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

Optimal. Leaf size=302 \[ -\frac {b \text {Li}_2\left (1-\frac {2}{\frac {\sqrt {1-c x}}{\sqrt {c x+1}}+1}\right ) \left (a+b \coth ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{c}+\frac {b \text {Li}_2\left (1-\frac {2 \sqrt {1-c x}}{\sqrt {c x+1} \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}+1\right )}\right ) \left (a+b \coth ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{c}-\frac {2 \coth ^{-1}\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \coth ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{c}-\frac {b^2 \text {Li}_3\left (1-\frac {2}{\frac {\sqrt {1-c x}}{\sqrt {c x+1}}+1}\right )}{2 c}+\frac {b^2 \text {Li}_3\left (1-\frac {2 \sqrt {1-c x}}{\sqrt {c x+1} \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}+1\right )}\right )}{2 c} \]

[Out]

-2*arccoth(1-2/(1-(-c*x+1)^(1/2)/(c*x+1)^(1/2)))*(a+b*arccoth((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^2/c-b*(a+b*arccot
h((-c*x+1)^(1/2)/(c*x+1)^(1/2)))*polylog(2,1-2/((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1))/c+b*(a+b*arccoth((-c*x+1)^(1/
2)/(c*x+1)^(1/2)))*polylog(2,1-2*(-c*x+1)^(1/2)/((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)/(c*x+1)^(1/2))/c-1/2*b^2*poly
log(3,1-2/((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1))/c+1/2*b^2*polylog(3,1-2*(-c*x+1)^(1/2)/((-c*x+1)^(1/2)/(c*x+1)^(1/
2)+1)/(c*x+1)^(1/2))/c

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Rubi [A]  time = 0.35, antiderivative size = 302, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {6681, 5915, 6053, 5949, 6057, 6610} \[ -\frac {b \text {PolyLog}\left (2,1-\frac {2}{\frac {\sqrt {1-c x}}{\sqrt {c x+1}}+1}\right ) \left (a+b \coth ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{c}+\frac {b \text {PolyLog}\left (2,1-\frac {2 \sqrt {1-c x}}{\sqrt {c x+1} \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}+1\right )}\right ) \left (a+b \coth ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )}{c}-\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{\frac {\sqrt {1-c x}}{\sqrt {c x+1}}+1}\right )}{2 c}+\frac {b^2 \text {PolyLog}\left (3,1-\frac {2 \sqrt {1-c x}}{\sqrt {c x+1} \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}+1\right )}\right )}{2 c}-\frac {2 \coth ^{-1}\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {c x+1}}}\right ) \left (a+b \coth ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )^2}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5915

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

Rule 5949

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcCoth[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 6053

Int[(ArcCoth[u_]*((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[
(Log[SimplifyIntegrand[1 + 1/u, x]]*(a + b*ArcCoth[c*x])^p)/(d + e*x^2), x], x] - Dist[1/2, Int[(Log[SimplifyI
ntegrand[1 - 1/u, x]]*(a + b*ArcCoth[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 6057

Int[(Log[u_]*((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[((a + b*ArcCo
th[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] - Dist[(b*p)/2, Int[((a + b*ArcCoth[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]

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]

Rubi steps

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

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Mathematica [F]  time = 0.54, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b \coth ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2}{1-c^2 x^2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {b^{2} \operatorname {arcoth}\left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right )^{2} + 2 \, a b \operatorname {arcoth}\left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a^{2}}{c^{2} x^{2} - 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (b \operatorname {arcoth}\left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a\right )}^{2}}{c^{2} x^{2} - 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.73, size = 696, normalized size = 2.30 \[ \frac {a^{2} \ln \left (c x +1\right )}{2 c}-\frac {a^{2} \ln \left (c x -1\right )}{2 c}+\frac {b^{2} \mathrm {arccoth}\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \ln \left (1+\frac {1}{\sqrt {\frac {\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-1}{\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1}}}\right )}{c}+\frac {2 b^{2} \mathrm {arccoth}\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (2, -\frac {1}{\sqrt {\frac {\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-1}{\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1}}}\right )}{c}-\frac {2 b^{2} \polylog \left (3, -\frac {1}{\sqrt {\frac {\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-1}{\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1}}}\right )}{c}+\frac {b^{2} \mathrm {arccoth}\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \ln \left (1-\frac {1}{\sqrt {\frac {\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-1}{\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1}}}\right )}{c}+\frac {2 b^{2} \mathrm {arccoth}\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (2, \frac {1}{\sqrt {\frac {\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-1}{\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1}}}\right )}{c}-\frac {2 b^{2} \polylog \left (3, \frac {1}{\sqrt {\frac {\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-1}{\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1}}}\right )}{c}-\frac {b^{2} \mathrm {arccoth}\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \ln \left (1+\frac {\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1}{\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-1}\right )}{c}-\frac {b^{2} \mathrm {arccoth}\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (2, -\frac {\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1}{\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-1}\right )}{c}+\frac {b^{2} \polylog \left (3, -\frac {\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1}{\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-1}\right )}{2 c}+\frac {2 a b \dilog \left (\frac {\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-1}{\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1}\right )}{c}-\frac {a b \dilog \left (\frac {\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}-1\right )^{2}}{\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1\right )^{2}}\right )}{2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccoth((-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*arccoth((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*ln(1+1/(((-c*x+1)^(1/2)/
(c*x+1)^(1/2)-1)/((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1))^(1/2))+2*b^2/c*arccoth((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylo
g(2,-1/(((-c*x+1)^(1/2)/(c*x+1)^(1/2)-1)/((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1))^(1/2))-2*b^2/c*polylog(3,-1/(((-c*x
+1)^(1/2)/(c*x+1)^(1/2)-1)/((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1))^(1/2))+b^2/c*arccoth((-c*x+1)^(1/2)/(c*x+1)^(1/2)
)^2*ln(1-1/(((-c*x+1)^(1/2)/(c*x+1)^(1/2)-1)/((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1))^(1/2))+2*b^2/c*arccoth((-c*x+1)
^(1/2)/(c*x+1)^(1/2))*polylog(2,1/(((-c*x+1)^(1/2)/(c*x+1)^(1/2)-1)/((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1))^(1/2))-2
*b^2/c*polylog(3,1/(((-c*x+1)^(1/2)/(c*x+1)^(1/2)-1)/((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1))^(1/2))-b^2/c*arccoth((-
c*x+1)^(1/2)/(c*x+1)^(1/2))^2*ln(1+1/((-c*x+1)^(1/2)/(c*x+1)^(1/2)-1)*((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1))-b^2/c*
arccoth((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(2,-1/((-c*x+1)^(1/2)/(c*x+1)^(1/2)-1)*((-c*x+1)^(1/2)/(c*x+1)^(1
/2)+1))+1/2*b^2/c*polylog(3,-1/((-c*x+1)^(1/2)/(c*x+1)^(1/2)-1)*((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1))+2*a*b/c*dilo
g(((-c*x+1)^(1/2)/(c*x+1)^(1/2)-1)/((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1))-1/2*a*b/c*dilog(((-c*x+1)^(1/2)/(c*x+1)^(
1/2)-1)^2/((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)^2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a^{2} {\left (\frac {\log \left (c x + 1\right )}{c} - \frac {\log \left (c x - 1\right )}{c}\right )} + \frac {{\left (b^{2} \log \left (c x + 1\right ) - b^{2} \log \left (-c x + 1\right )\right )} \log \left (-\sqrt {c x + 1} + \sqrt {-c x + 1}\right )^{2}}{8 \, c} + \int -\frac {2 \, {\left (\sqrt {c x + 1} b^{2} - \sqrt {-c x + 1} b^{2}\right )} \log \left (\sqrt {c x + 1} + \sqrt {-c x + 1}\right )^{2} + 8 \, {\left (\sqrt {c x + 1} a b - \sqrt {-c x + 1} a b\right )} \log \left (\sqrt {c x + 1} + \sqrt {-c x + 1}\right ) - {\left (4 \, {\left (\sqrt {c x + 1} b^{2} - \sqrt {-c x + 1} b^{2}\right )} \log \left (\sqrt {c x + 1} + \sqrt {-c x + 1}\right ) + {\left (8 \, a b - {\left (b^{2} c x - b^{2}\right )} \log \left (c x + 1\right ) + {\left (b^{2} c x - b^{2}\right )} \log \left (-c x + 1\right )\right )} \sqrt {c x + 1} - {\left (8 \, a b - {\left (b^{2} c x + b^{2}\right )} \log \left (c x + 1\right ) + {\left (b^{2} c x + b^{2}\right )} \log \left (-c x + 1\right )\right )} \sqrt {-c x + 1}\right )} \log \left (-\sqrt {c x + 1} + \sqrt {-c x + 1}\right )}{8 \, {\left ({\left (c^{2} x^{2} - 1\right )} \sqrt {c x + 1} - {\left (c^{2} x^{2} - 1\right )} \sqrt {-c x + 1}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccoth((-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) + sq
rt(-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))*
sqrt(-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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int -\frac {{\left (a+b\,\mathrm {acoth}\left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )\right )}^2}{c^2\,x^2-1} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {a^{2}}{c^{2} x^{2} - 1}\, dx - \int \frac {b^{2} \operatorname {acoth}^{2}{\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )}}{c^{2} x^{2} - 1}\, dx - \int \frac {2 a b \operatorname {acoth}{\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )}}{c^{2} x^{2} - 1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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