3.152 \(\int \frac{(a+b \sin ^{-1}(c x))^2}{x^2} \, dx\)

Optimal. Leaf size=81 \[ 2 i b^2 c \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )-2 i b^2 c \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x}-4 b c \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right ) \]

[Out]

-((a + b*ArcSin[c*x])^2/x) - 4*b*c*(a + b*ArcSin[c*x])*ArcTanh[E^(I*ArcSin[c*x])] + (2*I)*b^2*c*PolyLog[2, -E^
(I*ArcSin[c*x])] - (2*I)*b^2*c*PolyLog[2, E^(I*ArcSin[c*x])]

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Rubi [A]  time = 0.128171, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {4627, 4709, 4183, 2279, 2391} \[ 2 i b^2 c \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )-2 i b^2 c \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x}-4 b c \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[c*x])^2/x^2,x]

[Out]

-((a + b*ArcSin[c*x])^2/x) - 4*b*c*(a + b*ArcSin[c*x])*ArcTanh[E^(I*ArcSin[c*x])] + (2*I)*b^2*c*PolyLog[2, -E^
(I*ArcSin[c*x])] - (2*I)*b^2*c*PolyLog[2, E^(I*ArcSin[c*x])]

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4709

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

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, 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{\left (a+b \sin ^{-1}(c x)\right )^2}{x^2} \, dx &=-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x}+(2 b c) \int \frac{a+b \sin ^{-1}(c x)}{x \sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x}+(2 b c) \operatorname{Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x}-4 b c \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )-\left (2 b^2 c\right ) \operatorname{Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )+\left (2 b^2 c\right ) \operatorname{Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x}-4 b c \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )+\left (2 i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )-\left (2 i b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )\\ &=-\frac{\left (a+b \sin ^{-1}(c x)\right )^2}{x}-4 b c \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )+2 i b^2 c \text{Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )-2 i b^2 c \text{Li}_2\left (e^{i \sin ^{-1}(c x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.225321, size = 126, normalized size = 1.56 \[ -\frac{-i b^2 \left (2 c x \text{PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )-2 c x \text{PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )+i \sin ^{-1}(c x) \left (\sin ^{-1}(c x)+2 c x \left (\log \left (1+e^{i \sin ^{-1}(c x)}\right )-\log \left (1-e^{i \sin ^{-1}(c x)}\right )\right )\right )\right )+a^2+2 a b \left (c x \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )+\sin ^{-1}(c x)\right )}{x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcSin[c*x])^2/x^2,x]

[Out]

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

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Maple [A]  time = 0.056, size = 171, normalized size = 2.1 \begin{align*} -{\frac{{a}^{2}}{x}}-{\frac{{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{x}}+2\,c{b}^{2}\arcsin \left ( cx \right ) \ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) -2\,c{b}^{2}\arcsin \left ( cx \right ) \ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) -2\,i{b}^{2}c{\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) +2\,i{b}^{2}c{\it polylog} \left ( 2,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) -2\,{\frac{ab\arcsin \left ( cx \right ) }{x}}-2\,cab{\it Artanh} \left ({\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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