3.70 \(\int \frac{(a+b \tanh ^{-1}(c x^2))^2}{x^5} \, dx\)

Optimal. Leaf size=88 \[ \frac{1}{4} c^2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2-\frac{b c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )}{2 x^2}-\frac{\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{4 x^4}-\frac{1}{4} b^2 c^2 \log \left (1-c^2 x^4\right )+b^2 c^2 \log (x) \]

[Out]

-(b*c*(a + b*ArcTanh[c*x^2]))/(2*x^2) + (c^2*(a + b*ArcTanh[c*x^2])^2)/4 - (a + b*ArcTanh[c*x^2])^2/(4*x^4) +
b^2*c^2*Log[x] - (b^2*c^2*Log[1 - c^2*x^4])/4

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Rubi [C]  time = 1.05672, antiderivative size = 360, normalized size of antiderivative = 4.09, number of steps used = 46, number of rules used = 23, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 1.438, Rules used = {6099, 2454, 2398, 2411, 2347, 2344, 2301, 2316, 2315, 2314, 31, 2395, 44, 2439, 2416, 36, 29, 2392, 2391, 2394, 2393, 2410, 2390} \[ -\frac{1}{8} b^2 c^2 \text{PolyLog}\left (2,\frac{1}{2} \left (1-c x^2\right )\right )-\frac{1}{8} b^2 c^2 \text{PolyLog}\left (2,\frac{1}{2} \left (c x^2+1\right )\right )+\frac{1}{8} b c^2 \log \left (\frac{1}{2} \left (c x^2+1\right )\right ) \left (2 a-b \log \left (1-c x^2\right )\right )+\frac{1}{16} c^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac{b c \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )}{8 x^2}-\frac{b c \left (2 a-b \log \left (1-c x^2\right )\right )}{8 x^2}-\frac{b \log \left (c x^2+1\right ) \left (2 a-b \log \left (1-c x^2\right )\right )}{8 x^4}-\frac{\left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 x^4}+\frac{1}{16} b^2 c^2 \log ^2\left (c x^2+1\right )-\frac{1}{8} b^2 c^2 \log \left (1-c x^2\right )-\frac{1}{4} b^2 c^2 \log \left (c x^2+1\right )-\frac{1}{8} b^2 c^2 \log \left (\frac{1}{2} \left (1-c x^2\right )\right ) \log \left (c x^2+1\right )+b^2 c^2 \log (x)-\frac{b^2 \log ^2\left (c x^2+1\right )}{16 x^4}-\frac{b^2 c \log \left (c x^2+1\right )}{4 x^2} \]

Warning: Unable to verify antiderivative.

[In]

Int[(a + b*ArcTanh[c*x^2])^2/x^5,x]

[Out]

b^2*c^2*Log[x] - (b^2*c^2*Log[1 - c*x^2])/8 - (b*c*(2*a - b*Log[1 - c*x^2]))/(8*x^2) - (b*c*(1 - c*x^2)*(2*a -
b*Log[1 - c*x^2]))/(8*x^2) + (c^2*(2*a - b*Log[1 - c*x^2])^2)/16 - (2*a - b*Log[1 - c*x^2])^2/(16*x^4) + (b*c
^2*(2*a - b*Log[1 - c*x^2])*Log[(1 + c*x^2)/2])/8 - (b^2*c^2*Log[1 + c*x^2])/4 - (b^2*c*Log[1 + c*x^2])/(4*x^2
) - (b^2*c^2*Log[(1 - c*x^2)/2]*Log[1 + c*x^2])/8 - (b*(2*a - b*Log[1 - c*x^2])*Log[1 + c*x^2])/(8*x^4) + (b^2
*c^2*Log[1 + c*x^2]^2)/16 - (b^2*Log[1 + c*x^2]^2)/(16*x^4) - (b^2*c^2*PolyLog[2, (1 - c*x^2)/2])/8 - (b^2*c^2
*PolyLog[2, (1 + c*x^2)/2])/8

Rule 6099

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(d*x)^
m*(a + (b*Log[1 + c*x^n])/2 - (b*Log[1 - c*x^n])/2)^p, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && IGtQ[p, 0] &&
IntegerQ[m] && IntegerQ[n]

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2398

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((
f + g*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n])^p)/(g*(q + 1)), x] - Dist[(b*e*n*p)/(g*(q + 1)), Int[((f + g*x)^(q
+ 1)*(a + b*Log[c*(d + e*x)^n])^(p - 1))/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*
f - d*g, 0] && GtQ[p, 0] && NeQ[q, -1] && IntegersQ[2*p, 2*q] && ( !IGtQ[q, 0] || (EqQ[p, 2] && NeQ[q, 1]))

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2347

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_))/(x_), x_Symbol] :> Dist[1/d, Int[((
d + e*x)^(q + 1)*(a + b*Log[c*x^n])^p)/x, x], x] - Dist[e/d, Int[(d + e*x)^q*(a + b*Log[c*x^n])^p, x], x] /; F
reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && LtQ[q, -1] && IntegerQ[2*q]

Rule 2344

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

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2316

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[((a + b*Log[-((c*d)/e)])*Log[d + e*
x])/e, x] + Dist[b, Int[Log[-((e*x)/d)]/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[-((c*d)/e), 0]

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 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 31

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

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

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 2439

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + Log[(h_.)*((i_.) + (j_.)*(x_))^(m_.)]*
(g_.))*(x_)^(r_.), x_Symbol] :> Simp[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^p*(f + g*Log[h*(i + j*x)^m]))/(r +
1), x] + (-Dist[(g*j*m)/(r + 1), Int[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^p)/(i + j*x), x], x] - Dist[(b*e*n*
p)/(r + 1), Int[(x^(r + 1)*(a + b*Log[c*(d + e*x)^n])^(p - 1)*(f + g*Log[h*(i + j*x)^m]))/(d + e*x), x], x]) /
; FreeQ[{a, b, c, d, e, f, g, h, i, j, m, n}, x] && IGtQ[p, 0] && IntegerQ[r] && (EqQ[p, 1] || GtQ[r, 0]) && N
eQ[r, -1]

Rule 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2392

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

Rule 2391

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

Rule 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2393

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

Rule 2410

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

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rubi steps

\begin{align*} \int \frac{\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{x^5} \, dx &=\int \left (\frac{\left (2 a-b \log \left (1-c x^2\right )\right )^2}{4 x^5}-\frac{b \left (-2 a+b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{2 x^5}+\frac{b^2 \log ^2\left (1+c x^2\right )}{4 x^5}\right ) \, dx\\ &=\frac{1}{4} \int \frac{\left (2 a-b \log \left (1-c x^2\right )\right )^2}{x^5} \, dx-\frac{1}{2} b \int \frac{\left (-2 a+b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{x^5} \, dx+\frac{1}{4} b^2 \int \frac{\log ^2\left (1+c x^2\right )}{x^5} \, dx\\ &=\frac{1}{8} \operatorname{Subst}\left (\int \frac{(2 a-b \log (1-c x))^2}{x^3} \, dx,x,x^2\right )-\frac{1}{4} b \operatorname{Subst}\left (\int \frac{(-2 a+b \log (1-c x)) \log (1+c x)}{x^3} \, dx,x,x^2\right )+\frac{1}{8} b^2 \operatorname{Subst}\left (\int \frac{\log ^2(1+c x)}{x^3} \, dx,x,x^2\right )\\ &=-\frac{\left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 x^4}-\frac{b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 x^4}-\frac{b^2 \log ^2\left (1+c x^2\right )}{16 x^4}+\frac{1}{8} (b c) \operatorname{Subst}\left (\int \frac{2 a-b \log (1-c x)}{x^2 (1-c x)} \, dx,x,x^2\right )-\frac{1}{8} (b c) \operatorname{Subst}\left (\int \frac{-2 a+b \log (1-c x)}{x^2 (1+c x)} \, dx,x,x^2\right )+\frac{1}{8} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1+c x)}{x^2 (1-c x)} \, dx,x,x^2\right )+\frac{1}{8} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1+c x)}{x^2 (1+c x)} \, dx,x,x^2\right )\\ &=-\frac{\left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 x^4}-\frac{b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 x^4}-\frac{b^2 \log ^2\left (1+c x^2\right )}{16 x^4}-\frac{1}{8} b \operatorname{Subst}\left (\int \frac{2 a-b \log (x)}{x \left (\frac{1}{c}-\frac{x}{c}\right )^2} \, dx,x,1-c x^2\right )-\frac{1}{8} (b c) \operatorname{Subst}\left (\int \left (\frac{-2 a+b \log (1-c x)}{x^2}-\frac{c (-2 a+b \log (1-c x))}{x}+\frac{c^2 (-2 a+b \log (1-c x))}{1+c x}\right ) \, dx,x,x^2\right )+\frac{1}{8} \left (b^2 c\right ) \operatorname{Subst}\left (\int \left (\frac{\log (1+c x)}{x^2}+\frac{c \log (1+c x)}{x}-\frac{c^2 \log (1+c x)}{-1+c x}\right ) \, dx,x,x^2\right )+\frac{1}{8} \left (b^2 c\right ) \operatorname{Subst}\left (\int \left (\frac{\log (1+c x)}{x^2}-\frac{c \log (1+c x)}{x}+\frac{c^2 \log (1+c x)}{1+c x}\right ) \, dx,x,x^2\right )\\ &=-\frac{\left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 x^4}-\frac{b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 x^4}-\frac{b^2 \log ^2\left (1+c x^2\right )}{16 x^4}-\frac{1}{8} b \operatorname{Subst}\left (\int \frac{2 a-b \log (x)}{\left (\frac{1}{c}-\frac{x}{c}\right )^2} \, dx,x,1-c x^2\right )-\frac{1}{8} (b c) \operatorname{Subst}\left (\int \frac{2 a-b \log (x)}{x \left (\frac{1}{c}-\frac{x}{c}\right )} \, dx,x,1-c x^2\right )-\frac{1}{8} (b c) \operatorname{Subst}\left (\int \frac{-2 a+b \log (1-c x)}{x^2} \, dx,x,x^2\right )+2 \left (\frac{1}{8} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (1+c x)}{x^2} \, dx,x,x^2\right )\right )+\frac{1}{8} \left (b c^2\right ) \operatorname{Subst}\left (\int \frac{-2 a+b \log (1-c x)}{x} \, dx,x,x^2\right )-\frac{1}{8} \left (b c^3\right ) \operatorname{Subst}\left (\int \frac{-2 a+b \log (1-c x)}{1+c x} \, dx,x,x^2\right )-\frac{1}{8} \left (b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+c x)}{-1+c x} \, dx,x,x^2\right )+\frac{1}{8} \left (b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{\log (1+c x)}{1+c x} \, dx,x,x^2\right )\\ &=-\frac{1}{2} a b c^2 \log (x)-\frac{b c \left (2 a-b \log \left (1-c x^2\right )\right )}{8 x^2}-\frac{b c \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )}{8 x^2}-\frac{\left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 x^4}+\frac{1}{8} b c^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+c x^2\right )\right )-\frac{1}{8} b^2 c^2 \log \left (\frac{1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac{b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 x^4}-\frac{b^2 \log ^2\left (1+c x^2\right )}{16 x^4}-\frac{1}{8} (b c) \operatorname{Subst}\left (\int \frac{2 a-b \log (x)}{\frac{1}{c}-\frac{x}{c}} \, dx,x,1-c x^2\right )-\frac{1}{8} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{c}-\frac{x}{c}} \, dx,x,1-c x^2\right )-\frac{1}{8} \left (b c^2\right ) \operatorname{Subst}\left (\int \frac{2 a-b \log (x)}{x} \, dx,x,1-c x^2\right )+\frac{1}{8} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x (1-c x)} \, dx,x,x^2\right )+2 \left (-\frac{b^2 c \log \left (1+c x^2\right )}{8 x^2}+\frac{1}{8} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x (1+c x)} \, dx,x,x^2\right )\right )+\frac{1}{8} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\log (x)}{x} \, dx,x,1+c x^2\right )+\frac{1}{8} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-c x)}{x} \, dx,x,x^2\right )+\frac{1}{8} \left (b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{1}{2} (1-c x)\right )}{1+c x} \, dx,x,x^2\right )-\frac{1}{8} \left (b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{1}{2} (1+c x)\right )}{1-c x} \, dx,x,x^2\right )\\ &=\frac{1}{4} b^2 c^2 \log (x)-\frac{b c \left (2 a-b \log \left (1-c x^2\right )\right )}{8 x^2}-\frac{b c \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )}{8 x^2}+\frac{1}{16} c^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac{\left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 x^4}+\frac{1}{8} b c^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+c x^2\right )\right )-\frac{1}{8} b^2 c^2 \log \left (\frac{1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac{b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 x^4}+\frac{1}{16} b^2 c^2 \log ^2\left (1+c x^2\right )-\frac{b^2 \log ^2\left (1+c x^2\right )}{16 x^4}-\frac{1}{8} b^2 c^2 \text{Li}_2\left (c x^2\right )+\frac{1}{8} \left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{\log (x)}{\frac{1}{c}-\frac{x}{c}} \, dx,x,1-c x^2\right )+\frac{1}{8} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )+\frac{1}{8} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right )}{x} \, dx,x,1-c x^2\right )+\frac{1}{8} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{\log \left (1-\frac{x}{2}\right )}{x} \, dx,x,1+c x^2\right )+\frac{1}{8} \left (b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x} \, dx,x,x^2\right )+2 \left (-\frac{b^2 c \log \left (1+c x^2\right )}{8 x^2}+\frac{1}{8} \left (b^2 c^2\right ) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,x^2\right )-\frac{1}{8} \left (b^2 c^3\right ) \operatorname{Subst}\left (\int \frac{1}{1+c x} \, dx,x,x^2\right )\right )\\ &=\frac{1}{2} b^2 c^2 \log (x)-\frac{1}{8} b^2 c^2 \log \left (1-c x^2\right )-\frac{b c \left (2 a-b \log \left (1-c x^2\right )\right )}{8 x^2}-\frac{b c \left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )}{8 x^2}+\frac{1}{16} c^2 \left (2 a-b \log \left (1-c x^2\right )\right )^2-\frac{\left (2 a-b \log \left (1-c x^2\right )\right )^2}{16 x^4}+\frac{1}{8} b c^2 \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac{1}{2} \left (1+c x^2\right )\right )-\frac{1}{8} b^2 c^2 \log \left (\frac{1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac{b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{8 x^4}+\frac{1}{16} b^2 c^2 \log ^2\left (1+c x^2\right )-\frac{b^2 \log ^2\left (1+c x^2\right )}{16 x^4}+2 \left (\frac{1}{4} b^2 c^2 \log (x)-\frac{1}{8} b^2 c^2 \log \left (1+c x^2\right )-\frac{b^2 c \log \left (1+c x^2\right )}{8 x^2}\right )-\frac{1}{8} b^2 c^2 \text{Li}_2\left (\frac{1}{2} \left (1-c x^2\right )\right )-\frac{1}{8} b^2 c^2 \text{Li}_2\left (\frac{1}{2} \left (1+c x^2\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.083499, size = 111, normalized size = 1.26 \[ \frac{1}{4} \left (-\frac{a^2}{x^4}-b c^2 (a+b) \log \left (1-c x^2\right )+b c^2 (a-b) \log \left (c x^2+1\right )-\frac{2 a b c}{x^2}-\frac{2 b \tanh ^{-1}\left (c x^2\right ) \left (a+b c x^2\right )}{x^4}+\frac{b^2 \left (c^2 x^4-1\right ) \tanh ^{-1}\left (c x^2\right )^2}{x^4}+4 b^2 c^2 \log (x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^2])^2/x^5,x]

[Out]

(-(a^2/x^4) - (2*a*b*c)/x^2 - (2*b*(a + b*c*x^2)*ArcTanh[c*x^2])/x^4 + (b^2*(-1 + c^2*x^4)*ArcTanh[c*x^2]^2)/x
^4 + 4*b^2*c^2*Log[x] - b*(a + b)*c^2*Log[1 - c*x^2] + (a - b)*b*c^2*Log[1 + c*x^2])/4

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Maple [B]  time = 0.188, size = 257, normalized size = 2.9 \begin{align*}{\frac{{b}^{2} \left ({c}^{2}{x}^{4}-1 \right ) \left ( \ln \left ( c{x}^{2}+1 \right ) \right ) ^{2}}{16\,{x}^{4}}}-{\frac{b \left ({x}^{4}b\ln \left ( -c{x}^{2}+1 \right ){c}^{2}+2\,bc{x}^{2}-b\ln \left ( -c{x}^{2}+1 \right ) +2\,a \right ) \ln \left ( c{x}^{2}+1 \right ) }{8\,{x}^{4}}}+{\frac{{b}^{2}{c}^{2}{x}^{4} \left ( \ln \left ( -c{x}^{2}+1 \right ) \right ) ^{2}+4\,b{c}^{2}\ln \left ( c{x}^{2}+1 \right ){x}^{4}a-4\,{b}^{2}{c}^{2}\ln \left ( c{x}^{2}+1 \right ){x}^{4}-4\,b{c}^{2}\ln \left ( c{x}^{2}-1 \right ){x}^{4}a-4\,{b}^{2}{c}^{2}\ln \left ( c{x}^{2}-1 \right ){x}^{4}+16\,{b}^{2}{c}^{2}\ln \left ( x \right ){x}^{4}+4\,{b}^{2}c{x}^{2}\ln \left ( -c{x}^{2}+1 \right ) -8\,abc{x}^{2}-{b}^{2} \left ( \ln \left ( -c{x}^{2}+1 \right ) \right ) ^{2}+4\,b\ln \left ( -c{x}^{2}+1 \right ) a-4\,{a}^{2}}{16\,{x}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^2))^2/x^5,x)

[Out]

1/16*b^2*(c^2*x^4-1)/x^4*ln(c*x^2+1)^2-1/8*b*(x^4*b*ln(-c*x^2+1)*c^2+2*b*c*x^2-b*ln(-c*x^2+1)+2*a)/x^4*ln(c*x^
2+1)+1/16*(b^2*c^2*x^4*ln(-c*x^2+1)^2+4*b*c^2*ln(c*x^2+1)*x^4*a-4*b^2*c^2*ln(c*x^2+1)*x^4-4*b*c^2*ln(c*x^2-1)*
x^4*a-4*b^2*c^2*ln(c*x^2-1)*x^4+16*b^2*c^2*ln(x)*x^4+4*b^2*c*x^2*ln(-c*x^2+1)-8*a*b*c*x^2-b^2*ln(-c*x^2+1)^2+4
*b*ln(-c*x^2+1)*a-4*a^2)/x^4

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Maxima [B]  time = 1.00288, size = 236, normalized size = 2.68 \begin{align*} \frac{1}{4} \,{\left ({\left (c \log \left (c x^{2} + 1\right ) - c \log \left (c x^{2} - 1\right ) - \frac{2}{x^{2}}\right )} c - \frac{2 \, \operatorname{artanh}\left (c x^{2}\right )}{x^{4}}\right )} a b + \frac{1}{16} \,{\left ({\left (2 \,{\left (\log \left (c x^{2} - 1\right ) - 2\right )} \log \left (c x^{2} + 1\right ) - \log \left (c x^{2} + 1\right )^{2} - \log \left (c x^{2} - 1\right )^{2} - 4 \, \log \left (c x^{2} - 1\right ) + 16 \, \log \left (x\right )\right )} c^{2} + 4 \,{\left (c \log \left (c x^{2} + 1\right ) - c \log \left (c x^{2} - 1\right ) - \frac{2}{x^{2}}\right )} c \operatorname{artanh}\left (c x^{2}\right )\right )} b^{2} - \frac{b^{2} \operatorname{artanh}\left (c x^{2}\right )^{2}}{4 \, x^{4}} - \frac{a^{2}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))^2/x^5,x, algorithm="maxima")

[Out]

1/4*((c*log(c*x^2 + 1) - c*log(c*x^2 - 1) - 2/x^2)*c - 2*arctanh(c*x^2)/x^4)*a*b + 1/16*((2*(log(c*x^2 - 1) -
2)*log(c*x^2 + 1) - log(c*x^2 + 1)^2 - log(c*x^2 - 1)^2 - 4*log(c*x^2 - 1) + 16*log(x))*c^2 + 4*(c*log(c*x^2 +
1) - c*log(c*x^2 - 1) - 2/x^2)*c*arctanh(c*x^2))*b^2 - 1/4*b^2*arctanh(c*x^2)^2/x^4 - 1/4*a^2/x^4

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Fricas [A]  time = 2.16351, size = 324, normalized size = 3.68 \begin{align*} \frac{16 \, b^{2} c^{2} x^{4} \log \left (x\right ) + 4 \,{\left (a b - b^{2}\right )} c^{2} x^{4} \log \left (c x^{2} + 1\right ) - 4 \,{\left (a b + b^{2}\right )} c^{2} x^{4} \log \left (c x^{2} - 1\right ) - 8 \, a b c x^{2} +{\left (b^{2} c^{2} x^{4} - b^{2}\right )} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right )^{2} - 4 \, a^{2} - 4 \,{\left (b^{2} c x^{2} + a b\right )} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right )}{16 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))^2/x^5,x, algorithm="fricas")

[Out]

1/16*(16*b^2*c^2*x^4*log(x) + 4*(a*b - b^2)*c^2*x^4*log(c*x^2 + 1) - 4*(a*b + b^2)*c^2*x^4*log(c*x^2 - 1) - 8*
a*b*c*x^2 + (b^2*c^2*x^4 - b^2)*log(-(c*x^2 + 1)/(c*x^2 - 1))^2 - 4*a^2 - 4*(b^2*c*x^2 + a*b)*log(-(c*x^2 + 1)
/(c*x^2 - 1)))/x^4

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Sympy [A]  time = 26.0661, size = 175, normalized size = 1.99 \begin{align*} \begin{cases} - \frac{a^{2}}{4 x^{4}} + \frac{a b c^{2} \operatorname{atanh}{\left (c x^{2} \right )}}{2} - \frac{a b c}{2 x^{2}} - \frac{a b \operatorname{atanh}{\left (c x^{2} \right )}}{2 x^{4}} + b^{2} c^{2} \log{\left (x \right )} - \frac{b^{2} c^{2} \log{\left (x - i \sqrt{\frac{1}{c}} \right )}}{2} - \frac{b^{2} c^{2} \log{\left (x + i \sqrt{\frac{1}{c}} \right )}}{2} + \frac{b^{2} c^{2} \operatorname{atanh}^{2}{\left (c x^{2} \right )}}{4} + \frac{b^{2} c^{2} \operatorname{atanh}{\left (c x^{2} \right )}}{2} - \frac{b^{2} c \operatorname{atanh}{\left (c x^{2} \right )}}{2 x^{2}} - \frac{b^{2} \operatorname{atanh}^{2}{\left (c x^{2} \right )}}{4 x^{4}} & \text{for}\: c \neq 0 \\- \frac{a^{2}}{4 x^{4}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**2))**2/x**5,x)

[Out]

Piecewise((-a**2/(4*x**4) + a*b*c**2*atanh(c*x**2)/2 - a*b*c/(2*x**2) - a*b*atanh(c*x**2)/(2*x**4) + b**2*c**2
*log(x) - b**2*c**2*log(x - I*sqrt(1/c))/2 - b**2*c**2*log(x + I*sqrt(1/c))/2 + b**2*c**2*atanh(c*x**2)**2/4 +
b**2*c**2*atanh(c*x**2)/2 - b**2*c*atanh(c*x**2)/(2*x**2) - b**2*atanh(c*x**2)**2/(4*x**4), Ne(c, 0)), (-a**2
/(4*x**4), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c x^{2}\right ) + a\right )}^{2}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))^2/x^5,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x^2) + a)^2/x^5, x)