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

Optimal. Leaf size=205 \[ \frac{i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )}{c}-\frac{b^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right )}{2 c}+\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^3}{3 b c}-\frac{\log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^2}{c} \]

[Out]

((I/3)*(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3)/(b*c) - ((a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2*L
og[1 - E^((2*I)*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/c + (I*b*(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*Po
lyLog[2, E^((2*I)*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/c - (b^2*PolyLog[3, E^((2*I)*ArcSin[Sqrt[1 - c*x]/Sqr
t[1 + c*x]])])/(2*c)

________________________________________________________________________________________

Rubi [A]  time = 0.181683, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {6681, 4625, 3717, 2190, 2531, 2282, 6589} \[ \frac{i b \text{PolyLog}\left (2,e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )}{c}-\frac{b^2 \text{PolyLog}\left (3,e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right )}{2 c}+\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^3}{3 b c}-\frac{\log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )}\right ) \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^2}{c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2/(1 - c^2*x^2),x]

[Out]

((I/3)*(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3)/(b*c) - ((a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2*L
og[1 - E^((2*I)*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/c + (I*b*(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*Po
lyLog[2, E^((2*I)*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])])/c - (b^2*PolyLog[3, E^((2*I)*ArcSin[Sqrt[1 - c*x]/Sqr
t[1 + c*x]])])/(2*c)

Rule 6681

Int[((a_.) + (b_.)*(F_)[((c_.)*Sqrt[(d_.) + (e_.)*(x_)])/Sqrt[(f_.) + (g_.)*(x_)]])^(n_.)/((A_.) + (C_.)*(x_)^
2), x_Symbol] :> Dist[(2*e*g)/(C*(e*f - d*g)), Subst[Int[(a + b*F[c*x])^n/x, x], x, Sqrt[d + e*x]/Sqrt[f + g*x
]], x] /; FreeQ[{a, b, c, d, e, f, g, A, C, F}, x] && EqQ[C*d*f - A*e*g, 0] && EqQ[e*f + d*g, 0] && IGtQ[n, 0]

Rule 4625

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tan[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && 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 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}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2}{1-c^2 x^2} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \sin ^{-1}(x)\right )^2}{x} \, dx,x,\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{c}\\ &=-\frac{\operatorname{Subst}\left (\int (a+b x)^2 \cot (x) \, dx,x,\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )}{c}\\ &=\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^3}{3 b c}+\frac{(2 i) \operatorname{Subst}\left (\int \frac{e^{2 i x} (a+b x)^2}{1-e^{2 i x}} \, dx,x,\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )}{c}\\ &=\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^3}{3 b c}-\frac{\left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2 \log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}+\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )}{c}\\ &=\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^3}{3 b c}-\frac{\left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2 \log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}+\frac{i b \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}-\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_2\left (e^{2 i x}\right ) \, dx,x,\sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )}{c}\\ &=\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^3}{3 b c}-\frac{\left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2 \log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}+\frac{i b \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2(x)}{x} \, dx,x,e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{2 c}\\ &=\frac{i \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^3}{3 b c}-\frac{\left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2 \log \left (1-e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}+\frac{i b \left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \text{Li}_2\left (e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{c}-\frac{b^2 \text{Li}_3\left (e^{2 i \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}\right )}{2 c}\\ \end{align*}

Mathematica [F]  time = 0.690831, size = 0, normalized size = 0. \[ \int \frac{\left (a+b \sin ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2}{1-c^2 x^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2/(1 - c^2*x^2),x]

[Out]

Integrate[(a + b*ArcSin[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2/(1 - c^2*x^2), x]

________________________________________________________________________________________

Maple [B]  time = 0.008, size = 681, normalized size = 3.3 \begin{align*}{\frac{{a}^{2}\ln \left ( cx+1 \right ) }{2\,c}}-{\frac{{a}^{2}\ln \left ( cx-1 \right ) }{2\,c}}+{\frac{{\frac{i}{3}}{b}^{2}}{c} \left ( \arcsin \left ({\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}} \right ) \right ) ^{3}}-{\frac{{b}^{2}}{c} \left ( \arcsin \left ({\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}} \right ) \right ) ^{2}\ln \left ( 1-{i\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}}-\sqrt{1-{\frac{-cx+1}{cx+1}}} \right ) }+{\frac{2\,i{b}^{2}}{c}\arcsin \left ({\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}} \right ){\it polylog} \left ( 2,{i\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}}+\sqrt{1-{\frac{-cx+1}{cx+1}}} \right ) }-2\,{\frac{{b}^{2}}{c}{\it polylog} \left ( 3,{\frac{i\sqrt{-cx+1}}{\sqrt{cx+1}}}+\sqrt{1-{\frac{-cx+1}{cx+1}}} \right ) }-{\frac{{b}^{2}}{c} \left ( \arcsin \left ({\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}} \right ) \right ) ^{2}\ln \left ( 1+{i\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}}+\sqrt{1-{\frac{-cx+1}{cx+1}}} \right ) }+{\frac{2\,i{b}^{2}}{c}\arcsin \left ({\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}} \right ){\it polylog} \left ( 2,{-i\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}}-\sqrt{1-{\frac{-cx+1}{cx+1}}} \right ) }-2\,{\frac{{b}^{2}}{c}{\it polylog} \left ( 3,{\frac{-i\sqrt{-cx+1}}{\sqrt{cx+1}}}-\sqrt{1-{\frac{-cx+1}{cx+1}}} \right ) }+{\frac{iab}{c} \left ( \arcsin \left ({\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}} \right ) \right ) ^{2}}-2\,{\frac{ab}{c}\arcsin \left ({\frac{\sqrt{-cx+1}}{\sqrt{cx+1}}} \right ) \ln \left ( 1+{\frac{i\sqrt{-cx+1}}{\sqrt{cx+1}}}+\sqrt{1-{\frac{-cx+1}{cx+1}}} \right ) }+{\frac{2\,iab}{c}{\it polylog} \left ( 2,{-i\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}}-\sqrt{1-{\frac{-cx+1}{cx+1}}} \right ) }-2\,{\frac{ab}{c}\arcsin \left ({\frac{\sqrt{-cx+1}}{\sqrt{cx+1}}} \right ) \ln \left ( 1-{\frac{i\sqrt{-cx+1}}{\sqrt{cx+1}}}-\sqrt{1-{\frac{-cx+1}{cx+1}}} \right ) }+{\frac{2\,iab}{c}{\it polylog} \left ( 2,{i\sqrt{-cx+1}{\frac{1}{\sqrt{cx+1}}}}+\sqrt{1-{\frac{-cx+1}{cx+1}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^2/(-c^2*x^2+1),x)

[Out]

1/2*a^2/c*ln(c*x+1)-1/2*a^2/c*ln(c*x-1)+1/3*I*b^2/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^3-b^2/c*arcsin((-c*x+
1)^(1/2)/(c*x+1)^(1/2))^2*ln(1-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))+2*I*b^2/c*arcsin((-c
*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(2,I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))-2*b^2/c*polylo
g(3,I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))-b^2/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*ln
(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))+2*I*b^2/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))*p
olylog(2,-I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))-2*b^2/c*polylog(3,-I*(-c*x+1)^(1/2)/(c*x+
1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))+I*a*b/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2-2*a*b/c*arcsin((-c*x+1)^(1
/2)/(c*x+1)^(1/2))*ln(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*x+1)/(c*x+1))^(1/2))+2*I*a*b/c*polylog(2,-I*(-c*
x+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))-2*a*b/c*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2))*ln(1-I*(-c*x
+1)^(1/2)/(c*x+1)^(1/2)-(1-(-c*x+1)/(c*x+1))^(1/2))+2*I*a*b/c*polylog(2,I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)+(1-(-c*
x+1)/(c*x+1))^(1/2))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^2/(-c^2*x^2+1),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b^{2} \arcsin \left (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right )^{2} + 2 \, a b \arcsin \left (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right ) + a^{2}}{c^{2} x^{2} - 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^2/(-c^2*x^2+1),x, algorithm="fricas")

[Out]

integral(-(b^2*arcsin(sqrt(-c*x + 1)/sqrt(c*x + 1))^2 + 2*a*b*arcsin(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a^2)/(c^2
*x^2 - 1), x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asin((-c*x+1)**(1/2)/(c*x+1)**(1/2)))**2/(-c**2*x**2+1),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsin((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^2/(-c^2*x^2+1),x, algorithm="giac")

[Out]

Exception raised: TypeError