3.110 \(\int \frac {(a+b \tanh ^{-1}(c x))^2}{x^3 (d+c d x)^2} \, dx\)

Optimal. Leaf size=480 \[ -\frac {3 b c^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}+\frac {3 b c^2 \text {Li}_2\left (\frac {2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}-\frac {3 b c^2 \text {Li}_2\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}+\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 (c x+1)}-\frac {2 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}+\frac {b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^2 (c x+1)}+\frac {6 c^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}+\frac {3 c^2 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}-\frac {4 b c^2 \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 x}-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )}{d^2 x}+\frac {2 b^2 c^2 \text {Li}_2\left (\frac {2}{c x+1}-1\right )}{d^2}+\frac {3 b^2 c^2 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (\frac {2}{1-c x}-1\right )}{2 d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (1-\frac {2}{c x+1}\right )}{2 d^2}-\frac {b^2 c^2 \log \left (1-c^2 x^2\right )}{2 d^2}+\frac {b^2 c^2}{2 d^2 (c x+1)}+\frac {b^2 c^2 \log (x)}{d^2}-\frac {b^2 c^2 \tanh ^{-1}(c x)}{2 d^2} \]

[Out]

1/2*b^2*c^2/d^2/(c*x+1)-1/2*b^2*c^2*arctanh(c*x)/d^2-b*c*(a+b*arctanh(c*x))/d^2/x+b*c^2*(a+b*arctanh(c*x))/d^2
/(c*x+1)-2*c^2*(a+b*arctanh(c*x))^2/d^2-1/2*(a+b*arctanh(c*x))^2/d^2/x^2+2*c*(a+b*arctanh(c*x))^2/d^2/x+c^2*(a
+b*arctanh(c*x))^2/d^2/(c*x+1)-6*c^2*(a+b*arctanh(c*x))^2*arctanh(-1+2/(-c*x+1))/d^2+b^2*c^2*ln(x)/d^2+3*c^2*(
a+b*arctanh(c*x))^2*ln(2/(c*x+1))/d^2-1/2*b^2*c^2*ln(-c^2*x^2+1)/d^2-4*b*c^2*(a+b*arctanh(c*x))*ln(2-2/(c*x+1)
)/d^2-3*b*c^2*(a+b*arctanh(c*x))*polylog(2,1-2/(-c*x+1))/d^2+3*b*c^2*(a+b*arctanh(c*x))*polylog(2,-1+2/(-c*x+1
))/d^2-3*b*c^2*(a+b*arctanh(c*x))*polylog(2,1-2/(c*x+1))/d^2+2*b^2*c^2*polylog(2,-1+2/(c*x+1))/d^2+3/2*b^2*c^2
*polylog(3,1-2/(-c*x+1))/d^2-3/2*b^2*c^2*polylog(3,-1+2/(-c*x+1))/d^2-3/2*b^2*c^2*polylog(3,1-2/(c*x+1))/d^2

________________________________________________________________________________________

Rubi [A]  time = 0.95, antiderivative size = 480, normalized size of antiderivative = 1.00, number of steps used = 31, number of rules used = 22, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.000, Rules used = {5940, 5916, 5982, 266, 36, 29, 31, 5948, 5988, 5932, 2447, 5914, 6052, 6058, 6610, 5928, 5926, 627, 44, 207, 5918, 6056} \[ -\frac {3 b c^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}+\frac {3 b c^2 \text {PolyLog}\left (2,\frac {2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}-\frac {3 b c^2 \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}+\frac {2 b^2 c^2 \text {PolyLog}\left (2,\frac {2}{c x+1}-1\right )}{d^2}+\frac {3 b^2 c^2 \text {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{2 d^2}-\frac {3 b^2 c^2 \text {PolyLog}\left (3,\frac {2}{1-c x}-1\right )}{2 d^2}-\frac {3 b^2 c^2 \text {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{2 d^2}+\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 (c x+1)}-\frac {2 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}+\frac {b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^2 (c x+1)}+\frac {6 c^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}+\frac {3 c^2 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}-\frac {4 b c^2 \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 x}-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )}{d^2 x}-\frac {b^2 c^2 \log \left (1-c^2 x^2\right )}{2 d^2}+\frac {b^2 c^2}{2 d^2 (c x+1)}+\frac {b^2 c^2 \log (x)}{d^2}-\frac {b^2 c^2 \tanh ^{-1}(c x)}{2 d^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^2/(x^3*(d + c*d*x)^2),x]

[Out]

(b^2*c^2)/(2*d^2*(1 + c*x)) - (b^2*c^2*ArcTanh[c*x])/(2*d^2) - (b*c*(a + b*ArcTanh[c*x]))/(d^2*x) + (b*c^2*(a
+ b*ArcTanh[c*x]))/(d^2*(1 + c*x)) - (2*c^2*(a + b*ArcTanh[c*x])^2)/d^2 - (a + b*ArcTanh[c*x])^2/(2*d^2*x^2) +
 (2*c*(a + b*ArcTanh[c*x])^2)/(d^2*x) + (c^2*(a + b*ArcTanh[c*x])^2)/(d^2*(1 + c*x)) + (6*c^2*(a + b*ArcTanh[c
*x])^2*ArcTanh[1 - 2/(1 - c*x)])/d^2 + (b^2*c^2*Log[x])/d^2 + (3*c^2*(a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/
d^2 - (b^2*c^2*Log[1 - c^2*x^2])/(2*d^2) - (4*b*c^2*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)])/d^2 - (3*b*c^2*
(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 - c*x)])/d^2 + (3*b*c^2*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 - c
*x)])/d^2 - (3*b*c^2*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/d^2 + (2*b^2*c^2*PolyLog[2, -1 + 2/(1 +
 c*x)])/d^2 + (3*b^2*c^2*PolyLog[3, 1 - 2/(1 - c*x)])/(2*d^2) - (3*b^2*c^2*PolyLog[3, -1 + 2/(1 - c*x)])/(2*d^
2) - (3*b^2*c^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*d^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 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
 FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

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 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 5914

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

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 5918

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 5926

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

Rule 5928

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(
a + b*ArcTanh[c*x])^p)/(e*(q + 1)), x] - Dist[(b*c*p)/(e*(q + 1)), Int[ExpandIntegrand[(a + b*ArcTanh[c*x])^(p
 - 1), (d + e*x)^(q + 1)/(1 - c^2*x^2), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 1] && IntegerQ[q] &
& NeQ[q, -1]

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 5940

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

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 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 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 6052

Int[(ArcTanh[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/2, Int[
(Log[1 + u]*(a + b*ArcTanh[c*x])^p)/(d + e*x^2), x], x] - Dist[1/2, Int[(Log[1 - u]*(a + b*ArcTanh[c*x])^p)/(d
 + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[u^2 - (1 - 2/(1 - c*x
))^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 6058

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 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 )^2}{x^3 (d+c d x)^2} \, dx &=\int \left (\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 x^3}-\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 x^2}+\frac {3 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 x}-\frac {c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 (1+c x)^2}-\frac {3 c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 (1+c x)}\right ) \, dx\\ &=\frac {\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x^3} \, dx}{d^2}-\frac {(2 c) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2} \, dx}{d^2}+\frac {\left (3 c^2\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x} \, dx}{d^2}-\frac {c^3 \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^2} \, dx}{d^2}-\frac {\left (3 c^3\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{1+c x} \, dx}{d^2}\\ &=-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 x}+\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 (1+c x)}+\frac {6 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{d^2}+\frac {3 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{d^2}+\frac {(b c) \int \frac {a+b \tanh ^{-1}(c x)}{x^2 \left (1-c^2 x^2\right )} \, dx}{d^2}-\frac {\left (4 b c^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx}{d^2}-\frac {\left (2 b c^3\right ) \int \left (\frac {a+b \tanh ^{-1}(c x)}{2 (1+c x)^2}-\frac {a+b \tanh ^{-1}(c x)}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{d^2}-\frac {\left (6 b c^3\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d^2}-\frac {\left (12 b c^3\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d^2}\\ &=-\frac {2 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 x}+\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 (1+c x)}+\frac {6 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{d^2}+\frac {3 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{d^2}-\frac {3 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{d^2}+\frac {(b c) \int \frac {a+b \tanh ^{-1}(c x)}{x^2} \, dx}{d^2}-\frac {\left (4 b c^2\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx}{d^2}-\frac {\left (b c^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{d^2}+\frac {\left (b c^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{d^2}+\frac {\left (b c^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{d^2}+\frac {\left (6 b c^3\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d^2}-\frac {\left (6 b c^3\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}{d^2}+\frac {\left (3 b^2 c^3\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d^2}\\ &=-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )}{d^2 x}+\frac {b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^2 (1+c x)}-\frac {2 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 x}+\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 (1+c x)}+\frac {6 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{d^2}+\frac {3 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{d^2}-\frac {4 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )}{d^2}-\frac {3 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{d^2}+\frac {3 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{d^2}-\frac {3 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 d^2}+\frac {\left (b^2 c^2\right ) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx}{d^2}-\frac {\left (b^2 c^3\right ) \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{d^2}+\frac {\left (3 b^2 c^3\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d^2}-\frac {\left (3 b^2 c^3\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{d^2}+\frac {\left (4 b^2 c^3\right ) \int \frac {\log \left (2-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d^2}\\ &=-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )}{d^2 x}+\frac {b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^2 (1+c x)}-\frac {2 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 x}+\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 (1+c x)}+\frac {6 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{d^2}+\frac {3 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{d^2}-\frac {4 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )}{d^2}-\frac {3 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{d^2}+\frac {3 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{d^2}-\frac {3 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{d^2}+\frac {2 b^2 c^2 \text {Li}_2\left (-1+\frac {2}{1+c x}\right )}{d^2}+\frac {3 b^2 c^2 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (-1+\frac {2}{1-c x}\right )}{2 d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 d^2}+\frac {\left (b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )}{2 d^2}-\frac {\left (b^2 c^3\right ) \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{d^2}\\ &=-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )}{d^2 x}+\frac {b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^2 (1+c x)}-\frac {2 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 x}+\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 (1+c x)}+\frac {6 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{d^2}+\frac {3 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{d^2}-\frac {4 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )}{d^2}-\frac {3 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{d^2}+\frac {3 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{d^2}-\frac {3 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{d^2}+\frac {2 b^2 c^2 \text {Li}_2\left (-1+\frac {2}{1+c x}\right )}{d^2}+\frac {3 b^2 c^2 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (-1+\frac {2}{1-c x}\right )}{2 d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 d^2}+\frac {\left (b^2 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d^2}-\frac {\left (b^2 c^3\right ) \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{d^2}+\frac {\left (b^2 c^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )}{2 d^2}\\ &=\frac {b^2 c^2}{2 d^2 (1+c x)}-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )}{d^2 x}+\frac {b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^2 (1+c x)}-\frac {2 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 x}+\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 (1+c x)}+\frac {6 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{d^2}+\frac {b^2 c^2 \log (x)}{d^2}+\frac {3 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{d^2}-\frac {b^2 c^2 \log \left (1-c^2 x^2\right )}{2 d^2}-\frac {4 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )}{d^2}-\frac {3 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{d^2}+\frac {3 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{d^2}-\frac {3 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{d^2}+\frac {2 b^2 c^2 \text {Li}_2\left (-1+\frac {2}{1+c x}\right )}{d^2}+\frac {3 b^2 c^2 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (-1+\frac {2}{1-c x}\right )}{2 d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 d^2}+\frac {\left (b^2 c^3\right ) \int \frac {1}{-1+c^2 x^2} \, dx}{2 d^2}\\ &=\frac {b^2 c^2}{2 d^2 (1+c x)}-\frac {b^2 c^2 \tanh ^{-1}(c x)}{2 d^2}-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )}{d^2 x}+\frac {b c^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^2 (1+c x)}-\frac {2 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^2 x^2}+\frac {2 c \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 x}+\frac {c^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 (1+c x)}+\frac {6 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{d^2}+\frac {b^2 c^2 \log (x)}{d^2}+\frac {3 c^2 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{d^2}-\frac {b^2 c^2 \log \left (1-c^2 x^2\right )}{2 d^2}-\frac {4 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )}{d^2}-\frac {3 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{d^2}+\frac {3 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{d^2}-\frac {3 b c^2 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{d^2}+\frac {2 b^2 c^2 \text {Li}_2\left (-1+\frac {2}{1+c x}\right )}{d^2}+\frac {3 b^2 c^2 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (-1+\frac {2}{1-c x}\right )}{2 d^2}-\frac {3 b^2 c^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 d^2}\\ \end {align*}

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Mathematica [C]  time = 2.17, size = 452, normalized size = 0.94 \[ \frac {\frac {8 a^2 c^2}{c x+1}+24 a^2 c^2 \log (x)-24 a^2 c^2 \log (c x+1)+\frac {16 a^2 c}{x}-\frac {4 a^2}{x^2}+\frac {4 a b \left (-6 c^2 x^2 \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )+c x \left (-8 c x \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )-c x \sinh \left (2 \tanh ^{-1}(c x)\right )+c x \cosh \left (2 \tanh ^{-1}(c x)\right )-2\right )+2 \tanh ^{-1}(c x) \left (c^2 x^2+6 c^2 x^2 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-c^2 x^2 \sinh \left (2 \tanh ^{-1}(c x)\right )+c^2 x^2 \cosh \left (2 \tanh ^{-1}(c x)\right )+4 c x-1\right )\right )}{x^2}+b^2 c^2 \left (8 \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )-\frac {4 \tanh ^{-1}(c x)^2}{c^2 x^2}+24 \tanh ^{-1}(c x) \text {Li}_2\left (e^{2 \tanh ^{-1}(c x)}\right )+16 \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )-12 \text {Li}_3\left (e^{2 \tanh ^{-1}(c x)}\right )-16 \tanh ^{-1}(c x)^3+\frac {16 \tanh ^{-1}(c x)^2}{c x}-12 \tanh ^{-1}(c x)^2-\frac {8 \tanh ^{-1}(c x)}{c x}+24 \tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )-32 \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-4 \tanh ^{-1}(c x)^2 \sinh \left (2 \tanh ^{-1}(c x)\right )-4 \tanh ^{-1}(c x) \sinh \left (2 \tanh ^{-1}(c x)\right )-2 \sinh \left (2 \tanh ^{-1}(c x)\right )+4 \tanh ^{-1}(c x)^2 \cosh \left (2 \tanh ^{-1}(c x)\right )+4 \tanh ^{-1}(c x) \cosh \left (2 \tanh ^{-1}(c x)\right )+2 \cosh \left (2 \tanh ^{-1}(c x)\right )+i \pi ^3\right )}{8 d^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c*x])^2/(x^3*(d + c*d*x)^2),x]

[Out]

((-4*a^2)/x^2 + (16*a^2*c)/x + (8*a^2*c^2)/(1 + c*x) + 24*a^2*c^2*Log[x] - 24*a^2*c^2*Log[1 + c*x] + b^2*c^2*(
I*Pi^3 - (8*ArcTanh[c*x])/(c*x) - 12*ArcTanh[c*x]^2 - (4*ArcTanh[c*x]^2)/(c^2*x^2) + (16*ArcTanh[c*x]^2)/(c*x)
 - 16*ArcTanh[c*x]^3 + 2*Cosh[2*ArcTanh[c*x]] + 4*ArcTanh[c*x]*Cosh[2*ArcTanh[c*x]] + 4*ArcTanh[c*x]^2*Cosh[2*
ArcTanh[c*x]] - 32*ArcTanh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] + 24*ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])] +
 8*Log[(c*x)/Sqrt[1 - c^2*x^2]] + 16*PolyLog[2, E^(-2*ArcTanh[c*x])] + 24*ArcTanh[c*x]*PolyLog[2, E^(2*ArcTanh
[c*x])] - 12*PolyLog[3, E^(2*ArcTanh[c*x])] - 2*Sinh[2*ArcTanh[c*x]] - 4*ArcTanh[c*x]*Sinh[2*ArcTanh[c*x]] - 4
*ArcTanh[c*x]^2*Sinh[2*ArcTanh[c*x]]) + (4*a*b*(-6*c^2*x^2*PolyLog[2, E^(-2*ArcTanh[c*x])] + c*x*(-2 + c*x*Cos
h[2*ArcTanh[c*x]] - 8*c*x*Log[(c*x)/Sqrt[1 - c^2*x^2]] - c*x*Sinh[2*ArcTanh[c*x]]) + 2*ArcTanh[c*x]*(-1 + 4*c*
x + c^2*x^2 + c^2*x^2*Cosh[2*ArcTanh[c*x]] + 6*c^2*x^2*Log[1 - E^(-2*ArcTanh[c*x])] - c^2*x^2*Sinh[2*ArcTanh[c
*x]])))/x^2)/(8*d^2)

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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {artanh}\left (c x\right )^{2} + 2 \, a b \operatorname {artanh}\left (c x\right ) + a^{2}}{c^{2} d^{2} x^{5} + 2 \, c d^{2} x^{4} + d^{2} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/x^3/(c*d*x+d)^2,x, algorithm="fricas")

[Out]

integral((b^2*arctanh(c*x)^2 + 2*a*b*arctanh(c*x) + a^2)/(c^2*d^2*x^5 + 2*c*d^2*x^4 + d^2*x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2}}{{\left (c d x + d\right )}^{2} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/x^3/(c*d*x+d)^2,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^2/((c*d*x + d)^2*x^3), x)

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maple [C]  time = 2.34, size = 2009, normalized size = 4.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))^2/x^3/(c*d*x+d)^2,x)

[Out]

3*I*c^2*b^2/d^2*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*arctanh(c*x)^2-3/2*I*c^2
*b^2/d^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
*arctanh(c*x)^2-3/2*I*c^2*b^2/d^2*Pi*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(
c*x+1)^2/(-c^2*x^2+1)))^2*arctanh(c*x)^2+3/2*I*c^2*b^2/d^2*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*csgn(I*(c*x
+1)^2/(c^2*x^2-1))*arctanh(c*x)^2-3/2*I*c^2*b^2/d^2*Pi*csgn(I*(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*arctanh(c*x)^2+3/2*I*c^2*b^2/d^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*arctanh(c*x)^2+1/4*b^2*c^2/d^2/(c*x+1)-b^2*c^2*arctanh(c
*x)/d^2+4*c*a*b/d^2*arctanh(c*x)/x-1/2*c^3*b^2/d^2*arctanh(c*x)/(c*x+1)*x-3*c^2*a*b/d^2*ln(c*x)*ln(c*x+1)+2*c^
2*a*b/d^2*arctanh(c*x)/(c*x+1)-6*c^2*a*b/d^2*arctanh(c*x)*ln(c*x+1)-3*c^2*a*b/d^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+3
*c^2*a*b/d^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+6*c^2*a*b/d^2*arctanh(c*x)*ln(c*x)+3/2*I*c^2*b^2/d^2*Pi*csgn(I*(
(c*x+1)^2/(-c^2*x^2+1)-1))*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)))*arctanh(c*x)^2-3/2*I*c^2*b^2/d^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))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))*arctanh(c*x)^2+4*c^2*b^2/d^2*dilog((c*x+1)/(-c^2
*x^2+1)^(1/2))-3*c^2*a^2/d^2*ln(c*x+1)+3/2*I*c^2*b^2/d^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^
2+1)))^3*arctanh(c*x)^2+3/2*I*c^2*b^2/d^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
rctanh(c*x)^2+3/2*I*c^2*b^2/d^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*arctanh(c*x)^2+6*c^2*b^2/d^2*arctanh(c*x)*p
olylog(2,(c*x+1)/(-c^2*x^2+1)^(1/2))+2*c^2*a*b/d^2*ln(c*x+1)+2*c^2*a*b/d^2*ln(c*x-1)+3*c^2*b^2/d^2*arctanh(c*x
)^2*ln(c*x)-3*c^2*b^2/d^2*arctanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)+3*c^2*b^2/d^2*arctanh(c*x)^2*ln(1+(c*x+1
)/(-c^2*x^2+1)^(1/2))+6*c^2*b^2/d^2*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2))+3/2*c^2*a*b/d^2*ln(c*x
+1)^2-1/4*c^3*b^2/d^2/(c*x+1)*x-3*c^2*b^2/d^2*arctanh(c*x)^2*ln(c*x+1)-4*c^2*a*b/d^2*ln(c*x)-4*c^2*b^2/d^2*arc
tanh(c*x)*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+2*c*b^2/d^2*arctanh(c*x)^2/x+3*c^2*b^2/d^2*arctanh(c*x)^2*ln(1-(c*x
+1)/(-c^2*x^2+1)^(1/2))+c^2*b^2/d^2*arctanh(c*x)^2/(c*x+1)+1/2*c^2*b^2/d^2*arctanh(c*x)/(c*x+1)-c*b^2/d^2*arct
anh(c*x)/x+6*c^2*b^2/d^2*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))+3*c^2*a*b/d^2*dilog(1/2+1/2*c*x)+3*c^2*
b^2/d^2*arctanh(c*x)^2*ln(2)+c^2*a*b/d^2/(c*x+1)-c*a*b/d^2/x-a*b/d^2*arctanh(c*x)/x^2-3*c^2*a*b/d^2*dilog(c*x)
-3*c^2*a*b/d^2*dilog(c*x+1)-1/2*a^2/d^2/x^2+c^2*b^2/d^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2)-1)+c^2*b^2/d^2*ln(1+(c*x
+1)/(-c^2*x^2+1)^(1/2))+2*c^2*b^2/d^2*arctanh(c*x)^2-1/2*b^2/d^2*arctanh(c*x)^2/x^2+c^2*a^2/d^2/(c*x+1)+3*c^2*
a^2/d^2*ln(c*x)-6*c^2*b^2/d^2*polylog(3,(c*x+1)/(-c^2*x^2+1)^(1/2))-6*c^2*b^2/d^2*polylog(3,-(c*x+1)/(-c^2*x^2
+1)^(1/2))-4*c^2*b^2/d^2*dilog(1+(c*x+1)/(-c^2*x^2+1)^(1/2))-2*c^2*b^2/d^2*arctanh(c*x)^3+2*c*a^2/d^2/x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, a^{2} {\left (\frac {6 \, c^{2} \log \left (c x + 1\right )}{d^{2}} - \frac {6 \, c^{2} \log \relax (x)}{d^{2}} - \frac {6 \, c^{2} x^{2} + 3 \, c x - 1}{c d^{2} x^{3} + d^{2} x^{2}}\right )} + \frac {{\left (6 \, b^{2} c^{2} x^{2} + 3 \, b^{2} c x - b^{2} - 6 \, {\left (b^{2} c^{3} x^{3} + b^{2} c^{2} x^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{8 \, {\left (c d^{2} x^{3} + d^{2} x^{2}\right )}} + \int \frac {{\left (b^{2} c x - b^{2}\right )} \log \left (c x + 1\right )^{2} + 4 \, {\left (a b c x - a b\right )} \log \left (c x + 1\right ) - {\left (6 \, b^{2} c^{4} x^{4} + 9 \, b^{2} c^{3} x^{3} + 2 \, b^{2} c^{2} x^{2} - 4 \, a b + {\left (4 \, a b c - b^{2} c\right )} x - 2 \, {\left (3 \, b^{2} c^{5} x^{5} + 6 \, b^{2} c^{4} x^{4} + 3 \, b^{2} c^{3} x^{3} - b^{2} c x + b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{4 \, {\left (c^{3} d^{2} x^{6} + c^{2} d^{2} x^{5} - c d^{2} x^{4} - d^{2} x^{3}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^2/x^3/(c*d*x+d)^2,x, algorithm="maxima")

[Out]

-1/2*a^2*(6*c^2*log(c*x + 1)/d^2 - 6*c^2*log(x)/d^2 - (6*c^2*x^2 + 3*c*x - 1)/(c*d^2*x^3 + d^2*x^2)) + 1/8*(6*
b^2*c^2*x^2 + 3*b^2*c*x - b^2 - 6*(b^2*c^3*x^3 + b^2*c^2*x^2)*log(c*x + 1))*log(-c*x + 1)^2/(c*d^2*x^3 + d^2*x
^2) + integrate(1/4*((b^2*c*x - b^2)*log(c*x + 1)^2 + 4*(a*b*c*x - a*b)*log(c*x + 1) - (6*b^2*c^4*x^4 + 9*b^2*
c^3*x^3 + 2*b^2*c^2*x^2 - 4*a*b + (4*a*b*c - b^2*c)*x - 2*(3*b^2*c^5*x^5 + 6*b^2*c^4*x^4 + 3*b^2*c^3*x^3 - b^2
*c*x + b^2)*log(c*x + 1))*log(-c*x + 1))/(c^3*d^2*x^6 + c^2*d^2*x^5 - c*d^2*x^4 - d^2*x^3), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2}{x^3\,{\left (d+c\,d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))^2/(x^3*(d + c*d*x)^2),x)

[Out]

int((a + b*atanh(c*x))^2/(x^3*(d + c*d*x)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{2}}{c^{2} x^{5} + 2 c x^{4} + x^{3}}\, dx + \int \frac {b^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{c^{2} x^{5} + 2 c x^{4} + x^{3}}\, dx + \int \frac {2 a b \operatorname {atanh}{\left (c x \right )}}{c^{2} x^{5} + 2 c x^{4} + x^{3}}\, dx}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))**2/x**3/(c*d*x+d)**2,x)

[Out]

(Integral(a**2/(c**2*x**5 + 2*c*x**4 + x**3), x) + Integral(b**2*atanh(c*x)**2/(c**2*x**5 + 2*c*x**4 + x**3),
x) + Integral(2*a*b*atanh(c*x)/(c**2*x**5 + 2*c*x**4 + x**3), x))/d**2

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