3.1.32 \(\int \frac {(a+b \tanh ^{-1}(\frac {\sqrt {1-c x}}{\sqrt {1+c x}}))^2}{1-c^2 x^2} \, dx\) [32]

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

[Out]

2*arctanh(-1+2/(1-(-c*x+1)^(1/2)/(c*x+1)^(1/2)))*(a+b*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^2/c+b*(a+b*arctan
h((-c*x+1)^(1/2)/(c*x+1)^(1/2)))*polylog(2,1-2/(1-(-c*x+1)^(1/2)/(c*x+1)^(1/2)))/c-b*(a+b*arctanh((-c*x+1)^(1/
2)/(c*x+1)^(1/2)))*polylog(2,-1+2/(1-(-c*x+1)^(1/2)/(c*x+1)^(1/2)))/c-1/2*b^2*polylog(3,1-2/(1-(-c*x+1)^(1/2)/
(c*x+1)^(1/2)))/c+1/2*b^2*polylog(3,-1+2/(1-(-c*x+1)^(1/2)/(c*x+1)^(1/2)))/c

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Rubi [A]
time = 0.22, antiderivative size = 268, 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 = {6813, 6033, 6199, 6095, 6205, 6745} \begin {gather*} \frac {b \text {Li}_2\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 )}{c}-\frac {b \text {Li}_2\left (\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 {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}-\frac {b^2 \text {Li}_3\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {c x+1}}}\right )}{2 c}+\frac {b^2 \text {Li}_3\left (\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {c x+1}}}-1\right )}{2 c} \end {gather*}

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 6033

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 6095

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 6199

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 6205

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 6745

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 6813

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 \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2}{1-c^2 x^2} \, dx &=-\frac {\text {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) \text {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) \text {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) \text {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 \text {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 \text {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*}

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Mathematica [A]
time = 0.07, size = 324, normalized size = 1.21 \begin {gather*} -\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 )+b \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {PolyLog}\left (2,-\frac {\sqrt {1-c x}+\sqrt {1+c x}}{\sqrt {1-c x}-\sqrt {1+c x}}\right )-b \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {PolyLog}\left (2,\frac {\sqrt {1-c x}+\sqrt {1+c x}}{\sqrt {1-c x}-\sqrt {1+c x}}\right )-\frac {1}{2} b^2 \text {PolyLog}\left (3,-\frac {\sqrt {1-c x}+\sqrt {1+c x}}{\sqrt {1-c x}-\sqrt {1+c x}}\right )+\frac {1}{2} b^2 \text {PolyLog}\left (3,\frac {\sqrt {1-c x}+\sqrt {1+c x}}{\sqrt {1-c x}-\sqrt {1+c x}}\right )}{c} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(673\) vs. \(2(232)=464\).
time = 0.02, size = 674, normalized size = 2.51

method	result	size
default	\(\frac {a^{2} \ln \left (c x +1\right )}{2 c}-\frac {a^{2} \ln \left (c x -1\right )}{2 c}-\frac {b^{2} \arctanh \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \ln \left (1-\frac {\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1}{\sqrt {-\frac {-c x +1}{c x +1}+1}}\right )}{c}-\frac {2 b^{2} \arctanh \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (2, \frac {\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1}{\sqrt {-\frac {-c x +1}{c x +1}+1}}\right )}{c}+\frac {2 b^{2} \polylog \left (3, \frac {\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1}{\sqrt {-\frac {-c x +1}{c x +1}+1}}\right )}{c}-\frac {b^{2} \arctanh \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \ln \left (1+\frac {\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1}{\sqrt {-\frac {-c x +1}{c x +1}+1}}\right )}{c}-\frac {2 b^{2} \arctanh \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (2, -\frac {\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1}{\sqrt {-\frac {-c x +1}{c x +1}+1}}\right )}{c}+\frac {2 b^{2} \polylog \left (3, -\frac {\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1}{\sqrt {-\frac {-c x +1}{c x +1}+1}}\right )}{c}+\frac {b^{2} \arctanh \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right )^{2} \ln \left (\frac {\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1\right )^{2}}{-\frac {-c x +1}{c x +1}+1}+1\right )}{c}+\frac {b^{2} \arctanh \left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}\right ) \polylog \left (2, -\frac {\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1\right )^{2}}{-\frac {-c x +1}{c x +1}+1}\right )}{c}-\frac {b^{2} \polylog \left (3, -\frac {\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1\right )^{2}}{-\frac {-c x +1}{c x +1}+1}\right )}{2 c}-\frac {a b \left (4 \dilog \left (\frac {-\frac {-c x +1}{c x +1}+1}{\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1\right )^{2}}\right )-\dilog \left (\frac {\left (-\frac {-c x +1}{c x +1}+1\right )^{2}}{\left (\frac {\sqrt {-c x +1}}{\sqrt {c x +1}}+1\right )^{4}}\right )\right )}{2 c}\)	\(674\)

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,method=_RETURNVERBOSE)

[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(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))-1/2*a*b*(4*dilog(1/((-c*x
+1)^(1/2)/(c*x+1)^(1/2)+1)^2*(-(-c*x+1)/(c*x+1)+1))-dilog(1/((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)^4*(-(-c*x+1)/(c*x
+1)+1)^2))/c

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int -\frac {{\left (a+b\,\mathrm {atanh}\left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )\right )}^2}{c^2\,x^2-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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