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

Optimal. Leaf size=268 \[ \frac{b \text{PolyLog}\left (2,1-\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{c x+1}}}\right ) \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )}{c}-\frac{b \text{PolyLog}\left (2,\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{c x+1}}}-1\right ) \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )}{c}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{c x+1}}}\right )}{2 c}+\frac{b^2 \text{PolyLog}\left (3,\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{c x+1}}}-1\right )}{2 c}-\frac{2 \tanh ^{-1}\left (1-\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{c x+1}}}\right ) \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^2}{c} \]

[Out]

(-2*(a + b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2*ArcTanh[1 - 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])])/c + (b*(a
+ b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*PolyLog[2, 1 - 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])])/c - (b*(a + b*
ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*PolyLog[2, -1 + 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])])/c - (b^2*PolyLog[3
, 1 - 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])])/(2*c) + (b^2*PolyLog[3, -1 + 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])])
/(2*c)

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Rubi [A]  time = 0.31119, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {6681, 5914, 6052, 5948, 6058, 6610} \[ \frac{b \text{PolyLog}\left (2,1-\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{c x+1}}}\right ) \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )}{c}-\frac{b \text{PolyLog}\left (2,\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{c x+1}}}-1\right ) \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )}{c}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{c x+1}}}\right )}{2 c}+\frac{b^2 \text{PolyLog}\left (3,\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{c x+1}}}-1\right )}{2 c}-\frac{2 \tanh ^{-1}\left (1-\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{c x+1}}}\right ) \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^2}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2*ArcTanh[1 - 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])])/c + (b*(a
+ b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*PolyLog[2, 1 - 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])])/c - (b*(a + b*
ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*PolyLog[2, -1 + 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])])/c - (b^2*PolyLog[3
, 1 - 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])])/(2*c) + (b^2*PolyLog[3, -1 + 2/(1 - Sqrt[1 - c*x]/Sqrt[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 5914

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)/(x_), x_Symbol] :> Simp[2*(a + b*ArcTanh[c*x])^p*ArcTanh[1 - 2/(1
- c*x)], x] - Dist[2*b*c*p, Int[((a + b*ArcTanh[c*x])^(p - 1)*ArcTanh[1 - 2/(1 - c*x)])/(1 - c^2*x^2), x], x]
/; FreeQ[{a, b, c}, x] && IGtQ[p, 1]

Rule 6052

Int[(ArcTanh[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[
(Log[1 + u]*(a + b*ArcTanh[c*x])^p)/(d + e*x^2), x], x] - Dist[1/2, Int[(Log[1 - u]*(a + b*ArcTanh[c*x])^p)/(d
+ e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[u^2 - (1 - 2/(1 - c*x
))^2, 0]

Rule 5948

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rule 6058

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a + b*ArcT
anh[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p)/2, Int[((a + b*ArcTanh[c*x])^(p - 1)*PolyLog[2, 1 - u]
)/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1 -
2/(1 - c*x))^2, 0]

Rule 6610

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tanh ^{-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 \tanh ^{-1}(x)\right )^2}{x} \, dx,x,\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{c}\\ &=-\frac{2 \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{c}+\frac{(4 b) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}\left (1-\frac{2}{1-x}\right ) \left (a+b \tanh ^{-1}(x)\right )}{1-x^2} \, dx,x,\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{c}\\ &=-\frac{2 \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{c}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right ) \log \left (2-\frac{2}{1-x}\right )}{1-x^2} \, dx,x,\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{c}-\frac{(2 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right ) \log \left (\frac{2}{1-x}\right )}{1-x^2} \, dx,x,\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{c}\\ &=-\frac{2 \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{c}+\frac{b \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \text{Li}_2\left (1-\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{c}-\frac{b \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \text{Li}_2\left (-1+\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{c}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (1-\frac{2}{1-x}\right )}{1-x^2} \, dx,x,\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{c}+\frac{b^2 \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-1+\frac{2}{1-x}\right )}{1-x^2} \, dx,x,\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{c}\\ &=-\frac{2 \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2 \tanh ^{-1}\left (1-\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{c}+\frac{b \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \text{Li}_2\left (1-\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{c}-\frac{b \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \text{Li}_2\left (-1+\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{c}-\frac{b^2 \text{Li}_3\left (1-\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{2 c}+\frac{b^2 \text{Li}_3\left (-1+\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{2 c}\\ \end{align*}

Mathematica [A]  time = 0.100641, size = 324, normalized size = 1.21 \[ -\frac{b \text{PolyLog}\left (2,-\frac{\sqrt{1-c x}+\sqrt{c x+1}}{\sqrt{1-c x}-\sqrt{c x+1}}\right ) \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )-b \text{PolyLog}\left (2,\frac{\sqrt{1-c x}+\sqrt{c x+1}}{\sqrt{1-c x}-\sqrt{c x+1}}\right ) \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )-\frac{1}{2} b^2 \text{PolyLog}\left (3,-\frac{\sqrt{1-c x}+\sqrt{c x+1}}{\sqrt{1-c x}-\sqrt{c x+1}}\right )+\frac{1}{2} b^2 \text{PolyLog}\left (3,\frac{\sqrt{1-c x}+\sqrt{c x+1}}{\sqrt{1-c x}-\sqrt{c x+1}}\right )+2 \tanh ^{-1}\left (1-\frac{2}{1-\frac{\sqrt{1-c x}}{\sqrt{c x+1}}}\right ) \left (a+b \tanh ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^2}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*(a + b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2*ArcTanh[1 - 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])] + b*(a +
b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*PolyLog[2, -((Sqrt[1 - c*x] + Sqrt[1 + c*x])/(Sqrt[1 - c*x] - Sqrt[1 +
c*x]))] - b*(a + b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*PolyLog[2, (Sqrt[1 - c*x] + Sqrt[1 + c*x])/(Sqrt[1 -
c*x] - Sqrt[1 + c*x])] - (b^2*PolyLog[3, -((Sqrt[1 - c*x] + Sqrt[1 + c*x])/(Sqrt[1 - c*x] - Sqrt[1 + c*x]))])
/2 + (b^2*PolyLog[3, (Sqrt[1 - c*x] + Sqrt[1 + c*x])/(Sqrt[1 - c*x] - Sqrt[1 + c*x])])/2)/c)

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Maple [B]  time = 0.904, size = 676, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh((-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)-b^2/c*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*ln(1+((-c*x+1)^(1/2)/(c
*x+1)^(1/2)+1)/(-(-c*x+1)/(c*x+1)+1)^(1/2))-2*b^2/c*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(2,-((-c*x+1)
^(1/2)/(c*x+1)^(1/2)+1)/(-(-c*x+1)/(c*x+1)+1)^(1/2))+2*b^2/c*polylog(3,-((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)/(-(-c
*x+1)/(c*x+1)+1)^(1/2))+b^2/c*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*ln(((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)^2/(-
(-c*x+1)/(c*x+1)+1)+1)+b^2/c*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(2,-((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)
^2/(-(-c*x+1)/(c*x+1)+1))-1/2*b^2/c*polylog(3,-((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)^2/(-(-c*x+1)/(c*x+1)+1))-b^2/c
*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*ln(1-((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)/(-(-c*x+1)/(c*x+1)+1)^(1/2))-2*
b^2/c*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(2,((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)/(-(-c*x+1)/(c*x+1)+1)^(
1/2))+2*b^2/c*polylog(3,((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)/(-(-c*x+1)/(c*x+1)+1)^(1/2))-2*a*b/c*dilog(1/((-c*x+1
)^(1/2)/(c*x+1)^(1/2)+1)^2*(-(-c*x+1)/(c*x+1)+1))+1/2*a*b/c*dilog(1/((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)^4*(-(-c*x
+1)/(c*x+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 )}{c} - \frac{\log \left (c x - 1\right )}{c}\right )} + \frac{{\left (b^{2} \log \left (c x + 1\right ) - b^{2} \log \left (-c x + 1\right )\right )} \log \left (\sqrt{c x + 1} - \sqrt{-c x + 1}\right )^{2}}{8 \, c} + \int -\frac{2 \,{\left (\sqrt{c x + 1} b^{2} - \sqrt{-c x + 1} b^{2}\right )} \log \left (\sqrt{c x + 1} + \sqrt{-c x + 1}\right )^{2} + 8 \,{\left (\sqrt{c x + 1} a b - \sqrt{-c x + 1} a b\right )} \log \left (\sqrt{c x + 1} + \sqrt{-c x + 1}\right ) -{\left (4 \,{\left (\sqrt{c x + 1} b^{2} - \sqrt{-c x + 1} b^{2}\right )} \log \left (\sqrt{c x + 1} + \sqrt{-c x + 1}\right ) +{\left (8 \, a b -{\left (b^{2} c x - b^{2}\right )} \log \left (c x + 1\right ) +{\left (b^{2} c x - b^{2}\right )} \log \left (-c x + 1\right )\right )} \sqrt{c x + 1} -{\left (8 \, a b -{\left (b^{2} c x + b^{2}\right )} \log \left (c x + 1\right ) +{\left (b^{2} c x + b^{2}\right )} \log \left (-c x + 1\right )\right )} \sqrt{-c x + 1}\right )} \log \left (\sqrt{c x + 1} - \sqrt{-c x + 1}\right )}{8 \,{\left ({\left (c^{2} x^{2} - 1\right )} \sqrt{c x + 1} -{\left (c^{2} x^{2} - 1\right )} \sqrt{-c x + 1}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh((-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) + 1/8*(b^2*log(c*x + 1) - b^2*log(-c*x + 1))*log(sqrt(c*x + 1) - sqr
t(-c*x + 1))^2/c + integrate(-1/8*(2*(sqrt(c*x + 1)*b^2 - sqrt(-c*x + 1)*b^2)*log(sqrt(c*x + 1) + sqrt(-c*x +
1))^2 + 8*(sqrt(c*x + 1)*a*b - sqrt(-c*x + 1)*a*b)*log(sqrt(c*x + 1) + sqrt(-c*x + 1)) - (4*(sqrt(c*x + 1)*b^2
- sqrt(-c*x + 1)*b^2)*log(sqrt(c*x + 1) + sqrt(-c*x + 1)) + (8*a*b - (b^2*c*x - b^2)*log(c*x + 1) + (b^2*c*x
- b^2)*log(-c*x + 1))*sqrt(c*x + 1) - (8*a*b - (b^2*c*x + b^2)*log(c*x + 1) + (b^2*c*x + b^2)*log(-c*x + 1))*s
qrt(-c*x + 1))*log(sqrt(c*x + 1) - sqrt(-c*x + 1)))/((c^2*x^2 - 1)*sqrt(c*x + 1) - (c^2*x^2 - 1)*sqrt(-c*x + 1
)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b^{2} \operatorname{artanh}\left (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right )^{2} + 2 \, a b \operatorname{artanh}\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*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^2/(-c^2*x^2+1),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(a**2/(c**2*x**2 - 1), x) - Integral(b**2*atanh(sqrt(-c*x + 1)/sqrt(c*x + 1))**2/(c**2*x**2 - 1), x)
- Integral(2*a*b*atanh(sqrt(-c*x + 1)/sqrt(c*x + 1))/(c**2*x**2 - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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