3.2.8 \(\int \frac {(a+b \tanh ^{-1}(c x))^2}{x (d+c d x)^2} \, dx\) [108]

Optimal. Leaf size=295 \[ \frac {b^2}{2 d^2 (1+c x)}-\frac {b^2 \tanh ^{-1}(c x)}{2 d^2}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{d^2 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 d^2}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{d^2 (1+c x)}+\frac {2 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{d^2}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{d^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{d^2}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-1+\frac {2}{1-c x}\right )}{d^2}-\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{d^2}+\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{1-c x}\right )}{2 d^2}-\frac {b^2 \text {PolyLog}\left (3,-1+\frac {2}{1-c x}\right )}{2 d^2}-\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{1+c x}\right )}{2 d^2} \]

[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 = 0.47, 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 = {6087, 6033, 6199, 6095, 6205, 6745, 6065, 6063, 641, 46, 213, 6055, 6203} \begin {gather*} -\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} \end {gather*}

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 46

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] && Lt
Q[m + n + 2, 0])

Rule 213

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

Rule 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
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 6033

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 6055

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 6063

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 6065

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 6087

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 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 6199

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 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 6205

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 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 )^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] Result contains complex when optimal does not.
time = 0.48, size = 254, normalized size = 0.86 \begin {gather*} \frac {\frac {24 a^2}{1+c x}+24 a^2 \log (c x)-24 a^2 \log (1+c x)+12 a b \left (\cosh \left (2 \tanh ^{-1}(c x)\right )-2 \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )+2 \tanh ^{-1}(c x) \left (\cosh \left (2 \tanh ^{-1}(c x)\right )+2 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-\sinh \left (2 \tanh ^{-1}(c x)\right )\right )-\sinh \left (2 \tanh ^{-1}(c x)\right )\right )+b^2 \left (i \pi ^3-16 \tanh ^{-1}(c x)^3+6 \cosh \left (2 \tanh ^{-1}(c x)\right )+12 \tanh ^{-1}(c x) \cosh \left (2 \tanh ^{-1}(c x)\right )+12 \tanh ^{-1}(c x)^2 \cosh \left (2 \tanh ^{-1}(c x)\right )+24 \tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )+24 \tanh ^{-1}(c x) \text {PolyLog}\left (2,e^{2 \tanh ^{-1}(c x)}\right )-12 \text {PolyLog}\left (3,e^{2 \tanh ^{-1}(c x)}\right )-6 \sinh \left (2 \tanh ^{-1}(c x)\right )-12 \tanh ^{-1}(c x) \sinh \left (2 \tanh ^{-1}(c x)\right )-12 \tanh ^{-1}(c x)^2 \sinh \left (2 \tanh ^{-1}(c x)\right )\right )}{24 d^2} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 10.37, size = 1566, normalized size = 5.31

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1566\)
default	\(\text {Expression too large to display}\)	\(1566\)

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,method=_RETURNVERBOSE)

[Out]

a^2/d^2*ln(c*x)+1/2*I*b^2/d^2*Pi*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)))^3+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+1/2*I*b^2/d^2*Pi*arctanh(
c*x)^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3+1/2*I*b^2/d^2*Pi*arctanh(c*x)^2*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c
*x+1)^2/(-c^2*x^2+1)))^3+1/2*I*b^2/d^2*Pi*arctanh(c*x)^2*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)))-1/2*I*b^2/d^2*Pi*arctanh(c*x)^2
*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)))+I*b^2/d^2*Pi*arctanh(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+1/
2*I*b^2/d^2*Pi*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))-1/2*I*b^2/d^2
*Pi*arctanh(c*x)^2*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+1/2*I*b^2/d^2*Pi*arctanh(c*x)^2*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*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)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2-1/2*I*b^2/d^2*Pi*arctanh(c*x)^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csg
n(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2+1/4*b^2/d^2/(c*x+1)-1/2*b^2/d^2*arctanh(c*x)/(c*x+1)*c
*x+b^2/d^2*arctanh(c*x)^2/(c*x+1)-b^2/d^2*arctanh(c*x)^2*ln(c*x+1)+1/2*b^2/d^2*arctanh(c*x)/(c*x+1)+a*b/d^2/(c
*x+1)+a*b/d^2*dilog(1/2*c*x+1/2)+1/2*a*b/d^2*ln(c*x+1)^2+1/2*a*b/d^2*ln(c*x-1)-1/2*a*b/d^2*ln(c*x+1)+b^2/d^2*a
rctanh(c*x)^2*ln(2)+2*b^2/d^2*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))-a*b/d^2*dilog(c*x+1)-a*b/d^2*dilog
(c*x)+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)+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))+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*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(1/2*c*x+1/2)-a*b/d^2*ln(-1/2*c*x+1/2)*ln(c*x+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(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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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))^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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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}} \end {gather*}

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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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))^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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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 )}^2}{x\,{\left (d+c\,d\,x\right )}^2} \,d x \end {gather*}

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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