ^{-1}(c+d x)}\right )-96 b^2 \sin ^{-1}(c+d x) \text {Li}_2\left (e^{-2 i \sin ^{-1}(c+d x)}\right ) \left (2 a+b \sin ^{-1}(c+d x)\right )+96 i a b^2 \text {Li}_3\left (e^{-2 i \sin ^{-1}(c+d x)}\right )-64 a b^2 \sin ^{-1}(c+d x)^3+192 i a b^2 \sin ^{-1}(c+d x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c+d x)}\right )+8 \pi ^3 a b^2+96 i b^3 \sin ^{-1}(c+d x) \text {Li}_3\left (e^{-2 i \sin ^{-1}(c+d x)}\right )+48 b^3 \text {Li}_4\left (e^{-2 i \sin ^{-1}(c+d x)}\right )-16 b^3 \sin ^{-1}(c+d x)^4+64 i b^3 \sin ^{-1}(c+d x)^3 \log \left (1-e^{-2 i \sin ^{-1}(c+d x)}\right )+\pi ^4 b^3\right )}{64 d e} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-1/64*I)*(8*a*b^2*Pi^3 + b^3*Pi^4 + 96*a^2*b*ArcSin[c + d*x]^2 - 64*a*b^2*ArcSin[c + d*x]^3 - 16*b^3*ArcSin[
c + d*x]^4 + (192*I)*a*b^2*ArcSin[c + d*x]^2*Log[1 - E^((-2*I)*ArcSin[c + d*x])] + (64*I)*b^3*ArcSin[c + d*x]^
3*Log[1 - E^((-2*I)*ArcSin[c + d*x])] + (192*I)*a^2*b*ArcSin[c + d*x]*Log[1 - E^((2*I)*ArcSin[c + d*x])] + (64
*I)*a^3*Log[c + d*x] - 96*b^2*ArcSin[c + d*x]*(2*a + b*ArcSin[c + d*x])*PolyLog[2, E^((-2*I)*ArcSin[c + d*x])]
+ 96*a^2*b*PolyLog[2, E^((2*I)*ArcSin[c + d*x])] + (96*I)*a*b^2*PolyLog[3, E^((-2*I)*ArcSin[c + d*x])] + (96*
I)*b^3*ArcSin[c + d*x]*PolyLog[3, E^((-2*I)*ArcSin[c + d*x])] + 48*b^3*PolyLog[4, E^((-2*I)*ArcSin[c + d*x])])
)/(d*e)

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \arcsin \left (d x + c\right )^{3} + 3 \, a b^{2} \arcsin \left (d x + c\right )^{2} + 3 \, a^{2} b \arcsin \left (d x + c\right ) + a^{3}}{d e x + c e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.08, size = 828, normalized size = 4.90 \[ \frac {a^{3} \ln \left (d x +c \right )}{d e}-\frac {3 i b^{3} \arcsin \left (d x +c \right )^{2} \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}+\frac {b^{3} \arcsin \left (d x +c \right )^{3} \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}-\frac {6 i a \,b^{2} \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}+\frac {6 b^{3} \arcsin \left (d x +c \right ) \polylog \left (3, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}-\frac {3 i a^{2} b \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}+\frac {b^{3} \arcsin \left (d x +c \right )^{3} \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}-\frac {i a \,b^{2} \arcsin \left (d x +c \right )^{3}}{d e}+\frac {6 b^{3} \arcsin \left (d x +c \right ) \polylog \left (3, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}-\frac {3 i a^{2} b \arcsin \left (d x +c \right )^{2}}{2 d e}-\frac {3 i a^{2} b \polylog \left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}+\frac {3 a \,b^{2} \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 )}{d e}+\frac {6 i b^{3} \polylog \left (4, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}+\frac {6 a \,b^{2} \polylog \left (3, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}+\frac {3 a \,b^{2} \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 )}{d e}+\frac {6 i b^{3} \polylog \left (4, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}+\frac {6 a \,b^{2} \polylog \left (3, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e}-\frac {6 i a \,b^{2} \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}+\frac {3 a^{2} b \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}+\frac {3 a^{2} b \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}-\frac {i b^{3} \arcsin \left (d x +c \right )^{4}}{4 d e}-\frac {3 i b^{3} \arcsin \left (d x +c \right )^{2} \polylog \left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )}{d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a^{3} \log \left (d e x + c e\right )}{d e} + \int \frac {b^{3} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{3} + 3 \, a b^{2} \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )^{2} + 3 \, a^{2} b \arctan \left (d x + c, \sqrt {d x + c + 1} \sqrt {-d x - c + 1}\right )}{d e x + c e}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*log(d*e*x + c*e)/(d*e) + integrate((b^3*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))^3 + 3*a*b^2
*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d*x - c + 1))^2 + 3*a^2*b*arctan2(d*x + c, sqrt(d*x + c + 1)*sqrt(-d
*x - c + 1)))/(d*e*x + c*e), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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