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

Optimal. Leaf size=295 \[ -\frac {b \text {Li}_2\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}+\frac {b \text {Li}_2\left (\frac {2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}-\frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{d^2 (c x+1)}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 (c x+1)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^2}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}+\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{1-c x}\right )}{2 d^2}-\frac {b^2 \text {Li}_3\left (\frac {2}{1-c x}-1\right )}{2 d^2}-\frac {b^2 \text {Li}_3\left (1-\frac {2}{c x+1}\right )}{2 d^2}+\frac {b^2}{2 d^2 (c x+1)}-\frac {b^2 \tanh ^{-1}(c x)}{2 d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.64, antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 13, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.591, Rules used = {5940, 5914, 6052, 5948, 6058, 6610, 5928, 5926, 627, 44, 207, 5918, 6056} \[ -\frac {b \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}+\frac {b \text {PolyLog}\left (2,\frac {2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}-\frac {b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}+\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{2 d^2}-\frac {b^2 \text {PolyLog}\left (3,\frac {2}{1-c x}-1\right )}{2 d^2}-\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{2 d^2}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{d^2 (c x+1)}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 (c x+1)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^2}+\frac {2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}+\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d^2}+\frac {b^2}{2 d^2 (c x+1)}-\frac {b^2 \tanh ^{-1}(c x)}{2 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 0.86, size = 254, normalized size = 0.86 \[ \frac {\frac {24 a^2}{c x+1}+24 a^2 \log (c x)-24 a^2 \log (c x+1)+12 a b \left (-2 \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )-\sinh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (2 \tanh ^{-1}(c x)\right )+2 \tanh ^{-1}(c x) \left (2 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-\sinh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (2 \tanh ^{-1}(c x)\right )\right )\right )+b^2 \left (24 \tanh ^{-1}(c x) \text {Li}_2\left (e^{2 \tanh ^{-1}(c x)}\right )-12 \text {Li}_3\left (e^{2 \tanh ^{-1}(c x)}\right )-16 \tanh ^{-1}(c x)^3+24 \tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )-12 \tanh ^{-1}(c x)^2 \sinh \left (2 \tanh ^{-1}(c x)\right )-12 \tanh ^{-1}(c x) \sinh \left (2 \tanh ^{-1}(c x)\right )-6 \sinh \left (2 \tanh ^{-1}(c x)\right )+12 \tanh ^{-1}(c x)^2 \cosh \left (2 \tanh ^{-1}(c x)\right )+12 \tanh ^{-1}(c x) \cosh \left (2 \tanh ^{-1}(c x)\right )+6 \cosh \left (2 \tanh ^{-1}(c x)\right )+i \pi ^3\right )}{24 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((24*a^2)/(1 + c*x) + 24*a^2*Log[c*x] - 24*a^2*Log[1 + c*x] + 12*a*b*(Cosh[2*ArcTanh[c*x]] - 2*PolyLog[2, E^(-
2*ArcTanh[c*x])] + 2*ArcTanh[c*x]*(Cosh[2*ArcTanh[c*x]] + 2*Log[1 - E^(-2*ArcTanh[c*x])] - Sinh[2*ArcTanh[c*x]
]) - Sinh[2*ArcTanh[c*x]]) + b^2*(I*Pi^3 - 16*ArcTanh[c*x]^3 + 6*Cosh[2*ArcTanh[c*x]] + 12*ArcTanh[c*x]*Cosh[2
*ArcTanh[c*x]] + 12*ArcTanh[c*x]^2*Cosh[2*ArcTanh[c*x]] + 24*ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])] + 24*A
rcTanh[c*x]*PolyLog[2, E^(2*ArcTanh[c*x])] - 12*PolyLog[3, E^(2*ArcTanh[c*x])] - 6*Sinh[2*ArcTanh[c*x]] - 12*A
rcTanh[c*x]*Sinh[2*ArcTanh[c*x]] - 12*ArcTanh[c*x]^2*Sinh[2*ArcTanh[c*x]]))/(24*d^2)

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {artanh}\left (c x\right )^{2} + 2 \, a b \operatorname {artanh}\left (c x\right ) + a^{2}}{c^{2} d^{2} x^{3} + 2 \, c d^{2} x^{2} + d^{2} x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.60, size = 1566, normalized size = 5.31 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*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*arctanh(c*x)^2+1/2*I*b^2/d^2*
arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3+1/2*I*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)^2-1/2*b^2/d^2*arctanh(c*x)/(c*x+1)*c*x-a*b/d^2*dilog(c*x)-a*b/d^2*dilog(c*x+1)+b^
2/d^2*arctanh(c*x)^2*ln(c*x)-b^2/d^2*arctanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)+b^2/d^2*arctanh(c*x)^2*ln(1+(
c*x+1)/(-c^2*x^2+1)^(1/2))+2*b^2/d^2*arctanh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2))+2*b^2/d^2*arctanh(c*x
)*polylog(2,(c*x+1)/(-c^2*x^2+1)^(1/2))+b^2/d^2*arctanh(c*x)^2*ln(1-(c*x+1)/(-c^2*x^2+1)^(1/2))+a^2/d^2/(c*x+1
)-a^2/d^2*ln(c*x+1)-1/2*b^2/d^2*arctanh(c*x)^2-2/3*b^2/d^2*arctanh(c*x)^3+a^2/d^2*ln(c*x)-2*b^2/d^2*polylog(3,
(c*x+1)/(-c^2*x^2+1)^(1/2))-2*b^2/d^2*polylog(3,-(c*x+1)/(-c^2*x^2+1)^(1/2))+2*a*b/d^2*arctanh(c*x)*ln(c*x)-a*
b/d^2*ln(c*x)*ln(c*x+1)-1/4*b^2/d^2/(c*x+1)*c*x+2*a*b/d^2*arctanh(c*x)/(c*x+1)-2*a*b/d^2*arctanh(c*x)*ln(c*x+1
)-a*b/d^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+a*b/d^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+1/4*b^2/d^2/(c*x+1)+1/2*I*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)^2-1/2*I*b^2/d^2*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*c
sgn(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)))+1/2*I*b^2/d^2*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^
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+a*b/d^2/(c*x+1)+b^2/d^2*arctanh(c*x)^2/(c*x+1)+1/2*b^2/d^2*arctanh(c*x)/(c*x+1)+2*b^2/d^2*arctanh(c*x)^2*ln
((c*x+1)/(-c^2*x^2+1)^(1/2))+b^2/d^2*arctanh(c*x)^2*ln(2)-b^2/d^2*arctanh(c*x)^2*ln(c*x+1)+1/2*a*b/d^2*ln(c*x+
1)^2+a*b/d^2*dilog(1/2+1/2*c*x)-1/2*a*b/d^2*ln(c*x+1)+1/2*a*b/d^2*ln(c*x-1)-1/2*I*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+1/2*I*b^2/d^2
*arctanh(c*x)^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))-1/2*I*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+I*b
^2/d^2*arctanh(c*x)^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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*(1/(c*d^2*x + d^2) - log(c*x + 1)/d^2 + log(x)/d^2) + 1/4*(b^2 - (b^2*c*x + b^2)*log(c*x + 1))*log(-c*x +
1)^2/(c*d^2*x + d^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) - 2*(b^2
*c^2*x^2 - 2*a*b + (2*a*b*c + b^2*c)*x - (b^2*c^3*x^3 + 2*b^2*c^2*x^2 + b^2)*log(c*x + 1))*log(-c*x + 1))/(c^3
*d^2*x^4 + c^2*d^2*x^3 - c*d^2*x^2 - d^2*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2/(c**2*x**3 + 2*c*x**2 + x), x) + Integral(b**2*atanh(c*x)**2/(c**2*x**3 + 2*c*x**2 + x), x) + I
ntegral(2*a*b*atanh(c*x)/(c**2*x**3 + 2*c*x**2 + x), x))/d**2

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