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

Optimal. Leaf size=283 \[ \frac{i b \text{PolyLog}\left (2,1-\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{c x+1}}}\right ) \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )}{c}-\frac{i b \text{PolyLog}\left (2,-1+\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{c x+1}}}\right ) \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )}{c}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{c x+1}}}\right )}{2 c}-\frac{b^2 \text{PolyLog}\left (3,-1+\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{c x+1}}}\right )}{2 c}-\frac{2 \tanh ^{-1}\left (1-\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{c x+1}}}\right ) \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^2}{c} \]

[Out]

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

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Rubi [A]  time = 0.295276, antiderivative size = 283, 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, 4850, 4988, 4884, 4994, 6610} \[ \frac{i b \text{PolyLog}\left (2,1-\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{c x+1}}}\right ) \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )}{c}-\frac{i b \text{PolyLog}\left (2,-1+\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{c x+1}}}\right ) \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )}{c}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{c x+1}}}\right )}{2 c}-\frac{b^2 \text{PolyLog}\left (3,-1+\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{c x+1}}}\right )}{2 c}-\frac{2 \tanh ^{-1}\left (1-\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{c x+1}}}\right ) \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^2}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2*ArcTanh[1 - 2/(1 + (I*Sqrt[1 - c*x])/Sqrt[1 + c*x])])/c + (I
*b*(a + b*ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*PolyLog[2, 1 - 2/(1 + (I*Sqrt[1 - c*x])/Sqrt[1 + c*x])])/c - (I
*b*(a + b*ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*PolyLog[2, -1 + 2/(1 + (I*Sqrt[1 - c*x])/Sqrt[1 + c*x])])/c + (
b^2*PolyLog[3, 1 - 2/(1 + (I*Sqrt[1 - c*x])/Sqrt[1 + c*x])])/(2*c) - (b^2*PolyLog[3, -1 + 2/(1 + (I*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 4850

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

Rule 4988

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

Rule 4884

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

Rule 4994

Int[(Log[u_]*((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*Arc
Tan[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p*I)/2, Int[((a + b*ArcTan[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[e, c^2*d] && EqQ[(1 - u)^2 - (1 - (2*
I)/(I - 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 \tan ^{-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 \tan ^{-1}(x)\right )^2}{x} \, dx,x,\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{c}\\ &=-\frac{2 \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{c}+\frac{(4 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right ) \tanh ^{-1}\left (1-\frac{2}{1+i x}\right )}{1+x^2} \, dx,x,\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{c}\\ &=-\frac{2 \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{c}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tan ^{-1}(x)\right ) \log \left (2-\frac{2}{1+i 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 \tan ^{-1}(x)\right ) \log \left (\frac{2}{1+i x}\right )}{1+x^2} \, dx,x,\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{c}\\ &=-\frac{2 \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{c}+\frac{i b \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \text{Li}_2\left (1-\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{c}-\frac{i b \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \text{Li}_2\left (-1+\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{c}-\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (1-\frac{2}{1+i x}\right )}{1+x^2} \, dx,x,\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{c}+\frac{\left (i b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (-1+\frac{2}{1+i x}\right )}{1+x^2} \, dx,x,\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )}{c}\\ &=-\frac{2 \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right )^2 \tanh ^{-1}\left (1-\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{c}+\frac{i b \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \text{Li}_2\left (1-\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{c}-\frac{i b \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{1+c x}}\right )\right ) \text{Li}_2\left (-1+\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{c}+\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{2 c}-\frac{b^2 \text{Li}_3\left (-1+\frac{2}{1+\frac{i \sqrt{1-c x}}{\sqrt{1+c x}}}\right )}{2 c}\\ \end{align*}

Mathematica [A]  time = 0.106713, size = 354, normalized size = 1.25 \[ -\frac{2 i b \text{PolyLog}\left (2,-\frac{\sqrt{1-c x}+i \sqrt{c x+1}}{\sqrt{1-c x}-i \sqrt{c x+1}}\right ) \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )-2 i b \text{PolyLog}\left (2,\frac{\sqrt{1-c x}+i \sqrt{c x+1}}{\sqrt{1-c x}-i \sqrt{c x+1}}\right ) \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )+b^2 \text{PolyLog}\left (3,-\frac{\sqrt{1-c x}+i \sqrt{c x+1}}{\sqrt{1-c x}-i \sqrt{c x+1}}\right )-b^2 \text{PolyLog}\left (3,\frac{\sqrt{1-c x}+i \sqrt{c x+1}}{\sqrt{1-c x}-i \sqrt{c x+1}}\right )+4 \tanh ^{-1}\left (1-\frac{2 i}{-\frac{\sqrt{1-c x}}{\sqrt{c x+1}}+i}\right ) \left (a+b \tan ^{-1}\left (\frac{\sqrt{1-c x}}{\sqrt{c x+1}}\right )\right )^2}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 1.218, size = 947, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan((-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*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*ln(1-(1+I*(-c*x+1)^(1/2)/
(c*x+1)^(1/2))/((-c*x+1)/(c*x+1)+1)^(1/2))+2*I*b^2/c*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(2,(1+I*(-c*x
+1)^(1/2)/(c*x+1)^(1/2))/((-c*x+1)/(c*x+1)+1)^(1/2))-2*b^2/c*polylog(3,(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))/((-c
*x+1)/(c*x+1)+1)^(1/2))-b^2/c*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*ln(1+(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))/(
(-c*x+1)/(c*x+1)+1)^(1/2))-I*a*b/c*polylog(2,-(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))^2/((-c*x+1)/(c*x+1)+1))-2*b^2
/c*polylog(3,-(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))/((-c*x+1)/(c*x+1)+1)^(1/2))+b^2/c*arctan((-c*x+1)^(1/2)/(c*x+
1)^(1/2))^2*ln((1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))^2/((-c*x+1)/(c*x+1)+1)+1)-I*b^2/c*arctan((-c*x+1)^(1/2)/(c*x
+1)^(1/2))*polylog(2,-(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))^2/((-c*x+1)/(c*x+1)+1))+1/2*b^2/c*polylog(3,-(1+I*(-c
*x+1)^(1/2)/(c*x+1)^(1/2))^2/((-c*x+1)/(c*x+1)+1))-2*a*b/c*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2))*ln(1+(1+I*(-c*
x+1)^(1/2)/(c*x+1)^(1/2))/((-c*x+1)/(c*x+1)+1)^(1/2))+2*I*a*b/c*polylog(2,(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))/(
(-c*x+1)/(c*x+1)+1)^(1/2))-2*a*b/c*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2))*ln(1-(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2)
)/((-c*x+1)/(c*x+1)+1)^(1/2))+2*I*b^2/c*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(2,-(1+I*(-c*x+1)^(1/2)/(c
*x+1)^(1/2))/((-c*x+1)/(c*x+1)+1)^(1/2))+2*a*b/c*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2))*ln((1+I*(-c*x+1)^(1/2)/(
c*x+1)^(1/2))^2/((-c*x+1)/(c*x+1)+1)+1)+2*I*a*b/c*polylog(2,-(1+I*(-c*x+1)^(1/2)/(c*x+1)^(1/2))/((-c*x+1)/(c*x
+1)+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{b^{2} \log \left (2\right )^{2} \log \left (c x + 1\right ) - b^{2} \log \left (2\right )^{2} \log \left (-c x + 1\right ) - 4 \,{\left (b^{2} \log \left (c x + 1\right ) - b^{2} \log \left (-c x + 1\right )\right )} \arctan \left (\sqrt{-c x + 1}, \sqrt{c x + 1}\right )^{2} -{\left (12 \, b^{2}{\left (\frac{\log \left (c x + 1\right )}{c} - \frac{\log \left (c x - 1\right )}{c}\right )} \arctan \left (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right )^{2} + b^{2}{\left (\frac{\log \left (c x + 1\right )}{c} - \frac{\log \left (c x - 1\right )}{c}\right )} \log \left (2\right )^{2} - 4 \, b^{2} \int \frac{\sqrt{c x + 1} \sqrt{-c x + 1} \arctan \left (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right ) \log \left (c x + 1\right )}{c^{2} x^{2} - 1}\,{d x} + 4 \, b^{2} \int \frac{\sqrt{c x + 1} \sqrt{-c x + 1} \arctan \left (\frac{\sqrt{-c x + 1}}{\sqrt{c x + 1}}\right ) \log \left (-c x + 1\right )}{c^{2} x^{2} - 1}\,{d x} + \frac{32 \,{\left ({\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )} \arctan \left (\sqrt{-c x + 1}, \sqrt{c x + 1}\right ) - c \int \frac{e^{\left (\frac{1}{2} \, \log \left (c x + 1\right ) + \frac{1}{2} \, \log \left (-c x + 1\right )\right )} \log \left (c x + 1\right ) - e^{\left (\frac{1}{2} \, \log \left (c x + 1\right ) + \frac{1}{2} \, \log \left (-c x + 1\right )\right )} \log \left (-c x + 1\right )}{{\left (c^{2} x^{2} - 1\right )}{\left (c x + 1\right )} -{\left (c^{2} x^{2} - 1\right )}{\left (c x - 1\right )}}\,{d x}\right )} a b}{c}\right )} c}{32 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan((-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) - 1/32*(b^2*log(2)^2*log(c*x + 1) - b^2*log(2)^2*log(-c*x + 1) - 4*(
b^2*log(c*x + 1) - b^2*log(-c*x + 1))*arctan2(sqrt(-c*x + 1), sqrt(c*x + 1))^2 - (b^2*(log(c*x + 1)/c - log(c*
x - 1)/c)*log(2)^2 - 64*b^2*integrate(1/16*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan(sqrt(-c*x + 1)/sqrt(c*x + 1))*l
og(c*x + 1)/(c^2*x^2 - 1), x) + 64*b^2*integrate(1/16*sqrt(c*x + 1)*sqrt(-c*x + 1)*arctan(sqrt(-c*x + 1)/sqrt(
c*x + 1))*log(-c*x + 1)/(c^2*x^2 - 1), x) - 384*b^2*integrate(1/16*arctan(sqrt(-c*x + 1)/sqrt(c*x + 1))^2/(c^2
*x^2 - 1), x) - 1024*a*b*integrate(1/16*arctan(sqrt(-c*x + 1)/sqrt(c*x + 1))/(c^2*x^2 - 1), x))*c)/c

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

[Out]

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

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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*atan((-c*x+1)**(1/2)/(c*x+1)**(1/2)))**2/(-c**2*x**2+1),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{{\left (b \arctan \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*arctan((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^2/(-c^2*x^2+1),x, algorithm="giac")

[Out]

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