3.111 \(\int \frac {x^4 (a+b \tanh ^{-1}(c x))^2}{(d+c d x)^3} \, dx\)

Optimal. Leaf size=408 \[ \frac {6 b \text {Li}_2\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3}+\frac {15 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^5 d^3 (c x+1)}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^5 d^3 (c x+1)^2}+\frac {4 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^5 d^3 (c x+1)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^5 d^3 (c x+1)^2}-\frac {43 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^5 d^3}+\frac {6 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3}-\frac {6 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^5 d^3}+\frac {a b x}{c^4 d^3}-\frac {3 x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3}+\frac {3 b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^5 d^3}+\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{c x+1}\right )}{c^5 d^3}+\frac {29 b^2}{16 c^5 d^3 (c x+1)}-\frac {b^2}{16 c^5 d^3 (c x+1)^2}-\frac {29 b^2 \tanh ^{-1}(c x)}{16 c^5 d^3}+\frac {b^2 x \tanh ^{-1}(c x)}{c^4 d^3}+\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^5 d^3} \]

[Out]

a*b*x/c^4/d^3-1/16*b^2/c^5/d^3/(c*x+1)^2+29/16*b^2/c^5/d^3/(c*x+1)-29/16*b^2*arctanh(c*x)/c^5/d^3+b^2*x*arctan
h(c*x)/c^4/d^3-1/4*b*(a+b*arctanh(c*x))/c^5/d^3/(c*x+1)^2+15/4*b*(a+b*arctanh(c*x))/c^5/d^3/(c*x+1)-43/8*(a+b*
arctanh(c*x))^2/c^5/d^3-3*x*(a+b*arctanh(c*x))^2/c^4/d^3+1/2*x^2*(a+b*arctanh(c*x))^2/c^3/d^3-1/2*(a+b*arctanh
(c*x))^2/c^5/d^3/(c*x+1)^2+4*(a+b*arctanh(c*x))^2/c^5/d^3/(c*x+1)+6*b*(a+b*arctanh(c*x))*ln(2/(-c*x+1))/c^5/d^
3-6*(a+b*arctanh(c*x))^2*ln(2/(c*x+1))/c^5/d^3+1/2*b^2*ln(-c^2*x^2+1)/c^5/d^3+3*b^2*polylog(2,1-2/(-c*x+1))/c^
5/d^3+6*b*(a+b*arctanh(c*x))*polylog(2,1-2/(c*x+1))/c^5/d^3+3*b^2*polylog(3,1-2/(c*x+1))/c^5/d^3

________________________________________________________________________________________

Rubi [A]  time = 0.81, antiderivative size = 408, normalized size of antiderivative = 1.00, number of steps used = 37, number of rules used = 17, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.773, Rules used = {5940, 5910, 5984, 5918, 2402, 2315, 5916, 5980, 260, 5948, 5928, 5926, 627, 44, 207, 6056, 6610} \[ \frac {6 b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3}+\frac {3 b^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c^5 d^3}+\frac {3 b^2 \text {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{c^5 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3}+\frac {a b x}{c^4 d^3}+\frac {15 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^5 d^3 (c x+1)}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^5 d^3 (c x+1)^2}-\frac {3 x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3}+\frac {4 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^5 d^3 (c x+1)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^5 d^3 (c x+1)^2}-\frac {43 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^5 d^3}+\frac {6 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3}-\frac {6 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^5 d^3}+\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^5 d^3}+\frac {29 b^2}{16 c^5 d^3 (c x+1)}-\frac {b^2}{16 c^5 d^3 (c x+1)^2}+\frac {b^2 x \tanh ^{-1}(c x)}{c^4 d^3}-\frac {29 b^2 \tanh ^{-1}(c x)}{16 c^5 d^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcTanh[c*x])^2)/(d + c*d*x)^3,x]

[Out]

(a*b*x)/(c^4*d^3) - b^2/(16*c^5*d^3*(1 + c*x)^2) + (29*b^2)/(16*c^5*d^3*(1 + c*x)) - (29*b^2*ArcTanh[c*x])/(16
*c^5*d^3) + (b^2*x*ArcTanh[c*x])/(c^4*d^3) - (b*(a + b*ArcTanh[c*x]))/(4*c^5*d^3*(1 + c*x)^2) + (15*b*(a + b*A
rcTanh[c*x]))/(4*c^5*d^3*(1 + c*x)) - (43*(a + b*ArcTanh[c*x])^2)/(8*c^5*d^3) - (3*x*(a + b*ArcTanh[c*x])^2)/(
c^4*d^3) + (x^2*(a + b*ArcTanh[c*x])^2)/(2*c^3*d^3) - (a + b*ArcTanh[c*x])^2/(2*c^5*d^3*(1 + c*x)^2) + (4*(a +
b*ArcTanh[c*x])^2)/(c^5*d^3*(1 + c*x)) + (6*b*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)])/(c^5*d^3) - (6*(a + b*Ar
cTanh[c*x])^2*Log[2/(1 + c*x)])/(c^5*d^3) + (b^2*Log[1 - c^2*x^2])/(2*c^5*d^3) + (3*b^2*PolyLog[2, 1 - 2/(1 -
c*x)])/(c^5*d^3) + (6*b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/(c^5*d^3) + (3*b^2*PolyLog[3, 1 - 2/
(1 + c*x)])/(c^5*d^3)

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 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

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 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 5980

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2
/e, Int[(f*x)^(m - 2)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[(d*f^2)/e, 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] && GtQ[m, 1]

Rule 5984

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c
*x])^(p + 1)/(b*e*(p + 1)), x] + Dist[1/(c*d), Int[(a + b*ArcTanh[c*x])^p/(1 - c*x), x], x] /; FreeQ[{a, b, c,
d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[p, 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 {x^4 \left (a+b \tanh ^{-1}(c x)\right )^2}{(d+c d x)^3} \, dx &=\int \left (-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^3}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3 (1+c x)^3}-\frac {4 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3 (1+c x)^2}+\frac {6 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3 (1+c x)}\right ) \, dx\\ &=\frac {\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^3} \, dx}{c^4 d^3}-\frac {3 \int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{c^4 d^3}-\frac {4 \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{(1+c x)^2} \, dx}{c^4 d^3}+\frac {6 \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{1+c x} \, dx}{c^4 d^3}+\frac {\int x \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{c^3 d^3}\\ &=-\frac {3 x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^5 d^3 (1+c x)^2}+\frac {4 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^5 d^3 (1+c x)}-\frac {6 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {b \int \left (\frac {a+b \tanh ^{-1}(c x)}{2 (1+c x)^3}+\frac {a+b \tanh ^{-1}(c x)}{4 (1+c x)^2}-\frac {a+b \tanh ^{-1}(c x)}{4 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^4 d^3}-\frac {(8 b) \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}{c^4 d^3}+\frac {(12 b) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^4 d^3}+\frac {(6 b) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{c^3 d^3}-\frac {b \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{c^2 d^3}\\ &=-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^5 d^3}-\frac {3 x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^5 d^3 (1+c x)^2}+\frac {4 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^5 d^3 (1+c x)}-\frac {6 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {6 b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{4 c^4 d^3}-\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{4 c^4 d^3}+\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{2 c^4 d^3}+\frac {b \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c^4 d^3}-\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{c^4 d^3}-\frac {(4 b) \int \frac {a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{c^4 d^3}+\frac {(4 b) \int \frac {a+b \tanh ^{-1}(c x)}{-1+c^2 x^2} \, dx}{c^4 d^3}+\frac {(6 b) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{c^4 d^3}-\frac {\left (6 b^2\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^4 d^3}\\ &=\frac {a b x}{c^4 d^3}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^5 d^3 (1+c x)^2}+\frac {15 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^5 d^3 (1+c x)}-\frac {43 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^5 d^3}-\frac {3 x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^5 d^3 (1+c x)^2}+\frac {4 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^5 d^3 (1+c x)}+\frac {6 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^5 d^3}-\frac {6 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {6 b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {b^2 \int \frac {1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx}{4 c^4 d^3}+\frac {b^2 \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{4 c^4 d^3}+\frac {b^2 \int \tanh ^{-1}(c x) \, dx}{c^4 d^3}-\frac {\left (4 b^2\right ) \int \frac {1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{c^4 d^3}-\frac {\left (6 b^2\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{c^4 d^3}\\ &=\frac {a b x}{c^4 d^3}+\frac {b^2 x \tanh ^{-1}(c x)}{c^4 d^3}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^5 d^3 (1+c x)^2}+\frac {15 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^5 d^3 (1+c x)}-\frac {43 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^5 d^3}-\frac {3 x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^5 d^3 (1+c x)^2}+\frac {4 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^5 d^3 (1+c x)}+\frac {6 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^5 d^3}-\frac {6 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {6 b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {\left (6 b^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{c^5 d^3}+\frac {b^2 \int \frac {1}{(1-c x) (1+c x)^3} \, dx}{4 c^4 d^3}+\frac {b^2 \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{4 c^4 d^3}-\frac {\left (4 b^2\right ) \int \frac {1}{(1-c x) (1+c x)^2} \, dx}{c^4 d^3}-\frac {b^2 \int \frac {x}{1-c^2 x^2} \, dx}{c^3 d^3}\\ &=\frac {a b x}{c^4 d^3}+\frac {b^2 x \tanh ^{-1}(c x)}{c^4 d^3}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^5 d^3 (1+c x)^2}+\frac {15 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^5 d^3 (1+c x)}-\frac {43 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^5 d^3}-\frac {3 x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^5 d^3 (1+c x)^2}+\frac {4 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^5 d^3 (1+c x)}+\frac {6 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^5 d^3}-\frac {6 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^5 d^3}+\frac {3 b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^5 d^3}+\frac {6 b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {b^2 \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{4 c^4 d^3}+\frac {b^2 \int \left (\frac {1}{2 (1+c x)^3}+\frac {1}{4 (1+c x)^2}-\frac {1}{4 \left (-1+c^2 x^2\right )}\right ) \, dx}{4 c^4 d^3}-\frac {\left (4 b^2\right ) \int \left (\frac {1}{2 (1+c x)^2}-\frac {1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^4 d^3}\\ &=\frac {a b x}{c^4 d^3}-\frac {b^2}{16 c^5 d^3 (1+c x)^2}+\frac {29 b^2}{16 c^5 d^3 (1+c x)}+\frac {b^2 x \tanh ^{-1}(c x)}{c^4 d^3}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^5 d^3 (1+c x)^2}+\frac {15 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^5 d^3 (1+c x)}-\frac {43 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^5 d^3}-\frac {3 x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^5 d^3 (1+c x)^2}+\frac {4 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^5 d^3 (1+c x)}+\frac {6 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^5 d^3}-\frac {6 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^5 d^3}+\frac {3 b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^5 d^3}+\frac {6 b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{c^5 d^3}-\frac {b^2 \int \frac {1}{-1+c^2 x^2} \, dx}{16 c^4 d^3}-\frac {b^2 \int \frac {1}{-1+c^2 x^2} \, dx}{8 c^4 d^3}+\frac {\left (2 b^2\right ) \int \frac {1}{-1+c^2 x^2} \, dx}{c^4 d^3}\\ &=\frac {a b x}{c^4 d^3}-\frac {b^2}{16 c^5 d^3 (1+c x)^2}+\frac {29 b^2}{16 c^5 d^3 (1+c x)}-\frac {29 b^2 \tanh ^{-1}(c x)}{16 c^5 d^3}+\frac {b^2 x \tanh ^{-1}(c x)}{c^4 d^3}-\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^5 d^3 (1+c x)^2}+\frac {15 b \left (a+b \tanh ^{-1}(c x)\right )}{4 c^5 d^3 (1+c x)}-\frac {43 \left (a+b \tanh ^{-1}(c x)\right )^2}{8 c^5 d^3}-\frac {3 x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^3}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^3 d^3}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^5 d^3 (1+c x)^2}+\frac {4 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^5 d^3 (1+c x)}+\frac {6 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c^5 d^3}-\frac {6 \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^5 d^3}+\frac {3 b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^5 d^3}+\frac {6 b \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{c^5 d^3}+\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{c^5 d^3}\\ \end {align*}

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Mathematica [A]  time = 2.00, size = 420, normalized size = 1.03 \[ \frac {8 a^2 c^2 x^2-48 a^2 c x+\frac {64 a^2}{c x+1}-\frac {8 a^2}{(c x+1)^2}+96 a^2 \log (c x+1)+a b \left (-48 \log \left (1-c^2 x^2\right )+4 \tanh ^{-1}(c x) \left (4 c^2 x^2-24 c x-48 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-14 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )+14 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )-4\right )+96 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+16 c x-28 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )+28 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )\right )+16 b^2 \left (\frac {1}{64} \left (32 \log \left (1-c^2 x^2\right )+8 \tanh ^{-1}(c x)^2 \left (4 c^2 x^2-24 c x-48 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-14 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )+14 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )+20\right )+192 \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c x)}\right )-56 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )+56 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )+4 \tanh ^{-1}(c x) \left (16 c x+96 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-28 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )+28 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )\right )\right )+\left (6 \tanh ^{-1}(c x)-3\right ) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )\right )}{16 c^5 d^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*(a + b*ArcTanh[c*x])^2)/(d + c*d*x)^3,x]

[Out]

(-48*a^2*c*x + 8*a^2*c^2*x^2 - (8*a^2)/(1 + c*x)^2 + (64*a^2)/(1 + c*x) + 96*a^2*Log[1 + c*x] + a*b*(16*c*x +
28*Cosh[2*ArcTanh[c*x]] - Cosh[4*ArcTanh[c*x]] - 48*Log[1 - c^2*x^2] + 96*PolyLog[2, -E^(-2*ArcTanh[c*x])] - 2
8*Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]] + 4*ArcTanh[c*x]*(-4 - 24*c*x + 4*c^2*x^2 + 14*Cosh[2*ArcTanh[c*
x]] - Cosh[4*ArcTanh[c*x]] - 48*Log[1 + E^(-2*ArcTanh[c*x])] - 14*Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]])
) + 16*b^2*((-3 + 6*ArcTanh[c*x])*PolyLog[2, -E^(-2*ArcTanh[c*x])] + (56*Cosh[2*ArcTanh[c*x]] - Cosh[4*ArcTanh
[c*x]] + 32*Log[1 - c^2*x^2] + 192*PolyLog[3, -E^(-2*ArcTanh[c*x])] - 56*Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcTanh
[c*x]] + 4*ArcTanh[c*x]*(16*c*x + 28*Cosh[2*ArcTanh[c*x]] - Cosh[4*ArcTanh[c*x]] + 96*Log[1 + E^(-2*ArcTanh[c*
x])] - 28*Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]]) + 8*ArcTanh[c*x]^2*(20 - 24*c*x + 4*c^2*x^2 + 14*Cosh[2
*ArcTanh[c*x]] - Cosh[4*ArcTanh[c*x]] - 48*Log[1 + E^(-2*ArcTanh[c*x])] - 14*Sinh[2*ArcTanh[c*x]] + Sinh[4*Arc
Tanh[c*x]]))/64))/(16*c^5*d^3)

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fricas [F]  time = 1.06, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{4} \operatorname {artanh}\left (c x\right )^{2} + 2 \, a b x^{4} \operatorname {artanh}\left (c x\right ) + a^{2} x^{4}}{c^{3} d^{3} x^{3} + 3 \, c^{2} d^{3} x^{2} + 3 \, c d^{3} x + d^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x))^2/(c*d*x+d)^3,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x))^2/(c*d*x+d)^3,x, algorithm="giac")

[Out]

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

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maple [C]  time = 2.00, size = 1565, normalized size = 3.84 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-6*I/c^5*b^2/d^3*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))^2-3*I/c^5*
b^2/d^3*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x
^2+1)))^2+3*I/c^5*b^2/d^3*arctanh(c*x)^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)))^2-3*I/c^5*b^2/d^3*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))+a*b*x/c^4/d^3+b^2*x*arctanh(c*x)/c^4/d^3-1/64*b^2/c^5/d^3/(c*x+1)^2+7/8*b^2/c^5/d^3/(c*x+1)+b^2*
arctanh(c*x)/c^5/d^3+1/2/c^3*a^2/d^3*x^2-3/c^4*a^2/d^3*x+4/c^5*a^2/d^3/(c*x+1)-1/2/c^5*a^2/d^3/(c*x+1)^2+6/c^5
*a^2/d^3*ln(c*x+1)-43/8/c^5*b^2/d^3*arctanh(c*x)^2+3/c^5*b^2/d^3*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))-1/c^5*b^2/
d^3*ln(1+(c*x+1)^2/(-c^2*x^2+1))+6/c^5*b^2/d^3*dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))+6/c^5*b^2/d^3*dilog(1-I*(
c*x+1)/(-c^2*x^2+1)^(1/2))+4/c^5*b^2/d^3*arctanh(c*x)^3+1/c^5*a*b/d^3+3*I/c^5*b^2/d^3*arctanh(c*x)^2*Pi*csgn(I
/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2
+1)))+8/c^5*a*b/d^3*arctanh(c*x)/(c*x+1)-7/4/c^4*b^2/d^3*arctanh(c*x)/(c*x+1)*x-1/16/c^3*b^2/d^3*arctanh(c*x)/
(c*x+1)^2*x^2+1/8/c^4*b^2/d^3*arctanh(c*x)/(c*x+1)^2*x+1/c^3*a*b/d^3*arctanh(c*x)*x^2-6/c^4*a*b/d^3*arctanh(c*
x)*x-1/c^5*a*b/d^3*arctanh(c*x)/(c*x+1)^2+12/c^5*a*b/d^3*arctanh(c*x)*ln(c*x+1)+6/c^5*a*b/d^3*ln(-1/2*c*x+1/2)
*ln(c*x+1)-6/c^5*a*b/d^3*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)-6/c^5*b^2/d^3*arctanh(c*x)^2*ln(2)-3/c^5*a*b/d^3*ln(
c*x+1)^2-6/c^5*a*b/d^3*dilog(1/2+1/2*c*x)-43/8/c^5*a*b/d^3*ln(c*x+1)-5/8/c^5*a*b/d^3*ln(c*x-1)+1/2/c^3*b^2/d^3
*arctanh(c*x)^2*x^2-3/c^4*b^2/d^3*arctanh(c*x)^2*x+4/c^5*b^2/d^3*arctanh(c*x)^2/(c*x+1)-1/2/c^5*b^2/d^3*arctan
h(c*x)^2/(c*x+1)^2-6/c^5*b^2/d^3*arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))+6/c^5*b^2/d^3*arctanh(c*x)*ln
(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))+6/c^5*b^2/d^3*arctanh(c*x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))-12/c^5*b^2/d^3*
arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))-1/64/c^3*b^2/d^3/(c*x+1)^2*x^2+1/32/c^4*b^2/d^3/(c*x+1)^2*x-7/8/
c^4*b^2/d^3/(c*x+1)*x-1/4/c^5*a*b/d^3/(c*x+1)^2+15/4/c^5*a*b/d^3/(c*x+1)+6/c^5*b^2/d^3*arctanh(c*x)^2*ln(c*x+1
)+7/4/c^5*b^2/d^3*arctanh(c*x)/(c*x+1)-1/16/c^5*b^2/d^3*arctanh(c*x)/(c*x+1)^2-3*I/c^5*b^2/d^3*arctanh(c*x)^2*
Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3-3*I/c^5*b^2/d^3*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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a^{2} {\left (\frac {8 \, c x + 7}{c^{7} d^{3} x^{2} + 2 \, c^{6} d^{3} x + c^{5} d^{3}} + \frac {c x^{2} - 6 \, x}{c^{4} d^{3}} + \frac {12 \, \log \left (c x + 1\right )}{c^{5} d^{3}}\right )} + \frac {{\left (b^{2} c^{4} x^{4} - 4 \, b^{2} c^{3} x^{3} - 11 \, b^{2} c^{2} x^{2} + 2 \, b^{2} c x + 7 \, b^{2} + 12 \, {\left (b^{2} c^{2} x^{2} + 2 \, b^{2} c x + b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )^{2}}{8 \, {\left (c^{7} d^{3} x^{2} + 2 \, c^{6} d^{3} x + c^{5} d^{3}\right )}} - \int -\frac {{\left (b^{2} c^{5} x^{5} - b^{2} c^{4} x^{4}\right )} \log \left (c x + 1\right )^{2} + 4 \, {\left (a b c^{5} x^{5} - a b c^{4} x^{4}\right )} \log \left (c x + 1\right ) + {\left (15 \, b^{2} c^{3} x^{3} + 9 \, b^{2} c^{2} x^{2} - {\left (4 \, a b c^{5} + b^{2} c^{5}\right )} x^{5} + {\left (4 \, a b c^{4} + 3 \, b^{2} c^{4}\right )} x^{4} - 9 \, b^{2} c x - 7 \, b^{2} - 2 \, {\left (b^{2} c^{5} x^{5} - b^{2} c^{4} x^{4} + 6 \, b^{2} c^{3} x^{3} + 18 \, b^{2} c^{2} x^{2} + 18 \, b^{2} c x + 6 \, b^{2}\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{4 \, {\left (c^{8} d^{3} x^{4} + 2 \, c^{7} d^{3} x^{3} - 2 \, c^{5} d^{3} x - c^{4} d^{3}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x))^2/(c*d*x+d)^3,x, algorithm="maxima")

[Out]

1/2*a^2*((8*c*x + 7)/(c^7*d^3*x^2 + 2*c^6*d^3*x + c^5*d^3) + (c*x^2 - 6*x)/(c^4*d^3) + 12*log(c*x + 1)/(c^5*d^
3)) + 1/8*(b^2*c^4*x^4 - 4*b^2*c^3*x^3 - 11*b^2*c^2*x^2 + 2*b^2*c*x + 7*b^2 + 12*(b^2*c^2*x^2 + 2*b^2*c*x + b^
2)*log(c*x + 1))*log(-c*x + 1)^2/(c^7*d^3*x^2 + 2*c^6*d^3*x + c^5*d^3) - integrate(-1/4*((b^2*c^5*x^5 - b^2*c^
4*x^4)*log(c*x + 1)^2 + 4*(a*b*c^5*x^5 - a*b*c^4*x^4)*log(c*x + 1) + (15*b^2*c^3*x^3 + 9*b^2*c^2*x^2 - (4*a*b*
c^5 + b^2*c^5)*x^5 + (4*a*b*c^4 + 3*b^2*c^4)*x^4 - 9*b^2*c*x - 7*b^2 - 2*(b^2*c^5*x^5 - b^2*c^4*x^4 + 6*b^2*c^
3*x^3 + 18*b^2*c^2*x^2 + 18*b^2*c*x + 6*b^2)*log(c*x + 1))*log(-c*x + 1))/(c^8*d^3*x^4 + 2*c^7*d^3*x^3 - 2*c^5
*d^3*x - c^4*d^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4*(a + b*atanh(c*x))^2)/(d + c*d*x)^3,x)

[Out]

int((x^4*(a + b*atanh(c*x))^2)/(d + c*d*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a^{2} x^{4}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {b^{2} x^{4} \operatorname {atanh}^{2}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac {2 a b x^{4} \operatorname {atanh}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx}{d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2*x**4/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x) + Integral(b**2*x**4*atanh(c*x)**2/(c**3*x**3 +
3*c**2*x**2 + 3*c*x + 1), x) + Integral(2*a*b*x**4*atanh(c*x)/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x))/d**3

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