3.333 \(\int \frac{\sin ^{-1}(a+b x)^3}{(1-a^2-2 a b x-b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=128 \[ -\frac{3 i \sin ^{-1}(a+b x) \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(a+b x)}\right )}{b}+\frac{3 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(a+b x)}\right )}{2 b}+\frac{(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt{1-(a+b x)^2}}-\frac{i \sin ^{-1}(a+b x)^3}{b}+\frac{3 \sin ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sin ^{-1}(a+b x)}\right )}{b} \]

[Out]

((-I)*ArcSin[a + b*x]^3)/b + ((a + b*x)*ArcSin[a + b*x]^3)/(b*Sqrt[1 - (a + b*x)^2]) + (3*ArcSin[a + b*x]^2*Lo
g[1 + E^((2*I)*ArcSin[a + b*x])])/b - ((3*I)*ArcSin[a + b*x]*PolyLog[2, -E^((2*I)*ArcSin[a + b*x])])/b + (3*Po
lyLog[3, -E^((2*I)*ArcSin[a + b*x])])/(2*b)

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Rubi [A]  time = 0.206506, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {4807, 4651, 4675, 3719, 2190, 2531, 2282, 6589} \[ -\frac{3 i \sin ^{-1}(a+b x) \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(a+b x)}\right )}{b}+\frac{3 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(a+b x)}\right )}{2 b}+\frac{(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt{1-(a+b x)^2}}-\frac{i \sin ^{-1}(a+b x)^3}{b}+\frac{3 \sin ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sin ^{-1}(a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a + b*x]^3/(1 - a^2 - 2*a*b*x - b^2*x^2)^(3/2),x]

[Out]

((-I)*ArcSin[a + b*x]^3)/b + ((a + b*x)*ArcSin[a + b*x]^3)/(b*Sqrt[1 - (a + b*x)^2]) + (3*ArcSin[a + b*x]^2*Lo
g[1 + E^((2*I)*ArcSin[a + b*x])])/b - ((3*I)*ArcSin[a + b*x]*PolyLog[2, -E^((2*I)*ArcSin[a + b*x])])/b + (3*Po
lyLog[3, -E^((2*I)*ArcSin[a + b*x])])/(2*b)

Rule 4807

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + (C_.)*(x_)^2)^(p_.), x_Symbol] :> Di
st[1/d, Subst[Int[(-(C/d^2) + (C*x^2)/d^2)^p*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A,
B, C, n, p}, x] && EqQ[B*(1 - c^2) + 2*A*c*d, 0] && EqQ[2*c*C - B*d, 0]

Rule 4651

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSin[c
*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n)/Sqrt[d], Int[(x*(a + b*ArcSin[c*x])^(n - 1))/(d + e*x^2), x], x
] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && GtQ[d, 0]

Rule 4675

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

Rule 3719

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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*(e + f*x)))/(1 + E^(2*I*(e + f*x))), x], x] /; FreeQ[{c, d, e, f}, x] &&
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{\sin ^{-1}(a+b x)^3}{\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sin ^{-1}(x)^3}{\left (1-x^2\right )^{3/2}} \, dx,x,a+b x\right )}{b}\\ &=\frac{(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt{1-(a+b x)^2}}-\frac{3 \operatorname{Subst}\left (\int \frac{x \sin ^{-1}(x)^2}{1-x^2} \, dx,x,a+b x\right )}{b}\\ &=\frac{(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt{1-(a+b x)^2}}-\frac{3 \operatorname{Subst}\left (\int x^2 \tan (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac{i \sin ^{-1}(a+b x)^3}{b}+\frac{(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt{1-(a+b x)^2}}+\frac{(6 i) \operatorname{Subst}\left (\int \frac{e^{2 i x} x^2}{1+e^{2 i x}} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac{i \sin ^{-1}(a+b x)^3}{b}+\frac{(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt{1-(a+b x)^2}}+\frac{3 \sin ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sin ^{-1}(a+b x)}\right )}{b}-\frac{6 \operatorname{Subst}\left (\int x \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac{i \sin ^{-1}(a+b x)^3}{b}+\frac{(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt{1-(a+b x)^2}}+\frac{3 \sin ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sin ^{-1}(a+b x)}\right )}{b}-\frac{3 i \sin ^{-1}(a+b x) \text{Li}_2\left (-e^{2 i \sin ^{-1}(a+b x)}\right )}{b}+\frac{(3 i) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=-\frac{i \sin ^{-1}(a+b x)^3}{b}+\frac{(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt{1-(a+b x)^2}}+\frac{3 \sin ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sin ^{-1}(a+b x)}\right )}{b}-\frac{3 i \sin ^{-1}(a+b x) \text{Li}_2\left (-e^{2 i \sin ^{-1}(a+b x)}\right )}{b}+\frac{3 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(a+b x)}\right )}{2 b}\\ &=-\frac{i \sin ^{-1}(a+b x)^3}{b}+\frac{(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt{1-(a+b x)^2}}+\frac{3 \sin ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sin ^{-1}(a+b x)}\right )}{b}-\frac{3 i \sin ^{-1}(a+b x) \text{Li}_2\left (-e^{2 i \sin ^{-1}(a+b x)}\right )}{b}+\frac{3 \text{Li}_3\left (-e^{2 i \sin ^{-1}(a+b x)}\right )}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.572127, size = 144, normalized size = 1.12 \[ \frac{-6 i \sin ^{-1}(a+b x) \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(a+b x)}\right )+3 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(a+b x)}\right )+2 \sin ^{-1}(a+b x)^2 \left (\frac{\left (-i \sqrt{-a^2-2 a b x-b^2 x^2+1}+a+b x\right ) \sin ^{-1}(a+b x)}{\sqrt{-a^2-2 a b x-b^2 x^2+1}}+3 \log \left (1+e^{2 i \sin ^{-1}(a+b x)}\right )\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a + b*x]^3/(1 - a^2 - 2*a*b*x - b^2*x^2)^(3/2),x]

[Out]

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

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Maple [A]  time = 0.122, size = 235, normalized size = 1.8 \begin{align*}{\frac{ \left ( \arcsin \left ( bx+a \right ) \right ) ^{3}}{b \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}-1 \right ) } \left ( -\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}xb+i{x}^{2}{b}^{2}-\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}a+2\,ixab+i{a}^{2}-i \right ) }-{\frac{2\,i \left ( \arcsin \left ( bx+a \right ) \right ) ^{3}}{b}}+3\,{\frac{ \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}\ln \left ( 1+ \left ( i \left ( bx+a \right ) +\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) ^{2} \right ) }{b}}-{\frac{3\,i\arcsin \left ( bx+a \right ) }{b}{\it polylog} \left ( 2,- \left ( i \left ( bx+a \right ) +\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) ^{2} \right ) }+{\frac{3}{2\,b}{\it polylog} \left ( 3,- \left ( i \left ( bx+a \right ) +\sqrt{1- \left ( bx+a \right ) ^{2}} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsin(b*x+a)^3/(-b^2*x^2-2*a*b*x-a^2+1)^(3/2),x)

[Out]

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

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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(arcsin(b*x+a)^3/(-b^2*x^2-2*a*b*x-a^2+1)^(3/2),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{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} \arcsin \left (b x + a\right )^{3}}{b^{4} x^{4} + 4 \, a b^{3} x^{3} + 2 \,{\left (3 \, a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 4 \,{\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(b*x+a)^3/(-b^2*x^2-2*a*b*x-a^2+1)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*arcsin(b*x + a)^3/(b^4*x^4 + 4*a*b^3*x^3 + 2*(3*a^2 - 1)*b^2*x^2 +
a^4 + 4*(a^3 - a)*b*x - 2*a^2 + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asin}^{3}{\left (a + b x \right )}}{\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asin(b*x+a)**3/(-b**2*x**2-2*a*b*x-a**2+1)**(3/2),x)

[Out]

Integral(asin(a + b*x)**3/(-(a + b*x - 1)*(a + b*x + 1))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\arcsin \left (b x + a\right )^{3}}{{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(b*x+a)^3/(-b^2*x^2-2*a*b*x-a^2+1)^(3/2),x, algorithm="giac")

[Out]

integrate(arcsin(b*x + a)^3/(-b^2*x^2 - 2*a*b*x - a^2 + 1)^(3/2), x)