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

Optimal. Leaf size=361 \[ \frac {1}{2} c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+a b c^2 d^3 x+3 c^2 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2-3 b c d^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+3 b c d^3 \text {Li}_2\left (\frac {2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )+\frac {7}{2} c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+6 c d^3 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2-6 b c d^3 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+2 b c d^3 \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{2} b^2 c d^3 \log \left (1-c^2 x^2\right )+b^2 c^2 d^3 x \tanh ^{-1}(c x)-3 b^2 c d^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right )-b^2 c d^3 \text {Li}_2\left (\frac {2}{c x+1}-1\right )+\frac {3}{2} b^2 c d^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )-\frac {3}{2} b^2 c d^3 \text {Li}_3\left (\frac {2}{1-c x}-1\right ) \]

[Out]

a*b*c^2*d^3*x+b^2*c^2*d^3*x*arctanh(c*x)+7/2*c*d^3*(a+b*arctanh(c*x))^2-d^3*(a+b*arctanh(c*x))^2/x+3*c^2*d^3*x
*(a+b*arctanh(c*x))^2+1/2*c^3*d^3*x^2*(a+b*arctanh(c*x))^2-6*c*d^3*(a+b*arctanh(c*x))^2*arctanh(-1+2/(-c*x+1))
-6*b*c*d^3*(a+b*arctanh(c*x))*ln(2/(-c*x+1))+1/2*b^2*c*d^3*ln(-c^2*x^2+1)+2*b*c*d^3*(a+b*arctanh(c*x))*ln(2-2/
(c*x+1))-3*b^2*c*d^3*polylog(2,1-2/(-c*x+1))-3*b*c*d^3*(a+b*arctanh(c*x))*polylog(2,1-2/(-c*x+1))+3*b*c*d^3*(a
+b*arctanh(c*x))*polylog(2,-1+2/(-c*x+1))-b^2*c*d^3*polylog(2,-1+2/(c*x+1))+3/2*b^2*c*d^3*polylog(3,1-2/(-c*x+
1))-3/2*b^2*c*d^3*polylog(3,-1+2/(-c*x+1))

________________________________________________________________________________________

Rubi [A]  time = 0.78, antiderivative size = 361, normalized size of antiderivative = 1.00, number of steps used = 23, 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, 5988, 5932, 2447, 5914, 6052, 5948, 6058, 6610, 5980, 260} \[ -3 b c d^3 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+3 b c d^3 \text {PolyLog}\left (2,\frac {2}{1-c x}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )-3 b^2 c d^3 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )-b^2 c d^3 \text {PolyLog}\left (2,\frac {2}{c x+1}-1\right )+\frac {3}{2} b^2 c d^3 \text {PolyLog}\left (3,1-\frac {2}{1-c x}\right )-\frac {3}{2} b^2 c d^3 \text {PolyLog}\left (3,\frac {2}{1-c x}-1\right )+\frac {1}{2} c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+a b c^2 d^3 x+3 c^2 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {7}{2} c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+6 c d^3 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2-6 b c d^3 \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )+2 b c d^3 \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )+\frac {1}{2} b^2 c d^3 \log \left (1-c^2 x^2\right )+b^2 c^2 d^3 x \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

a*b*c^2*d^3*x + b^2*c^2*d^3*x*ArcTanh[c*x] + (7*c*d^3*(a + b*ArcTanh[c*x])^2)/2 - (d^3*(a + b*ArcTanh[c*x])^2)
/x + 3*c^2*d^3*x*(a + b*ArcTanh[c*x])^2 + (c^3*d^3*x^2*(a + b*ArcTanh[c*x])^2)/2 + 6*c*d^3*(a + b*ArcTanh[c*x]
)^2*ArcTanh[1 - 2/(1 - c*x)] - 6*b*c*d^3*(a + b*ArcTanh[c*x])*Log[2/(1 - c*x)] + (b^2*c*d^3*Log[1 - c^2*x^2])/
2 + 2*b*c*d^3*(a + b*ArcTanh[c*x])*Log[2 - 2/(1 + c*x)] - 3*b^2*c*d^3*PolyLog[2, 1 - 2/(1 - c*x)] - 3*b*c*d^3*
(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 - c*x)] + 3*b*c*d^3*(a + b*ArcTanh[c*x])*PolyLog[2, -1 + 2/(1 - c*x)]
- b^2*c*d^3*PolyLog[2, -1 + 2/(1 + c*x)] + (3*b^2*c*d^3*PolyLog[3, 1 - 2/(1 - c*x)])/2 - (3*b^2*c*d^3*PolyLog
[3, -1 + 2/(1 - c*x)])/2

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 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 2447

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

Rule 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 5914

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

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 5918

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

Rule 5932

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

Rule 5940

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

Rule 5948

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

Rule 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 5988

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

Rule 6052

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

Rule 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 {(d+c d x)^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x^2} \, dx &=\int \left (3 c^2 d^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x^2}+\frac {3 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+c^3 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2\right ) \, dx\\ &=d^3 \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x^2} \, dx+\left (3 c d^3\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x} \, dx+\left (3 c^2 d^3\right ) \int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx+\left (c^3 d^3\right ) \int x \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx\\ &=-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+3 c^2 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+6 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )+\left (2 b c d^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx-\left (12 b c^2 d^3\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx-\left (6 b c^3 d^3\right ) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx-\left (b c^4 d^3\right ) \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=4 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+3 c^2 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+6 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )+\left (2 b c d^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx+\left (b c^2 d^3\right ) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx-\left (b c^2 d^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx-\left (6 b c^2 d^3\right ) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx+\left (6 b c^2 d^3\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx-\left (6 b c^2 d^3\right ) \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=a b c^2 d^3 x+\frac {7}{2} c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+3 c^2 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+6 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )-6 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )+2 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )-3 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )+3 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )+\left (b^2 c^2 d^3\right ) \int \tanh ^{-1}(c x) \, dx-\left (2 b^2 c^2 d^3\right ) \int \frac {\log \left (2-\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx+\left (3 b^2 c^2 d^3\right ) \int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx-\left (3 b^2 c^2 d^3\right ) \int \frac {\text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx+\left (6 b^2 c^2 d^3\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx\\ &=a b c^2 d^3 x+b^2 c^2 d^3 x \tanh ^{-1}(c x)+\frac {7}{2} c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+3 c^2 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+6 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )-6 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )+2 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )-3 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )+3 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )-b^2 c d^3 \text {Li}_2\left (-1+\frac {2}{1+c x}\right )+\frac {3}{2} b^2 c d^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )-\frac {3}{2} b^2 c d^3 \text {Li}_3\left (-1+\frac {2}{1-c x}\right )-\left (6 b^2 c d^3\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )-\left (b^2 c^3 d^3\right ) \int \frac {x}{1-c^2 x^2} \, dx\\ &=a b c^2 d^3 x+b^2 c^2 d^3 x \tanh ^{-1}(c x)+\frac {7}{2} c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {d^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{x}+3 c^2 d^3 x \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {1}{2} c^3 d^3 x^2 \left (a+b \tanh ^{-1}(c x)\right )^2+6 c d^3 \left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )-6 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )+\frac {1}{2} b^2 c d^3 \log \left (1-c^2 x^2\right )+2 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )-3 b^2 c d^3 \text {Li}_2\left (1-\frac {2}{1-c x}\right )-3 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )+3 b c d^3 \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )-b^2 c d^3 \text {Li}_2\left (-1+\frac {2}{1+c x}\right )+\frac {3}{2} b^2 c d^3 \text {Li}_3\left (1-\frac {2}{1-c x}\right )-\frac {3}{2} b^2 c d^3 \text {Li}_3\left (-1+\frac {2}{1-c x}\right )\\ \end {align*}

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Mathematica [C]  time = 0.64, size = 479, normalized size = 1.33 \[ \frac {d^3 \left (4 a^2 c^3 x^3+24 a^2 c^2 x^2+24 a^2 c x \log (x)-8 a^2+8 a b c^3 x^3 \tanh ^{-1}(c x)+8 a b c^2 x^2+16 a b c x \log \left (1-c^2 x^2\right )+48 a b c^2 x^2 \tanh ^{-1}(c x)-24 a b c x \text {Li}_2(-c x)+24 a b c x \text {Li}_2(c x)+16 a b c x \log (c x)+4 a b c x \log (1-c x)-4 a b c x \log (c x+1)-16 a b \tanh ^{-1}(c x)+4 b^2 c^3 x^3 \tanh ^{-1}(c x)^2+4 b^2 c x \log \left (1-c^2 x^2\right )+24 b^2 c^2 x^2 \tanh ^{-1}(c x)^2+8 b^2 c^2 x^2 \tanh ^{-1}(c x)+24 b^2 c x \left (\tanh ^{-1}(c x)+1\right ) \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )-8 b^2 c x \text {Li}_2\left (e^{-2 \tanh ^{-1}(c x)}\right )+24 b^2 c x \tanh ^{-1}(c x) \text {Li}_2\left (e^{2 \tanh ^{-1}(c x)}\right )+12 b^2 c x \text {Li}_3\left (-e^{-2 \tanh ^{-1}(c x)}\right )-12 b^2 c x \text {Li}_3\left (e^{2 \tanh ^{-1}(c x)}\right )+i \pi ^3 b^2 c x-16 b^2 c x \tanh ^{-1}(c x)^3-20 b^2 c x \tanh ^{-1}(c x)^2-8 b^2 \tanh ^{-1}(c x)^2+16 b^2 c x \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-24 b^2 c x \tanh ^{-1}(c x)^2 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-48 b^2 c x \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+24 b^2 c x \tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )\right )}{8 x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(d^3*(-8*a^2 + I*b^2*c*Pi^3*x + 24*a^2*c^2*x^2 + 8*a*b*c^2*x^2 + 4*a^2*c^3*x^3 - 16*a*b*ArcTanh[c*x] + 48*a*b*
c^2*x^2*ArcTanh[c*x] + 8*b^2*c^2*x^2*ArcTanh[c*x] + 8*a*b*c^3*x^3*ArcTanh[c*x] - 8*b^2*ArcTanh[c*x]^2 - 20*b^2
*c*x*ArcTanh[c*x]^2 + 24*b^2*c^2*x^2*ArcTanh[c*x]^2 + 4*b^2*c^3*x^3*ArcTanh[c*x]^2 - 16*b^2*c*x*ArcTanh[c*x]^3
+ 16*b^2*c*x*ArcTanh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] - 48*b^2*c*x*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])]
- 24*b^2*c*x*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + 24*b^2*c*x*ArcTanh[c*x]^2*Log[1 - E^(2*ArcTanh[c*x
])] + 24*a^2*c*x*Log[x] + 16*a*b*c*x*Log[c*x] + 4*a*b*c*x*Log[1 - c*x] - 4*a*b*c*x*Log[1 + c*x] + 16*a*b*c*x*L
og[1 - c^2*x^2] + 4*b^2*c*x*Log[1 - c^2*x^2] + 24*b^2*c*x*(1 + ArcTanh[c*x])*PolyLog[2, -E^(-2*ArcTanh[c*x])]
- 8*b^2*c*x*PolyLog[2, E^(-2*ArcTanh[c*x])] + 24*b^2*c*x*ArcTanh[c*x]*PolyLog[2, E^(2*ArcTanh[c*x])] - 24*a*b*
c*x*PolyLog[2, -(c*x)] + 24*a*b*c*x*PolyLog[2, c*x] + 12*b^2*c*x*PolyLog[3, -E^(-2*ArcTanh[c*x])] - 12*b^2*c*x
*PolyLog[3, E^(2*ArcTanh[c*x])]))/(8*x)

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{2} c^{3} d^{3} x^{3} + 3 \, a^{2} c^{2} d^{3} x^{2} + 3 \, a^{2} c d^{3} x + a^{2} d^{3} + {\left (b^{2} c^{3} d^{3} x^{3} + 3 \, b^{2} c^{2} d^{3} x^{2} + 3 \, b^{2} c d^{3} x + b^{2} d^{3}\right )} \operatorname {artanh}\left (c x\right )^{2} + 2 \, {\left (a b c^{3} d^{3} x^{3} + 3 \, a b c^{2} d^{3} x^{2} + 3 \, a b c d^{3} x + a b d^{3}\right )} \operatorname {artanh}\left (c x\right )}{x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.94, size = 1270, normalized size = 3.52 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*d^3*a^2*c^3*x^2+3*x*a^2*c^2*d^3-d^3*b^2*arctanh(c*x)^2/x+3/2*c*d^3*b^2*arctanh(c*x)^2+c*d^3*b^2*arctanh(c*
x)-6*c*d^3*b^2*dilog(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))-6*c*d^3*b^2*dilog(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))+3/2*c*d
^3*b^2*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))-6*c*d^3*b^2*polylog(3,(c*x+1)/(-c^2*x^2+1)^(1/2))-6*c*d^3*b^2*polylo
g(3,-(c*x+1)/(-c^2*x^2+1)^(1/2))+3*c*d^3*a^2*ln(c*x)-c*d^3*b^2*ln(1+(c*x+1)^2/(-c^2*x^2+1))-2*c*d^3*b^2*dilog(
(c*x+1)/(-c^2*x^2+1)^(1/2))+2*c*d^3*b^2*dilog(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+a*b*c^2*d^3*x+b^2*c^2*d^3*x*arctan
h(c*x)+6*c*d^3*a*b*arctanh(c*x)*ln(c*x)-3*c*d^3*a*b*ln(c*x)*ln(c*x+1)+6*d^3*a*b*arctanh(c*x)*c^2*x+d^3*a*b*arc
tanh(c*x)*c^3*x^2+3/2*I*c*d^3*b^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)))*csg
n(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*d^3*b^2*arctanh(c*x)^2*c^3*x^2+5
/2*c*d^3*a*b*ln(c*x-1)+3/2*c*d^3*a*b*ln(c*x+1)-3*c*d^3*a*b*dilog(c*x+1)-3*c*d^3*a*b*dilog(c*x)+2*c*d^3*a*b*ln(
c*x)+3*c*d^3*b^2*arctanh(c*x)^2*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+3*c*d^3*b^2*arctanh(c*x)^2*ln(1-(c*x+1)/(-c^2
*x^2+1)^(1/2))+6*c*d^3*b^2*arctanh(c*x)*polylog(2,(c*x+1)/(-c^2*x^2+1)^(1/2))+3*c*d^3*b^2*arctanh(c*x)^2*ln(c*
x)-2*d^3*a*b*arctanh(c*x)/x-6*c*d^3*b^2*arctanh(c*x)*ln(1+I*(c*x+1)/(-c^2*x^2+1)^(1/2))-6*c*d^3*b^2*arctanh(c*
x)*ln(1-I*(c*x+1)/(-c^2*x^2+1)^(1/2))-3*c*d^3*b^2*arctanh(c*x)^2*ln((c*x+1)^2/(-c^2*x^2+1)-1)+6*c*d^3*b^2*arct
anh(c*x)*polylog(2,-(c*x+1)/(-c^2*x^2+1)^(1/2))-3*c*d^3*b^2*arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))+2*
c*d^3*b^2*arctanh(c*x)*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))+3*d^3*b^2*arctanh(c*x)^2*c^2*x+3/2*I*c*d^3*b^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-d^3*a^2/x-3/2*I*c*d^3*b^2*Pi*csgn(I
/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*((c*x+1)^2/(-c^2*x^2+1)-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2*arctanh(c*x)^2-3/
2*I*c*d^3*b^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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a^2*c^3*d^3*x^2 + 3*a^2*c^2*d^3*x + 3*(2*c*x*arctanh(c*x) + log(-c^2*x^2 + 1))*a*b*c*d^3 + 3*a^2*c*d^3*log
(x) - (c*(log(c^2*x^2 - 1) - log(x^2)) + 2*arctanh(c*x)/x)*a*b*d^3 - a^2*d^3/x + 1/8*(b^2*c^3*d^3*x^3 + 6*b^2*
c^2*d^3*x^2 - 2*b^2*d^3)*log(-c*x + 1)^2/x - integrate(-1/4*((b^2*c^4*d^3*x^4 + 2*b^2*c^3*d^3*x^3 - 2*b^2*c*d^
3*x - b^2*d^3)*log(c*x + 1)^2 + 4*(a*b*c^4*d^3*x^4 - a*b*c^3*d^3*x^3 + 3*a*b*c^2*d^3*x^2 - 3*a*b*c*d^3*x)*log(
c*x + 1) - (12*a*b*c^2*d^3*x^2 + (4*a*b*c^4*d^3 + b^2*c^4*d^3)*x^4 - 2*(2*a*b*c^3*d^3 - 3*b^2*c^3*d^3)*x^3 - 2
*(6*a*b*c*d^3 + b^2*c*d^3)*x + 2*(b^2*c^4*d^3*x^4 + 2*b^2*c^3*d^3*x^3 - 2*b^2*c*d^3*x - b^2*d^3)*log(c*x + 1))
*log(-c*x + 1))/(c*x^3 - x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ d^{3} \left (\int 3 a^{2} c^{2}\, dx + \int \frac {a^{2}}{x^{2}}\, dx + \int \frac {3 a^{2} c}{x}\, dx + \int a^{2} c^{3} x\, dx + \int 3 b^{2} c^{2} \operatorname {atanh}^{2}{\left (c x \right )}\, dx + \int \frac {b^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{x^{2}}\, dx + \int 6 a b c^{2} \operatorname {atanh}{\left (c x \right )}\, dx + \int \frac {2 a b \operatorname {atanh}{\left (c x \right )}}{x^{2}}\, dx + \int \frac {3 b^{2} c \operatorname {atanh}^{2}{\left (c x \right )}}{x}\, dx + \int b^{2} c^{3} x \operatorname {atanh}^{2}{\left (c x \right )}\, dx + \int \frac {6 a b c \operatorname {atanh}{\left (c x \right )}}{x}\, dx + \int 2 a b c^{3} x \operatorname {atanh}{\left (c x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d**3*(Integral(3*a**2*c**2, x) + Integral(a**2/x**2, x) + Integral(3*a**2*c/x, x) + Integral(a**2*c**3*x, x) +
Integral(3*b**2*c**2*atanh(c*x)**2, x) + Integral(b**2*atanh(c*x)**2/x**2, x) + Integral(6*a*b*c**2*atanh(c*x
), x) + Integral(2*a*b*atanh(c*x)/x**2, x) + Integral(3*b**2*c*atanh(c*x)**2/x, x) + Integral(b**2*c**3*x*atan
h(c*x)**2, x) + Integral(6*a*b*c*atanh(c*x)/x, x) + Integral(2*a*b*c**3*x*atanh(c*x), x))

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