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

Optimal. Leaf size=179 \[ -\frac{i \sqrt{1-a^2 x^2} \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(a x)}\right )}{a c \sqrt{c-a^2 c x^2}}+\frac{x \sin ^{-1}(a x)^2}{c \sqrt{c-a^2 c x^2}}-\frac{i \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{a c \sqrt{c-a^2 c x^2}}+\frac{2 \sqrt{1-a^2 x^2} \sin ^{-1}(a x) \log \left (1+e^{2 i \sin ^{-1}(a x)}\right )}{a c \sqrt{c-a^2 c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.128756, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4653, 4675, 3719, 2190, 2279, 2391} \[ -\frac{i \sqrt{1-a^2 x^2} \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(a x)}\right )}{a c \sqrt{c-a^2 c x^2}}+\frac{x \sin ^{-1}(a x)^2}{c \sqrt{c-a^2 c x^2}}-\frac{i \sqrt{1-a^2 x^2} \sin ^{-1}(a x)^2}{a c \sqrt{c-a^2 c x^2}}+\frac{2 \sqrt{1-a^2 x^2} \sin ^{-1}(a x) \log \left (1+e^{2 i \sin ^{-1}(a x)}\right )}{a c \sqrt{c-a^2 c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

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

Mathematica [A]  time = 0.221492, size = 108, normalized size = 0.6 \[ \frac{\sin ^{-1}(a x) \left (a x \sin ^{-1}(a x)+\sqrt{1-a^2 x^2} \left (2 \log \left (1+e^{2 i \sin ^{-1}(a x)}\right )-i \sin ^{-1}(a x)\right )\right )-i \sqrt{1-a^2 x^2} \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(a x)}\right )}{a c \sqrt{c \left (1-a^2 x^2\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.102, size = 169, normalized size = 0.9 \begin{align*} -{\frac{ \left ( \arcsin \left ( ax \right ) \right ) ^{2}}{a{c}^{2} \left ({a}^{2}{x}^{2}-1 \right ) }\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) } \left ( i\sqrt{-{a}^{2}{x}^{2}+1}+ax \right ) }+{\frac{i}{a{c}^{2} \left ({a}^{2}{x}^{2}-1 \right ) }\sqrt{-{a}^{2}{x}^{2}+1}\sqrt{-c \left ({a}^{2}{x}^{2}-1 \right ) } \left ( 2\,i\arcsin \left ( ax \right ) \ln \left ( 1+ \left ( iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) ^{2} \right ) +2\, \left ( \arcsin \left ( ax \right ) \right ) ^{2}+{\it polylog} \left ( 2,- \left ( iax+\sqrt{-{a}^{2}{x}^{2}+1} \right ) ^{2} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{-a^{2} c x^{2} + c} \arcsin \left (a x\right )^{2}}{a^{4} c^{2} x^{4} - 2 \, a^{2} c^{2} x^{2} + c^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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