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 \[ -\frac {3 i \sin ^{-1}(a+b x) \text {Li}_2\left (-e^{2 i \sin ^{-1}(a+b x)}\right )}{b}+\frac {3 \text {Li}_3\left (-e^{2 i \sin ^{-1}(a+b x)}\right )}{2 b}+\frac {(a+b x) \sin ^{-1}(a+b x)^3}{b \sqrt {1-(a+b x)^2}}-\frac {i \sin ^{-1}(a+b x)^3}{b}+\frac {3 \sin ^{-1}(a+b x)^2 \log \left (1+e^{2 i \sin ^{-1}(a+b x)}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 128, normalized size of antiderivative = 1.00, 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 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 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 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 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 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*}

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Mathematica [A]  time = 0.63, size = 144, normalized size = 1.12 \[ \frac {2 \sin ^{-1}(a+b x)^2 \left (\frac {\left (-i \sqrt {-a^2-2 a b x-b^2 x^2+1}+a+b x\right ) \sin ^{-1}(a+b x)}{\sqrt {-a^2-2 a b x-b^2 x^2+1}}+3 \log \left (1+e^{2 i \sin ^{-1}(a+b x)}\right )\right )-6 i \sin ^{-1}(a+b x) \text {Li}_2\left (-e^{2 i \sin ^{-1}(a+b x)}\right )+3 \text {Li}_3\left (-e^{2 i \sin ^{-1}(a+b x)}\right )}{2 b} \]

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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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} \arcsin \left (b x + a\right )^{3}}{b^{4} x^{4} + 4 \, a b^{3} x^{3} + 2 \, {\left (3 \, a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\arcsin \left (b x + a\right )^{3}}{{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [A]  time = 0.40, size = 235, normalized size = 1.84 \[ \frac {\left (-\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, x b +i x^{2} b^{2}-\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}\, a +2 i x a b +i a^{2}-i\right ) \arcsin \left (b x +a \right )^{3}}{b \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )}-\frac {2 i \arcsin \left (b x +a \right )^{3}}{b}-\frac {3 i \arcsin \left (b x +a \right ) \polylog \left (2, -\left (i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )^{2}\right )}{b}+\frac {3 \arcsin \left (b x +a \right )^{2} \ln \left (1+\left (i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )^{2}\right )}{b}+\frac {3 \polylog \left (3, -\left (i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )^{2}\right )}{2 b} \]

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*I*arcsin(b*x+a)*polylog(2,-(I*(b*x+a)+(1-(b*x+a)^2)^(1/
2))^2)/b+3*arcsin(b*x+a)^2*ln(1+(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 [A]  time = 0.67, size = 137, normalized size = 1.07 \[ \frac {3}{2} \, b {\left (\frac {\log \left (b x + a + 1\right )}{b^{2}} + \frac {\log \left (b x + a - 1\right )}{b^{2}}\right )} \arcsin \left (b x + a\right )^{2} + {\left (\frac {b^{2} x}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}} + \frac {a b}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}\right )} \arcsin \left (b x + a\right )^{3} \]

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]

3/2*b*(log(b*x + a + 1)/b^2 + log(b*x + a - 1)/b^2)*arcsin(b*x + a)^2 + (b^2*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt
(-b^2*x^2 - 2*a*b*x - a^2 + 1)) + a*b/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)))*arcsin(b
*x + a)^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {asin}\left (a+b\,x\right )}^3}{{\left (-a^2-2\,a\,b\,x-b^2\,x^2+1\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asin}^{3}{\left (a + b x \right )}}{\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac {3}{2}}}\, dx \]

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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