3.1.33 \(\int \frac {(a+b \tanh ^{-1}(c x))^3}{x^4} \, dx\) [33]

Optimal. Leaf size=200 \[ -\frac {b^2 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {1}{2} b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}+\frac {1}{3} c^3 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 x^3}+b^3 c^3 \log (x)-\frac {1}{2} b^3 c^3 \log \left (1-c^2 x^2\right )+b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )-b^2 c^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-1+\frac {2}{1+c x}\right )-\frac {1}{2} b^3 c^3 \text {PolyLog}\left (3,-1+\frac {2}{1+c x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.34, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.786, Rules used = {6037, 6129, 272, 36, 29, 31, 6095, 6135, 6079, 6203, 6745} \begin {gather*} -b^2 c^3 \text {Li}_2\left (\frac {2}{c x+1}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac {b^2 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {1}{3} c^3 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac {1}{2} b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2+b c^3 \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 x^3}-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac {1}{2} b^3 c^3 \text {Li}_3\left (\frac {2}{c x+1}-1\right )+b^3 c^3 \log (x)-\frac {1}{2} b^3 c^3 \log \left (1-c^2 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^3/x^4,x]

[Out]

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

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 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 x)\right )^3}{x^4} \, dx &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 x^3}+(b c) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x^3 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 x^3}+(b c) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x^3} \, dx+\left (b c^3\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x \left (1-c^2 x^2\right )} \, dx\\ &=-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}+\frac {1}{3} c^3 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 x^3}+\left (b^2 c^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x^2 \left (1-c^2 x^2\right )} \, dx+\left (b c^3\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x (1+c x)} \, dx\\ &=-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}+\frac {1}{3} c^3 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 x^3}+b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )+\left (b^2 c^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x^2} \, dx+\left (b^2 c^4\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx-\left (2 b^2 c^4\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx\\ &=-\frac {b^2 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {1}{2} b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}+\frac {1}{3} c^3 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 x^3}+b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )-b^2 c^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c x}\right )+\left (b^3 c^3\right ) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx+\left (b^3 c^4\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx\\ &=-\frac {b^2 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {1}{2} b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}+\frac {1}{3} c^3 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 x^3}+b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )-b^2 c^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c x}\right )-\frac {1}{2} b^3 c^3 \text {Li}_3\left (-1+\frac {2}{1+c x}\right )+\frac {1}{2} \left (b^3 c^3\right ) \text {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {b^2 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {1}{2} b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}+\frac {1}{3} c^3 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 x^3}+b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )-b^2 c^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c x}\right )-\frac {1}{2} b^3 c^3 \text {Li}_3\left (-1+\frac {2}{1+c x}\right )+\frac {1}{2} \left (b^3 c^3\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} \left (b^3 c^5\right ) \text {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )\\ &=-\frac {b^2 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac {1}{2} b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}+\frac {1}{3} c^3 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{3 x^3}+b^3 c^3 \log (x)-\frac {1}{2} b^3 c^3 \log \left (1-c^2 x^2\right )+b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )-b^2 c^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1+c x}\right )-\frac {1}{2} b^3 c^3 \text {Li}_3\left (-1+\frac {2}{1+c x}\right )\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.64, size = 323, normalized size = 1.62 \begin {gather*} -\frac {8 a^3+12 a^2 b c x+24 a^2 b \tanh ^{-1}(c x)-24 a^2 b c^3 x^3 \log (x)+12 a^2 b c^3 x^3 \log \left (1-c^2 x^2\right )+24 a b^2 \left (c^2 x^2+\left (1-c^3 x^3\right ) \tanh ^{-1}(c x)^2-c x \tanh ^{-1}(c x) \left (-1+c^2 x^2+2 c^2 x^2 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )\right )+c^3 x^3 \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )\right )+b^3 \left (-i c^3 \pi ^3 x^3+24 c^2 x^2 \tanh ^{-1}(c x)+12 c x \tanh ^{-1}(c x)^2-12 c^3 x^3 \tanh ^{-1}(c x)^2+8 \tanh ^{-1}(c x)^3+8 c^3 x^3 \tanh ^{-1}(c x)^3-24 c^3 x^3 \tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )-24 c^3 x^3 \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )-24 c^3 x^3 \tanh ^{-1}(c x) \text {PolyLog}\left (2,e^{2 \tanh ^{-1}(c x)}\right )+12 c^3 x^3 \text {PolyLog}\left (3,e^{2 \tanh ^{-1}(c x)}\right )\right )}{24 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(8*a^3 + 12*a^2*b*c*x + 24*a^2*b*ArcTanh[c*x] - 24*a^2*b*c^3*x^3*Log[x] + 12*a^2*b*c^3*x^3*Log[1 - c^2*x
^2] + 24*a*b^2*(c^2*x^2 + (1 - c^3*x^3)*ArcTanh[c*x]^2 - c*x*ArcTanh[c*x]*(-1 + c^2*x^2 + 2*c^2*x^2*Log[1 - E^
(-2*ArcTanh[c*x])]) + c^3*x^3*PolyLog[2, E^(-2*ArcTanh[c*x])]) + b^3*((-I)*c^3*Pi^3*x^3 + 24*c^2*x^2*ArcTanh[c
*x] + 12*c*x*ArcTanh[c*x]^2 - 12*c^3*x^3*ArcTanh[c*x]^2 + 8*ArcTanh[c*x]^3 + 8*c^3*x^3*ArcTanh[c*x]^3 - 24*c^3
*x^3*ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])] - 24*c^3*x^3*Log[(c*x)/Sqrt[1 - c^2*x^2]] - 24*c^3*x^3*ArcTanh
[c*x]*PolyLog[2, E^(2*ArcTanh[c*x])] + 12*c^3*x^3*PolyLog[3, E^(2*ArcTanh[c*x])]))/x^3

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 2.55, size = 1716, normalized size = 8.58

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1716\)
default	\(\text {Expression too large to display}\)	\(1716\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3*(-1/3*a^3/c^3/x^3-1/2*I*b^3*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*((c*x+1)^2/(-c^2*x
^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2+1/4*I*b^3*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(
c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2-1/2*I*b^3*arctanh(c*x)^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-
1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))-1/4*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)
^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))+1/2*I*b^3*arctanh(c*x)^2*Pi*csgn(I*
(c*x+1)^2/(c^2*x^2-1))^2*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))+1/4*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*
x^2-1))*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2+a*b^2*dilog(1/2*c*x+1/2)+b^3*ln(2)*arctanh(c*x)^2+2*a*b^2*arctanh
(c*x)*ln(c*x)-a*b^2*ln(c*x)*ln(c*x+1)-1/2*a^2*b/c^2/x^2+1/2*I*b^3*arctanh(c*x)^2*Pi-1/2*b^3/c^2/x^2*arctanh(c*
x)^2-b^3*arctanh(c*x)/c/x-a*b^2/c/x-1/2*a^2*b*ln(c*x-1)-1/2*a^2*b*ln(c*x+1)+b^3*arctanh(c*x)^2*ln((c*x+1)/(-c^
2*x^2+1)^(1/2))-1/2*b^3*arctanh(c*x)^2*ln(c*x-1)-1/2*b^3*arctanh(c*x)^2*ln(c*x+1)-1/4*a*b^2*ln(c*x-1)^2+1/4*a*
b^2*ln(c*x+1)^2-1/2*a*b^2*ln(c*x-1)+1/2*a*b^2*ln(c*x+1)-a*b^2*arctanh(c*x)*ln(c*x+1)+1/2*a*b^2*ln(c*x-1)*ln(1/
2*c*x+1/2)+b^3*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+b^3*ln((c*x+1)/(-c^2*x^2+1)^(1/2)-1)-1/2*a*b^2*ln(-1/2*c*x+1/2
)*ln(c*x+1)+1/2*a*b^2*ln(-1/2*c*x+1/2)*ln(1/2*c*x+1/2)-a*b^2*arctanh(c*x)*ln(c*x-1)-a*b^2*dilog(c*x)-a*b^2*dil
og(c*x+1)+a^2*b*ln(c*x)-b^3*arctanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)+2*b^3*arctanh(c*x)*polylog(2,(c*x+1)/(
-c^2*x^2+1)^(1/2))+b^3*arctanh(c*x)^2*ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2))+2*b^3*arctanh(c*x)*polylog(2,-(c*x+1)/(
-c^2*x^2+1)^(1/2))+b^3*arctanh(c*x)^2*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+b^3*ln(c*x)*arctanh(c*x)^2-2*b^3*polylo
g(3,-(c*x+1)/(-c^2*x^2+1)^(1/2))-2*b^3*polylog(3,(c*x+1)/(-c^2*x^2+1)^(1/2))-b^3*arctanh(c*x)+1/2*b^3*arctanh(
c*x)^2-1/3*b^3*arctanh(c*x)^3+1/2*I*b^3*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*((c*x+1)^2
/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))-1/4*I*b^3*arctanh(c*x)^2*Pi*cs
gn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(c*x+1)^2/(c^
2*x^2-1))-a*b^2/c^3/x^3*arctanh(c*x)^2-a^2*b/c^3/x^3*arctanh(c*x)+1/4*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2
/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3+1/4*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3-1/2*I*b
^3*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))^2+1/2*I*b^3*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^
2*x^2+1)))^3+1/2*I*b^3*arctanh(c*x)^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3-a*b^2
/c^2/x^2*arctanh(c*x)-1/3*b^3/c^3/x^3*arctanh(c*x)^3)

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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(c*x))^3/x^4,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atanh(c*x))**3/x**4, 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(c*x))^3/x^4,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^3/x^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\,x\right )\right )}^3}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))^3/x^4,x)

[Out]

int((a + b*atanh(c*x))^3/x^4, x)

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