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

Optimal. Leaf size=331 \[ \frac {3 b \text {Li}_2\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^2}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^2 (c x+1)}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^2 (c x+1)}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^2}+\frac {4 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^2}-\frac {3 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^2}+\frac {a b x}{c^3 d^2}-\frac {2 x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^2}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d^2}+\frac {2 b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c^4 d^2}+\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{c x+1}\right )}{2 c^4 d^2}+\frac {b^2}{2 c^4 d^2 (c x+1)}-\frac {b^2 \tanh ^{-1}(c x)}{2 c^4 d^2}+\frac {b^2 x \tanh ^{-1}(c x)}{c^3 d^2}+\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.63, antiderivative size = 331, normalized size of antiderivative = 1.00, number of steps used = 24, 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} \[ \frac {3 b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^2}+\frac {2 b^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c^4 d^2}+\frac {3 b^2 \text {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{2 c^4 d^2}+\frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d^2}+\frac {a b x}{c^3 d^2}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^2 (c x+1)}-\frac {2 x \left (a+b \tanh ^{-1}(c x)\right )^2}{c^3 d^2}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^2 (c x+1)}-\frac {3 \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^2}+\frac {4 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^2}-\frac {3 \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^4 d^2}+\frac {b^2 \log \left (1-c^2 x^2\right )}{2 c^4 d^2}+\frac {b^2}{2 c^4 d^2 (c x+1)}+\frac {b^2 x \tanh ^{-1}(c x)}{c^3 d^2}-\frac {b^2 \tanh ^{-1}(c x)}{2 c^4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 1.27, size = 354, normalized size = 1.07 \[ \frac {2 a^2 c^2 x^2-8 a^2 c x+\frac {4 a^2}{c x+1}+12 a^2 \log (c x+1)+2 a b \left (-4 \log \left (1-c^2 x^2\right )+2 \tanh ^{-1}(c x) \left (c^2 x^2-4 c x-6 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-\sinh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (2 \tanh ^{-1}(c x)\right )-1\right )+6 \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+2 c x-\sinh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (2 \tanh ^{-1}(c x)\right )\right )+b^2 \left (2 \log \left (1-c^2 x^2\right )+2 c^2 x^2 \tanh ^{-1}(c x)^2+4 \left (3 \tanh ^{-1}(c x)-2\right ) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+6 \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c x)}\right )-8 c x \tanh ^{-1}(c x)^2+6 \tanh ^{-1}(c x)^2+4 c x \tanh ^{-1}(c x)-12 \tanh ^{-1}(c x)^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+16 \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-2 \tanh ^{-1}(c x)^2 \sinh \left (2 \tanh ^{-1}(c x)\right )-2 \tanh ^{-1}(c x) \sinh \left (2 \tanh ^{-1}(c x)\right )-\sinh \left (2 \tanh ^{-1}(c x)\right )+2 \tanh ^{-1}(c x)^2 \cosh \left (2 \tanh ^{-1}(c x)\right )+2 \tanh ^{-1}(c x) \cosh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (2 \tanh ^{-1}(c x)\right )\right )}{4 c^4 d^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-8*a^2*c*x + 2*a^2*c^2*x^2 + (4*a^2)/(1 + c*x) + 12*a^2*Log[1 + c*x] + 2*a*b*(2*c*x + Cosh[2*ArcTanh[c*x]] -
4*Log[1 - c^2*x^2] + 6*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 2*ArcTanh[c*x]*(-1 - 4*c*x + c^2*x^2 + Cosh[2*ArcTan
h[c*x]] - 6*Log[1 + E^(-2*ArcTanh[c*x])] - Sinh[2*ArcTanh[c*x]]) - Sinh[2*ArcTanh[c*x]]) + b^2*(4*c*x*ArcTanh[
c*x] + 6*ArcTanh[c*x]^2 - 8*c*x*ArcTanh[c*x]^2 + 2*c^2*x^2*ArcTanh[c*x]^2 + Cosh[2*ArcTanh[c*x]] + 2*ArcTanh[c
*x]*Cosh[2*ArcTanh[c*x]] + 2*ArcTanh[c*x]^2*Cosh[2*ArcTanh[c*x]] + 16*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])
] - 12*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + 2*Log[1 - c^2*x^2] + 4*(-2 + 3*ArcTanh[c*x])*PolyLog[2, -
E^(-2*ArcTanh[c*x])] + 6*PolyLog[3, -E^(-2*ArcTanh[c*x])] - Sinh[2*ArcTanh[c*x]] - 2*ArcTanh[c*x]*Sinh[2*ArcTa
nh[c*x]] - 2*ArcTanh[c*x]^2*Sinh[2*ArcTanh[c*x]]))/(4*c^4*d^2)

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x^{3} \operatorname {artanh}\left (c x\right )^{2} + 2 \, a b x^{3} \operatorname {artanh}\left (c x\right ) + a^{2} x^{3}}{c^{2} d^{2} x^{2} + 2 \, c d^{2} x + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.74, size = 1354, normalized size = 4.09 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*(2/(c^5*d^2*x + c^4*d^2) + (c*x^2 - 4*x)/(c^3*d^2) + 6*log(c*x + 1)/(c^4*d^2)) + 1/8*(b^2*c^3*x^3 - 3*
b^2*c^2*x^2 - 4*b^2*c*x + 2*b^2 + 6*(b^2*c*x + b^2)*log(c*x + 1))*log(-c*x + 1)^2/(c^5*d^2*x + c^4*d^2) - inte
grate(-1/4*((b^2*c^4*x^4 - b^2*c^3*x^3)*log(c*x + 1)^2 + 4*(a*b*c^4*x^4 - a*b*c^3*x^3)*log(c*x + 1) + (7*b^2*c
^2*x^2 - (4*a*b*c^4 + b^2*c^4)*x^4 + 2*b^2*c*x + 2*(2*a*b*c^3 + b^2*c^3)*x^3 - 2*b^2 - 2*(b^2*c^4*x^4 - b^2*c^
3*x^3 + 3*b^2*c^2*x^2 + 6*b^2*c*x + 3*b^2)*log(c*x + 1))*log(-c*x + 1))/(c^6*d^2*x^3 + c^5*d^2*x^2 - c^4*d^2*x
- c^3*d^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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