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

Optimal. Leaf size=117 \[ -b \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+b \text{PolyLog}\left (2,\frac{2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} b^2 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )-\frac{1}{2} b^2 \text{PolyLog}\left (3,\frac{2}{1-c x}-1\right )+2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2 \]

[Out]

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

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Rubi [A]  time = 0.263671, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {5914, 6052, 5948, 6058, 6610} \[ -b \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+b \text{PolyLog}\left (2,\frac{2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )+\frac{1}{2} b^2 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )-\frac{1}{2} b^2 \text{PolyLog}\left (3,\frac{2}{1-c x}-1\right )+2 \tanh ^{-1}\left (1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^2/x,x]

[Out]

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

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

Mathematica [A]  time = 0.0715302, size = 120, normalized size = 1.03 \[ \frac{1}{2} b \left (2 \text{PolyLog}\left (2,\frac{c x+1}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )-2 \text{PolyLog}\left (2,\frac{c x+1}{c x-1}\right ) \left (a+b \tanh ^{-1}(c x)\right )+b \left (\text{PolyLog}\left (3,\frac{c x+1}{c x-1}\right )-\text{PolyLog}\left (3,\frac{c x+1}{1-c x}\right )\right )\right )+2 \tanh ^{-1}\left (\frac{c x+1}{c x-1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2 \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])^2/x,x]

[Out]

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

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Maple [C]  time = 0.214, size = 701, normalized size = 6. \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))^2/x,x)

[Out]

a^2*ln(c*x)+b^2*ln(c*x)*arctanh(c*x)^2-b^2*arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))+1/2*b^2*polylog(3,-
(c*x+1)^2/(-c^2*x^2+1))-b^2*arctanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)+b^2*arctanh(c*x)^2*ln(1-(c*x+1)/(-c^2*
x^2+1)^(1/2))+2*b^2*arctanh(c*x)*polylog(2,(c*x+1)/(-c^2*x^2+1)^(1/2))-2*b^2*polylog(3,(c*x+1)/(-c^2*x^2+1)^(1
/2))+b^2*arctanh(c*x)^2*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+2*b^2*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1
/2))-2*b^2*polylog(3,-(c*x+1)/(-c^2*x^2+1)^(1/2))+1/2*I*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)^2/(-
c^2*x^2+1)+1))^3*arctanh(c*x)^2-1/2*I*b^2*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)
-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c*x)^2+1/2*I*b^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I/((c*x+
1)^2/(-c^2*x^2+1)+1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))*arctanh(c*x)^2-1/2*I*b^2*P
i*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*arctanh(c
*x)^2+2*a*b*ln(c*x)*arctanh(c*x)-a*b*ln(c*x)*ln(c*x+1)-a*b*dilog(c*x)-a*b*dilog(c*x+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \log \left (x\right ) + \int \frac{b^{2}{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}^{2}}{4 \, x} + \frac{a b{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*log(x) + integrate(1/4*b^2*(log(c*x + 1) - log(-c*x + 1))^2/x + a*b*(log(c*x + 1) - log(-c*x + 1))/x, 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 (c x\right )^{2} + 2 \, a b \operatorname{artanh}\left (c x\right ) + a^{2}}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atanh(c*x))**2/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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