3.1.28 \(\int \frac {(a+b \tanh ^{-1}(c+d x))^3}{(c e+d e x)^4} \, dx\) [28]

Optimal. Leaf size=269 \[ -\frac {b^2 \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}+\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b^3 \log (c+d x)}{d e^4}-\frac {b^3 \log \left (1-(c+d x)^2\right )}{2 d e^4}+\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1+c+d x}\right )}{d e^4}-\frac {b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {PolyLog}\left (2,-1+\frac {2}{1+c+d x}\right )}{d e^4}-\frac {b^3 \text {PolyLog}\left (3,-1+\frac {2}{1+c+d x}\right )}{2 d e^4} \]

[Out]

-b^2*(a+b*arctanh(d*x+c))/d/e^4/(d*x+c)+1/2*b*(a+b*arctanh(d*x+c))^2/d/e^4-1/2*b*(a+b*arctanh(d*x+c))^2/d/e^4/
(d*x+c)^2+1/3*(a+b*arctanh(d*x+c))^3/d/e^4-1/3*(a+b*arctanh(d*x+c))^3/d/e^4/(d*x+c)^3+b^3*ln(d*x+c)/d/e^4-1/2*
b^3*ln(1-(d*x+c)^2)/d/e^4+b*(a+b*arctanh(d*x+c))^2*ln(2-2/(d*x+c+1))/d/e^4-b^2*(a+b*arctanh(d*x+c))*polylog(2,
-1+2/(d*x+c+1))/d/e^4-1/2*b^3*polylog(3,-1+2/(d*x+c+1))/d/e^4

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Rubi [A]
time = 0.36, antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {6242, 12, 6037, 6129, 272, 36, 31, 29, 6095, 6135, 6079, 6203, 6745} \begin {gather*} -\frac {b^2 \text {Li}_2\left (\frac {2}{c+d x+1}-1\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^4}-\frac {b^2 \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4}+\frac {b \log \left (2-\frac {2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e^4}-\frac {b^3 \text {Li}_3\left (\frac {2}{c+d x+1}-1\right )}{2 d e^4}+\frac {b^3 \log (c+d x)}{d e^4}-\frac {b^3 \log \left (1-(c+d x)^2\right )}{2 d e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c + d*x])^3/(c*e + d*e*x)^4,x]

[Out]

-((b^2*(a + b*ArcTanh[c + d*x]))/(d*e^4*(c + d*x))) + (b*(a + b*ArcTanh[c + d*x])^2)/(2*d*e^4) - (b*(a + b*Arc
Tanh[c + d*x])^2)/(2*d*e^4*(c + d*x)^2) + (a + b*ArcTanh[c + d*x])^3/(3*d*e^4) - (a + b*ArcTanh[c + d*x])^3/(3
*d*e^4*(c + d*x)^3) + (b^3*Log[c + d*x])/(d*e^4) - (b^3*Log[1 - (c + d*x)^2])/(2*d*e^4) + (b*(a + b*ArcTanh[c
+ d*x])^2*Log[2 - 2/(1 + c + d*x)])/(d*e^4) - (b^2*(a + b*ArcTanh[c + d*x])*PolyLog[2, -1 + 2/(1 + c + d*x)])/
(d*e^4) - (b^3*PolyLog[3, -1 + 2/(1 + c + d*x)])/(2*d*e^4)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 6037

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

Rule 6079

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[(a + b*ArcTanh[c*x
])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Dist[b*c*(p/d), Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2 - 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 6129

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

Rule 6135

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

Rule 6203

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan
h[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 6242

Int[((a_.) + ArcTanh[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[(f*(x/d))^m*(a + b*ArcTanh[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f,
0] && 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 \frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{(c e+d e x)^4} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^3}{e^4 x^4} \, dx,x,c+d x\right )}{d}\\ &=\frac {\text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^3}{x^4} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{x^3 \left (1-x^2\right )} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{x^3} \, dx,x,c+d x\right )}{d e^4}+\frac {b \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{x \left (1-x^2\right )} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^2}{x (1+x)} \, dx,x,c+d x\right )}{d e^4}+\frac {b^2 \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{x^2 \left (1-x^2\right )} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1+c+d x}\right )}{d e^4}+\frac {b^2 \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{x^2} \, dx,x,c+d x\right )}{d e^4}+\frac {b^2 \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(x)}{1-x^2} \, dx,x,c+d x\right )}{d e^4}-\frac {\left (2 b^2\right ) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right ) \log \left (2-\frac {2}{1+x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d e^4}\\ &=-\frac {b^2 \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}+\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1+c+d x}\right )}{d e^4}-\frac {b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c+d x}\right )}{d e^4}+\frac {b^3 \text {Subst}\left (\int \frac {1}{x \left (1-x^2\right )} \, dx,x,c+d x\right )}{d e^4}+\frac {b^3 \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 e^4}\\ &=-\frac {b^2 \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}+\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1+c+d x}\right )}{d e^4}-\frac {b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c+d x}\right )}{d e^4}-\frac {b^3 \text {Li}_3\left (-1+\frac {2}{1+c+d x}\right )}{2 d e^4}+\frac {b^3 \text {Subst}\left (\int \frac {1}{(1-x) x} \, dx,x,(c+d x)^2\right )}{2 d e^4}\\ &=-\frac {b^2 \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}+\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1+c+d x}\right )}{d e^4}-\frac {b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c+d x}\right )}{d e^4}-\frac {b^3 \text {Li}_3\left (-1+\frac {2}{1+c+d x}\right )}{2 d e^4}+\frac {b^3 \text {Subst}\left (\int \frac {1}{1-x} \, dx,x,(c+d x)^2\right )}{2 d e^4}+\frac {b^3 \text {Subst}\left (\int \frac {1}{x} \, dx,x,(c+d x)^2\right )}{2 d e^4}\\ &=-\frac {b^2 \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^4 (c+d x)}+\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4}-\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 d e^4 (c+d x)^2}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4}-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{3 d e^4 (c+d x)^3}+\frac {b^3 \log (c+d x)}{d e^4}-\frac {b^3 \log \left (1-(c+d x)^2\right )}{2 d e^4}+\frac {b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \log \left (2-\frac {2}{1+c+d x}\right )}{d e^4}-\frac {b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c+d x}\right )}{d e^4}-\frac {b^3 \text {Li}_3\left (-1+\frac {2}{1+c+d x}\right )}{2 d e^4}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.91, size = 393, normalized size = 1.46 \begin {gather*} \frac {-\frac {2 a^3}{(c+d x)^3}-\frac {3 a^2 b}{(c+d x)^2}-\frac {6 a^2 b \tanh ^{-1}(c+d x)}{(c+d x)^3}+6 a^2 b \log (c+d x)-3 a^2 b \log \left (1-c^2-2 c d x-d^2 x^2\right )+6 a b^2 \left (-\frac {(c+d x)^2+\tanh ^{-1}(c+d x)^2}{(c+d x)^3}+\tanh ^{-1}(c+d x) \left (-\frac {1-(c+d x)^2}{(c+d x)^2}+\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 )+6 b^3 \left (\frac {i \pi ^3}{24}-\frac {\tanh ^{-1}(c+d x)}{c+d x}-\frac {\left (1-(c+d x)^2\right ) \tanh ^{-1}(c+d x)^2}{2 (c+d x)^2}-\frac {1}{3} \tanh ^{-1}(c+d x)^3-\frac {\tanh ^{-1}(c+d x)^3}{3 (c+d x)}-\frac {\left (1-(c+d x)^2\right ) \tanh ^{-1}(c+d x)^3}{3 (c+d x)^3}+\tanh ^{-1}(c+d x)^2 \log \left (1-e^{2 \tanh ^{-1}(c+d x)}\right )+\log \left (\frac {c+d x}{\sqrt {1-(c+d x)^2}}\right )+\tanh ^{-1}(c+d x) \text {PolyLog}\left (2,e^{2 \tanh ^{-1}(c+d x)}\right )-\frac {1}{2} \text {PolyLog}\left (3,e^{2 \tanh ^{-1}(c+d x)}\right )\right )}{6 d e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c + d*x])^3/(c*e + d*e*x)^4,x]

[Out]

((-2*a^3)/(c + d*x)^3 - (3*a^2*b)/(c + d*x)^2 - (6*a^2*b*ArcTanh[c + d*x])/(c + d*x)^3 + 6*a^2*b*Log[c + d*x]
- 3*a^2*b*Log[1 - c^2 - 2*c*d*x - d^2*x^2] + 6*a*b^2*(-(((c + d*x)^2 + ArcTanh[c + d*x]^2)/(c + d*x)^3) + ArcT
anh[c + d*x]*(-((1 - (c + d*x)^2)/(c + d*x)^2) + ArcTanh[c + d*x] + 2*Log[1 - E^(-2*ArcTanh[c + d*x])]) - Poly
Log[2, E^(-2*ArcTanh[c + d*x])]) + 6*b^3*((I/24)*Pi^3 - ArcTanh[c + d*x]/(c + d*x) - ((1 - (c + d*x)^2)*ArcTan
h[c + d*x]^2)/(2*(c + d*x)^2) - ArcTanh[c + d*x]^3/3 - ArcTanh[c + d*x]^3/(3*(c + d*x)) - ((1 - (c + d*x)^2)*A
rcTanh[c + d*x]^3)/(3*(c + d*x)^3) + ArcTanh[c + d*x]^2*Log[1 - E^(2*ArcTanh[c + d*x])] + Log[(c + d*x)/Sqrt[1
- (c + d*x)^2]] + ArcTanh[c + d*x]*PolyLog[2, E^(2*ArcTanh[c + d*x])] - PolyLog[3, E^(2*ArcTanh[c + d*x])]/2)
)/(6*d*e^4)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 6.38, size = 2080, normalized size = 7.73

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2080\)
default	\(\text {Expression too large to display}\)	\(2080\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(d*x+c))^3/(d*e*x+c*e)^4,x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/3*a^3/e^4/(d*x+c)^3+1/2*I*b^3/e^4*Pi*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)
))^3*arctanh(d*x+c)^2+1/2*I*b^3/e^4*Pi*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1))*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^
2)))*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*arctanh(d*x+c)^2-1/4*I*b^3/e^4*Pi*arc
tanh(d*x+c)^2*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1+(d*x+c+1)^2/(1-(d*x+c)
^2)))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))+1/4*I*b^3/e^4*Pi*arctanh(d*x+c)^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))*cs
gn(I*(d*x+c+1)/(1-(d*x+c)^2)^(1/2))^2-1/2*I*b^3/e^4*Pi*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*csgn(I*((d*x+c+1)
^2/(1-(d*x+c)^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^2*arctanh(d*x+c)^2-1/2*I*b^3/e^4*Pi*csgn(I*((d*x+c+1)^2/(1-
(d*x+c)^2)-1))*csgn(I*((d*x+c+1)^2/(1-(d*x+c)^2)-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^2*arctanh(d*x+c)^2+1/4*I*b^
3/e^4*Pi*arctanh(d*x+c)^2*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1+(d*x+c+1)^
2/(1-(d*x+c)^2)))^2-1/4*I*b^3/e^4*Pi*arctanh(d*x+c)^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1+(d*x+c+1)^2/(1-(d*x+
c)^2)))^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1))+1/2*I*b^3/e^4*Pi*arctanh(d*x+c)^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)
)^2*csgn(I*(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-1/2*I*b^3/e^4*Pi*arctanh(d*x+c)^2*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2
)))^2+1/2*I*b^3/e^4*Pi*arctanh(d*x+c)^2*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^3+1/4*I*b^3/e^4*Pi*arctanh(d*x+c
)^2*csgn(I*(d*x+c+1)^2/((d*x+c)^2-1)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^3+1/4*I*b^3/e^4*Pi*arctanh(d*x+c)^2*csgn(I
*(d*x+c+1)^2/((d*x+c)^2-1))^3-a^2*b/e^4/(d*x+c)^3*arctanh(d*x+c)-a*b^2/e^4/(d*x+c)^3*arctanh(d*x+c)^2-a*b^2/e^
4*arctanh(d*x+c)*ln(d*x+c+1)-a*b^2/e^4/(d*x+c)^2*arctanh(d*x+c)+2*a*b^2/e^4*ln(d*x+c)*arctanh(d*x+c)-a*b^2/e^4
*arctanh(d*x+c)*ln(d*x+c-1)+1/2*a*b^2/e^4*ln(d*x+c-1)*ln(1/2*d*x+1/2*c+1/2)-1/2*a*b^2/e^4*ln(-1/2*d*x-1/2*c+1/
2)*ln(d*x+c+1)+1/2*a*b^2/e^4*ln(-1/2*d*x-1/2*c+1/2)*ln(1/2*d*x+1/2*c+1/2)-a*b^2/e^4*ln(d*x+c)*ln(d*x+c+1)+1/2*
I*b^3/e^4*Pi*arctanh(d*x+c)^2-a*b^2/e^4/(d*x+c)+1/2*a*b^2/e^4*ln(d*x+c+1)-1/2*a*b^2/e^4*ln(d*x+c-1)+a*b^2/e^4*
dilog(1/2*d*x+1/2*c+1/2)-1/4*a*b^2/e^4*ln(d*x+c-1)^2+1/4*a*b^2/e^4*ln(d*x+c+1)^2-a*b^2/e^4*dilog(d*x+c+1)-a*b^
2/e^4*dilog(d*x+c)-1/2*a^2*b/e^4/(d*x+c)^2-1/2*a^2*b/e^4*ln(d*x+c+1)+a^2*b/e^4*ln(d*x+c)-1/2*a^2*b/e^4*ln(d*x+
c-1)-b^3/e^4/(d*x+c)*arctanh(d*x+c)+b^3/e^4*ln(d*x+c)*arctanh(d*x+c)^2+2*b^3/e^4*arctanh(d*x+c)*polylog(2,(d*x
+c+1)/(1-(d*x+c)^2)^(1/2))+b^3/e^4*arctanh(d*x+c)^2*ln(1-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+b^3/e^4*arctanh(d*x+c)
^2*ln(1+(d*x+c+1)/(1-(d*x+c)^2)^(1/2))+2*b^3/e^4*arctanh(d*x+c)*polylog(2,-(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-b^3/
e^4*arctanh(d*x+c)^2*ln((d*x+c+1)^2/(1-(d*x+c)^2)-1)-1/2*b^3/e^4/(d*x+c)^2*arctanh(d*x+c)^2-1/2*b^3/e^4*arctan
h(d*x+c)^2*ln(d*x+c-1)-1/2*b^3/e^4*arctanh(d*x+c)^2*ln(d*x+c+1)+b^3/e^4*ln(2)*arctanh(d*x+c)^2+b^3/e^4*arctanh
(d*x+c)^2*ln((d*x+c+1)/(1-(d*x+c)^2)^(1/2))-1/3*b^3/e^4/(d*x+c)^3*arctanh(d*x+c)^3-b^3/e^4*arctanh(d*x+c)+1/2*
b^3/e^4*arctanh(d*x+c)^2-2*b^3/e^4*polylog(3,(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-2*b^3/e^4*polylog(3,-(d*x+c+1)/(1-
(d*x+c)^2)^(1/2))+b^3/e^4*ln(1+(d*x+c+1)/(1-(d*x+c)^2)^(1/2))-1/3*b^3/e^4*arctanh(d*x+c)^3+b^3/e^4*ln((d*x+c+1
)/(1-(d*x+c)^2)^(1/2)-1))

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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/(d*e*x+c*e)^4,x, algorithm="maxima")

[Out]

-1/2*(d*(e^(-4)*log(d*x + c + 1)/d^2 - 2*e^(-4)*log(d*x + c)/d^2 + e^(-4)*log(d*x + c - 1)/d^2 + 1/(d^4*x^2*e^
4 + 2*c*d^3*x*e^4 + c^2*d^2*e^4)) + 2*arctanh(d*x + c)/(d^4*x^3*e^4 + 3*c*d^3*x^2*e^4 + 3*c^2*d^2*x*e^4 + c^3*
d*e^4))*a^2*b - 1/3*a^3/(d^4*x^3*e^4 + 3*c*d^3*x^2*e^4 + 3*c^2*d^2*x*e^4 + c^3*d*e^4) - 1/24*((b^3*d^3*x^3 + 3
*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + (c^3 - 1)*b^3)*log(-d*x - c + 1)^3 + 3*(b^3*d*x + b^3*c + 2*a*b^2 + (b^3*d^3*
x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + (c^3 + 1)*b^3)*log(d*x + c + 1))*log(-d*x - c + 1)^2)/(d^4*x^3*e^4 + 3
*c*d^3*x^2*e^4 + 3*c^2*d^2*x*e^4 + c^3*d*e^4) - 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 + (2*b^3*d^2*x^2 + 2*b^3*c^2 + 4*a*b^2*c - 3*(b^3*d*x + b^3*(c
- 1))*log(d*x + c + 1)^2 + 4*(b^3*c*d + a*b^2*d)*x + 2*(b^3*d^4*x^4 + 4*b^3*c*d^3*x^3 + 6*b^3*c^2*d^2*x^2 + (
c^4 + c)*b^3 - 6*a*b^2*(c - 1) + ((4*c^3*d + d)*b^3 - 6*a*b^2*d)*x)*log(d*x + c + 1))*log(-d*x - c + 1))/(d^5*
x^5*e^4 + (5*c*d^4 - d^4)*x^4*e^4 + 2*(5*c^2*d^3 - 2*c*d^3)*x^3*e^4 + 2*(5*c^3*d^2 - 3*c^2*d^2)*x^2*e^4 + (5*c
^4*d - 4*c^3*d)*x*e^4 + (c^5 - c^4)*e^4), 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/(d*e*x+c*e)^4,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)*e^(-4)/(d^4*x^
4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{3}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b^{3} \operatorname {atanh}^{3}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {3 a b^{2} \operatorname {atanh}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {3 a^{2} b \operatorname {atanh}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(d*x+c))**3/(d*e*x+c*e)**4,x)

[Out]

(Integral(a**3/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(b**3*atanh(c
+ d*x)**3/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(3*a*b**2*atanh(c +
d*x)**2/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x) + Integral(3*a**2*b*atanh(c +
d*x)/(c**4 + 4*c**3*d*x + 6*c**2*d**2*x**2 + 4*c*d**3*x**3 + d**4*x**4), x))/e**4

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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/(d*e*x+c*e)^4,x, algorithm="giac")

[Out]

integrate((b*arctanh(d*x + c) + a)^3/(d*e*x + c*e)^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c + d*x))^3/(c*e + d*e*x)^4,x)

[Out]

int((a + b*atanh(c + d*x))^3/(c*e + d*e*x)^4, x)

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