3.299 \(\int \frac {\sin ^{-1}(a x)^3}{(c-a^2 c x^2)^{3/2}} \, dx\)

Optimal. Leaf size=238 \[ -\frac {3 i \sqrt {1-a^2 x^2} \sin ^{-1}(a x) \text {Li}_2\left (-e^{2 i \sin ^{-1}(a x)}\right )}{a c \sqrt {c-a^2 c x^2}}+\frac {3 \sqrt {1-a^2 x^2} \text {Li}_3\left (-e^{2 i \sin ^{-1}(a x)}\right )}{2 a c \sqrt {c-a^2 c x^2}}+\frac {x \sin ^{-1}(a x)^3}{c \sqrt {c-a^2 c x^2}}-\frac {i \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{a c \sqrt {c-a^2 c x^2}}+\frac {3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2 \log \left (1+e^{2 i \sin ^{-1}(a x)}\right )}{a c \sqrt {c-a^2 c x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 238, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {4653, 4675, 3719, 2190, 2531, 2282, 6589} \[ -\frac {3 i \sqrt {1-a^2 x^2} \sin ^{-1}(a x) \text {PolyLog}\left (2,-e^{2 i \sin ^{-1}(a x)}\right )}{a c \sqrt {c-a^2 c x^2}}+\frac {3 \sqrt {1-a^2 x^2} \text {PolyLog}\left (3,-e^{2 i \sin ^{-1}(a x)}\right )}{2 a c \sqrt {c-a^2 c x^2}}+\frac {x \sin ^{-1}(a x)^3}{c \sqrt {c-a^2 c x^2}}-\frac {i \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{a c \sqrt {c-a^2 c x^2}}+\frac {3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2 \log \left (1+e^{2 i \sin ^{-1}(a x)}\right )}{a c \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Int[ArcSin[a*x]^3/(c - a^2*c*x^2)^(3/2),x]

[Out]

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

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 4653

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[1 - c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSin[c*x
])^(n - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 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 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 x)^3}{\left (c-a^2 c x^2\right )^{3/2}} \, dx &=\frac {x \sin ^{-1}(a x)^3}{c \sqrt {c-a^2 c x^2}}-\frac {\left (3 a \sqrt {1-a^2 x^2}\right ) \int \frac {x \sin ^{-1}(a x)^2}{1-a^2 x^2} \, dx}{c \sqrt {c-a^2 c x^2}}\\ &=\frac {x \sin ^{-1}(a x)^3}{c \sqrt {c-a^2 c x^2}}-\frac {\left (3 \sqrt {1-a^2 x^2}\right ) \operatorname {Subst}\left (\int x^2 \tan (x) \, dx,x,\sin ^{-1}(a x)\right )}{a c \sqrt {c-a^2 c x^2}}\\ &=\frac {x \sin ^{-1}(a x)^3}{c \sqrt {c-a^2 c x^2}}-\frac {i \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{a c \sqrt {c-a^2 c x^2}}+\frac {\left (6 i \sqrt {1-a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{2 i x} x^2}{1+e^{2 i x}} \, dx,x,\sin ^{-1}(a x)\right )}{a c \sqrt {c-a^2 c x^2}}\\ &=\frac {x \sin ^{-1}(a x)^3}{c \sqrt {c-a^2 c x^2}}-\frac {i \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{a c \sqrt {c-a^2 c x^2}}+\frac {3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2 \log \left (1+e^{2 i \sin ^{-1}(a x)}\right )}{a c \sqrt {c-a^2 c x^2}}-\frac {\left (6 \sqrt {1-a^2 x^2}\right ) \operatorname {Subst}\left (\int x \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{a c \sqrt {c-a^2 c x^2}}\\ &=\frac {x \sin ^{-1}(a x)^3}{c \sqrt {c-a^2 c x^2}}-\frac {i \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{a c \sqrt {c-a^2 c x^2}}+\frac {3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2 \log \left (1+e^{2 i \sin ^{-1}(a x)}\right )}{a c \sqrt {c-a^2 c x^2}}-\frac {3 i \sqrt {1-a^2 x^2} \sin ^{-1}(a x) \text {Li}_2\left (-e^{2 i \sin ^{-1}(a x)}\right )}{a c \sqrt {c-a^2 c x^2}}+\frac {\left (3 i \sqrt {1-a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(a x)\right )}{a c \sqrt {c-a^2 c x^2}}\\ &=\frac {x \sin ^{-1}(a x)^3}{c \sqrt {c-a^2 c x^2}}-\frac {i \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{a c \sqrt {c-a^2 c x^2}}+\frac {3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2 \log \left (1+e^{2 i \sin ^{-1}(a x)}\right )}{a c \sqrt {c-a^2 c x^2}}-\frac {3 i \sqrt {1-a^2 x^2} \sin ^{-1}(a x) \text {Li}_2\left (-e^{2 i \sin ^{-1}(a x)}\right )}{a c \sqrt {c-a^2 c x^2}}+\frac {\left (3 \sqrt {1-a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(a x)}\right )}{2 a c \sqrt {c-a^2 c x^2}}\\ &=\frac {x \sin ^{-1}(a x)^3}{c \sqrt {c-a^2 c x^2}}-\frac {i \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^3}{a c \sqrt {c-a^2 c x^2}}+\frac {3 \sqrt {1-a^2 x^2} \sin ^{-1}(a x)^2 \log \left (1+e^{2 i \sin ^{-1}(a x)}\right )}{a c \sqrt {c-a^2 c x^2}}-\frac {3 i \sqrt {1-a^2 x^2} \sin ^{-1}(a x) \text {Li}_2\left (-e^{2 i \sin ^{-1}(a x)}\right )}{a c \sqrt {c-a^2 c x^2}}+\frac {3 \sqrt {1-a^2 x^2} \text {Li}_3\left (-e^{2 i \sin ^{-1}(a x)}\right )}{2 a c \sqrt {c-a^2 c x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.28, size = 157, normalized size = 0.66 \[ \frac {-6 i \sqrt {1-a^2 x^2} \sin ^{-1}(a x) \text {Li}_2\left (-e^{2 i \sin ^{-1}(a x)}\right )+3 \sqrt {1-a^2 x^2} \text {Li}_3\left (-e^{2 i \sin ^{-1}(a x)}\right )+2 \sin ^{-1}(a x)^2 \left (\left (a x-i \sqrt {1-a^2 x^2}\right ) \sin ^{-1}(a x)+3 \sqrt {1-a^2 x^2} \log \left (1+e^{2 i \sin ^{-1}(a x)}\right )\right )}{2 a c \sqrt {c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSin[a*x]^3/(c - a^2*c*x^2)^(3/2),x]

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-a^{2} c x^{2} + c} \arcsin \left (a x\right )^{3}}{a^{4} c^{2} x^{4} - 2 \, a^{2} c^{2} x^{2} + c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.21, size = 203, normalized size = 0.85 \[ -\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \left (i \sqrt {-a^{2} x^{2}+1}+a x \right ) \arcsin \left (a x \right )^{3}}{a \,c^{2} \left (a^{2} x^{2}-1\right )}-\frac {\sqrt {-c \left (a^{2} x^{2}-1\right )}\, \sqrt {-a^{2} x^{2}+1}\, \left (-4 i \arcsin \left (a x \right )^{3}+6 \arcsin \left (a x \right )^{2} \ln \left (1+\left (i a x +\sqrt {-a^{2} x^{2}+1}\right )^{2}\right )-6 i \arcsin \left (a x \right ) \polylog \left (2, -\left (i a x +\sqrt {-a^{2} x^{2}+1}\right )^{2}\right )+3 \polylog \left (3, -\left (i a x +\sqrt {-a^{2} x^{2}+1}\right )^{2}\right )\right )}{2 a \,c^{2} \left (a^{2} x^{2}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.40, size = 49, normalized size = 0.21 \[ \frac {x \arcsin \left (a x\right )^{3}}{\sqrt {-a^{2} c x^{2} + c} c} - \frac {3 \, \arcsin \left (a x\right )^{2} \log \left (x^{2} - \frac {1}{a^{2}}\right )}{2 \, a c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsin(a*x)^3/(-a^2*c*x^2+c)^(3/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asin(a*x)^3/(c - a^2*c*x^2)^(3/2),x)

[Out]

int(asin(a*x)^3/(c - a^2*c*x^2)^(3/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________