3.1.47 \(\int (a+b \tanh ^{-1}(c+d x))^3 \, dx\) [47]

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

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 132, normalized size of antiderivative = 1.00, 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 = {6238, 6021, 6131, 6055, 6095, 6205, 6745} \begin {gather*} -\frac {3 b^2 \text {Li}_2\left (1-\frac {2}{-c-d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{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}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{-c-d x+1}\right )}{2 d} \end {gather*}

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 6021

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

Rule 6055

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 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 6131

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 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 6238

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 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]

Rubi steps

\begin {align*} \int \left (a+b \tanh ^{-1}(c+d x)\right )^3 \, dx &=\frac {\text {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) \text {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) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.18, size = 205, normalized size = 1.55 \begin {gather*} \frac {2 a^3 d x+6 a^2 b d x \tanh ^{-1}(c+d x)-3 a^2 b (-1+c) \log (1-c-d x)+3 a^2 b (1+c) \log (1+c+d x)+6 a b^2 \left (\tanh ^{-1}(c+d x) \left ((-1+c+d x) \tanh ^{-1}(c+d x)-2 \log \left (1+e^{-2 \tanh ^{-1}(c+d x)}\right )\right )+\text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c+d x)}\right )\right )+2 b^3 \left (\tanh ^{-1}(c+d x)^2 \left ((-1+c+d x) \tanh ^{-1}(c+d x)-3 \log \left (1+e^{-2 \tanh ^{-1}(c+d x)}\right )\right )+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 )\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*d*x + 6*a^2*b*d*x*ArcTanh[c + d*x] - 3*a^2*b*(-1 + c)*Log[1 - c - d*x] + 3*a^2*b*(1 + c)*Log[1 + c + d*
x] + 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])]) + PolyLo
g[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*ArcTanh[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] Leaf count of result is larger than twice the leaf count of optimal. \(283\) vs. \(2(130)=260\).
time = 0.50, size = 284, normalized size = 2.15

method	result	size
derivativedivides	\(\frac {\left (d x +c \right ) a^{3}+\left (d x +c \right ) b^{3} \arctanh \left (d x +c \right )^{3}+b^{3} \arctanh \left (d x +c \right )^{3}-3 b^{3} \arctanh \left (d x +c \right )^{2} \ln \left (1+\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}\right )-3 b^{3} \arctanh \left (d x +c \right ) \polylog \left (2, -\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}\right )+\frac {3 b^{3} \polylog \left (3, -\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}\right )}{2}+3 \left (d x +c \right ) a \,b^{2} \arctanh \left (d x +c \right )^{2}+3 a \,b^{2} \arctanh \left (d x +c \right )^{2}-6 \arctanh \left (d x +c \right ) \ln \left (1+\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}\right ) a \,b^{2}-3 \polylog \left (2, -\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}\right ) a \,b^{2}+3 \left (d x +c \right ) a^{2} b \arctanh \left (d x +c \right )+\frac {3 a^{2} b \ln \left (1-\left (d x +c \right )^{2}\right )}{2}}{d}\)	\(284\)
default	\(\frac {\left (d x +c \right ) a^{3}+\left (d x +c \right ) b^{3} \arctanh \left (d x +c \right )^{3}+b^{3} \arctanh \left (d x +c \right )^{3}-3 b^{3} \arctanh \left (d x +c \right )^{2} \ln \left (1+\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}\right )-3 b^{3} \arctanh \left (d x +c \right ) \polylog \left (2, -\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}\right )+\frac {3 b^{3} \polylog \left (3, -\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}\right )}{2}+3 \left (d x +c \right ) a \,b^{2} \arctanh \left (d x +c \right )^{2}+3 a \,b^{2} \arctanh \left (d x +c \right )^{2}-6 \arctanh \left (d x +c \right ) \ln \left (1+\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}\right ) a \,b^{2}-3 \polylog \left (2, -\frac {\left (d x +c +1\right )^{2}}{1-\left (d x +c \right )^{2}}\right ) a \,b^{2}+3 \left (d x +c \right ) a^{2} b \arctanh \left (d x +c \right )+\frac {3 a^{2} b \ln \left (1-\left (d x +c \right )^{2}\right )}{2}}{d}\)	\(284\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(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 = 0.00, 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(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \operatorname {atanh}{\left (c + d x \right )}\right )^{3}\, dx \end {gather*}

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 = 0.00, 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(d*x+c))^3,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^3 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c + d*x))^3,x)

[Out]

int((a + b*atanh(c + d*x))^3, x)

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