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

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

[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

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Rubi [A]  time = 0.498341, antiderivative size = 200, normalized size of antiderivative = 1., 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 = {5916, 5982, 266, 36, 29, 31, 5948, 5988, 5932, 6056, 6610} \[ -b^2 c^3 \text{PolyLog}\left (2,\frac{2}{c x+1}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac{1}{2} b^3 c^3 \text{PolyLog}\left (3,\frac{2}{c x+1}-1\right )-\frac{b^2 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{x}+\frac{1}{3} c^3 \left (a+b \tanh ^{-1}(c x)\right )^3+\frac{1}{2} b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2+b c^3 \log \left (2-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2-\frac{b c \left (a+b \tanh ^{-1}(c x)\right )^2}{2 x^2}-\frac{\left (a+b \tanh ^{-1}(c x)\right )^3}{3 x^3}-\frac{1}{2} b^3 c^3 \log \left (1-c^2 x^2\right )+b^3 c^3 \log (x) \]

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 5916

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

Rule 5982

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 266

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 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 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 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 5988

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 5932

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 6056

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[((a + b*ArcTa
nh[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 \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 ) \operatorname{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 ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{2} \left (b^3 c^5\right ) \operatorname{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*}

Mathematica [C]  time = 0.861358, size = 323, normalized size = 1.62 \[ -\frac{24 a b^2 \left (c^3 x^3 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )+c^2 x^2+\left (1-c^3 x^3\right ) \tanh ^{-1}(c x)^2-c x \tanh ^{-1}(c x) \left (c^2 x^2+2 c^2 x^2 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-1\right )\right )+b^3 \left (-24 c^3 x^3 \tanh ^{-1}(c x) \text{PolyLog}\left (2,e^{2 \tanh ^{-1}(c x)}\right )+12 c^3 x^3 \text{PolyLog}\left (3,e^{2 \tanh ^{-1}(c x)}\right )-i \pi ^3 c^3 x^3-24 c^3 x^3 \log \left (\frac{c x}{\sqrt{1-c^2 x^2}}\right )+8 c^3 x^3 \tanh ^{-1}(c x)^3-12 c^3 x^3 \tanh ^{-1}(c x)^2+24 c^2 x^2 \tanh ^{-1}(c x)-24 c^3 x^3 \tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )+12 c x \tanh ^{-1}(c x)^2+8 \tanh ^{-1}(c x)^3\right )-24 a^2 b c^3 x^3 \log (x)+12 a^2 b c^3 x^3 \log \left (1-c^2 x^2\right )+12 a^2 b c x+24 a^2 b \tanh ^{-1}(c x)+8 a^3}{24 x^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(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*A
rcTanh[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])]))/(24*x^3)

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Maple [C]  time = 0.849, size = 1838, normalized size = 9.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c^3*a*b^2*ln(c*x-1)+1/2*c^3*a*b^2*ln(c*x+1)-1/2*c^3*a^2*b*ln(c*x-1)-1/2*c^3*a^2*b*ln(c*x+1)-1/4*c^3*a*b^2
*ln(c*x-1)^2+1/4*c^3*a*b^2*ln(c*x+1)^2-a^2*b/x^3*arctanh(c*x)-a*b^2/x^3*arctanh(c*x)^2-1/4*I*c^3*b^3*arctanh(c
*x)^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2
/(-c^2*x^2+1)+1))*Pi+1/2*I*c^3*b^3*arctanh(c*x)^2*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*((c*x+1)^2/(-c^2*x
^2+1)-1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))*Pi-1/3*a^3/x^3-1/3*b^3/x^3*arctanh(c*x
)^3+c^3*b^3*ln((c*x+1)/(-c^2*x^2+1)^(1/2)-1)+c^3*b^3*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))-2*c^3*b^3*polylog(3,-(c*
x+1)/(-c^2*x^2+1)^(1/2))-2*c^3*b^3*polylog(3,(c*x+1)/(-c^2*x^2+1)^(1/2))-c^3*b^3*arctanh(c*x)-1/3*c^3*b^3*arct
anh(c*x)^3+1/2*c^3*b^3*arctanh(c*x)^2+1/4*I*c^3*b^3*arctanh(c*x)^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*Pi+1/2*I*c^
3*b^3*arctanh(c*x)^2*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^3*Pi+1/2*I*c^3*b^3*arctanh(c*x)^2*csgn(I*((c*x+1)^2/(-
c^2*x^2+1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^3*Pi+1/4*I*c^3*b^3*arctanh(c*x)^2*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x
+1)^2/(-c^2*x^2+1)+1))^3*Pi-1/2*I*c^3*b^3*arctanh(c*x)^2*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))^2*Pi+1/4*I*c^3*b^3
*arctanh(c*x)^2*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*
Pi-c^2*a*b^2/x-1/2*c*a^2*b/x^2-c^2*b^3*arctanh(c*x)/x-1/2*c*b^3*arctanh(c*x)^2/x^2+c^3*b^3*arctanh(c*x)^2*ln(2
)+c^3*a*b^2*dilog(1/2+1/2*c*x)+c^3*b^3*arctanh(c*x)^2*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+c^3*b^3*ln(c*x)*arctanh
(c*x)^2-c^3*b^3*arctanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)+2*c^3*b^3*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^
2+1)^(1/2))+c^3*b^3*arctanh(c*x)^2*ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2))+2*c^3*b^3*arctanh(c*x)*polylog(2,(c*x+1)/(
-c^2*x^2+1)^(1/2))-1/2*c^3*b^3*arctanh(c*x)^2*ln(c*x-1)-1/2*c^3*b^3*arctanh(c*x)^2*ln(c*x+1)+c^3*b^3*arctanh(c
*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))-c^3*a*b^2*dilog(c*x)-c^3*a*b^2*dilog(c*x+1)+c^3*a^2*b*ln(c*x)-c^3*a*b^2*a
rctanh(c*x)*ln(c*x-1)+2*c^3*a*b^2*arctanh(c*x)*ln(c*x)-c^3*a*b^2*ln(c*x)*ln(c*x+1)-c^3*a*b^2*arctanh(c*x)*ln(c
*x+1)+1/2*c^3*a*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)-1/2*c^3*a*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+1/2*c^3*a*b^2*ln(-1/2*c
*x+1/2)*ln(1/2+1/2*c*x)-c*a*b^2*arctanh(c*x)/x^2+1/2*I*c^3*b^3*arctanh(c*x)^2*Pi-1/2*I*c^3*b^3*arctanh(c*x)^2*
csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*Pi+1/4*I*c^
3*b^3*arctanh(c*x)^2*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))*Pi+1/2*I*c^3*b^3*arcta
nh(c*x)^2*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*Pi-1/4*I*c^3*b^3*arctanh(c*x)^2*c
sgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*Pi-1/2*I*c^3*b^3*arcta
nh(c*x)^2*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/((c*x+1)^2/(-c^2*x^2+1)+1))^2*P
i

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{1}{2} \,{\left ({\left (c^{2} \log \left (c^{2} x^{2} - 1\right ) - c^{2} \log \left (x^{2}\right ) + \frac{1}{x^{2}}\right )} c + \frac{2 \, \operatorname{artanh}\left (c x\right )}{x^{3}}\right )} a^{2} b - \frac{a^{3}}{3 \, x^{3}} - \frac{{\left (b^{3} c^{3} x^{3} - b^{3}\right )} \log \left (-c x + 1\right )^{3} + 3 \,{\left (b^{3} c x + 2 \, a b^{2} +{\left (b^{3} c^{3} x^{3} + b^{3}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{24 \, x^{3}} - \int -\frac{{\left (b^{3} c x - b^{3}\right )} \log \left (c x + 1\right )^{3} + 6 \,{\left (a b^{2} c x - a b^{2}\right )} \log \left (c x + 1\right )^{2} +{\left (2 \, b^{3} c^{2} x^{2} + 4 \, a b^{2} c x - 3 \,{\left (b^{3} c x - b^{3}\right )} \log \left (c x + 1\right )^{2} + 2 \,{\left (b^{3} c^{4} x^{4} + 6 \, a b^{2} -{\left (6 \, a b^{2} c - b^{3} c\right )} x\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{8 \,{\left (c x^{5} - x^{4}\right )}}\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{3} \operatorname{artanh}\left (c x\right )^{3} + 3 \, a b^{2} \operatorname{artanh}\left (c x\right )^{2} + 3 \, a^{2} b \operatorname{artanh}\left (c x\right ) + a^{3}}{x^{4}}, x\right ) \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atanh}{\left (c x \right )}\right )^{3}}{x^{4}}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{3}}{x^{4}}\,{d x} \end{align*}

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)