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

Optimal. Leaf size=132 \[ -\frac {3 b^2 \text {Li}_2\left (1-\frac {2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{d}+\frac {(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^3}{d}+\frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{d}-\frac {3 b \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d}+\frac {3 b^3 \text {Li}_3\left (1-\frac {2}{-c-d x+1}\right )}{2 d} \]

[Out]

(a+b*arccoth(d*x+c))^3/d+(d*x+c)*(a+b*arccoth(d*x+c))^3/d-3*b*(a+b*arccoth(d*x+c))^2*ln(2/(-d*x-c+1))/d-3*b^2*
(a+b*arccoth(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 = 0.23, 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 = {6104, 5911, 5985, 5919, 5949, 6059, 6610} \[ -\frac {3 b^2 \text {PolyLog}\left (2,1-\frac {2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{d}+\frac {3 b^3 \text {PolyLog}\left (3,1-\frac {2}{-c-d x+1}\right )}{2 d}+\frac {(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^3}{d}+\frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{d}-\frac {3 b \log \left (\frac {2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCoth[c + d*x])^3,x]

[Out]

(a + b*ArcCoth[c + d*x])^3/d + ((c + d*x)*(a + b*ArcCoth[c + d*x])^3)/d - (3*b*(a + b*ArcCoth[c + d*x])^2*Log[
2/(1 - c - d*x)])/d - (3*b^2*(a + b*ArcCoth[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 5911

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

Rule 5919

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCoth[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcCoth[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 5949

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcCoth[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 5985

Int[(((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcCoth[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcCoth[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 6059

Int[(Log[u_]*((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a + b*ArcC
oth[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p)/2, Int[((a + b*ArcCoth[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 6104

Int[((a_.) + ArcCoth[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCoth[x])^p, x
], x, c + d*x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 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 \coth ^{-1}(c+d x)\right )^3 \, dx &=\frac {\operatorname {Subst}\left (\int \left (a+b \coth ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac {(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^3}{d}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {x \left (a+b \coth ^{-1}(x)\right )^2}{1-x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{d}+\frac {(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^3}{d}-\frac {(3 b) \operatorname {Subst}\left (\int \frac {\left (a+b \coth ^{-1}(x)\right )^2}{1-x} \, dx,x,c+d x\right )}{d}\\ &=\frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{d}+\frac {(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^3}{d}-\frac {3 b \left (a+b \coth ^{-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 \coth ^{-1}(x)\right ) \log \left (\frac {2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{d}+\frac {(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^3}{d}-\frac {3 b \left (a+b \coth ^{-1}(c+d x)\right )^2 \log \left (\frac {2}{1-c-d x}\right )}{d}-\frac {3 b^2 \left (a+b \coth ^{-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 \coth ^{-1}(c+d x)\right )^3}{d}+\frac {(c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^3}{d}-\frac {3 b \left (a+b \coth ^{-1}(c+d x)\right )^2 \log \left (\frac {2}{1-c-d x}\right )}{d}-\frac {3 b^2 \left (a+b \coth ^{-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 [C]  time = 0.33, size = 208, normalized size = 1.58 \[ \frac {2 a^3 (c+d x)+3 a^2 b \log \left (1-(c+d x)^2\right )+6 a^2 b (c+d x) \coth ^{-1}(c+d x)+6 a b^2 \left (\text {Li}_2\left (e^{-2 \coth ^{-1}(c+d x)}\right )+\coth ^{-1}(c+d x) \left ((c+d x-1) \coth ^{-1}(c+d x)-2 \log \left (1-e^{-2 \coth ^{-1}(c+d x)}\right )\right )\right )+2 b^3 \left (-3 \coth ^{-1}(c+d x) \text {Li}_2\left (e^{2 \coth ^{-1}(c+d x)}\right )+\frac {3}{2} \text {Li}_3\left (e^{2 \coth ^{-1}(c+d x)}\right )+(c+d x) \coth ^{-1}(c+d x)^3+\coth ^{-1}(c+d x)^3-3 \coth ^{-1}(c+d x)^2 \log \left (1-e^{2 \coth ^{-1}(c+d x)}\right )-\frac {i \pi ^3}{8}\right )}{2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*a^3*(c + d*x) + 6*a^2*b*(c + d*x)*ArcCoth[c + d*x] + 3*a^2*b*Log[1 - (c + d*x)^2] + 6*a*b^2*(ArcCoth[c + d*
x]*((-1 + c + d*x)*ArcCoth[c + d*x] - 2*Log[1 - E^(-2*ArcCoth[c + d*x])]) + PolyLog[2, E^(-2*ArcCoth[c + d*x])
]) + 2*b^3*((-1/8*I)*Pi^3 + ArcCoth[c + d*x]^3 + (c + d*x)*ArcCoth[c + d*x]^3 - 3*ArcCoth[c + d*x]^2*Log[1 - E
^(2*ArcCoth[c + d*x])] - 3*ArcCoth[c + d*x]*PolyLog[2, E^(2*ArcCoth[c + d*x])] + (3*PolyLog[3, E^(2*ArcCoth[c
+ d*x])])/2))/(2*d)

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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{3} \operatorname {arcoth}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname {arcoth}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname {arcoth}\left (d x + c\right ) + a^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \operatorname {arcoth}\left (d x + c\right ) + a\right )}^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.30, size = 485, normalized size = 3.67 \[ a^{3} x +\frac {a^{3} c}{d}+\mathrm {arccoth}\left (d x +c \right )^{3} x \,b^{3}+\frac {\mathrm {arccoth}\left (d x +c \right )^{3} b^{3} c}{d}+\frac {b^{3} \mathrm {arccoth}\left (d x +c \right )^{3}}{d}-\frac {3 \mathrm {arccoth}\left (d x +c \right )^{2} \ln \left (1+\frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) b^{3}}{d}-\frac {3 \mathrm {arccoth}\left (d x +c \right )^{2} \ln \left (1-\frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) b^{3}}{d}-\frac {6 \,\mathrm {arccoth}\left (d x +c \right ) \polylog \left (2, -\frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) b^{3}}{d}-\frac {6 \,\mathrm {arccoth}\left (d x +c \right ) \polylog \left (2, \frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) b^{3}}{d}+\frac {6 \polylog \left (3, -\frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) b^{3}}{d}+\frac {6 \polylog \left (3, \frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) b^{3}}{d}+3 \mathrm {arccoth}\left (d x +c \right )^{2} x a \,b^{2}+\frac {3 \mathrm {arccoth}\left (d x +c \right )^{2} a \,b^{2} c}{d}+\frac {3 a \,b^{2} \mathrm {arccoth}\left (d x +c \right )^{2}}{d}-\frac {6 \,\mathrm {arccoth}\left (d x +c \right ) \ln \left (1+\frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) a \,b^{2}}{d}-\frac {6 \,\mathrm {arccoth}\left (d x +c \right ) \ln \left (1-\frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) a \,b^{2}}{d}-\frac {6 \polylog \left (2, -\frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) a \,b^{2}}{d}-\frac {6 \polylog \left (2, \frac {1}{\sqrt {\frac {d x +c -1}{d x +c +1}}}\right ) a \,b^{2}}{d}+3 \,\mathrm {arccoth}\left (d x +c \right ) x \,a^{2} b +\frac {3 \,\mathrm {arccoth}\left (d x +c \right ) a^{2} b c}{d}+\frac {3 a^{2} b \ln \left (\left (d x +c \right )^{2}-1\right )}{2 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} x + \frac {3 \, {\left (2 \, {\left (d x + c\right )} \operatorname {arcoth}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} a^{2} b}{2 \, d} + \frac {{\left (b^{3} d x + b^{3} {\left (c + 1\right )}\right )} \log \left (d x + c + 1\right )^{3} + 3 \, {\left (2 \, a b^{2} d x - {\left (b^{3} d x + b^{3} {\left (c - 1\right )}\right )} \log \left (d x + c - 1\right )\right )} \log \left (d x + c + 1\right )^{2}}{8 \, d} + \int -\frac {{\left (b^{3} d x + b^{3} {\left (c + 1\right )}\right )} \log \left (d x + c - 1\right )^{3} - 6 \, {\left (a b^{2} d x + a b^{2} {\left (c + 1\right )}\right )} \log \left (d x + c - 1\right )^{2} + 3 \, {\left (4 \, a b^{2} d x - {\left (b^{3} d x + b^{3} {\left (c + 1\right )}\right )} \log \left (d x + c - 1\right )^{2} + 2 \, {\left (2 \, a b^{2} {\left (c + 1\right )} - b^{3} {\left (c - 1\right )} + {\left (2 \, a b^{2} d - b^{3} d\right )} x\right )} \log \left (d x + c - 1\right )\right )} \log \left (d x + c + 1\right )}{8 \, {\left (d x + c + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3*x + 3/2*(2*(d*x + c)*arccoth(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 + integrate(
-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*(2*a*b^2*(c + 1) - b^3*(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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,\mathrm {acoth}\left (c+d\,x\right )\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acoth(c + d*x))^3,x)

[Out]

int((a + b*acoth(c + d*x))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \operatorname {acoth}{\left (c + d x \right )}\right )^{3}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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