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

Optimal. Leaf size=132 \[ -\frac{3 b^2 \text{PolyLog}\left (2,1-\frac{2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d}+\frac{3 b^3 \text{PolyLog}\left (3,1-\frac{2}{-c-d x+1}\right )}{2 d}+\frac{(c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{d}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d}-\frac{3 b \log \left (\frac{2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d} \]

[Out]

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

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Rubi [A]  time = 0.229602, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {6103, 5910, 5984, 5918, 5948, 6058, 6610} \[ -\frac{3 b^2 \text{PolyLog}\left (2,1-\frac{2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d}+\frac{3 b^3 \text{PolyLog}\left (3,1-\frac{2}{-c-d x+1}\right )}{2 d}+\frac{(c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{d}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d}-\frac{3 b \log \left (\frac{2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c + d*x])^3,x]

[Out]

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

Rule 6103

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

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+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \left (a+b \tanh ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{(c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{d}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{x \left (a+b \tanh ^{-1}(x)\right )^2}{1-x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d}+\frac{(c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{d}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right )^2}{1-x} \, dx,x,c+d x\right )}{d}\\ &=\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d}+\frac{(c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{d}-\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1-c-d x}\right )}{d}+\frac{\left (6 b^2\right ) \operatorname{Subst}\left (\int \frac{\left (a+b \tanh ^{-1}(x)\right ) \log \left (\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d}+\frac{(c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{d}-\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1-c-d x}\right )}{d}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{d}+\frac{\left (3 b^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (1-\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^3}{d}+\frac{(c+d x) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{d}-\frac{3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1-c-d x}\right )}{d}-\frac{3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{d}+\frac{3 b^3 \text{Li}_3\left (1-\frac{2}{1-c-d x}\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.148561, size = 194, normalized size = 1.47 \[ \frac{6 a b^2 \left (\text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c+d x)}\right )+\tanh ^{-1}(c+d x) \left ((c+d x-1) \tanh ^{-1}(c+d x)-2 \log \left (e^{-2 \tanh ^{-1}(c+d x)}+1\right )\right )\right )+2 b^3 \left (3 \tanh ^{-1}(c+d x) \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c+d x)}\right )+\frac{3}{2} \text{PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c+d x)}\right )+\tanh ^{-1}(c+d x)^2 \left ((c+d x-1) \tanh ^{-1}(c+d x)-3 \log \left (e^{-2 \tanh ^{-1}(c+d x)}+1\right )\right )\right )+3 a^2 b \log \left (1-(c+d x)^2\right )+6 a^2 b (c+d x) \tanh ^{-1}(c+d x)+2 a^3 (c+d x)}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c + d*x])^3,x]

[Out]

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

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Maple [B]  time = 0.106, size = 346, normalized size = 2.6 \begin{align*} x{a}^{3}+{\frac{{a}^{3}c}{d}}+ \left ({\it Artanh} \left ( dx+c \right ) \right ) ^{3}x{b}^{3}+{\frac{ \left ({\it Artanh} \left ( dx+c \right ) \right ) ^{3}{b}^{3}c}{d}}+{\frac{ \left ({\it Artanh} \left ( dx+c \right ) \right ) ^{3}{b}^{3}}{d}}-3\,{\frac{{b}^{3} \left ({\it Artanh} \left ( dx+c \right ) \right ) ^{2}}{d}\ln \left ({\frac{ \left ( dx+c+1 \right ) ^{2}}{1- \left ( dx+c \right ) ^{2}}}+1 \right ) }-3\,{\frac{{b}^{3}{\it Artanh} \left ( dx+c \right ) }{d}{\it polylog} \left ( 2,-{\frac{ \left ( dx+c+1 \right ) ^{2}}{1- \left ( dx+c \right ) ^{2}}} \right ) }+{\frac{3\,{b}^{3}}{2\,d}{\it polylog} \left ( 3,-{\frac{ \left ( dx+c+1 \right ) ^{2}}{1- \left ( dx+c \right ) ^{2}}} \right ) }+3\, \left ({\it Artanh} \left ( dx+c \right ) \right ) ^{2}xa{b}^{2}+3\,{\frac{ \left ({\it Artanh} \left ( dx+c \right ) \right ) ^{2}a{b}^{2}c}{d}}+3\,{\frac{ \left ({\it Artanh} \left ( dx+c \right ) \right ) ^{2}a{b}^{2}}{d}}-6\,{\frac{{\it Artanh} \left ( dx+c \right ) a{b}^{2}}{d}\ln \left ({\frac{ \left ( dx+c+1 \right ) ^{2}}{1- \left ( dx+c \right ) ^{2}}}+1 \right ) }-3\,{\frac{a{b}^{2}}{d}{\it polylog} \left ( 2,-{\frac{ \left ( dx+c+1 \right ) ^{2}}{1- \left ( dx+c \right ) ^{2}}} \right ) }+3\,{\it Artanh} \left ( dx+c \right ) x{a}^{2}b+3\,{\frac{{\it Artanh} \left ( dx+c \right ){a}^{2}bc}{d}}+{\frac{3\,{a}^{2}b\ln \left ( 1- \left ( dx+c \right ) ^{2} \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x + 3/2*(2*(d*x + c)*arctanh(d*x + c) + log(-(d*x + c)^2 + 1))*a^2*b/d - 1/8*((b^3*d*x + b^3*(c - 1))*log(
-d*x - c + 1)^3 - 3*(2*a*b^2*d*x + (b^3*d*x + b^3*(c + 1))*log(d*x + c + 1))*log(-d*x - c + 1)^2)/d - integrat
e(-1/8*((b^3*d*x + b^3*(c - 1))*log(d*x + c + 1)^3 + 6*(a*b^2*d*x + a*b^2*(c - 1))*log(d*x + c + 1)^2 - 3*(4*a
*b^2*d*x + (b^3*d*x + b^3*(c - 1))*log(d*x + c + 1)^2 + 2*(b^3*(c + 1) + 2*a*b^2*(c - 1) + (2*a*b^2*d + b^3*d)
*x)*log(d*x + c + 1))*log(-d*x - c + 1))/(d*x + c - 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 (d x + c\right )^{3} + 3 \, a b^{2} \operatorname{artanh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname{artanh}\left (d x + c\right ) + a^{3}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(b^3*arctanh(d*x + c)^3 + 3*a*b^2*arctanh(d*x + c)^2 + 3*a^2*b*arctanh(d*x + c) + 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 + d x \right )}\right )^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atanh(c + d*x))**3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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