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

Optimal. Leaf size=148 \[ -\frac{b \text{PolyLog}\left (2,1-\frac{2}{1-c x^n}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )}{n}+\frac{b \text{PolyLog}\left (2,\frac{2}{1-c x^n}-1\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )}{n}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1-c x^n}\right )}{2 n}-\frac{b^2 \text{PolyLog}\left (3,\frac{2}{1-c x^n}-1\right )}{2 n}+\frac{2 \tanh ^{-1}\left (1-\frac{2}{1-c x^n}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2}{n} \]

[Out]

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

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Rubi [A]  time = 0.313463, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {6095, 5914, 6052, 5948, 6058, 6610} \[ -\frac{b \text{PolyLog}\left (2,1-\frac{2}{1-c x^n}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )}{n}+\frac{b \text{PolyLog}\left (2,\frac{2}{1-c x^n}-1\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )}{n}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{1-c x^n}\right )}{2 n}-\frac{b^2 \text{PolyLog}\left (3,\frac{2}{1-c x^n}-1\right )}{2 n}+\frac{2 \tanh ^{-1}\left (1-\frac{2}{1-c x^n}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2}{n} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^n])^2/x,x]

[Out]

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

Rule 6095

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*ArcTanh[c*x])
^p/x, x], x, x^n], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 0]

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

Mathematica [A]  time = 0.128001, size = 146, normalized size = 0.99 \[ \frac{\frac{1}{2} b \left (2 \text{PolyLog}\left (2,\frac{c x^n+1}{1-c x^n}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )-2 \text{PolyLog}\left (2,\frac{c x^n+1}{c x^n-1}\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )+b \left (\text{PolyLog}\left (3,\frac{c x^n+1}{c x^n-1}\right )-\text{PolyLog}\left (3,\frac{c x^n+1}{1-c x^n}\right )\right )\right )+2 \tanh ^{-1}\left (\frac{2}{c x^n-1}+1\right ) \left (a+b \tanh ^{-1}\left (c x^n\right )\right )^2}{n} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^n])^2/x,x]

[Out]

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

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Maple [C]  time = 0.33, size = 880, 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^n))^2/x,x)

[Out]

1/n*a^2*ln(c*x^n)+1/n*b^2*ln(c*x^n)*arctanh(c*x^n)^2-1/n*b^2*arctanh(c*x^n)*polylog(2,-(c*x^n+1)^2/(-c^2*(x^n)
^2+1))+1/2/n*b^2*polylog(3,-(c*x^n+1)^2/(-c^2*(x^n)^2+1))-1/n*b^2*arctanh(c*x^n)^2*ln((c*x^n+1)^2/(-c^2*(x^n)^
2+1)-1)+1/n*b^2*arctanh(c*x^n)^2*ln(1-(c*x^n+1)/(-c^2*(x^n)^2+1)^(1/2))+2/n*b^2*arctanh(c*x^n)*polylog(2,(c*x^
n+1)/(-c^2*(x^n)^2+1)^(1/2))-2/n*b^2*polylog(3,(c*x^n+1)/(-c^2*(x^n)^2+1)^(1/2))+1/n*b^2*arctanh(c*x^n)^2*ln(1
+(c*x^n+1)/(-c^2*(x^n)^2+1)^(1/2))+2/n*b^2*arctanh(c*x^n)*polylog(2,-(c*x^n+1)/(-c^2*(x^n)^2+1)^(1/2))-2/n*b^2
*polylog(3,-(c*x^n+1)/(-c^2*(x^n)^2+1)^(1/2))+1/2*I/n*b^2*Pi*csgn(I*((c*x^n+1)^2/(-c^2*(x^n)^2+1)-1)/((c*x^n+1
)^2/(-c^2*(x^n)^2+1)+1))^3*arctanh(c*x^n)^2+1/2*I/n*b^2*Pi*csgn(I*((c*x^n+1)^2/(-c^2*(x^n)^2+1)-1))*csgn(I/((c
*x^n+1)^2/(-c^2*(x^n)^2+1)+1))*csgn(I*((c*x^n+1)^2/(-c^2*(x^n)^2+1)-1)/((c*x^n+1)^2/(-c^2*(x^n)^2+1)+1))*arcta
nh(c*x^n)^2-1/2*I/n*b^2*Pi*csgn(I/((c*x^n+1)^2/(-c^2*(x^n)^2+1)+1))*csgn(I*((c*x^n+1)^2/(-c^2*(x^n)^2+1)-1)/((
c*x^n+1)^2/(-c^2*(x^n)^2+1)+1))^2*arctanh(c*x^n)^2-1/2*I/n*b^2*Pi*csgn(I*((c*x^n+1)^2/(-c^2*(x^n)^2+1)-1))*csg
n(I*((c*x^n+1)^2/(-c^2*(x^n)^2+1)-1)/((c*x^n+1)^2/(-c^2*(x^n)^2+1)+1))^2*arctanh(c*x^n)^2-1/n*a*b*ln(c*x^n)*ln
(c*x^n+1)+2/n*a*b*ln(c*x^n)*arctanh(c*x^n)-1/n*a*b*dilog(c*x^n)-1/n*a*b*dilog(c*x^n+1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*b^2*log(-c*x^n + 1)^2*log(x) + a^2*log(x) - integrate(-1/4*((b^2*c*x^n - b^2)*log(c*x^n + 1)^2 + 4*(a*b*c*
x^n - a*b)*log(c*x^n + 1) + 2*(2*a*b - (b^2*c*n*log(x) + 2*a*b*c)*x^n - (b^2*c*x^n - b^2)*log(c*x^n + 1))*log(
-c*x^n + 1))/(c*x*x^n - 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^{n}\right )^{2} + 2 \, a b \operatorname{artanh}\left (c x^{n}\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^n))^2/x,x, algorithm="fricas")

[Out]

integral((b^2*arctanh(c*x^n)^2 + 2*a*b*arctanh(c*x^n) + 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^{n} \right )}\right )^{2}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atanh(c*x**n))**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^{n}\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^n))^2/x,x, algorithm="giac")

[Out]

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