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

Optimal. Leaf size=131 \[ \frac{i b \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac{i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac{b^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )}{2 d}+\frac{b^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )}{2 d}-\frac{2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d} \]

[Out]

(-2*(a + b*ArcSin[c*x])^2*ArcTanh[E^((2*I)*ArcSin[c*x])])/d + (I*b*(a + b*ArcSin[c*x])*PolyLog[2, -E^((2*I)*Ar
cSin[c*x])])/d - (I*b*(a + b*ArcSin[c*x])*PolyLog[2, E^((2*I)*ArcSin[c*x])])/d - (b^2*PolyLog[3, -E^((2*I)*Arc
Sin[c*x])])/(2*d) + (b^2*PolyLog[3, E^((2*I)*ArcSin[c*x])])/(2*d)

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Rubi [A]  time = 0.196594, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {4679, 4419, 4183, 2531, 2282, 6589} \[ \frac{i b \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac{i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d}-\frac{b^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )}{2 d}+\frac{b^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}(c x)}\right )}{2 d}-\frac{2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*ArcSin[c*x])^2*ArcTanh[E^((2*I)*ArcSin[c*x])])/d + (I*b*(a + b*ArcSin[c*x])*PolyLog[2, -E^((2*I)*Ar
cSin[c*x])])/d - (I*b*(a + b*ArcSin[c*x])*PolyLog[2, E^((2*I)*ArcSin[c*x])])/d - (b^2*PolyLog[3, -E^((2*I)*Arc
Sin[c*x])])/(2*d) + (b^2*PolyLog[3, E^((2*I)*ArcSin[c*x])])/(2*d)

Rule 4679

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

Rule 4419

Int[Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sec[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dist[
2^n, Int[(c + d*x)^m*Csc[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d, m}, x] && IntegerQ[n] && RationalQ[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 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{\left (a+b \sin ^{-1}(c x)\right )^2}{x \left (d-c^2 d x^2\right )} \, dx &=\frac{\operatorname{Subst}\left (\int (a+b x)^2 \csc (x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{d}\\ &=\frac{2 \operatorname{Subst}\left (\int (a+b x)^2 \csc (2 x) \, dx,x,\sin ^{-1}(c x)\right )}{d}\\ &=-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d}-\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}+\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}\\ &=-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d}+\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d}-\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d}-\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d}\\ &=-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d}+\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d}-\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 d}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 d}\\ &=-\frac{2 \left (a+b \sin ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d}+\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d}-\frac{i b \left (a+b \sin ^{-1}(c x)\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d}-\frac{b^2 \text{Li}_3\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 d}+\frac{b^2 \text{Li}_3\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.194788, size = 254, normalized size = 1.94 \[ \frac{24 i b \text{PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-24 i a b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )+24 i b^2 \sin ^{-1}(c x) \text{PolyLog}\left (2,e^{-2 i \sin ^{-1}(c x)}\right )+12 b^2 \text{PolyLog}\left (3,e^{-2 i \sin ^{-1}(c x)}\right )-12 b^2 \text{PolyLog}\left (3,-e^{2 i \sin ^{-1}(c x)}\right )-12 a^2 \log \left (1-c^2 x^2\right )+24 a^2 \log (c x)+48 a b \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-48 a b \sin ^{-1}(c x) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )+16 i b^2 \sin ^{-1}(c x)^3+24 b^2 \sin ^{-1}(c x)^2 \log \left (1-e^{-2 i \sin ^{-1}(c x)}\right )-24 b^2 \sin ^{-1}(c x)^2 \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )-i \pi ^3 b^2}{24 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I)*b^2*Pi^3 + (16*I)*b^2*ArcSin[c*x]^3 + 24*b^2*ArcSin[c*x]^2*Log[1 - E^((-2*I)*ArcSin[c*x])] + 48*a*b*ArcS
in[c*x]*Log[1 - E^((2*I)*ArcSin[c*x])] - 48*a*b*ArcSin[c*x]*Log[1 + E^((2*I)*ArcSin[c*x])] - 24*b^2*ArcSin[c*x
]^2*Log[1 + E^((2*I)*ArcSin[c*x])] + 24*a^2*Log[c*x] - 12*a^2*Log[1 - c^2*x^2] + (24*I)*b^2*ArcSin[c*x]*PolyLo
g[2, E^((-2*I)*ArcSin[c*x])] + (24*I)*b*(a + b*ArcSin[c*x])*PolyLog[2, -E^((2*I)*ArcSin[c*x])] - (24*I)*a*b*Po
lyLog[2, E^((2*I)*ArcSin[c*x])] + 12*b^2*PolyLog[3, E^((-2*I)*ArcSin[c*x])] - 12*b^2*PolyLog[3, -E^((2*I)*ArcS
in[c*x])])/(24*d)

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Maple [B]  time = 0.092, size = 529, normalized size = 4. \begin{align*} -{\frac{{a}^{2}\ln \left ( cx-1 \right ) }{2\,d}}-{\frac{{a}^{2}\ln \left ( cx+1 \right ) }{2\,d}}+{\frac{{a}^{2}\ln \left ( cx \right ) }{d}}+{\frac{{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{d}\ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{2\,i{b}^{2}\arcsin \left ( cx \right ) }{d}{\it polylog} \left ( 2,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+2\,{\frac{{b}^{2}{\it polylog} \left ( 3,-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) }{d}}+{\frac{{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{d}\ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{2\,i{b}^{2}\arcsin \left ( cx \right ) }{d}{\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }+2\,{\frac{{b}^{2}{\it polylog} \left ( 3,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }{d}}-{\frac{{b}^{2} \left ( \arcsin \left ( cx \right ) \right ) ^{2}}{d}\ln \left ( 1+ \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }-{\frac{2\,iab}{d}{\it polylog} \left ( 2,icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }-{\frac{{b}^{2}}{2\,d}{\it polylog} \left ( 3,- \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+2\,{\frac{ab\arcsin \left ( cx \right ) \ln \left ( 1+icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) }{d}}+{\frac{i{b}^{2}\arcsin \left ( cx \right ) }{d}{\it polylog} \left ( 2,- \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }+2\,{\frac{ab\arcsin \left ( cx \right ) \ln \left ( 1-icx-\sqrt{-{c}^{2}{x}^{2}+1} \right ) }{d}}+{\frac{iab}{d}{\it polylog} \left ( 2,- \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }-2\,{\frac{ab\arcsin \left ( cx \right ) \ln \left ( 1+ \left ( icx+\sqrt{-{c}^{2}{x}^{2}+1} \right ) ^{2} \right ) }{d}}-{\frac{2\,iab}{d}{\it polylog} \left ( 2,-icx-\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/(-c^2*d*x^2+d),x)

[Out]

-1/2*a^2/d*ln(c*x-1)-1/2*a^2/d*ln(c*x+1)+a^2/d*ln(c*x)+b^2/d*arcsin(c*x)^2*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))-2*I*
b^2/d*arcsin(c*x)*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))+2*b^2/d*polylog(3,-I*c*x-(-c^2*x^2+1)^(1/2))+b^2/d*arcs
in(c*x)^2*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))-2*I*b^2/d*arcsin(c*x)*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))+2*b^2/d*pol
ylog(3,I*c*x+(-c^2*x^2+1)^(1/2))-b^2/d*arcsin(c*x)^2*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)-2*I*a*b/d*polylog(2,I*
c*x+(-c^2*x^2+1)^(1/2))-1/2*b^2*polylog(3,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)/d+2*a*b/d*arcsin(c*x)*ln(1+I*c*x+(-c^
2*x^2+1)^(1/2))+I*b^2/d*arcsin(c*x)*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)+2*a*b/d*arcsin(c*x)*ln(1-I*c*x-(-
c^2*x^2+1)^(1/2))+I*a*b/d*polylog(2,-(I*c*x+(-c^2*x^2+1)^(1/2))^2)-2*a*b/d*arcsin(c*x)*ln(1+(I*c*x+(-c^2*x^2+1
)^(1/2))^2)-2*I*a*b/d*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a^2*(log(c*x + 1)/d + log(c*x - 1)/d - 2*log(x)/d) - integrate((b^2*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x
+ 1))^2 + 2*a*b*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))/(c^2*d*x^3 - d*x), 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}}{c^{2} d x^{3} - d x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)/(c^2*d*x^3 - d*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}{{\left (c^{2} d x^{2} - d\right )} x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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