3.29 \(\int (a+b \tanh ^{-1}(c x))^3 \, dx\)

Optimal. Leaf size=108 \[ -\frac{3 b^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac{3 b^3 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{2 c}+x \left (a+b \tanh ^{-1}(c x)\right )^3+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{c}-\frac{3 b \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c} \]

[Out]

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

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Rubi [A]  time = 0.217234, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {5910, 5984, 5918, 5948, 6058, 6610} \[ -\frac{3 b^2 \text{PolyLog}\left (2,1-\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c}+\frac{3 b^3 \text{PolyLog}\left (3,1-\frac{2}{1-c x}\right )}{2 c}+x \left (a+b \tanh ^{-1}(c x)\right )^3+\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{c}-\frac{3 b \log \left (\frac{2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^3,x]

[Out]

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

Rule 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 5984

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 0]

Rule 5918

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

Mathematica [A]  time = 0.260217, size = 161, normalized size = 1.49 \[ \frac{6 a b^2 \left (\text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+\tanh ^{-1}(c x) \left ((c x-1) \tanh ^{-1}(c x)-2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )\right )+b^3 \left (6 \tanh ^{-1}(c x) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+3 \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )+2 \tanh ^{-1}(c x)^2 \left ((c x-1) \tanh ^{-1}(c x)-3 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )\right )+3 a^2 b \log \left (1-c^2 x^2\right )+6 a^2 b c x \tanh ^{-1}(c x)+2 a^3 c x}{2 c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c*x])^3,x]

[Out]

(2*a^3*c*x + 6*a^2*b*c*x*ArcTanh[c*x] + 3*a^2*b*Log[1 - c^2*x^2] + 6*a*b^2*(ArcTanh[c*x]*((-1 + c*x)*ArcTanh[c
*x] - 2*Log[1 + E^(-2*ArcTanh[c*x])]) + PolyLog[2, -E^(-2*ArcTanh[c*x])]) + b^3*(2*ArcTanh[c*x]^2*((-1 + c*x)*
ArcTanh[c*x] - 3*Log[1 + E^(-2*ArcTanh[c*x])]) + 6*ArcTanh[c*x]*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 3*PolyLog[3
, -E^(-2*ArcTanh[c*x])]))/(2*c)

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Maple [B]  time = 0.082, size = 261, normalized size = 2.4 \begin{align*} x{a}^{3}+{b}^{3}x \left ({\it Artanh} \left ( cx \right ) \right ) ^{3}+{\frac{{b}^{3} \left ({\it Artanh} \left ( cx \right ) \right ) ^{3}}{c}}-3\,{\frac{{b}^{3} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}}{c}\ln \left ({\frac{ \left ( cx+1 \right ) ^{2}}{-{c}^{2}{x}^{2}+1}}+1 \right ) }-3\,{\frac{{b}^{3}{\it Artanh} \left ( cx \right ) }{c}{\it polylog} \left ( 2,-{\frac{ \left ( cx+1 \right ) ^{2}}{-{c}^{2}{x}^{2}+1}} \right ) }+{\frac{3\,{b}^{3}}{2\,c}{\it polylog} \left ( 3,-{\frac{ \left ( cx+1 \right ) ^{2}}{-{c}^{2}{x}^{2}+1}} \right ) }+3\,xa{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}+3\,{\frac{a{b}^{2} \left ({\it Artanh} \left ( cx \right ) \right ) ^{2}}{c}}-6\,{\frac{{\it Artanh} \left ( cx \right ) a{b}^{2}}{c}\ln \left ({\frac{ \left ( cx+1 \right ) ^{2}}{-{c}^{2}{x}^{2}+1}}+1 \right ) }-3\,{\frac{a{b}^{2}}{c}{\it polylog} \left ( 2,-{\frac{ \left ( cx+1 \right ) ^{2}}{-{c}^{2}{x}^{2}+1}} \right ) }+3\,x{a}^{2}b{\it Artanh} \left ( cx \right ) +{\frac{3\,{a}^{2}b\ln \left ( -{c}^{2}{x}^{2}+1 \right ) }{2\,c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))^3,x)

[Out]

x*a^3+b^3*x*arctanh(c*x)^3+1/c*b^3*arctanh(c*x)^3-3/c*b^3*arctanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)+1)-3/c*b^3*
arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))+3/2/c*b^3*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))+3*x*a*b^2*arctanh
(c*x)^2+3/c*a*b^2*arctanh(c*x)^2-6/c*arctanh(c*x)*ln((c*x+1)^2/(-c^2*x^2+1)+1)*a*b^2-3/c*polylog(2,-(c*x+1)^2/
(-c^2*x^2+1))*a*b^2+3*x*a^2*b*arctanh(c*x)+3/2/c*a^2*b*ln(-c^2*x^2+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{3} x + \frac{3 \,{\left (2 \, c x \operatorname{artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} a^{2} b}{2 \, c} - \frac{{\left (b^{3} c x - b^{3}\right )} \log \left (-c x + 1\right )^{3} - 3 \,{\left (2 \, a b^{2} c x +{\left (b^{3} c x + b^{3}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{8 \, c} - \int -\frac{{\left (b^{3} c x - b^{3}\right )} \log \left (c x + 1\right )^{3} + 6 \,{\left (a b^{2} c x - a b^{2}\right )} \log \left (c x + 1\right )^{2} - 3 \,{\left (4 \, a b^{2} c x +{\left (b^{3} c x - b^{3}\right )} \log \left (c x + 1\right )^{2} - 2 \,{\left (2 \, a b^{2} - b^{3} -{\left (2 \, a b^{2} c + b^{3} c\right )} x\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{8 \,{\left (c x - 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^3,x, algorithm="maxima")

[Out]

a^3*x + 3/2*(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*a^2*b/c - 1/8*((b^3*c*x - b^3)*log(-c*x + 1)^3 - 3*(2*a*b
^2*c*x + (b^3*c*x + b^3)*log(c*x + 1))*log(-c*x + 1)^2)/c - integrate(-1/8*((b^3*c*x - b^3)*log(c*x + 1)^3 + 6
*(a*b^2*c*x - a*b^2)*log(c*x + 1)^2 - 3*(4*a*b^2*c*x + (b^3*c*x - b^3)*log(c*x + 1)^2 - 2*(2*a*b^2 - b^3 - (2*
a*b^2*c + b^3*c)*x)*log(c*x + 1))*log(-c*x + 1))/(c*x - 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b^{3} \operatorname{artanh}\left (c x\right )^{3} + 3 \, a b^{2} \operatorname{artanh}\left (c x\right )^{2} + 3 \, a^{2} b \operatorname{artanh}\left (c x\right ) + a^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^3,x, algorithm="fricas")

[Out]

integral(b^3*arctanh(c*x)^3 + 3*a*b^2*arctanh(c*x)^2 + 3*a^2*b*arctanh(c*x) + a^3, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{atanh}{\left (c x \right )}\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))**3,x)

[Out]

Integral((a + b*atanh(c*x))**3, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^3,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^3, x)