3.1.31 \(\int \frac {(a+b \tanh ^{-1}(\frac {\sqrt {1-c x}}{\sqrt {1+c x}}))^3}{1-c^2 x^2} \, dx\) [31]

Optimal. Leaf size=409 \[ -\frac {2 \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {3 b \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {PolyLog}\left (2,1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {3 b \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {PolyLog}\left (2,-1+\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {3 b^2 \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {PolyLog}\left (3,1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}+\frac {3 b^2 \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {PolyLog}\left (3,-1+\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}+\frac {3 b^3 \text {PolyLog}\left (4,1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{4 c}-\frac {3 b^3 \text {PolyLog}\left (4,-1+\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{4 c} \]

[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)))^3/c+3/2*b*(a+b*ar
ctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2)))^2*polylog(2,1-2/(1-(-c*x+1)^(1/2)/(c*x+1)^(1/2)))/c-3/2*b*(a+b*arctanh((-
c*x+1)^(1/2)/(c*x+1)^(1/2)))^2*polylog(2,-1+2/(1-(-c*x+1)^(1/2)/(c*x+1)^(1/2)))/c-3/2*b^2*(a+b*arctanh((-c*x+1
)^(1/2)/(c*x+1)^(1/2)))*polylog(3,1-2/(1-(-c*x+1)^(1/2)/(c*x+1)^(1/2)))/c+3/2*b^2*(a+b*arctanh((-c*x+1)^(1/2)/
(c*x+1)^(1/2)))*polylog(3,-1+2/(1-(-c*x+1)^(1/2)/(c*x+1)^(1/2)))/c+3/4*b^3*polylog(4,1-2/(1-(-c*x+1)^(1/2)/(c*
x+1)^(1/2)))/c-3/4*b^3*polylog(4,-1+2/(1-(-c*x+1)^(1/2)/(c*x+1)^(1/2)))/c

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Rubi [A]
time = 0.36, antiderivative size = 409, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {6813, 6033, 6199, 6095, 6205, 6209, 6745} \begin {gather*} -\frac {3 b^2 \text {Li}_3\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 {3 b^2 \text {Li}_3\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 )}{2 c}+\frac {3 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 )^2}{2 c}-\frac {3 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 )^2}{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 )^3}{c}+\frac {3 b^3 \text {Li}_4\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {c x+1}}}\right )}{4 c}-\frac {3 b^3 \text {Li}_4\left (\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {c x+1}}}-1\right )}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3*ArcTanh[1 - 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])])/c + (3*b*
(a + b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2*PolyLog[2, 1 - 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])])/(2*c) - (3
*b*(a + b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2*PolyLog[2, -1 + 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])])/(2*c)
- (3*b^2*(a + b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*PolyLog[3, 1 - 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])])/(2*
c) + (3*b^2*(a + b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*PolyLog[3, -1 + 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])])
/(2*c) + (3*b^3*PolyLog[4, 1 - 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])])/(4*c) - (3*b^3*PolyLog[4, -1 + 2/(1 - Sqr
t[1 - c*x]/Sqrt[1 + c*x])])/(4*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 6209

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*PolyLog[k_, u_])/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a +
b*ArcTanh[c*x])^p*(PolyLog[k + 1, u]/(2*c*d)), x] - Dist[b*(p/2), Int[(a + b*ArcTanh[c*x])^(p - 1)*(PolyLog[k
+ 1, u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, k}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[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 )^3}{1-c^2 x^2} \, dx &=-\frac {\text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^3}{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 )^3 \tanh ^{-1}\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {(6 b) \text {Subst}\left (\int \frac {\tanh ^{-1}\left (1-\frac {2}{1-x}\right ) \left (a+b \tanh ^{-1}(x)\right )^2}{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 )^3 \tanh ^{-1}\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {(3 b) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2 \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 {(3 b) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2 \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 )^3 \tanh ^{-1}\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {3 b \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {3 b \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right ) \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 {\left (3 b^2\right ) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right ) \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 )^3 \tanh ^{-1}\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {3 b \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {3 b \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {3 b^2 \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_3\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}+\frac {3 b^2 \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_3\left (-1+\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}+\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_3\left (1-\frac {2}{1-x}\right )}{1-x^2} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{2 c}-\frac {\left (3 b^3\right ) \text {Subst}\left (\int \frac {\text {Li}_3\left (-1+\frac {2}{1-x}\right )}{1-x^2} \, dx,x,\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )}{2 c}\\ &=-\frac {2 \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{c}+\frac {3 b \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {3 b \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}-\frac {3 b^2 \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_3\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}+\frac {3 b^2 \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {Li}_3\left (-1+\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{2 c}+\frac {3 b^3 \text {Li}_4\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{4 c}-\frac {3 b^3 \text {Li}_4\left (-1+\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )}{4 c}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 482, normalized size = 1.18 \begin {gather*} -\frac {8 \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^3 \tanh ^{-1}\left (1-\frac {2}{1-\frac {\sqrt {1-c x}}{\sqrt {1+c x}}}\right )+6 b \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {PolyLog}\left (2,-\frac {\sqrt {1-c x}+\sqrt {1+c x}}{\sqrt {1-c x}-\sqrt {1+c x}}\right )-6 b \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2 \text {PolyLog}\left (2,\frac {\sqrt {1-c x}+\sqrt {1+c x}}{\sqrt {1-c x}-\sqrt {1+c x}}\right )-6 b^2 \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {PolyLog}\left (3,-\frac {\sqrt {1-c x}+\sqrt {1+c x}}{\sqrt {1-c x}-\sqrt {1+c x}}\right )+6 b^2 \left (a+b \tanh ^{-1}\left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right ) \text {PolyLog}\left (3,\frac {\sqrt {1-c x}+\sqrt {1+c x}}{\sqrt {1-c x}-\sqrt {1+c x}}\right )+3 b^3 \text {PolyLog}\left (4,-\frac {\sqrt {1-c x}+\sqrt {1+c x}}{\sqrt {1-c x}-\sqrt {1+c x}}\right )-3 b^3 \text {PolyLog}\left (4,\frac {\sqrt {1-c x}+\sqrt {1+c x}}{\sqrt {1-c x}-\sqrt {1+c x}}\right )}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(8*(a + b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^3*ArcTanh[1 - 2/(1 - Sqrt[1 - c*x]/Sqrt[1 + c*x])] + 6*b*
(a + b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2*PolyLog[2, -((Sqrt[1 - c*x] + Sqrt[1 + c*x])/(Sqrt[1 - c*x] - S
qrt[1 + c*x]))] - 6*b*(a + b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])^2*PolyLog[2, (Sqrt[1 - c*x] + Sqrt[1 + c*x]
)/(Sqrt[1 - c*x] - Sqrt[1 + c*x])] - 6*b^2*(a + b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*PolyLog[3, -((Sqrt[1 -
c*x] + Sqrt[1 + c*x])/(Sqrt[1 - c*x] - Sqrt[1 + c*x]))] + 6*b^2*(a + b*ArcTanh[Sqrt[1 - c*x]/Sqrt[1 + c*x]])*
PolyLog[3, (Sqrt[1 - c*x] + Sqrt[1 + c*x])/(Sqrt[1 - c*x] - Sqrt[1 + c*x])] + 3*b^3*PolyLog[4, -((Sqrt[1 - c*x
] + Sqrt[1 + c*x])/(Sqrt[1 - c*x] - Sqrt[1 + c*x]))] - 3*b^3*PolyLog[4, (Sqrt[1 - c*x] + Sqrt[1 + c*x])/(Sqrt[
1 - c*x] - Sqrt[1 + c*x])])/c

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1444\) vs. \(2(349)=698\).
time = 0.31, size = 1445, normalized size = 3.53

method	result	size
default	\(\text {Expression too large to display}\)	\(1445\)

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)))^3/(-c^2*x^2+1),x,method=_RETURNVERBOSE)

[Out]

1/2*a^3/c*ln(c*x+1)-1/2*a^3/c*ln(c*x-1)-b^3/c*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2))^3*ln(1-((-c*x+1)^(1/2)/(c*
x+1)^(1/2)+1)/(-(-c*x+1)/(c*x+1)+1)^(1/2))-3*b^3/c*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*polylog(2,((-c*x+1)
^(1/2)/(c*x+1)^(1/2)+1)/(-(-c*x+1)/(c*x+1)+1)^(1/2))+6*b^3/c*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(3,(
(-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)/(-(-c*x+1)/(c*x+1)+1)^(1/2))-6*b^3/c*polylog(4,((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1
)/(-(-c*x+1)/(c*x+1)+1)^(1/2))-b^3/c*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2))^3*ln(1+((-c*x+1)^(1/2)/(c*x+1)^(1/2
)+1)/(-(-c*x+1)/(c*x+1)+1)^(1/2))-3*b^3/c*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*polylog(2,-((-c*x+1)^(1/2)/(
c*x+1)^(1/2)+1)/(-(-c*x+1)/(c*x+1)+1)^(1/2))+6*b^3/c*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(3,-((-c*x+1
)^(1/2)/(c*x+1)^(1/2)+1)/(-(-c*x+1)/(c*x+1)+1)^(1/2))-6*b^3/c*polylog(4,-((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)/(-(-
c*x+1)/(c*x+1)+1)^(1/2))+b^3/c*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2))^3*ln(((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)^2/(
-(-c*x+1)/(c*x+1)+1)+1)+3/2*b^3/c*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2))^2*polylog(2,-((-c*x+1)^(1/2)/(c*x+1)^(
1/2)+1)^2/(-(-c*x+1)/(c*x+1)+1))-3/2*b^3/c*arctanh((-c*x+1)^(1/2)/(c*x+1)^(1/2))*polylog(3,-((-c*x+1)^(1/2)/(c
*x+1)^(1/2)+1)^2/(-(-c*x+1)/(c*x+1)+1))+3/4*b^3/c*polylog(4,-((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)^2/(-(-c*x+1)/(c*
x+1)+1))-3*a*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))-6*a*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))+6*a*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))-3*
a*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))-6*a*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))+6*a*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))+3*a*b^2/c*a
rctanh((-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)+3*a*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))-3
/2*a*b^2/c*polylog(3,-((-c*x+1)^(1/2)/(c*x+1)^(1/2)+1)^2/(-(-c*x+1)/(c*x+1)+1))-3/4*a^2*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)))^3/(-c^2*x^2+1),x, algorithm="maxima")

[Out]

1/2*a^3*(log(c*x + 1)/c - log(c*x - 1)/c) - 1/16*(b^3*log(c*x + 1) - b^3*log(-c*x + 1))*log(sqrt(c*x + 1) - sq
rt(-c*x + 1))^3/c - integrate(1/32*(4*(sqrt(c*x + 1)*b^3 - sqrt(-c*x + 1)*b^3)*log(sqrt(c*x + 1) + sqrt(-c*x +
1))^3 + 24*(sqrt(c*x + 1)*a*b^2 - sqrt(-c*x + 1)*a*b^2)*log(sqrt(c*x + 1) + sqrt(-c*x + 1))^2 + 3*(4*(sqrt(c*
x + 1)*b^3 - sqrt(-c*x + 1)*b^3)*log(sqrt(c*x + 1) + sqrt(-c*x + 1)) + (8*a*b^2 - (b^3*c*x - b^3)*log(c*x + 1)
+ (b^3*c*x - b^3)*log(-c*x + 1))*sqrt(c*x + 1) - (8*a*b^2 - (b^3*c*x + b^3)*log(c*x + 1) + (b^3*c*x + b^3)*lo
g(-c*x + 1))*sqrt(-c*x + 1))*log(sqrt(c*x + 1) - sqrt(-c*x + 1))^2 + 48*(sqrt(c*x + 1)*a^2*b - sqrt(-c*x + 1)*
a^2*b)*log(sqrt(c*x + 1) + sqrt(-c*x + 1)) - 12*(4*sqrt(c*x + 1)*a^2*b - 4*sqrt(-c*x + 1)*a^2*b + (sqrt(c*x +
1)*b^3 - sqrt(-c*x + 1)*b^3)*log(sqrt(c*x + 1) + sqrt(-c*x + 1))^2 + 4*(sqrt(c*x + 1)*a*b^2 - sqrt(-c*x + 1)*a
*b^2)*log(sqrt(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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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)))^3/(-c^2*x^2+1),x, algorithm="fricas")

[Out]

integral(-(b^3*arctanh(sqrt(-c*x + 1)/sqrt(c*x + 1))^3 + 3*a*b^2*arctanh(sqrt(-c*x + 1)/sqrt(c*x + 1))^2 + 3*a
^2*b*arctanh(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a^3)/(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^{3}}{c^{2} x^{2} - 1}\, dx - \int \frac {b^{3} \operatorname {atanh}^{3}{\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )}}{c^{2} x^{2} - 1}\, dx - \int \frac {3 a b^{2} \operatorname {atanh}^{2}{\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )}}{c^{2} x^{2} - 1}\, dx - \int \frac {3 a^{2} 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)))**3/(-c**2*x**2+1),x)

[Out]

-Integral(a**3/(c**2*x**2 - 1), x) - Integral(b**3*atanh(sqrt(-c*x + 1)/sqrt(c*x + 1))**3/(c**2*x**2 - 1), x)
- Integral(3*a*b**2*atanh(sqrt(-c*x + 1)/sqrt(c*x + 1))**2/(c**2*x**2 - 1), x) - Integral(3*a**2*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)))^3/(-c^2*x^2+1),x, algorithm="giac")

[Out]

integrate(-(b*arctanh(sqrt(-c*x + 1)/sqrt(c*x + 1)) + a)^3/(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 )}^3}{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)))^3/(c^2*x^2 - 1),x)

[Out]

int(-(a + b*atanh((1 - c*x)^(1/2)/(c*x + 1)^(1/2)))^3/(c^2*x^2 - 1), x)

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